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A005070
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Sum of primes == 1 (mod 3) dividing n.
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6
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0, 0, 0, 0, 0, 0, 7, 0, 0, 0, 0, 0, 13, 7, 0, 0, 0, 0, 19, 0, 7, 0, 0, 0, 0, 13, 0, 7, 0, 0, 31, 0, 0, 0, 7, 0, 37, 19, 13, 0, 0, 7, 43, 0, 0, 0, 0, 0, 7, 0, 0, 13, 0, 0, 0, 7, 19, 0, 0, 0, 61, 31, 7, 0, 13, 0, 67, 0, 0, 7, 0, 0, 73, 37, 0, 19, 7, 13, 79, 0, 0, 0, 0, 7, 0, 43, 0, 0, 0, 0, 20, 0, 31, 0, 19, 0, 97, 7, 0, 0, 0, 0, 103, 13, 7, 0, 0, 0, 109, 0, 37
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OFFSET
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1,7
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LINKS
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FORMULA
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Additive with a(p^e) = p if p == 1 (mod 3), 0 otherwise.
a(1) = 0; for n > 1, a(n) = a(A028234(n)) + A020639(n)*[A020639(n) == 1 (mod 3)]. (Here [] is the Iverson bracket, giving in this case 1 whenever the smallest prime is of the form 3k+1, and 0 otherwise.) - Antti Karttunen, May 12 2017
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EXAMPLE
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For n = 5, a(5) = 0 as 5 modulo 3 = 2.
For n = 49 = 7*7, a(49) = 7 as 7 modulo 3 = 1, and each such prime is counted only once.
For n = 91 = 7*13, a(91) = 7+13 = 20, as both primes are of the form 3k+1.
For n = 10001 = 73*137, only 73 is of the form 3k+1, thus a(10001) = 73.
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MATHEMATICA
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Table[DivisorSum[n, # &, And[PrimeQ@ #, Mod[#, 3] == 1] &], {n, 111}] (* Michael De Vlieger, May 12 2017 *)
f[p_, e_] := If[Mod[p, 3] == 1, p, 0]; a[n_] := Plus @@ f @@@ FactorInteger[n]; a[1] = 0; Array[a, 100] (* Amiram Eldar, Jun 21 2022 *)
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PROG
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(Python)
from sympy import factorint, primefactors
def a028234(n):
f = factorint(n)
m = min(f)
return 1 if n==1 else n/(m**f[m])
def a020639(n): return min(primefactors(n)) if n>1 else 1
def a(n): return 0 if n==1 else a(a028234(n)) + a020639(n)*(1*(a020639(n)%3==1)) # Indranil Ghosh, May 12 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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EXTENSIONS
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STATUS
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approved
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