A175778 Length of the longest positive arithmetic progression, with nonzero common difference, that sums to n.

1, 1, 2, 2, 2, 3, 2, 2, 3, 4, 2, 3, 2, 4, 5, 4, 2, 4, 2, 5, 6, 4, 2, 4, 5, 4, 6, 7, 2, 5, 2, 4, 6, 4, 7, 8, 2, 4, 6, 5, 2, 7, 2, 8, 9, 4, 2, 6, 7, 5, 6, 8, 2, 9, 10, 7, 6, 4, 2, 8, 2, 4, 9, 8, 10, 11, 2, 8, 6, 7, 2, 9, 2, 4, 10, 8, 11, 12, 2, 8, 9, 4, 2, 8, 10, 4, 6, 11, 2, 12, 13, 8, 6, 4, 10, 8, 2, 7, 11

Offset

1,3

Comments

If n is an odd prime then a(n)=2.

It appears that if n is at least 10 and twice a prime, then a(n)=4.

If n is the triangular number k*(k+1)/2 (A000217), then a(n)=k.

As a follow-up to previous comment, it appears that, if sequence is displayed as a regular triangle, then the diagonals starting at rows 1, 2, 4, 7, 11, 16, 22, ...(likely A000124) are constituted of consecutive integers. - Michel Marcus, Jul 13 2014

Links

Antti Karttunen, Table of n, a(n) for n = 1..65537

Example

The arithmetic progressions {1,3,5} and {2,3,4}, of length 3, sum to 9. No longer arithmetic progression sums to 9, so a(9)=3.
As a triangle, sequence starts:
1;
1, 2;
2, 2, 3;
2, 2, 3, 4;
2, 3, 2, 4, 5;
4, 2, 4, 2, 5, 6;
4, 2, 4, 5, 4, 6, 7;
2, 5, 2, 4, 6, 4, 7, 8;
2, 4, 6, 5, 2, 7, 2, 8, 9;
4, 2, 6, 7, 5, 6, 8, 2, 9, 10;

Prog

(PARI) isokap(sumap, nbap) = {for (d=1, sumap, fap = (sumap - (nbap-1)*nbap*d/2)/nbap; if (fap <= 0, return (0)); if (type(fap) == "t_INT", return (1)); ); return (0); }
a(n) = {if (n == 1, return (1)); forstep(len=n-1, 1, -1, if (isokap(n, len), return (len)); ); } \\ Michel Marcus, Jul 13 2014

Crossrefs

Sequence in context: A104011 A242879 A176775 * A357039 A226182 A099774

Adjacent sequences: A175775 A175776 A175777 * A175779 A175780 A175781

Keyword

nonn,look

Author

John W. Layman, Sep 03 2010

Status

approved