A115561 a(n) = lpf((n/lpf(n))/lpf(n/lpf(n))), where lpf=A020639, least prime factor.
1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 1, 3, 1, 5, 1, 1, 1, 2, 1, 1, 3, 7, 1, 5, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 1, 7, 1, 11, 5, 1, 1, 2, 1, 5, 1, 13, 1, 3, 1, 2, 1, 1, 1, 3, 1, 1, 7, 2, 1, 11, 1, 17, 1, 7, 1, 2, 1, 1, 5, 19, 1, 13, 1, 2, 3, 1, 1, 3, 1, 1, 1, 2, 1, 3, 1, 23, 1, 1, 1, 2, 1, 7, 11, 5, 1
Offset: 1
Keywords
Links
- Antti Karttunen, Table of n, a(n) for n = 1..10000
- Eric Weisstein's World of Mathematics, Least Prime Factor
Crossrefs
Programs
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Mathematica
f[n_] := FactorInteger[n][[1, 1]]; Table[f[#/f@ #] &[n/f@ n], {n, 101}] (* Michael De Vlieger, Aug 14 2017 *)
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PARI
a020639(n) = if(n>1, if(n>n=factor(n, 0)[1, 1], n, factor(n)[1, 1]), 1) \\ after M. F. Hasler in A020639 a(n) = a020639((n/a020639(n))/a020639(n/a020639(n))) \\ Felix Fröhlich, Jul 15 2019
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Python
from sympy import divisors, primefactors def a032752(n): return 1 if n==1 else divisors(n)[-2] def a020639(n): return 1 if n==1 else primefactors(n)[0] def a(n): return a020639(a032752(a032752(n))) print([a(n) for n in range(1, 102)]) # Indranil Ghosh, Aug 12 2017
Comments