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a(n) = (binomial(p^n, p^(n-1)) - binomial(p^(n-1), p^(n-2))) / p^(3n-2) for p = 3.
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%I #17 Oct 26 2024 23:00:15

%S 1,2143,39057044954221855,

%T 507249004999029430448035076427591041390649615630234312261967

%N a(n) = (binomial(p^n, p^(n-1)) - binomial(p^(n-1), p^(n-2))) / p^(3n-2) for p = 3.

%C Consider the difference between two binomials f(p,k) = binomial(p^k, p^(k-1)) - binomial(p^(k-1), p^(k-2)).

%C A theorem from the A. I. Shirshov paper (in Russian) states:

%C p^(3k - 3) divides f(p,k) for prime p = 2 and k > 2.

%C p^(3k - 2) divides f(p,k) for prime p = 3 and k > 1.

%C p^(3k - 1) divides f(p,k) for prime p > 3 and k > 1.

%D D. B. Fuks and Serge Tabachnikov, Mathematical Omnibus: Thirty Lectures on Classic Mathematics, American Mathematical Society, 2007. Lecture 2. Arithmetical Properties of Binomial Coefficients, pages 27-44

%H D. B. Fuks and M. B. Fuks, <a href="http://kvant.mccme.ru/1970/06/arifmetika_binomialnyh_koeffic.htm">Arithmetics of binomial coefficients</a>, Kvant 6 (1970), 17-25. (in Russian)

%H A. I. Shirshov, <a href="http://kvant.mccme.ru/1971/10/ob_odnom_svojstve_binomialnyh.htm">On one property of binomial coefficients</a>, Kvant 10 (1971), 16-20. (in Russian)

%F a(n) = (binomial(3^n, 3^(n-1)) - binomial(3^(n-1), 3^(n-2))) / 3^(3*n-2).

%t p = 3; Table[(Binomial[p^n, p^(n - 1)] - Binomial[p^(n - 1), p^(n - 2)]) / 3^(3n - 2), {n, 2, 6}]

%Y Cf. A211600, A211602.

%K nonn,easy

%O 2,2

%A _Alexander Adamchuk_, Apr 16 2012