OFFSET
1,3
COMMENTS
In other words: a(1) = 0, and for any n > 0, if a(1) + ... + a(n) is a square number then a(n+1) = n, otherwise a(n+1) = a(n).
This sequence is unbounded: if a(1) + ... + a(n) = u^2, then a(n+1) = n, u^2 + n * (2*u+n) = (u+n)^2, so a(1) + ... + a(m) is a square number for some m > n, and a(m+1) = m > a(n+1).
LINKS
Rémy Sigrist, Table of n, a(n) for n = 1..10000
EXAMPLE
Sequence begins:
n a(n) a(1)+...+a(n) Square?
-- ---- ------------- -------
1 0 0 Yes
2 1 1 Yes
3 2 3 No
4 2 5 No
5 2 7 No
6 2 9 Yes
7 6 15 No
8 6 21 No
9 6 27 No
10 6 33 No
MATHEMATICA
Module[{s = 0, a = 0}, Table[If[IntegerQ[Sqrt[s += a]], a = n-1]; a, {n, 100}]] (* Paolo Xausa, Jul 29 2025, after Rémy Sigrist *)
PROG
(PARI) { t = 0; v = 0; for (n = 1, 70, print1 (v", "); t += v; if (issquare(t), v = n; ); ); }
CROSSREFS
KEYWORD
nonn
AUTHOR
Rémy Sigrist, Jul 19 2025
STATUS
approved
