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A037992
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Smallest number with 2^n divisors.
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38
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1, 2, 6, 24, 120, 840, 7560, 83160, 1081080, 17297280, 294053760, 5587021440, 128501493120, 3212537328000, 93163582512000, 2888071057872000, 106858629141264000, 4381203794791824000, 188391763176048432000, 8854412869274276304000, 433866230594439538896000
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OFFSET
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0,2
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COMMENTS
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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
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LINKS
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FORMULA
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MATHEMATICA
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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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PROG
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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<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 */
(Haskell)
a037992 n = head [x | x <- [1..], a000005 x == 2 ^ n]
(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
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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