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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.
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%I #12 Aug 28 2018 00:21:00

%S 4,1,5,8,9,12,6,20,10,3,39,65,52,11,7,42,147,441,294,366,222,35,514,

%T 257,1285,771,3084,672,99,925,608,291,2061,229,495,140,81,288,12088,

%U 1511,750,476,209,2603,752,7645,1079,5816,2210,2830,1996,1162,328,3690

%N 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.

%C Is this sequence infinite?

%C For any v > 0, let b_v be the variant of this sequence starting with v:

%C - b_4 = a (this sequence),

%C - b_1 = A000027 (and indeed, A000217(n) divides A000537(n) for any n > 0),

%C - for v in {1, 2, 3, 5}: b_v(n) = n for all n > 0 except a finite number.

%C This sequence is a variant of A318358.

%H Rémy Sigrist, <a href="/A318359/b318359.txt">Table of n, a(n) for n = 1..500</a>

%e For n = 3:

%e - (4^3 + 1^3 + 2^3) == 3 mod (4 + 1 + 2),

%e - (4^3 + 1^3 + 3^3) == 4 mod (4 + 1 + 3),

%e - (4^3 + 1^3 + 5^3) == 0 mod (4 + 1 + 5),

%e - hence a(3) = 5.

%o (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)))

%Y Cf. A000027, A000217, A000537, A318358.

%K nonn

%O 1,1

%A _Rémy Sigrist_, Aug 24 2018