A037992 Smallest number with 2^n divisors.
1, 2, 6, 24, 120, 840, 7560, 83160, 1081080, 17297280, 294053760, 5587021440, 128501493120, 3212537328000, 93163582512000, 2888071057872000, 106858629141264000, 4381203794791824000, 188391763176048432000, 8854412869274276304000, 433866230594439538896000
Offset: 0
Links
- Danny Rorabaugh, Table of n, a(n) for n = 0..350 (first 101 terms from T. D. Noe)
- Project Euler, Problem 500!!!
Programs
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Haskell
a037992 n = head [x | x <- [1..], a000005 x == 2 ^ n] -- Reinhard Zumkeller, Apr 08 2015
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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 *)
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PARI
{a(n)= local(A,m,c,k,p); if(n<1, n==0, c=0; A=1; m=1; while( c
Michael 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
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
Comments