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A217280
a(n) = Sum_{k=0..n} binomial(n,k)^3 * 2^k * 3^(n-k).
1
1, 5, 61, 845, 12865, 206525, 3434005, 58542125, 1016757025, 17915319125, 319316172781, 5744962350605, 104168557434241, 1901265562745645, 34897397882019685, 643662507367299725, 11922610595971162945, 221673143774064873125, 4135241181985049252125
OFFSET
0,2
COMMENTS
5 divides a(n) except when a(n) == 1 (mod 5), which occurs only when the base 5 representation of n consists solely of digits '0' and '2' (conjecture). Positions n at which a(n) == 1 (mod 5) begin: [0, 2, 10,12, 50,52,60,62, 250,252,260,262,300,302,310,312, 1250,1252,1260,1262,1300,1302,1310,1312,1500,1502,1510,1512, 1550,1552,1560,1562, ...].
FORMULA
a(n) ~ 3^n*(1+(2/3)^(1/3))^(3*n+2)/(2^(4/3)*3^(1/6)*Pi*n)*(1+((2/3)^(1/3)+(3/2)^(1/3)-5)/(9*n)). - Vaclav Kotesovec, Sep 30 2012
G.f.: hypergeom([1/3, 2/3],[1],162*x^2/(1-5*x)^3)/(1-5*x). - Mark van Hoeij, Apr 11 2014
D-finite with recurrence 2*(n^2)*a(n) -5*n*(5*n-3)*a(n-1) +3*(-83*n^2+216*n-146)*a(n-2) +5*(-137*n^2+585*n-633)*a(n-3) -625*(n-3)^2*a(n-4)=0. - R. J. Mathar, Jul 27 2022
MATHEMATICA
Table[Sum[Binomial[n, k]^3*2^k*3^(n-k), {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Sep 30 2012 *)
PROG
(PARI) {a(n)=sum(k=0, n, binomial(n, k)^3*2^k*3^(n-k))}
for(n=0, 21, print1(a(n), ", "))
CROSSREFS
Sequence in context: A189444 A373106 A217811 * A236955 A064328 A363867
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Sep 29 2012
STATUS
approved