OFFSET
0,2
LINKS
Robert Israel, Table of n, a(n) for n = 0..960
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)
EXAMPLE
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 + ...
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 *)
PROG
(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), ", "))
(Magma) m:=30; R<x>:=PowerSeriesRing(Rationals(), m); Coefficients(R!( 1/Sqrt(1 - 10*x - 11*x^2) )); // G. C. Greubel, Dec 09 2018
(Sage) s=(1/sqrt(1 - 10*x - 11*x^2)).series(x, 30); s.coefficients(x, sparse=False) # G. C. Greubel, Dec 09 2018
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Dec 09 2018
STATUS
approved