A355750 Sum of the divisors of 2n minus the number of divisors of 2n.

1, 4, 8, 11, 14, 22, 20, 26, 33, 36, 32, 52, 38, 50, 64, 57, 50, 82, 56, 82, 88, 78, 68, 114, 87, 92, 112, 112, 86, 156, 92, 120, 136, 120, 136, 183, 110, 134, 160, 176, 122, 212, 128, 172, 222, 162, 140, 240, 165, 208, 208, 202, 158, 268, 208, 238, 232, 204, 176, 344, 182

Offset

1,2

Comments

Consider the partitions of 2n into 2 parts (s,t), where s <= t. a(n) gives the sum of all the quotients t/s such that t/s is an integer. (See example.)

Links

Table of n, a(n) for n=1..61.

Formula

a(n) = sigma(2n) - tau(2n).

a(n) = Sum_{d|2n} (2n-d)/d.

a(n) = A065608(2n) = A000203(2n) - A000005(2n).

a(n) = A062731(n) - A099777(n).

a(n) = Sum_{k=1..n} m*c(m), where m=(2n-k)/k and c(m)=1-ceiling(m)+floor(m).

Example

a(7) = 20; the partitions of 2*7 = 14 into two parts (s,t) where s <= t are: (1,13), (2,12), (3,11), (4,10), (5,9), (6,8), and (7,7). The sum of the quotients t/s such that each t/s is an integer is then: 13/1 + 12/2 + 7/7 = 13 + 6 + 1 = 20.

Mathematica

Table[DivisorSigma[1, 2 n] - DivisorSigma[0, 2 n], {n, 80}]

Prog

(PARI) a(n) = my(f=factor(2*n)); sigma(f) - numdiv(f); \\ Michel Marcus, Jul 16 2022

Crossrefs

Cf. A000005 (tau), A000203 (sigma), A062731, A099777.

Bisection of A065608.

Sequence in context: A311046 A311047 A311048 * A311049 A073302 A191105

Adjacent sequences: A355747 A355748 A355749 * A355751 A355752 A355753

Keyword

nonn,easy

Author

Wesley Ivan Hurt, Jul 15 2022

Status

approved