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A241919
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If n is a prime power, p_i^e, a(n) = i, (with a(1)=0), otherwise difference (i-j) of the indices of the two largest distinct primes p_i, p_j, i > j in the prime factorization of n: a(n) = A061395(n) - A061395(A051119(n)).
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10
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0, 1, 2, 1, 3, 1, 4, 1, 2, 2, 5, 1, 6, 3, 1, 1, 7, 1, 8, 2, 2, 4, 9, 1, 3, 5, 2, 3, 10, 1, 11, 1, 3, 6, 1, 1, 12, 7, 4, 2, 13, 2, 14, 4, 1, 8, 15, 1, 4, 2, 5, 5, 16, 1, 2, 3, 6, 9, 17, 1, 18, 10, 2, 1, 3, 3, 19, 6, 7, 1, 20, 1, 21, 11, 1, 7, 1, 4, 22, 2, 2, 12, 23
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OFFSET
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1,3
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COMMENTS
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LINKS
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FORMULA
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PROG
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(Haskell)
a241919 1 = 0
a241919 n = i - j where
(i:j:_) = map a049084 $ reverse (1 : a027748_row n)
(Python)
from sympy import factorint, primefactors, primepi
def a061395(n): return 0 if n==1 else primepi(primefactors(n)[-1])
def a053585(n):
if n==1: return 1
p = primefactors(n)[-1]
return p**factorint(n)[p]
def a051119(n): return n/a053585(n)
def a(n): return a061395(n) - a061395(a051119(n)) # Indranil Ghosh, May 19 2017
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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