A367628 Sum of the divisors of n <= tau(n).
1, 3, 1, 3, 1, 6, 1, 7, 4, 3, 1, 16, 1, 3, 4, 7, 1, 12, 1, 12, 4, 3, 1, 24, 1, 3, 4, 7, 1, 17, 1, 7, 4, 3, 1, 25, 1, 3, 4, 20, 1, 19, 1, 7, 9, 3, 1, 24, 1, 8, 4, 7, 1, 12, 1, 22, 4, 3, 1, 43, 1, 3, 4, 7, 1, 12, 1, 7, 4, 15, 1, 45, 1, 3, 9, 7, 1, 12, 1, 30, 4, 3, 1, 35, 1
Offset: 1
Keywords
Examples
a(12) = 16. The sum of the divisors of 12 <= tau(12) = 6 are 1 + 2 + 3 + 4 + 6 = 16.
Programs
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Mathematica
Table[Sum[k(1-Ceiling[n/k]+Floor[n/k]), {k, DivisorSigma[0, n]}], {n, 100}] -
PARI
a(n) = my(t=numdiv(n)); sumdiv(n, d, if (d <=t, d)); \\ Michel Marcus, Nov 25 2023
Formula
a(n) = Sum_{d|n, d<=tau(n)} d.
Comments