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A238529
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a(0) = a(1) = 0, and for n > 1, a(n) = number of iterations of A238525 (n modulo sopfr(n)) needed to reach either 0 or 1. Here sopfr(n) is the sum of the prime factors of n, with multiplicity, A001414.
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5
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0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 1, 3, 1, 2, 2, 2, 1, 2, 1, 1, 1, 2, 2, 3, 2, 2, 1, 2, 2, 2, 1, 2, 1, 3, 1, 2, 1, 2, 2, 2, 2, 1, 1, 3, 2, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 1, 2, 2, 1, 1, 1, 1, 3, 3, 2, 2, 2, 1, 2, 3, 3, 1, 1, 2, 2, 2, 2, 1, 3, 2, 2, 3, 2, 2, 2, 1, 2, 3, 2, 1, 3, 1, 3, 1
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OFFSET
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0,9
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COMMENTS
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Previous name was: Recursive depth of n modulo sopfr(n), where sopfr(n) is the sum of the prime factors of n, with multiplicity.
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LINKS
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FORMULA
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EXAMPLE
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a(2) = 1, because 2 mod sopfr(2) = 2 mod 2 = 0, and further recursion (0 mod sopfr(0)) is undefined.
a(8) = 2, because 8 mod sopfr(8) = 8 mod 6 = 2, and 2 mod sopfr(2) is defined as above, giving 8 a recursive depth of 2.
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MATHEMATICA
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Array[-1 + Length@ NestWhileList[Mod[#, Total@ Flatten@ Map[ConstantArray[#1, #2] & @@ # &, FactorInteger@ #]] &, #, # > 1 &] &, 105, 0] (* Michael De Vlieger, Oct 20 2017 *)
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PROG
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(Sage)
def a(n):
d = 0
while n>1:
n = n % sum([f[0]*f[1] for f in factor(n)])
d = d+1
return d
(PARI)
A001414(n) = { my(f=factor(n)); sum(k=1, matsize(f)[1], f[k, 1]*f[k, 2]); };
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Terms a(0) = a(1) = 0 prepended and name changed by Antti Karttunen, Oct 20 2017
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STATUS
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approved
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