A158119 Unsigned bisection of A157308 and A157310.
1, 1, 3, 38, 947, 37394, 2120190, 162980012, 16330173251, 2070201641498, 324240251016266, 61525045423103316, 13913915097436287598, 3698477457114061621492, 1141824214469896983332508
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 3*x^2 + 38*x^3 + 947*x^4 + 37394*x^5 + 2120190*x^6 + 162980012*x^7 + 16330173251*x^8 + 2070201641498*x^9 + 324240251016266*x^10 +... RELATED FUNCTIONS. G.f. of A157308, B(x) = x + A(-x^2), satisfies the condition that both B(x) and F(x) = B(x*F(x)) = o.g.f. of A155585 have zeros for every other coefficient after initial terms: A157308 = [1,1,-1,0,3,0,-38,0,947,0,-37394,0,2120190,0,...]; A155585 = [1,1,0,-2,0,16,0,-272,0,7936,0,-353792,0,...]. ... G.f. of A157310, C(x) = 2+x - A(-x^2), satisfies the condition that both C(x) and G(x) = C(x/G(x)) = o.g.f. of A157309 have zeros for every other coefficient after initial terms: A157310 = [1,1,1,0,-3,0,38,0,-947,0,37394,0,-2120190,0,...]; A157309 = [1,1,0,-1,0,9,0,-176,0,5693,0,-272185,0,...]. ...
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..200 (terms 0..50 from Paul D. Hanna)
- M. Bozejko and T. Hasebe, On free infinite divisibility for classical Meixner distributions, arXiv preprint arXiv:1302.4885 [math.PR], 2013-2014.
Programs
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Mathematica
terms = 30; F[x_] = Sum[n! x^n/Product[(1 + 2 k x), {k, 1, n}], {n, 0, terms+1}] + O[x]^(terms+1); A[x_] = x/InverseSeries[x F[x]]; Partition[CoefficientList[A[x], x][[1 ;; terms]], 2][[All, 1]] // Abs (* Jean-François Alcover, Jul 27 2018 *) -
PARI
{a(n)=local(A=[1, 1]); for(i=1, 2*n, if(#A%2==0, A=concat(A, 0);); if(#A%2==1, A=concat(A, t); A[ #A]=-subst(Vec(x/serreverse(x*Ser(A)))[ #A], t, 0))); (-1)^n*Vec(x/serreverse(x*Ser(A)))[2*n+1]} -
PARI
{a(n) = my(A=[1],CF=1); for(i=1,n, A=concat(A,0); for(i=1,#A, CF = Ser(A) - (#A-i+1)^2*x/CF ); A[#A] = -polcoeff(CF,#A-1) );A[n+1] } for(n=0,20,print1(a(n),", ")) \\ Paul D. Hanna, Nov 04 2020
Formula
G.f. A(x) satisfies: 1 = A(x) - x/(A(x) - 2^2*x/(A(x) - 3^2*x/(A(x) - 4^2*x/(A(x) - 5^2*x/(A(x) - 6^2*x/(A(x) - ...)))))), a continued fraction. - Paul D. Hanna, Nov 04 2020
Conjecture: a(m) == 1 (mod 2) iff m is a power of 2 or m=0. [Paul D. Hanna, Mar 16 2009]
a(n) ~ 2^(4*n + 3) * n^(2*n + 1/2) / (Pi^(2*n + 1/2) * exp(2*n)). - Vaclav Kotesovec, Nov 12 2020