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

%I A333562 #13 Oct 06 2021 14:24:32
%S A333562 1,15,769,47103,3080193,208470015,14413725697,1011196362751,
%T A333562 71695889072129,5124481173422079,368599603785760769,
%U A333562 26648859989512290303,1934777421539431153665,140966705275001764839423,10301634747725237826093057,754776795329691207916847103
%N A333562 a(n) = Sum_{j = 0..3*n} binomial(n+j-1,j)*2^j.
%C A333562 Column 3 of the square array A333560. Compare with A119259(n) = Sum_{j = 0..n} binomial(n+j-1,j)*2^j.
%C A333562 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.
%F A333562 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.
%F A333562 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.
%F A333562 a(n) ~ 2^(11*n + 3/2) / (5*sqrt(Pi*n) * 3^(3*n + 1/2)). - _Vaclav Kotesovec_, Mar 28 2020
%e A333562 Examples of congruences:
%e A333562 a(11) - a(1) = 26648859989512290303 - 15 = (2^4)*3*(11^3)*417118394526551 == 0 ( mod 11^3 ).
%e A333562 a(3*7) - a(3) = 121414496850169263529624169428526563327 - 47103 = (2^11)*(7^4)*24691554473186884926207539141513 == 0 ( mod 7^3 ).
%e A333562 a(5^2) - a(5) = 3682696038139661781421472944275523824848470015 - 208470015 = (2^16)*(5^7)*71*1315737187*37481160881*205425986821331 == 0 ( mod 5^6 ).
%p A333562 seq(add( binomial(n+j-1,j)*2^j, j = 0..3*n), n = 0..25);
%t A333562 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 *)
%o A333562 (PARI) a(n) = sum(j = 0, 3*n, binomial(n+j-1,j)*2^j); \\ _Michel Marcus_, Mar 28 2020
%Y A333562 Cf. A002293, A062752, A119259, A333560, A333561.
%K A333562 nonn,easy
%O A333562 0,2
%A A333562 _Peter Bala_, Mar 27 2020