A318108 a(n) = Sum_{k=0..n} (3*n-2*k)!/((n-k)!^3*k!)*(-3)^k, n >= 0.
1, 3, 27, 303, 3771, 49653, 677979, 9496791, 135572859, 1963940073, 28783474677, 425872190241, 6350923156059, 95341185353781, 1439433069482547, 21839152342265703, 332769145298428539, 5089688869615075521, 78108038975852093889, 1202268428203687094493, 18555675891246972931221
Offset: 0
Examples
A(x) = 1 + 3*x + 27*x^2 + 303*x^3 + 3771*x^4 + 49653*x^5 + 677979*x^6 + ...
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..829 (terms 0..100 from Gheorghe Coserea)
Programs
-
Mathematica
Table[Sum[(3n-2k)!/(((n-k)!)^3 k!) (-3)^k,{k,0,n}],{n,0,20}] (* Harvey P. Dale, Mar 01 2019 *) -
PARI
a(n) = sum(k=0, n, (3*n-2*k)!/((n-k)!^3*k!)*(-3)^k); vector(21, n, a(n-1))
Formula
G.f. y=A(x) satisfies: 0 = x*(6*x - 1)*(27*x^3 + 27*x^2 - 18*x + 1)*y'' + (486*x^4 + 216*x^3 - 189*x^2 + 36*x - 1)*y' + 3*(3*x + 1)*(18*x^2 - 6*x + 1)*y.
Recurrence: n^2*(3*n - 4)*a(n) = 3*(3*n - 2)*(6*n^2 - 10*n + 3)*a(n-1) - 9*(9*n^3 - 30*n^2 + 29*n - 6)*a(n-2) - 27*(n-2)^2*(3*n - 1)*a(n-3). - Vaclav Kotesovec, Mar 01 2019
From Peter Bala, Mar 16 2023: (Start)
a(n) = Sum_{k = 0..n} (-3)^(n-k)*binomial(n,k)*binomial(n + 2*k,n)* binomial(2*k,k).
Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for positive integers n and r and all primes p >= 5. (End)
G.f.: hypergeom([1/3, 2/3],[1],27*x/(1+3*x)^3)/(1+3*x). - Mark van Hoeij, Nov 28 2024
Comments