

A100253


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


7



8, 3, 241, 97, 26, 49, 8, 0, 55, 31, 23, 97, 274, 15, 721, 49, 120, 0, 305, 161, 1681, 89, 24, 577, 1126, 53, 244, 3361, 146, 241, 528, 97, 991, 35, 351, 6049, 223, 191, 65521, 1921, 288, 127, 213281, 485, 10081, 323, 48, 721, 1520, 0, 2449
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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



FORMULA



EXAMPLE

a(3)=241 since 1 + 4 + 7 +...+ 241 = 99^2 = 9801 and no other arithmetic series with initial term 1, difference 3 and fewer terms sums to a perfect square.


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



