OFFSET
0,2
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..200
FORMULA
a(n) = (n+1)*Sum_{k=0..n} (binomial(k+1, n-k) * binomial(n+2*k+2, n+k+1)) / (n+k+2)).
A(x) = B'(x)/B(x) where B(x) = (-1-sqrt(1-8*x) + sqrt(2+2*sqrt(1-8*x)+8*x)) / (4*x) is the g.f. of A186997.
a(n) ~ 2^(3*n+2)/sqrt(3*Pi*n). - Vaclav Kotesovec, Apr 12 2014
Conjecture: -(3*n-1)*(2*n+1)*(n+1)*a(n) +2*(21*n^3+14*n^2-2*n-1)*a(n-1) +8*n*(3*n+2)*(2*n-1)*a(n-2)=0. - R. J. Mathar, Jun 14 2016
From Peter Bala, Jul 27 2024: (Start)
The following remarks assume an offset of 1.
a(n) = [x^n] c(x + x^2)^n, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108.
It follows that 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.
Calculation suggests that the stronger supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) hold for all primes p >= 3. (End)
MATHEMATICA
CoefficientList[Series[(1/(16*x*(1 + x)*(-1 + 8*x)))*(8 - Sqrt[2]*Sqrt[1 + Sqrt[1 - 8*x] + 4*x] - 3*Sqrt[2 - 16*x]*Sqrt[1 + Sqrt[1 - 8*x] + 4*x] + 8*x*(-7 - 8*x + Sqrt[2]*Sqrt[1 + Sqrt[1 - 8*x] + 4*x])), {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 12 2014 *)
PROG
(Maxima)
a(n):=(n+1)*sum((binomial(k+1, n-k)*binomial(n+2*k+2, n+k+1))/(n+k+2), k, 0, n);
(PARI) for(n=0, 30, print1((n+1)*sum(k=0, n, (binomial(k+1, n-k)*binomial(n +2*k +2, n+k+1))/(n+k+2)), ", ")) \\ G. C. Greubel, Sep 30 2018
(Magma) [(n+1)*(&+[Binomial(k+1, n-k)*Binomial(n+2*k+2, n+k+1)/(n+k+2): k in [0..n]]): n in [0..30]]; // G. C. Greubel, Sep 30 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vladimir Kruchinin, Apr 11 2014
STATUS
approved