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a(n) = 2*binomial(3*n,n) - 9*binomial(2*n,n).
9

%I #26 Nov 07 2022 17:01:03

%S -7,-12,-24,-12,360,3738,28812,201672,1355112,8936070,58427226,

%T 380724552,2479017996,16151245488,105359408760,688338793488,

%U 4504288103784,29521135717470,193771020939510,1273649831269200,8382448392851610,55234026483856110,364347399072847320

%N a(n) = 2*binomial(3*n,n) - 9*binomial(2*n,n).

%C For integers j and k, not necessarily positive, define u(n) = k^2*(k - 1)*binomial(j*n,n) - j^2*(j - 1)*binomial(k*n,n). We conjecture that u(p) == u(1) (mod p^5) for all primes p >= 7. This is essentially the case (j, k) = (3, 2). [Follows from Helou and Terjanian (2008), Section 3, Proposition 2.]

%C Conjecture: for r >= 2, u(p^r) == u(p^(r-1)) ( mod p^(3*r+3) ) for all primes p >= 5. - _Peter Bala_, Oct 13 2022

%H R. R. Aidagulov and M. A. Alekseyev, <a href="https://doi.org/10.1007/s10958-018-3948-0">On p-adic approximation of sums of binomial coefficients</a>, Journal of Mathematical Sciences 233:5 (2018), 626-634; <a href="https://arxiv.org/abs/1602.02632">arXiv:1602.02632 [math.NT]</a>, 2018.

%H C. Helou and G. Terjanian, <a href="https://doi.org/10.1016/j.jnt.2007.06.008">On Wolstenholme’s theorem and its converse</a>, J. Number Theory 128 (2008), 475-499.

%H 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 preprint arXiv:1111.3057 [math.NT], 2011.

%F a(n) = 2*A005809(n) - 9*A000984(n).

%F a(p) == a(1) (mod p^4) for all primes p >= 5 by Meštrović, Section 3, equation 15.

%F Conjecture: the stronger supercongruence a(p) == a(1) (mod p^5) holds for all primes p >= 7.

%F The conjecture is true: apply Helou and Terjanian, Section 3, Proposition 2. - _Peter Bala_, Oct 22 2022

%F The conjecture was proved by Aidagulov and Alekseyev; see the remarks following Corollary 2. - _Peter Bala_, Oct 29 2022

%p seq(2*binomial(3*n,n) - 9*binomial(2*n,n), n = 0..20);

%Y Cf. A000984, A005809, A268589, A357508.

%K sign,easy

%O 0,1

%A _Peter Bala_, Oct 01 2022