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 A175778 Length of the longest positive arithmetic progression, with nonzero common difference, that sums to n. 0
 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 (list; graph; refs; listen; history; text; internal format)
 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 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 * A226182 A099774 A305973 Adjacent sequences:  A175775 A175776 A175777 * A175779 A175780 A175781 KEYWORD nonn AUTHOR John W. Layman, Sep 03 2010 STATUS approved

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Last modified September 29 06:29 EDT 2020. Contains 337425 sequences. (Running on oeis4.)