A333562 a(n) = Sum_{j = 0..3*n} binomial(n+j-1,j)*2^j.

1, 15, 769, 47103, 3080193, 208470015, 14413725697, 1011196362751, 71695889072129, 5124481173422079, 368599603785760769, 26648859989512290303, 1934777421539431153665, 140966705275001764839423, 10301634747725237826093057, 754776795329691207916847103

Offset

0,2

Comments

Column 3 of the square array A333560. Compare with A119259(n) = Sum_{j = 0..n} binomial(n+j-1,j)*2^j.

We conjecture that this sequence satisfies the congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Some examples are given below.

Links

Table of n, a(n) for n=0..15.

Formula

Conjectural o.g.f.: 1/(1 + x) + 16*x*f'(8*x)/(2*f(8*x) - 1), where f(x) = 1 + x + 4*x^2 + 22*x^3 + 140*x^4 + ... is the o.g.f. of A002293.

exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 15*x + 497*x^2 + 22031*x^3 + ... appears to be the o.g.f. of A062752.

a(n) ~ 2^(11*n + 3/2) / (5*sqrt(Pi*n) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020

Example

Examples of congruences:
a(11) - a(1) = 26648859989512290303 - 15 = (2^4)*3*(11^3)*417118394526551 == 0 ( mod 11^3 ).
a(3*7) - a(3) = 121414496850169263529624169428526563327 - 47103 = (2^11)*(7^4)*24691554473186884926207539141513 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 3682696038139661781421472944275523824848470015 - 208470015 = (2^16)*(5^7)*71*1315737187*37481160881*205425986821331 == 0 ( mod 5^6 ).

Maple

seq(add( binomial(n+j-1, j)*2^j, j = 0..3*n), n = 0..25);

Mathematica

Table[(-1)^n - 2^(3*n+1) * Binomial[4*n, 3*n+1] * Hypergeometric2F1[1, 4*n+1, 3*n+2, 2], {n, 0, 15}] (* Vaclav Kotesovec, Mar 28 2020 *)

Prog

(PARI) a(n) = sum(j = 0, 3*n, binomial(n+j-1, j)*2^j); \\ Michel Marcus, Mar 28 2020

Crossrefs

Cf. A002293, A062752, A119259, A333560, A333561.

Sequence in context: A055683 A196465 A177598 * A116094 A267754 A317798

Adjacent sequences: A333559 A333560 A333561 * A333563 A333564 A333565

Keyword

nonn,easy

Author

Peter Bala, Mar 27 2020

Status

approved