This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A333561 #20 Mar 07 2022 03:49:50
%S A333561 1,7,129,2815,65537,1579007,38862849,970522623,24494735361,
%T A333561 623210135551,15956734640129,410649406472191,10612705274626049,
%U A333561 275241225206890495,7159857331658817537,186731505521384226815,4880983719142471237633,127836403093194475044863
%N A333561 a(n) = Sum_{j = 0..2*n} binomial(n+j-1,j)*2^j.
%C A333561 Column 2 of the square array A333560. Compare with A119259(n) = Sum_{j = 0..n} binomial(n+j-1,j)*2^j.
%C A333561 We conjecture that this sequence satisfies the supercongruences 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 A333561 Conjectural o.g.f.: 1/(1 + x) + 8*x*f'(4*x)/(2*f(4*x) - 1), where f(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + ... is the o.g.f. of A001764.
%F A333561 exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 7*x + 89*x^2 + 1447*x^3 + ... appears to be the o.g.f. of A062747.
%F A333561 Conjectural recurrence: n*(n - 1)*(2*n - 1)*(3098*n - 6455)*a(n) = (n - 1)*(172988*n^3 - 585840*n^2 + 550321*n - 169824)*a(n-1) - 12*(11825*n^4 - 168518*n^3 + 627675*n^2 - 853766*n + 350744)*a(n-2) - 36*(n - 3)*(3*n - 7)*(3*n - 8)*(991*n - 724)*a(n-3) with a(1) = 7, a(2) = 129, a(3) = 2815.
%F A333561 From _Vaclav Kotesovec_, Mar 28 2020: (Start)
%F A333561 a(n) ~ 3^(3*n + 1/2) / (4*sqrt(Pi*n)).
%F A333561 Recurrence: n*(2*n - 1)*(7*n^2 - 20*n + 14)*a(n) = (364*n^4 - 1411*n^3 + 1818*n^2 - 868*n + 120)*a(n-1) + 6*(3*n - 5)*(3*n - 4)*(7*n^2 - 6*n + 1)*a(n-2). (End)
%F A333561 From _Peter Bala_, Mar 05 2022: (Start)
%F A333561 a(n) = Sum_{k = 0..2*n} binomial(3*n, 2*n-k)*binomial(n+k-1,k).
%F A333561 a(n) = [x^(2*n)] ( (1 + x^3)/(1 - x) )^n.
%F A333561 The o.g.f. satisfies the algebraic equation (108*x^3 + 212*x^2 + 100*x - 4)*A(x)^3 - (216*x^2 + 208*x - 8)*A(x)^2 + (48*x^2 + 155*x - 5)*A(x) + 8*x^2 - 40*x + 1 = 0. (End)
%F A333561 a(n) = binomial(3*n, 2*n)*hypergeom([-2*n, n], [n + 1], -1). - _Peter Luschny_, Mar 07 2022
%e A333561 Examples of supercongruences:
%e A333561 a(11) - a(1) = 410649406472191 - 7 = (2^3)*3*(11^3)*12855290711 == 0 ( mod 11^3 ).
%e A333561 a(3*7) - a(3) = 61103847305642669128888090623 - 2815 = (2^8)*(7^5)* 87326419*162627033103121 == 0 ( mod 7^3 ).
%e A333561 a(5^2) - a(5) = 29754989698128108780761000609579007 - 1579007 = (2^11)*(5^6)*179*751*10267*673710468794491483 == 0 ( mod 5^6 ).
%p A333561 seq(add( binomial(n+j-1,j)*2^j, j = 0..2*n), n = 0..25);
%t A333561 Table[(-1)^n - 2^(2*n+1) * Binomial[3*n, 2*n+1] * Hypergeometric2F1[1, 3*n+1, 2*n+2, 2], {n, 0, 20}] (* _Vaclav Kotesovec_, Mar 28 2020 *)
%o A333561 (PARI) a(n) = sum(j = 0, 2*n, binomial(n+j-1,j)*2^j); \\ _Michel Marcus_, Mar 28 2020
%Y A333561 Cf. A001764, A062747, A119259, A333560, A333562.
%K A333561 nonn,easy
%O A333561 0,2
%A A333561 _Peter Bala_, Mar 27 2020