A322246 Expansion of g.f. 1/sqrt(1 - 10*x - 11*x^2).
1, 5, 43, 395, 3811, 37775, 381205, 3895925, 40193395, 417697775, 4366043473, 45852847265, 483447391309, 5114115365585, 54252753665083, 576948203182475, 6148667240501395, 65651351673108575, 702154850931542305, 7520927108084780225, 80666557496061224281, 866249916689104887005, 9312623039533986068863, 100216202771039576006495, 1079454220008183284872861
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + 5*x + 43*x^2 + 395*x^3 + 3811*x^4 + 37775*x^5 + 381205*x^6 + 3895925*x^7 + 40193395*x^8 + 417697775*x^9 + 4366043473*x^10 + ...
such that A(x)^2 = 1/(1 - 10*x - 11*x^2).
RELATED SERIES.
exp( Sum_{n>=1} a(n)*x^n/n ) = 1 + 5*x + 34*x^2 + 260*x^3 + 2137*x^4 + 18425*x^5 + 164395*x^6 + 1505075*x^7 + 14058979*x^8 + 133459055*x^9 + 1283753308*x^10 + ...
Links
- Robert Israel, Table of n, a(n) for n = 0..960
Programs
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GAP
List([0..30], n -> Sum([0..n], k-> (-1)^(n-k)*3^k*Binomial(n,k) *Binomial(2*k,k))); # G. C. Greubel, Dec 09 2018
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Magma
m:=30; R
:=PowerSeriesRing(Rationals(), m); Coefficients(R!( 1/Sqrt(1 - 10*x - 11*x^2) )); // G. C. Greubel, Dec 09 2018 -
Maple
f:= gfun:-rectoproc({{(11*n+11)*a(n)+(15+10*n)*a(n+1)+(-n-2)*a(n+2), a(0) = 1, a(1) = 5},a(n),remember): map(f, [$0..40]); # Robert Israel, Dec 09 2018 -
Mathematica
CoefficientList[Series[1/Sqrt[1 - 10*x - 11*x^2], {x,0,30}], x] (* G. C. Greubel, Dec 09 2018 *) -
PARI
/* Using generating function: */ {a(n) = polcoeff( 1/sqrt(1 - 10*x - 11*x^2 +x*O(x^n)),n)} for(n=0,30,print1(a(n),", ")) -
PARI
/* Using binomial formula: */ {a(n) = sum(k=0,n, (-1)^(n-k)*3^k*binomial(n,k)*binomial(2*k,k))} for(n=0,30,print1(a(n),", ")) -
PARI
/* Using binomial formula: */ {a(n) = sum(k=0,n, 11^(n-k)*(-3)^k*binomial(n,k)*binomial(2*k,k))} for(n=0,30,print1(a(n),", ")) -
PARI
/* a(n) is central coefficient in (1 + 5*x + 9*x^2)^n */ {a(n) = polcoeff( (1 + 5*x + 9*x^2 +x*O(x^n))^n, n)} for(n=0,30,print1(a(n),", ")) -
Sage
s=(1/sqrt(1 - 10*x - 11*x^2)).series(x, 30); s.coefficients(x, sparse=False) # G. C. Greubel, Dec 09 2018
Formula
a(n) = Sum_{k=0..n} 11^(n-k) * (-3)^k * binomial(n,k)*binomial(2*k,k).
a(n) = Sum_{k=0..n} (-1)^(n-k) * 3^k * binomial(n,k)*binomial(2*k,k).
a(n) equals the (central) coefficient of x^n in (1 + 5*x + 9*x^2)^n.
D-finite with recurrence: (11*n+11)*a(n)+(15+10*n)*a(n+1)+(-n-2)*a(n+2)=0. - Robert Israel, Dec 09 2018
a(n) ~ 11^(n + 1/2) / (2*sqrt(3*Pi*n)). - Vaclav Kotesovec, Dec 13 2018
E.g.f.: exp(5*x) * BesselI(0,6*x). - Ilya Gutkovskiy, Feb 02 2021
a(n) = 11^n*2F1([1/2, -n], [1], 12/11), where 2F1 is the hypergeometric function. - Stefano Spezia, Feb 02 2021
From Peter Bala, Oct 13 2024: (Start)
a(n) = Integral_{x = -1..11} x^n * w(x) dx, where w(x) = 1/( Pi*sqrt((1 + x)*(11 - x)) ) is positive on the interval (-1, 11). The weight function w(x) is singular at x = -1 and at x = 11 and is the solution of the Hausdorff moment problem.
Inverse binomial transform of A098658.
The Gauss congruences hold: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r. (End)
a(n) = (1/4)^n * Sum_{k=0..n} (-1)^k * 11^(n-k) * binomial(2*k,k) * binomial(2*(n-k),n-k). - Seiichi Manyama, Aug 18 2025
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