cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A242032 A sequence related to lower bounds for the number of distinct differentiable structures on spheres of the form S^(4*k-1).

Original entry on oeis.org

2, 2, 7, 31, 127, 73, 1414477, 8191, 16931177, 5749691557, 91546277357, 3324754717, 1982765468311237, 22076500342261, 65053034220152267, 925118910976041358111, 16555640865486520478399, 8089941578146657681, 29167285342563717499865628061
Offset: 0

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Author

Peter Luschny, Aug 12 2014

Keywords

Comments

Definition: Divide the prime factorization of an integer M into two parts: L(k, M) = product{p^v(M) the highest power of a prime p which divides M and p < k} and H(k, M) = M / L(k, M). Then a(n) = H(2*floor((n+1)/2), A246053(n)).
For some n the a(n) are lower bounds for the number of distinct differentiable structures on spheres. Compare theorem 2 of the Milnor 1959 paper which asserts that a(2) and a(4) through a(8) are lower bounds for the spheres S^(4*k-1) for k = 2 and 4,..,8.
Let b(n) = numerator(B(2*n)/(2*n)!*(4^n-2)*(-1)^(n-1)), B(n) Bernoulli number. Apart from n=0 and n=1 the first n such that a(n) != b(n) is n = 1437. Thus in the range [2..1436] the a(n) are the numerators in the Taylor series for x*cosec(x), A036280.

Examples

			a( 0) = 2
a( 1) = 2
a( 2) = 7
a( 3) = 31
a( 4) = 127
a( 5) = 73
a( 6) = 23 * 89 * 691
a( 7) = 8191
a( 8) = 31 * 151 * 3617
a( 9) = 43867 * 131071
a(10) = 283 * 617 * 524287
a(11) = 127 * 131 * 337 * 593
a(12) = 47 * 103 * 178481 * 2294797
a(13) = 31 * 601 * 1801 * 657931
		

Crossrefs

Programs

  • Mathematica
    h[x_] := Zeta[2x] (4^x-2);
    a[n_] := Module[{M, k, p}, M = Denominator[h[Quotient[n+1, 2]] h[Quotient[ n, 2]]/h[n]]; k = 2 Quotient[n+1, 2]; p = 2; While[p < k, While[ Divisible[M, p], M = M/p]; p = NextPrime[p]]; M];
    Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jul 12 2019, from Sage code *)
  • Sage
    def A242032(n):
        h = lambda x: zeta(2*x)*(4^x-2)
        M = Integer((h((n+1)//2)*h(n//2)/h(n)).denominator())
        k = 2*((n+1)//2)
        P = Primes()
        p = P.first()
        while p < k:
            while p.divides(M):
                M /= p
            p = P.next(p)
        return M
    [A242032(n) for n in (0..30)]