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Re: [math-fun] proportion of bernoulli numbers that are = 1/6.
by N. J. A. Sloane 13 Feb '07

13 Feb '07

Simon said: Subject : The proportion of Bernoulli numbers that are = to 1/6. I made a calculation for the numbers that are = 1/6, I went up to B(5285000). Me: did you mean those that have DENOMINATOR 6 ? If so, see A051222 in the OEIS, which has references to the work of H. Cohen and G. Tenenbaum. Neil

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[math-fun] Comment re Allan Weshler's 1996 posting to the math-fun mailing list.
by RAMSEYKk2＠aol.com 12 Feb '07

12 Feb '07

Sloane's ONline encyclopedia of integer sequences" lists Allan Weshler's J-Determinant sequence as sequence A022344. Until I rediscovered this sequence and added my own comments no information on this sequence was given except a reference to "Posting to math-fun mailing list dated December 4, 1996". Was this your mailing list? If so members may be interested in my following conjectures: Background: Every integer based Fibonacci sequence ultimately appears as a row, n, of Wtthoffs array (Sloane's sequence no A035513) and can be characterized by their J-Determinant as given by Allan Weshler's J-Determinant sequence (A022344 where a(n) = (the absolute value of F(n,i-1)*F(n,i+1) - F(n,i)^2, which is invariant with i varying. Also, in 1961 Horadam, Amer. Math. Monthly 68(1961) pp 751-753) showed that for any Fibonacci sequence, F, that the triple (2F(n+1)*F(n+2), F(n)*F(n+3),2F(n+1)*F(n+2) + F(n)^2) is a Pythagorean number triple (a,b,c) for all n, i.e. a^2 + b^2 = c^2. Conjecture 1: Let any two integer based Fibonacci sequences, F1 and F2, be sequences appearing in rows "n" and "m" respectively and thus have characteristic values, a(n) and a(m) respectively. Then there is a third Fibonacci sequence that is defined by F3(i) = F1(i)*F2(j+1) - F1(i+1)*F2(j) where j is a constant. This third Fibonacci sequence appears in a row, p, of Wythoffs array and has the characteristic value a(p) = a(n)*a(m). Conjecture 2: For every Fibonacci sequence having coprime adjacent terms and appearing in a row (n) of Wythoff's array, the "c" values of Horadam's Pythagorean triples, 2F(n+1)*F(n+2) -F(n)^2), where the c values are reduced by dividing through by the greatest common divisor, form the bijection of a Fibonacci sequence having the characteristic value of (a(n))^2. Anyone interested in this may refer to my further comments to sequences A035513, A022344 in Sloane's online encyclopedia and my own sequence (Sloane's A127561) which is a tabulation of terms, a(n) from A022344, as a table where the row and column positions dictate the starting values of a Fibonacci sequence having the characteristic value a(n), and which has very interesting prime modulus properties.

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[math-fun] proportion of bernoulli numbers that are = 1/6.
by Simon Plouffe 12 Feb '07

12 Feb '07

Hello everybody, Subject : The proportion of Bernoulli numbers that are = to 1/6. I made a calculation for the numbers that are = 1/6, I went up to B(5285000). - The proportion is NOT 1/6, after B(5285000) the ratio is .152616880782547821315725 and still slowly decreasing. It is hard to tell where it goes, 0.15 ? 1/7 ? I don't know, your guess is certainly as good as mine. any idea ? I believe somebody did such a calculation in the past, Wagstaff ? SimonPlouffe

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[math-fun] more pi methods
by R. William Gosper 10 Feb '07

10 Feb '07

(Thought funsters might enjoy eavesdropping.) Jason, I'm still occupied by your pi question. I believe we've already rejected montecarloing or integrating Buffon's needle. How about integrating the area under sqrt(1-x^2)? Besides the numerous numerical methods, one can expand via the binomial thm and termwise integrate. Expanding at x=0 gives the slow series i 1/2 inf (- 1) ( ) ==== i %pi \ --- = > ------------, 4 / 2 i + 1 ==== i = 0 but expanding at x=1 gives one bit/term: i 1/2 inf (- 1) ( ) ==== i %pi \ --- = > ------------------. 4 / i - 3/2 ==== (2 i + 3) 2 i = 0 Better yet, only integrate from sqrt(3)/2 to 1: inf ==== i + 3/2 \ (2 - sqrt(3)) 1/2 3 sqrt(3) %pi = 12 > -------------------- ( ) + --------- / i i 2 === (2 i + 3) (- 4) i = 0 gives > 1 decimal digit/term (ratio = .06699). Expansion at points other than 0 and 1 produces a powerseries whose coefficients are themselves sums. But these obey a three term recurrence, so the double sum reduces to a 4x4 matrix product. Strangely, integrate(expansion at 1/sqrt(2),x,0,1) produces a divergent series. How about inf ==== 2 \ 1 %pi > -- = ---- ? / 2 6 ==== n n = 1 (I have an Euleroid derivation of this in the form of an annotated Macsyma notebook, starting with the atan x series.) Here's a neat one: the average value of gcd([k sqrt(2)],[k sqrt(3)])^-2 should approach pi^2/15, for large intervals of k. ([]:=floor) Too bad you don't like logs. I have *really* magic ways to get pi/ln 2 and (pi/ln 2)^2, the latter using a random bit stream. Heck, pi is basically log(-1). Ah, how about (d5) t(n) := if n = 0 then .2d1 else (t(n - 1), 2 %% ---------------------------------) %% (--, sqrt(%% + 1) + sqrt(1 - %%)) n 2 (c6) t(24) (d6) 3.14159265358979d0 or (d7) s(n) := if n = 0 then .4d1 else (s(n - 1), 2 %% --------------------) 2 %% 1 + sqrt(1 + ------) n + 1 4 (c8) s(25) (d8) 3.14159265358979d0 ? --rwg POINT SOURCE PROSECUTION

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[math-fun] Exact formula for this series?
by Mike Stay 09 Feb '07

09 Feb '07

0.38490017945975 0.29101243968677 0.21666469449485 0.15928582101343 0.11593604442994 0.08373599185262 0.06012913678590 0.04299237598573 This is the magnitude of the nonzero entries in log(G_n)/(i pi), where G_n is the unitary matrix implementing Grover's algorithm for an n-qubit quantum computer. More specifically, T_1 = |1 1| / sqrt(2) |1 -1| T_n = T_1 tensor T_{n-1} S_{n,k} = the 2^n x 2^n identity matrix except entry (k,k)=-1 Grover's algorithm is then G_n = T_n S_{n,1} T_n S_{n,m} where m is the marked state. They decrease roughly like sqrt(2) at each step, but not exactly. Does anyone know how I could figure out a closed form for the terms, or a reference if someone else already has? -- Mike Stay metaweta(a)gmail.com http://math.ucr.edu/~mike

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[math-fun] New proof of Euclid's Theorem
by Paul R. Pudaite 09 Feb '07

09 Feb '07

In the December 2006 issue of Am. Math. Monthly , Filip Saidak presents "A New Proof of Euclid's Theorem." Here's a concise rendition: Let f(x) = x^2 + x, let g[0] be a positive integer, and for n = 1, 2, 3, ..., let g[n] = f(g[n-1]). Then g[n] has at least n different prime factors. I omit the proof for those who might enjoy working it out for themselves. The case of g[0] = 1 seems somewhat canonical. I'm curious about which primes appear as factors in the sequence of g[n] (2, 6, 42, 1807, ...). This sequence begins: 2, 3, 7, 13, 43, 73, 139, 181, 547, 607, 1033, 1171, ... Paul

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[math-fun] Inequality puzzle
by Daniel Asimov 08 Feb '07

08 Feb '07

Given fixed real numbers c_1 < c_2 < . . . < c_n, define f(sigma) = Sum_{1 <= j <= n} c_j c_sigma(j) for any permutation sigma of {1,...,n}. Prove that sigma = id_{1,...n} is the unique permutation that maximizes f(sigma). --Dan

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[math-fun] # of non-isomorphic polyomino graphs
by Thane Plambeck 06 Feb '07

06 Feb '07

I'm not finding the following sequence in the OEIS, probably because I'm blind to a mistake I've made in counting them. Consider the set of graphs G(n) constructed in the following way. Take the set of polyominos with n cells. For each polyomino P in turn, put a vertex at the center of each cell, and draw edges between two vertices if they share an edge in the polyomino, obtaining a plain old unidirected graph G with n vertices. I'm interested in the # of nonisomorphic graphs that are obtained in this way, for each n. Starting at n=1, I get the sequence 1,1,1,3,4,10 and hits in the OEIS that don't look like they fit for larger n -- Thane Plambeck tplambeck(a)gmail.com http://www.plambeck.org/ehome.htm

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[math-fun] Groups with exactly two conjugacy classes
by Eugene Salamin 05 Feb '07

05 Feb '07

Can some give an example of a group having exactly two conjugacy classes, other than the group with two elements? Necessarily, the non-identity class consists of elements of infinite order. Gene ____________________________________________________________________________________ Want to start your own business? Learn how on Yahoo! Small Business. http://smallbusiness.yahoo.com/r-index

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[math-fun] predict Pochhammer untuplication?
by R. William Gosper 02 Feb '07

02 Feb '07

Define oo /===\ i | | 1 - a q (a; q) := | | ------------, x | | x + i i = 0 1 - a q a product of |x| binomials when x is an integer. By Mathematica, 2 3 3 3 (a; q) = (a q ; q ) (a; q ) (a q; q ) 3 n n - 1/3 n n + 1/3 3 5 4 5 5 (a; q) = (a q ; q ) (a q ; q ) (a; q ) 5 n n - 2/5 n - 1/5 n 5 2 5 (a q; q ) (a q ; q ) n + 1/5 n + 2/5 4 5 2 5 5 3 5 = (a q ; q ) (a q ; q ) (a; q ) (a q ; q ) n - 2/5 n - 1/5 n n + 1/5 5 (a q; q ) n + 2/5 4 5 3 5 5 2 5 = (a q ; q ) (a q ; q ) (a; q ) (a q ; q ) n - 3/5 n - 1/5 n n + 1/5 5 (a q; q ) n + 3/5 What's likely to be the case for 7? 2, 4, and 6? --rwg BRAINSTEM TRIBESMAN b jdb(a)math.arizona.edu aka(a)pu.ac.in hic(a)iname.com aadler(a)sbcglobal.net misha(a)nasledov.com aaronl(a)vitelus.com srinivasamu rthyb(a)yahoo.com rjpetti(a)rcn.com jeffrey.golden(a)verizon.net fateman(a)cs.berkeley.edu djeffrey(a)uwo.ca Rob.Corless(a)uwo.ca ch ongo(a)mellis.com mikeh(a)maths.unsw.edu.au AnneStrass(a)aol.com cshapiro(a)panix.com whitfield.diffie(a)Sun.COM gregaleg909@yahoo .com ono(a)math.wisc.edu tjhuber(a)math.uiuc.edu ruiming_zhang(a)hotmail.com b elh(a)osots.com rudy(a)rudyrucker.com silver(a)pilgrim.com sk(a)pilgrim.com peter7387(a)aol.com rwellessprint(a)earthlink.net b danny(a)appliedminds.com meryt(a)worldnet.att.net djeffrey(a)uwo.ca Rob.Corless(a)uwo.ca smj(a)cirr.com thomasmrodgers(a)yahoo.com mtrott(a)wolfram.com oleg(a)wolfram.com pavlyk(a)wolfram.com countolaf148(a)hotmail.com

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