login
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)^2*A108625(n, k).
1

%I #11 Sep 29 2024 09:19:17

%S 1,11,559,42923,3996751,416148761,46607742091,5494953548459,

%T 673006663951183,84887862132737261,10960384987436995309,

%U 1442356991928315362009,192832695330486199076011,26126080107002322789528197,3580244300364122205875141059,495479786625880934937914634923,69162655828624756613259127413583

%N a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)^2*A108625(n, k).

%C Cf. A005258(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)^2.

%C The sequence of Apéry numbers A005258 satisfies the pair of supercongruences

%C 1) A005258(n*p^r) == A005258(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r

%C and

%C 2) A005258(n*p^r - 1) == A005258(n*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r.

%C We conjecture that the present sequence satisfies the same pair of supercongruences. Some examples are given below.

%e Examples of supercongruences:

%e (1a) a(13) - a(1) = 26126080107002322789528197 - 11 = 2*3*(7^3)*11*(13^3)*101* 5200975019533651 == 0 (mod 13^3);

%e (1b) a(13 - 1) - a(1 - 1) = 192832695330486199076011 - 1 = 2*3*5*(13^3)*107*457* 1366649*43779661 == 0 (mod 13^3).

%e (2a) a(5^2) - a(5) = 1806832316206138774124773283611717478292644188023761 - 416148761 = (2^3)*(3^3)*(5^8)*953*22470418607359387809254496551076506667197 = 0 (mod 5^8);

%e (2b) a(5^2 - 1) - a(5 - 1) = 12360782465873293068286987426696670402529985934251 - 3996751 = (2^2)*3*(5^6)*11779*1268093*9320371*634752259*746015256807592939 == 0 (mod 5^6).

%p A108625(n, k) := add(binomial(n, i)^2 * binomial(n+k-i, k-i), i = 0..k):

%p a(n) := add((-1)^(n+k)*binomial(n, k)*binomial(n+k, k)^2*A108625(n, k), k = 0..n):

%p seq(a(n), n = 0..25);

%Y Cf. A005258, A376458 - A376466.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Sep 24 2024