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A192015
Arithmetic derivative of prime powers: a(n) = A003415(A000961(n)).
7
0, 1, 1, 4, 1, 1, 12, 6, 1, 1, 32, 1, 1, 1, 10, 27, 1, 1, 80, 1, 1, 1, 1, 14, 1, 1, 1, 192, 1, 1, 1, 1, 108, 1, 1, 1, 1, 1, 1, 1, 1, 22, 75, 1, 448, 1, 1, 1, 1, 1, 1, 1, 1, 26, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 405, 1, 1024, 1, 1, 1, 1, 1, 1, 1, 34
OFFSET
1,4
COMMENTS
a(A000040(n)) = 1; a(A002808(n)) > 1;
A001787, A027471, A100484, A079705 and A051674 are subsequences;
A001787 and A024622 give record values and where they occur;
A192016(n) = A003415(a(n)).
LINKS
Eric Weisstein's World of Mathematics, Prime Power
FORMULA
a(n) = A025474(n) * A025473(n)^(A025474(n) - 1).
MATHEMATICA
Join[{0}, Reap[For[n = 1, n <= 300, n++, f = FactorInteger[n]; If[Length[f] == 1, Sow[n*Total[Apply[#2/#1&, f, {1}]]]]]][[2, 1]]] (* Jean-François Alcover, Feb 21 2014 *)
PROG
(Haskell)
a192015 = a003415 . a000961 -- Reinhard Zumkeller, Apr 16 2014
(Python)
from sympy import primepi, integer_nthroot, factorint
def A192015(n):
if n == 1: return 0
def f(x): return int(n+x-1-sum(primepi(integer_nthroot(x, k)[0]) for k in range(1, x.bit_length())))
m, k = n, f(n)
while m != k:
m, k = k, f(k)
return sum((m*e//p for p, e in factorint(m).items())) # Chai Wah Wu, Aug 15 2024
CROSSREFS
Sequence in context: A146967 A173152 A156049 * A205946 A101919 A055106
KEYWORD
nonn
AUTHOR
Reinhard Zumkeller, Jun 26 2011
STATUS
approved