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A105560
a(1) = 1, and for n >= 2, a(n) = prime(bigomega(n)), where prime(n) = A000040(n) and bigomega(n) = A001222(n).
16
1, 2, 2, 3, 2, 3, 2, 5, 3, 3, 2, 5, 2, 3, 3, 7, 2, 5, 2, 5, 3, 3, 2, 7, 3, 3, 5, 5, 2, 5, 2, 11, 3, 3, 3, 7, 2, 3, 3, 7, 2, 5, 2, 5, 5, 3, 2, 11, 3, 5, 3, 5, 2, 7, 3, 7, 3, 3, 2, 7, 2, 3, 5, 13, 3, 5, 2, 5, 3, 5, 2, 11, 2, 3, 5, 5, 3, 5, 2, 11, 7, 3, 2, 7, 3, 3, 3, 7, 2, 7, 3, 5, 3, 3, 3, 13, 2, 5, 5, 7
OFFSET
1,2
COMMENTS
From Antti Karttunen, Jul 21 2014: (Start)
a(n) divides A122111(n), A242424(n), A243072(n), A243073(n) because a(n) divides all the terms in column n of A243070.
a(2n-1) divides A243505(n) and a(2n-1)^2 divides A122111(2n-1).
(End)
LINKS
FORMULA
a(1) = 1, and for n >= 2, a(n) = A000040(A001222(n)).
From Antti Karttunen, Jul 21 2014: (Start)
a(n) = A008578(1 + A001222(n)).
a(n) = A006530(A122111(n)).
a(n) = A122111(n) / A122111(A064989(n)).
a(2n-1) = A122111(2n-1) / A243505(n).
a(n) = A242424(n) / A064989(n).
(End)
MATHEMATICA
Table[Prime[Sum[FactorInteger[n][[i, 2]], {i, 1, Length[FactorInteger[n]]}]], {n, 2, 40}] (* Stefan Steinerberger, May 16 2007 *)
PROG
(PARI) d(n) = for(x=2, n, print1(prime(bigomega(x))", "))
(Python)
from sympy import prime, primefactors
def a001222(n): return 0 if n==1 else a001222(n/primefactors(n)[0]) + 1
def a(n): return 1 if n==1 else prime(a001222(n)) # Indranil Ghosh, Jun 15 2017
KEYWORD
easy,nonn
AUTHOR
Cino Hilliard, May 03 2005
EXTENSIONS
a(1) = 1 prepended by Antti Karttunen, Jul 21 2014
STATUS
approved