A211600 - OEIS

%I #20 Jun 06 2021 09:04:37

%S 1,25,146745,55927250376633,91366371314728099305354933301689,

%T 2750710880016902131123422793322699970110063817946068739768171777481145

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

%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(2^n, 2^(n-1) - binomial(2^(n-1), 2^(n-2))) / 2^(3n-3).

%F a(n) = (A037293(n) - A037293(n-1)) / 2^(3n - 3).

%e a(3) = 1 is the difference between central binomials C(8,4) - C(4,2) = 70 - 6 = 64 divided by 2^(3*2 - 3) = 64.

%p A211600:=n->(binomial(2^n, 2^(n - 1)) - binomial(2^(n - 1), 2^(n - 2))) / 2^(3*n - 3): seq(A211600(n), n=3..9); # _Wesley Ivan Hurt_, Apr 25 2017

%t p = 2; Table[(Binomial[p^n, p^(n - 1)] - Binomial[p^(n - 1), p^(n - 2)]) / 2^(3n - 3), {n, 3, 9}]

%Y Cf. A037293, A211601, A211602.

%K nonn,easy

%O 3,2

%A _Alexander Adamchuk_, Apr 16 2012