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Expansion of g.f.: 1/sqrt(1 - 10*x - 11*x^2).
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%I #27 Oct 16 2024 09:22:52

%S 1,5,43,395,3811,37775,381205,3895925,40193395,417697775,4366043473,

%T 45852847265,483447391309,5114115365585,54252753665083,

%U 576948203182475,6148667240501395,65651351673108575,702154850931542305,7520927108084780225,80666557496061224281,866249916689104887005,9312623039533986068863,100216202771039576006495,1079454220008183284872861

%N Expansion of g.f.: 1/sqrt(1 - 10*x - 11*x^2).

%H Robert Israel, <a href="/A322246/b322246.txt">Table of n, a(n) for n = 0..960</a>

%F a(n) = Sum_{k=0..n} 11^(n-k) * (-3)^k * binomial(n,k)*binomial(2*k,k).

%F a(n) = Sum_{k=0..n} (-1)^(n-k) * 3^k * binomial(n,k)*binomial(2*k,k).

%F a(n) equals the (central) coefficient of x^n in (1 + 5*x + 9*x^2)^n.

%F 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

%F a(n) ~ 11^(n + 1/2) / (2*sqrt(3*Pi*n)). - _Vaclav Kotesovec_, Dec 13 2018

%F E.g.f.: exp(5*x) * BesselI(0,6*x). - _Ilya Gutkovskiy_, Feb 02 2021

%F a(n) = 11^n*2F1([1/2, -n], [1], 12/11), where 2F1 is the hypergeometric function. - _Stefano Spezia_, Feb 02 2021

%F From _Peter Bala_, Oct 13 2024: (Start)

%F 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.

%F Inverse binomial transform of A098658.

%F 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)

%e 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 + ...

%e such that A(x)^2 = 1/(1 - 10*x - 11*x^2).

%e RELATED SERIES.

%e 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 + ...

%p 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):

%p map(f, [$0..40]); # _Robert Israel_, Dec 09 2018

%t CoefficientList[Series[1/Sqrt[1 - 10*x - 11*x^2], {x,0,30}], x] (* _G. C. Greubel_, Dec 09 2018 *)

%o (PARI) /* Using generating function: */

%o {a(n) = polcoeff( 1/sqrt(1 - 10*x - 11*x^2 +x*O(x^n)),n)}

%o for(n=0,30,print1(a(n),", "))

%o (PARI) /* Using binomial formula: */

%o {a(n) = sum(k=0,n, (-1)^(n-k)*3^k*binomial(n,k)*binomial(2*k,k))}

%o for(n=0,30,print1(a(n),", "))

%o (PARI) /* Using binomial formula: */

%o {a(n) = sum(k=0,n, 11^(n-k)*(-3)^k*binomial(n,k)*binomial(2*k,k))}

%o for(n=0,30,print1(a(n),", "))

%o (PARI) /* a(n) is central coefficient in (1 + 5*x + 9*x^2)^n */

%o {a(n) = polcoeff( (1 + 5*x + 9*x^2 +x*O(x^n))^n, n)}

%o for(n=0,30,print1(a(n),", "))

%o (Magma) m:=30; R<x>:=PowerSeriesRing(Rationals(), m); Coefficients(R!( 1/Sqrt(1 - 10*x - 11*x^2) )); // _G. C. Greubel_, Dec 09 2018

%o (Sage) s=(1/sqrt(1 - 10*x - 11*x^2)).series(x, 30); s.coefficients(x, sparse=False) # _G. C. Greubel_, Dec 09 2018

%o (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

%Y Cf. A098658, A322247.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Dec 09 2018