A100254 Let j be the smallest integer for which 1 + (1+1*n) + (1+2*n) + ... + (1+j*n) = k^2 = s. Then a(n)=j; if no such j exists, then a(n)=0.

7, 1, 80, 24, 5, 8, 1, 0, 6, 3, 2, 8, 21, 1, 48, 3, 7, 0, 16, 8, 80, 4, 1, 24, 45, 2, 9, 120, 5, 8, 17, 3, 30, 1, 10, 168, 6, 5, 1680, 48, 7, 3, 4960, 11, 224, 7, 1, 15, 31, 0, 48, 24, 16, 12, 288, 8, 48, 6, 26, 80, 117, 1, 136160, 195, 13, 3, 9, 840, 1520, 24, 49, 8, 70, 2, 1680

Offset

1,1

Comments

Basis for sequence is shortest arithmetic series with initial term 1 and difference n that sums to a perfect square.

Links

Table of n, a(n) for n=1..75.

Formula

1 + (1+1*n) + (1+2*n) + ... + (1+a(n)*n) = 1 + (1+1*n) + (1+2*n) + ... + A100253(n) = A100251(n)^2 = A100252(n).

Example

a(3)=80 since 1 + 4 + 7 +...+ (1+80*3) = 99^2 = 9801 and no other arithmetic series with initial term 1, difference 3 and fewer terms sums to a perfect square.

Mathematica

a[n_] := Block[{k = 1}, While[ !IntegerQ[ Sqrt[(k + 1)(1 + k*n/2)]], k++ ]; k]; a[18] = a[50] = 0; Table[ a[n], {n, 75}] (* Robert G. Wilson v, Nov 27 2004 *)

Crossrefs

Sequence in context: A321187 A221367 A110788 * A119935 A027447 A027517

Adjacent sequences: A100251 A100252 A100253 * A100255 A100256 A100257

Keyword

nonn

Author

Charlie Marion, Nov 21 2004

Extensions

More terms from Robert G. Wilson v, Nov 27 2004

Status

approved