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A318359
a(1) = 4; for n > 1, a(n) is the least positive number not yet in the sequence such that Sum_{k=1..n} a(k) divides Sum_{k=1..n} a(k)^3.
2
4, 1, 5, 8, 9, 12, 6, 20, 10, 3, 39, 65, 52, 11, 7, 42, 147, 441, 294, 366, 222, 35, 514, 257, 1285, 771, 3084, 672, 99, 925, 608, 291, 2061, 229, 495, 140, 81, 288, 12088, 1511, 750, 476, 209, 2603, 752, 7645, 1079, 5816, 2210, 2830, 1996, 1162, 328, 3690
OFFSET
1,1
COMMENTS
Is this sequence infinite?
For any v > 0, let b_v be the variant of this sequence starting with v:
- b_4 = a (this sequence),
- b_1 = A000027 (and indeed, A000217(n) divides A000537(n) for any n > 0),
- for v in {1, 2, 3, 5}: b_v(n) = n for all n > 0 except a finite number.
This sequence is a variant of A318358.
LINKS
EXAMPLE
For n = 3:
- (4^3 + 1^3 + 2^3) == 3 mod (4 + 1 + 2),
- (4^3 + 1^3 + 3^3) == 4 mod (4 + 1 + 3),
- (4^3 + 1^3 + 5^3) == 0 mod (4 + 1 + 5),
- hence a(3) = 5.
PROG
(PARI) s=0; s3=0; p=0; v=4; for (n=1, 54, print1 (v ", "); s+=v; s3+=v^3; p+=2^v; for (w=1, oo, if (!bittest(p, w) && (s3+w^3)%(s+w)==0, v=w; break)))
CROSSREFS
KEYWORD
nonn
AUTHOR
Rémy Sigrist, Aug 24 2018
STATUS
approved