A351112 Number of balanced numbers dividing n.
1, 2, 2, 2, 1, 4, 1, 2, 2, 2, 1, 5, 1, 3, 3, 2, 1, 4, 1, 2, 2, 2, 1, 5, 1, 2, 2, 3, 1, 6, 1, 2, 2, 2, 2, 5, 1, 2, 2, 2, 1, 6, 1, 2, 3, 2, 1, 5, 1, 2, 2, 2, 1, 4, 1, 4, 2, 2, 1, 7, 1, 2, 2, 2, 1, 4, 1, 2, 2, 5, 1, 5, 1, 2, 3, 2, 1, 5, 1, 2, 2, 2, 1, 7, 1, 2, 2, 2, 1, 6, 1, 2
Offset: 1
Keywords
Examples
a(4) = 2; the balanced divisors of 4 are 1 and 2. a(5) = 1; 1 is the only balanced divisor of 5. a(6) = 4; the balanced divisors of 6 are 1,2,3,6.
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Maple
f:= proc(n) uses numtheory; nops(select(t -> sigma(t) mod phi(t) = 0, divisors(n))) end proc: map(f, [$1..100]); # Robert Israel, Nov 28 2023
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Mathematica
a[n_] := DivisorSum[n, 1 &, Divisible[DivisorSigma[1, #], EulerPhi[#]] &]; Array[a, 100] (* Amiram Eldar, Feb 01 2022 *)
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PARI
a(n) = sumdiv(n, d, if (!(sigma(d) % eulerphi(d)), 1)); \\ Michel Marcus, Feb 01 2022
Formula
a(n) = Sum_{d|n, phi(d)|sigma(d)} 1.
a(n) = Sum_{d|n} A351114(d).
a(n) = tau(n) - Sum_{d|n} sign(sigma(d) mod phi(d)).
Conjecture: asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=1} 1/A020492(k) = 2.4343... (assuming empirically that this sum of reciprocals converges). - Amiram Eldar, Dec 27 2024
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