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A383403
Partial sums of the sum of the divisors of the numbers of the form 6*k + 3, k >= 0.
3
4, 17, 41, 73, 113, 161, 217, 295, 367, 447, 551, 647, 771, 892, 1012, 1140, 1296, 1488, 1640, 1822, 1990, 2166, 2406, 2598, 2826, 3060, 3276, 3564, 3824, 4064, 4312, 4632, 4968, 5240, 5552, 5840, 6136, 6539, 6923, 7243, 7607, 7943, 8375, 8765, 9125, 9573, 9989, 10469, 10861
OFFSET
0,1
COMMENTS
Partial sums of the sum of the divisors of A016945.
See the illustration of a(3) and a(10) as the total area (or total number of cells) in the diagram of the symmetric representation of sigma in the Links section.
Also consider a spiral similar to the spiral described in A239660 but with six wedges on the triangular grid. A "diamond" formed by two adjacent triangles has area 1. a(n) is the total number of diamonds (or the total area) in the third wedge after n + 1 turns. The spiral can be visualized from the top view of the stepped pyramid described in A274536. The graph is named W3 in the Plot 6 of the Links section.
FORMULA
a(n) = Sum_{k=0..n} sigma(6*k+3).
a(n) = (11*Pi^2/24) * n^2 + O(n*log(n)). - Amiram Eldar, Apr 28 2025
EXAMPLE
For n = 3 the first four terms of the numbers of the form 6*k + 3, k >= 0, are [3, 9, 15, 21]. The divisors of them are [1, 3], [1, 3, 9], [1, 3, 5, 15], [1, 3, 7, 21]. The sum of the divisors of them are [4, 13, 24, 32] respectively, and the sum of all divisors of them are 4 + 13 + 24 + 32 = 73, so a(3) = 73.
MATHEMATICA
Accumulate@ Array[DivisorSigma[1, 6 # + 3] &, 55, 0]
PROG
(PARI) a(n) = sum(k=0, n, sigma(6*k+3));
CROSSREFS
Sequences of the same family are A363161, A365442, this sequence, A365444, A383405, A365446.
Sequence in context: A145995 A275815 A018973 * A212575 A162148 A166781
KEYWORD
nonn,easy
AUTHOR
Omar E. Pol, Apr 27 2025
STATUS
approved