[math-fun] Greedy Queens question
Dear Math Fun, Take an infinite chessboard, and starting at the top left corner, scan it by upwards anti-diagonals, placing a Queen whenever you reach a square that is not attacked by an existing Queen, The top left portion of the board is Qxxxxx... xxxQxx... xQxxxx... xxxxQx... xxQxxx... xxxxxx... xxxxxx... xxxxxQ... ...... The n-th term of sequence https://oeis.org/A065188 tells which row contains the Queen in column n. (It is easy to show that every column eventually contains a Queen.) If you click the "graph" button in A065188 you will see that the points appear to lie on two roughly straight lines, of slopes phi and 1/phi, phi being the golden ratio! The question is, why? How do these queens know about the golden ratio? Presumably this is related to budding sequences and sunflower seeds. Notes. Let c(n) = A065188(n). A199134 lists n such that c(n)<n (the points on the line of slope 1/phi), and A275884 lists n such that c(n)>=n (the line of slope phi). A275885 gives lengths of runs of consecutive terms in A193134. Initially I thought these lengths were always 1 or 2, which might have been significant, but that is false. A275888 gives first differences of A275884. Initially I thought these terms were always 1, 2, or 3, but that is also false. I can't correct these errors until Monday, because the OEIS is in read-only mode for a server upgrade. I can send a larger picture of the chess-board to anyone who is interested.
I think there is a missing row of x's in the image. It shows a Q in the 6th column of row 8, but that should be row 9. George http://georgehart.com On 8/19/2016 4:02 PM, Neil Sloane wrote:
Dear Math Fun, Take an infinite chessboard, and starting at the top left corner, scan it by upwards anti-diagonals, placing a Queen whenever you reach a square that is not attacked by an existing Queen, The top left portion of the board is Qxxxxx... xxxQxx... xQxxxx... xxxxQx... xxQxxx... xxxxxx... xxxxxx... xxxxxQ... ...... The n-th term of sequence https://oeis.org/A065188 tells which row contains the Queen in column n. (It is easy to show that every column eventually contains a Queen.) If you click the "graph" button in A065188 you will see that the points appear to lie on two roughly straight lines, of slopes phi and 1/phi, phi being the golden ratio! The question is, why? How do these queens know about the golden ratio? Presumably this is related to budding sequences and sunflower seeds.
Notes. Let c(n) = A065188(n). A199134 lists n such that c(n)<n (the points on the line of slope 1/phi), and A275884 lists n such that c(n)>=n (the line of slope phi). A275885 gives lengths of runs of consecutive terms in A193134. Initially I thought these lengths were always 1 or 2, which might have been significant, but that is false. A275888 gives first differences of A275884. Initially I thought these terms were always 1, 2, or 3, but that is also false. I can't correct these errors until Monday, because the OEIS is in read-only mode for a server upgrade.
I can send a larger picture of the chess-board to anyone who is interested.
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What happens if the queens are also knightriders (i.e. they can also attack on lines of slope +-1/2 and +-2?) - Cris
On Aug 19, 2016, at 2:02 PM, Neil Sloane <njasloane@gmail.com> wrote:
Dear Math Fun, Take an infinite chessboard, and starting at the top left corner, scan it by upwards anti-diagonals, placing a Queen whenever you reach a square that is not attacked by an existing Queen, The top left portion of the board is Qxxxxx... xxxQxx... xQxxxx... xxxxQx... xxQxxx... xxxxxx... xxxxxx... xxxxxQ... ...... The n-th term of sequence https://oeis.org/A065188 tells which row contains the Queen in column n. (It is easy to show that every column eventually contains a Queen.) If you click the "graph" button in A065188 you will see that the points appear to lie on two roughly straight lines, of slopes phi and 1/phi, phi being the golden ratio! The question is, why? How do these queens know about the golden ratio? Presumably this is related to budding sequences and sunflower seeds.
Notes. Let c(n) = A065188(n). A199134 lists n such that c(n)<n (the points on the line of slope 1/phi), and A275884 lists n such that c(n)>=n (the line of slope phi). A275885 gives lengths of runs of consecutive terms in A193134. Initially I thought these lengths were always 1 or 2, which might have been significant, but that is false. A275888 gives first differences of A275884. Initially I thought these terms were always 1, 2, or 3, but that is also false. I can't correct these errors until Monday, because the OEIS is in read-only mode for a server upgrade.
I can send a larger picture of the chess-board to anyone who is interested.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Oh hey, I remember working this out on a chess board in my grandmother's house in 1992! Your queens are marking the winning positions in the nim variant where you have only up to two piles and you're also allowed to remove the same number from each pile (=move diagonally). Surely this is enough for NJAS to fill in all the details :-). --Michael On Fri, Aug 19, 2016 at 2:41 PM, Cris Moore <moore@santafe.edu> wrote:
What happens if the queens are also knightriders (i.e. they can also attack on lines of slope +-1/2 and +-2?)
- Cris
On Aug 19, 2016, at 2:02 PM, Neil Sloane <njasloane@gmail.com> wrote:
Dear Math Fun, Take an infinite chessboard, and starting at the top left corner, scan it by upwards anti-diagonals, placing a Queen whenever you reach a square that is not attacked by an existing Queen, The top left portion of the board is Qxxxxx... xxxQxx... xQxxxx... xxxxQx... xxQxxx... xxxxxx... xxxxxx... xxxxxQ... ...... The n-th term of sequence https://oeis.org/A065188 tells which row contains the Queen in column n. (It is easy to show that every column eventually contains a Queen.) If you click the "graph" button in A065188 you will see that the points appear to lie on two roughly straight lines, of slopes phi and 1/phi, phi being the golden ratio! The question is, why? How do these queens know about the golden ratio? Presumably this is related to budding sequences and sunflower seeds.
Notes. Let c(n) = A065188(n). A199134 lists n such that c(n)<n (the points on the line of slope 1/phi), and A275884 lists n such that c(n)>=n (the line of slope phi). A275885 gives lengths of runs of consecutive terms in A193134. Initially I thought these lengths were always 1 or 2, which might have been significant, but that is false. A275888 gives first differences of A275884. Initially I thought these terms were always 1, 2, or 3, but that is also false. I can't correct these errors until Monday, because the OEIS is in read-only mode for a server upgrade.
I can send a larger picture of the chess-board to anyone who is interested.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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-- Forewarned is worth an octopus in the bush.
This is a restatement of Wythoff's variant of Nim, which has been well-studied. Here are a couple references: http://www.wisdom.weizmann.ac.il/~fraenkel/Papers/WythoffWisdomJune62016.pdf http://www.msri.org/people/staff/levy/files/Book56/43nivasch.pdf J.P. On Sat, Aug 20, 2016 at 1:13 AM, Michael Kleber <michael.kleber@gmail.com> wrote:
Oh hey, I remember working this out on a chess board in my grandmother's house in 1992! Your queens are marking the winning positions in the nim variant where you have only up to two piles and you're also allowed to remove the same number from each pile (=move diagonally). Surely this is enough for NJAS to fill in all the details :-).
--Michael
On Fri, Aug 19, 2016 at 2:41 PM, Cris Moore <moore@santafe.edu> wrote:
What happens if the queens are also knightriders (i.e. they can also attack on lines of slope +-1/2 and +-2?)
- Cris
On Aug 19, 2016, at 2:02 PM, Neil Sloane <njasloane@gmail.com> wrote:
Dear Math Fun, Take an infinite chessboard, and starting at the top left corner, scan it by upwards anti-diagonals, placing a Queen whenever you reach a square that is not attacked by an existing Queen, The top left portion of the board is Qxxxxx... xxxQxx... xQxxxx... xxxxQx... xxQxxx... xxxxxx... xxxxxx... xxxxxQ... ...... The n-th term of sequence https://oeis.org/A065188 tells which row contains the Queen in column n. (It is easy to show that every column eventually contains a Queen.) If you click the "graph" button in A065188 you will see that the points appear to lie on two roughly straight lines, of slopes phi and 1/phi, phi being the golden ratio! The question is, why? How do these queens know about the golden ratio? Presumably this is related to budding sequences and sunflower seeds.
Notes. Let c(n) = A065188(n). A199134 lists n such that c(n)<n (the points on the line of slope 1/phi), and A275884 lists n such that c(n)>=n (the line of slope phi). A275885 gives lengths of runs of consecutive terms in A193134.
Initially
I thought these lengths were always 1 or 2, which might have been significant, but that is false. A275888 gives first differences of A275884. Initially I thought these terms were always 1, 2, or 3, but that is also false. I can't correct these errors until Monday, because the OEIS is in read-only mode for a server upgrade.
I can send a larger picture of the chess-board to anyone who is interested.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Very nice! Wythoff’s nim allows players to remove coins from the left pile, the right pile, or the same amount from both piles. This corresponds to moving down towards the x-axis, left towards the y-axis, or diagonally. As one of your references says, this is like moving a queen south, west, or southwest, where the first player to reach the origin wins. …But since Neil’s procedure takes the first available position in each antidiagonal, is seems that his queens mark the losing positions in a variant of Nim where we can, in addition, transfer coins from the left pile to the right pile, corresponding to also letting the queen move (say) northwest. Has this variant of Nim been studied? - Cris
On Aug 20, 2016, at 5:53 AM, J.P. Grossman <jpg@alum.mit.edu> wrote:
This is a restatement of Wythoff's variant of Nim, which has been well-studied. Here are a couple references:
http://www.wisdom.weizmann.ac.il/~fraenkel/Papers/WythoffWisdomJune62016.pdf http://www.msri.org/people/staff/levy/files/Book56/43nivasch.pdf
J.P.
On Sat, Aug 20, 2016 at 1:13 AM, Michael Kleber <michael.kleber@gmail.com> wrote:
Oh hey, I remember working this out on a chess board in my grandmother's house in 1992! Your queens are marking the winning positions in the nim variant where you have only up to two piles and you're also allowed to remove the same number from each pile (=move diagonally). Surely this is enough for NJAS to fill in all the details :-).
--Michael
On Fri, Aug 19, 2016 at 2:41 PM, Cris Moore <moore@santafe.edu> wrote:
What happens if the queens are also knightriders (i.e. they can also attack on lines of slope +-1/2 and +-2?)
- Cris
On Aug 19, 2016, at 2:02 PM, Neil Sloane <njasloane@gmail.com> wrote:
Dear Math Fun, Take an infinite chessboard, and starting at the top left corner, scan it by upwards anti-diagonals, placing a Queen whenever you reach a square that is not attacked by an existing Queen, The top left portion of the board is Qxxxxx... xxxQxx... xQxxxx... xxxxQx... xxQxxx... xxxxxx... xxxxxx... xxxxxQ... ...... The n-th term of sequence https://oeis.org/A065188 tells which row contains the Queen in column n. (It is easy to show that every column eventually contains a Queen.) If you click the "graph" button in A065188 you will see that the points appear to lie on two roughly straight lines, of slopes phi and 1/phi, phi being the golden ratio! The question is, why? How do these queens know about the golden ratio? Presumably this is related to budding sequences and sunflower seeds.
Notes. Let c(n) = A065188(n). A199134 lists n such that c(n)<n (the points on the line of slope 1/phi), and A275884 lists n such that c(n)>=n (the line of slope phi). A275885 gives lengths of runs of consecutive terms in A193134.
Initially
I thought these lengths were always 1 or 2, which might have been significant, but that is false. A275888 gives first differences of A275884. Initially I thought these terms were always 1, 2, or 3, but that is also false. I can't correct these errors until Monday, because the OEIS is in read-only mode for a server upgrade.
I can send a larger picture of the chess-board to anyone who is interested.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Cristopher Moore Professor, Santa Fe Institute The Nature of Computation Cristopher Moore and Stephan Mertens Available now at all good bookstores, or through Oxford University Press http://www.nature-of-computation.org/
In particular, the arrangement of greedy queens is asymmetric [about reflection in the diagonal], whereas the winning positions in Wythoff Nim are symmetric.
Sent: Saturday, August 20, 2016 at 8:27 PM From: "Cris Moore" <moore@santafe.edu> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Greedy Queens question
Very nice! Wythoff’s nim allows players to remove coins from the left pile, the right pile, or the same amount from both piles. This corresponds to moving down towards the x-axis, left towards the y-axis, or diagonally. As one of your references says, this is like moving a queen south, west, or southwest, where the first player to reach the origin wins.
…But since Neil’s procedure takes the first available position in each antidiagonal, is seems that his queens mark the losing positions in a variant of Nim where we can, in addition, transfer coins from the left pile to the right pile, corresponding to also letting the queen move (say) northwest. Has this variant of Nim been studied?
- Cris
On Aug 20, 2016, at 5:53 AM, J.P. Grossman <jpg@alum.mit.edu> wrote:
This is a restatement of Wythoff's variant of Nim, which has been well-studied. Here are a couple references:
http://www.wisdom.weizmann.ac.il/~fraenkel/Papers/WythoffWisdomJune62016.pdf http://www.msri.org/people/staff/levy/files/Book56/43nivasch.pdf
J.P.
On Sat, Aug 20, 2016 at 1:13 AM, Michael Kleber <michael.kleber@gmail.com> wrote:
Oh hey, I remember working this out on a chess board in my grandmother's house in 1992! Your queens are marking the winning positions in the nim variant where you have only up to two piles and you're also allowed to remove the same number from each pile (=move diagonally). Surely this is enough for NJAS to fill in all the details :-).
--Michael
On Fri, Aug 19, 2016 at 2:41 PM, Cris Moore <moore@santafe.edu> wrote:
What happens if the queens are also knightriders (i.e. they can also attack on lines of slope +-1/2 and +-2?)
- Cris
On Aug 19, 2016, at 2:02 PM, Neil Sloane <njasloane@gmail.com> wrote:
Dear Math Fun, Take an infinite chessboard, and starting at the top left corner, scan it by upwards anti-diagonals, placing a Queen whenever you reach a square that is not attacked by an existing Queen, The top left portion of the board is Qxxxxx... xxxQxx... xQxxxx... xxxxQx... xxQxxx... xxxxxx... xxxxxx... xxxxxQ... ...... The n-th term of sequence https://oeis.org/A065188 tells which row contains the Queen in column n. (It is easy to show that every column eventually contains a Queen.) If you click the "graph" button in A065188 you will see that the points appear to lie on two roughly straight lines, of slopes phi and 1/phi, phi being the golden ratio! The question is, why? How do these queens know about the golden ratio? Presumably this is related to budding sequences and sunflower seeds.
Notes. Let c(n) = A065188(n). A199134 lists n such that c(n)<n (the points on the line of slope 1/phi), and A275884 lists n such that c(n)>=n (the line of slope phi). A275885 gives lengths of runs of consecutive terms in A193134.
Initially
I thought these lengths were always 1 or 2, which might have been significant, but that is false. A275888 gives first differences of A275884. Initially I thought these terms were always 1, 2, or 3, but that is also false. I can't correct these errors until Monday, because the OEIS is in read-only mode for a server upgrade.
I can send a larger picture of the chess-board to anyone who is interested.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Cristopher Moore Professor, Santa Fe Institute
The Nature of Computation Cristopher Moore and Stephan Mertens Available now at all good bookstores, or through Oxford University Press http://www.nature-of-computation.org/
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Quite right! I saw "queens" and "golden ratio" and assumed they were the same, but I didn't read Neil's definition carefully enough. J.P. On Sat, Aug 20, 2016 at 2:42 PM, Adam P. Goucher <apgoucher@gmx.com> wrote:
In particular, the arrangement of greedy queens is asymmetric [about reflection in the diagonal], whereas the winning positions in Wythoff Nim are symmetric.
Sent: Saturday, August 20, 2016 at 8:27 PM From: "Cris Moore" <moore@santafe.edu> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Greedy Queens question
Very nice! Wythoff’s nim allows players to remove coins from the left pile, the right pile, or the same amount from both piles. This corresponds to moving down towards the x-axis, left towards the y-axis, or diagonally. As one of your references says, this is like moving a queen south, west, or southwest, where the first player to reach the origin wins.
…But since Neil’s procedure takes the first available position in each antidiagonal, is seems that his queens mark the losing positions in a variant of Nim where we can, in addition, transfer coins from the left pile to the right pile, corresponding to also letting the queen move (say) northwest. Has this variant of Nim been studied?
- Cris
On Aug 20, 2016, at 5:53 AM, J.P. Grossman <jpg@alum.mit.edu> wrote:
This is a restatement of Wythoff's variant of Nim, which has been well-studied. Here are a couple references:
http://www.wisdom.weizmann.ac.il/~fraenkel/Papers/ WythoffWisdomJune62016.pdf http://www.msri.org/people/staff/levy/files/Book56/43nivasch.pdf
J.P.
On Sat, Aug 20, 2016 at 1:13 AM, Michael Kleber < michael.kleber@gmail.com> wrote:
Oh hey, I remember working this out on a chess board in my grandmother's house in 1992! Your queens are marking the winning positions in the nim variant where you have only up to two piles and you're also allowed to remove the same number from each pile (=move diagonally). Surely this is enough for NJAS to fill in all the details :-).
--Michael
On Fri, Aug 19, 2016 at 2:41 PM, Cris Moore <moore@santafe.edu> wrote:
What happens if the queens are also knightriders (i.e. they can also attack on lines of slope +-1/2 and +-2?)
- Cris
On Aug 19, 2016, at 2:02 PM, Neil Sloane <njasloane@gmail.com>
wrote:
Dear Math Fun, Take an infinite chessboard, and starting at the top left corner, scan it by upwards anti-diagonals, placing a Queen whenever you reach a square that is not attacked by an existing
Queen,
The top left portion of the board is Qxxxxx... xxxQxx... xQxxxx... xxxxQx... xxQxxx... xxxxxx... xxxxxx... xxxxxQ... ...... The n-th term of sequence https://oeis.org/A065188 tells which row contains the Queen in column n. (It is easy to show that every column eventually contains a Queen.) If you click the "graph" button in A065188 you will see that the points appear to lie on two roughly straight lines, of slopes phi and 1/phi, phi being the golden ratio! The question is, why? How do these queens know about the golden ratio? Presumably this is related to budding sequences and sunflower seeds.
Notes. Let c(n) = A065188(n). A199134 lists n such that c(n)<n (the points on the line of slope 1/phi), and A275884 lists n such that c(n)>=n (the line of slope phi). A275885 gives lengths of runs of consecutive terms in A193134. Initially I thought these lengths were always 1 or 2, which might have been significant, but that is false. A275888 gives first differences of A275884. Initially I thought these terms were always 1, 2, or 3, but that is also false. I can't correct these errors until Monday, because the OEIS is in read-only mode for a server upgrade.
I can send a larger picture of the chess-board to anyone who is interested.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Cristopher Moore Professor, Santa Fe Institute
The Nature of Computation Cristopher Moore and Stephan Mertens Available now at all good bookstores, or through Oxford University Press http://www.nature-of-computation.org/
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Indeed, while both fall asymptotically along the lines of slope phi and 1/phi, the set of Neil’s queens and Wythoff’s --- and even the antidiagonals on which they live --- are incomparable. For instance, Neil has queens at (3,1), (2,4), and (4,3), while Wythoff has none on those antidiagonals. Conversely, Wythoff has quees at (3,5) and (5,3) and Neil has none on that antidiagonal. - Cris p.s. the origin (and the first or rather zeroth queen) is at (0,0).
On Aug 20, 2016, at 3:52 PM, J.P. Grossman <jpg@alum.mit.edu> wrote:
Quite right! I saw "queens" and "golden ratio" and assumed they were the same, but I didn't read Neil's definition carefully enough.
J.P.
On Sat, Aug 20, 2016 at 2:42 PM, Adam P. Goucher <apgoucher@gmx.com> wrote:
In particular, the arrangement of greedy queens is asymmetric [about reflection in the diagonal], whereas the winning positions in Wythoff Nim are symmetric.
Sent: Saturday, August 20, 2016 at 8:27 PM From: "Cris Moore" <moore@santafe.edu> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Greedy Queens question
Very nice! Wythoff’s nim allows players to remove coins from the left pile, the right pile, or the same amount from both piles. This corresponds to moving down towards the x-axis, left towards the y-axis, or diagonally. As one of your references says, this is like moving a queen south, west, or southwest, where the first player to reach the origin wins.
…But since Neil’s procedure takes the first available position in each antidiagonal, is seems that his queens mark the losing positions in a variant of Nim where we can, in addition, transfer coins from the left pile to the right pile, corresponding to also letting the queen move (say) northwest. Has this variant of Nim been studied?
- Cris
On Aug 20, 2016, at 5:53 AM, J.P. Grossman <jpg@alum.mit.edu> wrote:
This is a restatement of Wythoff's variant of Nim, which has been well-studied. Here are a couple references:
http://www.wisdom.weizmann.ac.il/~fraenkel/Papers/ WythoffWisdomJune62016.pdf http://www.msri.org/people/staff/levy/files/Book56/43nivasch.pdf
J.P.
On Sat, Aug 20, 2016 at 1:13 AM, Michael Kleber < michael.kleber@gmail.com> wrote:
Oh hey, I remember working this out on a chess board in my grandmother's house in 1992! Your queens are marking the winning positions in the nim variant where you have only up to two piles and you're also allowed to remove the same number from each pile (=move diagonally). Surely this is enough for NJAS to fill in all the details :-).
--Michael
On Fri, Aug 19, 2016 at 2:41 PM, Cris Moore <moore@santafe.edu> wrote:
What happens if the queens are also knightriders (i.e. they can also attack on lines of slope +-1/2 and +-2?)
- Cris
> On Aug 19, 2016, at 2:02 PM, Neil Sloane <njasloane@gmail.com>
wrote:
> > Dear Math Fun, Take an infinite chessboard, and starting at the > top left corner, scan it by upwards anti-diagonals, placing a Queen > whenever you reach a square that is not attacked by an existing Queen, > The top left portion of the board is > Qxxxxx... > xxxQxx... > xQxxxx... > xxxxQx... > xxQxxx... > xxxxxx... > xxxxxx... > xxxxxQ... > ...... > The n-th term of sequence https://oeis.org/A065188 tells > which row contains the Queen in column n. (It is easy to > show that every column eventually contains a Queen.) > If you click the "graph" button in A065188 you will > see that the points appear to lie on two roughly straight > lines, of slopes phi and 1/phi, phi being the golden ratio! > The question is, why? > How do these queens know about the golden ratio? > Presumably this is related to budding sequences and sunflower seeds. > > Notes. > Let c(n) = A065188(n). A199134 lists n such that c(n)<n (the points on the > line of slope 1/phi), and A275884 lists n such that c(n)>=n (the line > of slope phi). > A275885 gives lengths of runs of consecutive terms in A193134. Initially > I thought these lengths were always 1 or 2, which might have been > significant, but that is false. A275888 gives first differences of A275884. > Initially I thought these terms were always 1, 2, or 3, but that is also false. > I can't correct these errors until Monday, because the OEIS is in read-only > mode for a server upgrade. > > I can send a larger picture of the chess-board to anyone who is interested. > > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Cristopher Moore Professor, Santa Fe Institute
The Nature of Computation Cristopher Moore and Stephan Mertens Available now at all good bookstores, or through Oxford University Press http://www.nature-of-computation.org/
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participants (6)
-
Adam P. Goucher -
Cris Moore -
George Hart -
J.P. Grossman -
Michael Kleber -
Neil Sloane