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 A333592 #31 Nov 04 2024 22:22:33
%S A333592 1,2,14,146,1742,22252,296438,4063866,56884430,808970960,11649069764,
%T A333592 169444272692,2485268015414,36707034407396,545386280953262,
%U A333592 8144809577111146,122177689609022670,1839933272106181720,27804610617723365072,421476329309967621504,6406685024966332359492
%N A333592 a(n) = Sum_{k = 0..n} binomial(n+k-1, k)^2.
%C A333592 Compare with the closed-form evaluation Sum_{k = 0..n} C(n+k-1,k) = C(2*n,n) = A000984(n) (see the first comment in A001700 for a proof).
%C A333592 It is well-known that Sum_{k = 0..n} C(n,k)^2 = C(2*n,n). Here, we consider by analogy Sum_{k = 0..n} C(-n,k)^2, where C(-n,k) = (-1)^k*C(n+k-1,k) for integer n and nonnegative integer k.
%C A333592 The sequence b(n) = C(2*n,n) of central binomial coefficients satisfies the supercongruences b(n*p^k) = b(n*p^(k-1)) ( mod p^(3*k) ) for all primes p >= 5 and positive integers n and k - see Meštrović. We conjecture that the present sequence also satisfies these congruences. Some examples of the congruences are given below.
%C A333592 More generally, calculation suggests that for positive integer A and integer B, the sequence a(A,B;n) := Sum_{k = 0..A*n} C(B*n+k-1,k)^2 may satisfy the same congruences.
%C A333592 The sequence (a(p) - 2)/(2*p^3) for prime p >= 5 begins [89, 5924, 63652995, 8353899501, 187251503369243, 30724327840061789, 937835335872800013431, ...]. Cf. A034602.
%H A333592 Seiichi Manyama, <a href="/A333592/b333592.txt">Table of n, a(n) for n = 0..500</a>
%H A333592 Romeo Meštrović, <a href="http://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011)</a>, arXiv:1111.3057 [math.NT], 2011.
%F A333592 a(n) ~ 2^(4*n) / (3*Pi*n). - _Vaclav Kotesovec_, Mar 28 2020
%F A333592 For n >= 1, a(n) = 2 * Sum_{k = 0..n-1} binomial(n+k, k)*binomial(n+k-1, k) = 2 * Sum_{k = 0..n-1} (n + k)/n *binomial(n+k-1, k)^2. - _Peter Bala_, Nov 02 2024
%e A333592 Examples of supercongruences:
%e A333592 a(11) - a(1) = 169444272692 - 2 = 2*(3^2)*5*7*(11^3)*397*509 == 0 ( mod 11^3 ).
%e A333592 a(2*7) - a(2) = 545386280953262 - 14 = (2^5)*(3^2)*(7^4)*788714021 == 0 ( mod 7^3 ).
%e A333592 a(5^2) - a(5) = 5375188503768783714940459752 - 22252 = (2^2)*(5^6)*(31^2)* 89493252924350197127 == 0 ( mod 5^6 ).
%p A333592 seq( add( binomial(n+k-1,k)^2, k = 0..n ), n = 0..25);
%t A333592 Table[Binomial[2*n-1, n]^2 * HypergeometricPFQ[{1, -n, -n}, {1 - 2*n, 1 - 2*n}, 1], {n, 1, 20}] (* _Vaclav Kotesovec_, Mar 28 2020 *)
%o A333592 (PARI) a(n) = sum(k=0, n, binomial(n+k-1, k)^2); \\ _Michel Marcus_, Mar 29 2020.
%o A333592 (Python)
%o A333592 from math import comb
%o A333592 def A333592(n): return sum(comb(n+k-1,k)**2 for k in range(n+1)) if n else 1 # _Chai Wah Wu_, Oct 28 2022
%Y A333592 Cf. A000984, A034602, A333593.
%K A333592 nonn,easy
%O A333592 0,2
%A A333592 _Peter Bala_, Mar 27 2020