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A293899
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Number of proper divisors of the form 3k+1 minus number of proper divisors of the form 3k+2.
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4
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0, 1, 1, 0, 1, 0, 1, 1, 1, -1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 2, -1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 1, 0, -1, 1, 1, 1, 1, 2, -1, 1, 0, 1, 1, 0, -1, 1, 1, 2, 1, 0, 1, 1, 0, -1, 1, 2, -1, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 0, -1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, -1, 1, 2, -1, 1, 0, -1, 1, 0, 3, 1, 2, -1, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0
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OFFSET
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1,21
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LINKS
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FORMULA
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When n = 3k, a(n) = A002324(n), when n = 3k+1, a(n) = A002324(n) - 1, when n = 3k+2, a(n) = A002324(n) + 1.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Pi/(3*sqrt(3)) = 0.604599... (A073010). - Amiram Eldar, Nov 25 2023
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MATHEMATICA
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Table[DivisorSum[n, 1 &, And[Mod[#, 3] == 1, # != n] &] - DivisorSum[n, 1 &, And[Mod[#, 3] == 2, # != n] &], {n, 105}] (* Michael De Vlieger, Nov 08 2017 *)
Table[Total[Which[Mod[#, 3]==1, 1, Mod[#, 3]==2, -1, True, 0]&/@Most[ Divisors[ n]]], {n, 110}] (* Harvey P. Dale, Nov 26 2021 *)
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PROG
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(PARI)
A293895(n) = sumdiv(n, d, (d<n)*(1==(d%3)));
A293896(n) = sumdiv(n, d, (d<n)*(2==(d%3)));
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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