A211601 -id:A211601 - cp's OEIS frontend

A211600 a(n) = (binomial(p^n, p^(n-1)) - binomial(p^(n-1), p^(n-2))) / p^(3n-3) for p = 2.

1, 25, 146745, 55927250376633, 91366371314728099305354933301689, 2750710880016902131123422793322699970110063817946068739768171777481145

Offset: 3

Alexander Adamchuk, Apr 16 2012

Consider the difference between two binomials f(p,k) = binomial(p^k, p^(k-1)) - binomial(p^(k-1), p^(k-2)).
A theorem from the A. I. Shirshov paper (in Russian) states:
p^(3k - 3) divides f(p,k) for prime p = 2 and k > 2.
p^(3k - 2) divides f(p,k) for prime p = 3 and k > 1.
p^(3k - 1) divides f(p,k) for prime p > 3 and k > 1.

			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.

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

D. B. Fuks and M. B. Fuks, Arithmetics of binomial coefficients, Kvant 6 (1970), 17-25. (in Russian)
A. I. Shirshov, On one property of binomial coefficients, Kvant 10 (1971), 16-20. (in Russian)

Cf. A037293, A211601, A211602.

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

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

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

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

cp's OEIS Frontend

A211600 a(n) = (binomial(p^n, p^(n-1)) - binomial(p^(n-1), p^(n-2))) / p^(3n-3) for p = 2.

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