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A100253
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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.
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3
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| Basis for sequence is shortest arithmetic series with initial term 1 and difference n that sums to a perfect square.
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FORMULA
| 1+(1+1*n)+(1+2*n)+...+(1+A100254(n)*n)= 1+(1+1*n)+(1+2*n)+...+a(n)=A100251(n)^2=A100252(n)
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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.
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CROSSREFS
| Sequence in context: A197245 A038281 A131502 * A175040 A200108 A088397
Adjacent sequences: A100250 A100251 A100252 * A100254 A100255 A100256
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KEYWORD
| nonn
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AUTHOR
| Charlie Marion (charliemath(AT)optonline.net), Nov 21 2004
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