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Smallest number with 2^n divisors.
42

%I #116 Aug 07 2024 06:03:24

%S 1,2,6,24,120,840,7560,83160,1081080,17297280,294053760,5587021440,

%T 128501493120,3212537328000,93163582512000,2888071057872000,

%U 106858629141264000,4381203794791824000,188391763176048432000,8854412869274276304000,433866230594439538896000

%N Smallest number with 2^n divisors.

%C 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

%C 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

%H Danny Rorabaugh, <a href="/A037992/b037992.txt">Table of n, a(n) for n = 0..350</a> (first 101 terms from T. D. Noe)

%H Project Euler, <a href="https://projecteuler.net/problem=500">Problem 500!!!</a>

%F A000005(a(n)) = A000079(n).

%F a(n) = Product_{k=1..n} A050376(k), product of the first n terms of A050376. - _Lekraj Beedassy_, Jun 30 2004

%F a(n) = A052330(2^n -1). - _Thomas Ordowski_, Jun 29 2005

%F 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

%F a(n) = A005179(2^n). - _Ivan N. Ianakiev_, Apr 01 2015

%F a(n+1)/a(n) = A050376(n+1). - _Jinyuan Wang_, Oct 14 2018

%t 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 *)

%o (PARI) {a(n)= local(A,m,c,k,p); if(n<1, n==0, c=0; A=1; m=1; while( c<n, m++; if( isprime(m) || ( (k=ispower(m,,&p)) && isprime(p)& k ==2^valuation(k,2) ), A*=m; c++)); A)} /* _Michael Somos_, Apr 15 2005 */

%o (Haskell)

%o a037992 n = head [x | x <- [1..], a000005 x == 2 ^ n]

%o -- _Reinhard Zumkeller_, Apr 08 2015

%o (Python)

%o def a(n):

%o product = 1

%o k = 1

%o for i in range(n+1):

%o product *= k # k=A050376(i), for i>=1

%o while product % k == 0:

%o k += 1

%o return product

%o # _Jason L. Miller_, Mar 20 2024

%Y Cf. A000005, A000079, A001221, A005179, A007947, A050376, A052330, A074239.

%K nonn,nice

%O 0,2

%A _J. Lowell_

%E a(18) from _Don Reble_, Aug 20 2002