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.

A037992 Smallest number with 2^n divisors.

Original entry on oeis.org

1, 2, 6, 24, 120, 840, 7560, 83160, 1081080, 17297280, 294053760, 5587021440, 128501493120, 3212537328000, 93163582512000, 2888071057872000, 106858629141264000, 4381203794791824000, 188391763176048432000, 8854412869274276304000, 433866230594439538896000
Offset: 0

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Comments

Positions where the number of infinitary divisors of n (A037445), increases to a record (cf. A002182), or infinitary analog of highly composite numbers (A002182). - Vladimir Shevelev, May 13-22 2016
Infinitary superabundant numbers: numbers m with record values of the infinitary abundancy index, A049417(m)/m > A049417(k)/k for all k < m. - Amiram Eldar, Sep 20 2019

Crossrefs

Programs

  • Haskell
    a037992 n = head [x | x <- [1..], a000005 x == 2 ^ n]
    -- Reinhard Zumkeller, Apr 08 2015
    
  • Mathematica
    a[0] = 1; a[n_] := a[n] = Catch[ For[ k = 2, True, k++, If[ an = k*a[n-1]; DivisorSigma[0, an] == 2^n, Throw[an]]]]; Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Apr 16 2012 *)
  • PARI
    {a(n)= local(A,m,c,k,p); if(n<1, n==0, c=0; A=1; m=1; while( cMichael Somos, Apr 15 2005 */
    
  • Python
    def a(n):
      product = 1
      k = 1
      for i in range(n+1):
        product *= k   # k=A050376(i), for i>=1
        while product % k == 0:
          k += 1
      return product
    # Jason L. Miller, Mar 20 2024

Formula

A000005(a(n)) = A000079(n).
a(n) = Product_{k=1..n} A050376(k), product of the first n terms of A050376. - Lekraj Beedassy, Jun 30 2004
a(n) = A052330(2^n -1). - Thomas Ordowski, Jun 29 2005
A001221(a(n+1)) <= A001221(a(n))+1, see also A074239; A007947(a(n)) gives a sequence of primorials (A002110) in nondecreasing order. - Reinhard Zumkeller, Apr 16 2006, corrected: Apr 09 2015
a(n) = A005179(2^n). - Ivan N. Ianakiev, Apr 01 2015
a(n+1)/a(n) = A050376(n+1). - Jinyuan Wang, Oct 14 2018

Extensions

a(18) from Don Reble, Aug 20 2002