[math-fun] You wouldn't believe (fwd)
Funsters & fansters might be amused by this somewhat unlikely coincidence. I had calculated a couple of hundred terms of a fourth order recurring sequence [it's A005178 in OEIS, if you want details] and was looking for the ranks of apparition of various primes. A 19-digit prime factor of the 53rd term turned up as a substring of the 103rd term! Have a prime time in 7^2 x 41. R. ---------- Forwarded message ---------- Date: Wed, 17 Dec 2008 09:24:06 -0700 (MST) From: Richard Guy <rkg@cpsc.ucalgary.ca> To: Hugh Cowie Williams <williams@math.ucalgary.ca> Subject: You wouldn't believe Hugh, before sending the final (??) version of the quadric file, I decided I'd do a search for larger primes just to see if there were any double ranks. While searching with 3140540902719737029 which is a factor of a(53), I discovered that it's a substring of a(103): 93118232931779128686097911301602920314054090271973702981658870701 ^^^^^^^^^^^^^^^^^^^ R.
The next year also could be presented using just 7's and ones 7*7*(7*(7-1) -1)
7*(7*7*(7-7/7)-7) Hugo 2008/12/20 Alexander Povolotsky <apovolot@gmail.com>
The next year also could be presented using just 7's and ones
7*7*(7*(7-1) -1)
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Greetings, Professor Guy, You're right in your subject line; I wouldn't believe it. I am nevertheless disappointed to find out that it isn't actually true; you seem to have been misled by some kind of editing error. Details below. At 12:45 PM 12/20/2008, you wrote:
Funsters & fansters might be amused by this somewhat unlikely coincidence. I had calculated a couple of hundred terms of a fourth order recurring sequence [it's A005178 in OEIS, if you want details] and was looking for the ranks of apparition of various primes. A 19-digit prime factor of the 53rd term turned up as a substring of the 103rd term!
any double ranks. While searching with
3140540902719737029
which is a factor of a(53),
With the indexing used in the OEIS, it's actually a(52) which equals 63389 * 3140540902719737029; no problem there.
I discovered that it's a substring of a(103):
93118232931779128686097911301602920314054090271973702981658870701 ^^^^^^^^^^^^^^^^^^^
But alas, a(102) is only 9311823293177912868609791130160292081658870701; the 19-digit prime seems to have been accidentally inserted into your copy of the number. (I'm not on all the lists you submitted your note to, so I can't reply to them all directly; thus I'm sending this only to you.) Best wishes for the holidays and the new year. -- Fred W. Helenius fredh@ix.netcom.com
At 01:43 PM 12/20/2008, I wrote:
(I'm not on all the lists you submitted your note to, so I can't reply to them all directly; thus I'm sending this only to you.)
At least that was my intention, naively thinking that "Reply" would reply to the sender, not the list The seqfan list is still out of the loop, though. -- Fred W. Helenius fredh@ix.netcom.com
Dear Richard and Seqfans, A005178 is periodic modulo 30 (pay attention that start from 0 not 1): %C A005178 congruent to 0 mod 100 if n is congruent to 14 or 29 mod 30 %C A005178 congruent to 1 mod 100 if n is congruent to 0 or 1 or 12 or 16 or 27 or 28 mod 30 %C A005178 congruent to 5 mod 100 if n is congruent to 2 or 11 or 17 or 26 mod 30 %C A005178 congruent to 11 mod 100 if n is congruent to 3 or 25 mod 30 %C A005178 congruent to 36 mod 100 if n is congruent to 4 or 9 or 19 or 24 mod 30 %C A005178 congruent to 45 mod 100 if n is congruent to 8 or 20 mod 30 %C A005178 congruent to 51 mod 100 if n is congruent to 13 or 15 mod 30 %C A005178 congruent to 61 mod 100 if n is congruent to 10 or 18 mod 30 %C A005178 congruent to 81 mod 100 if n is congruent to 6 or 7 or 21 or 22 mod 30 %C A005178 congruent to 95 mod 100 if n is congruent to 5 or 23 mod 30 Together 10 different rests modulo 30. Best wishes Artur
Sorry about this, which turns out to be an unintentional hoax. Fred Helenius observes that I managed accidentally to insert the 19-digit prime into the term a(103), which is only 9311823293177912868609791130160292081658870701 Note that the offset in A005178 is ``wrong''. This is a ``divisibility sequence'', i.e., with the ``right'' offset a(m) divides a(n) just if m divides n, as in the Fibonacci sequence. The way to put it ``right'' is to define the n-th member to be the number of domino tilings of a 4 by n-1 rectangle [# of matchings of the graph P4 x P(n-1)] giving a(2)=1, a(1)=1 (the empty matching of the empty graph), a(3)=5: --- | | --- --- | | --- | | --- | | --- --- and | | --- --- --- a(n) = a(n-1) + 5a(n-2) + a(n-3) - a(n-4) with a(0) = 0 and a(-n) = a(n). Hugh Williams & I may have a paper within a finite time about such sequences (see also A003757) in which some primes have two ranks of apparition. R. On Sat, 20 Dec 2008, Richard Guy wrote:
Funsters & fansters might be amused by this somewhat unlikely coincidence. I had calculated a couple of hundred terms of a fourth order recurring sequence [it's A005178 in OEIS, if you want details] and was looking for the ranks of apparition of various primes. A 19-digit prime factor of the 53rd term turned up as a substring of the 103rd term!
Have a prime time in 7^2 x 41. R.
---------- Forwarded message ---------- Date: Wed, 17 Dec 2008 09:24:06 -0700 (MST) From: Richard Guy <rkg@cpsc.ucalgary.ca> To: Hugh Cowie Williams <williams@math.ucalgary.ca> Subject: You wouldn't believe
Hugh, before sending the final (??) version of the quadric file, I decided I'd do a search for larger primes just to see if there were any double ranks. While searching with
3140540902719737029
which is a factor of a(53), I discovered that it's a substring of a(103):
93118232931779128686097911301602920314054090271973702981658870701 ^^^^^^^^^^^^^^^^^^^ R.
participants (5)
-
Alexander Povolotsky -
Artur -
Fred W. Helenius -
Hugo van der Sanden -
Richard Guy