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Search: id:A100254
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| A100254 |
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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)=j; if no such j exists, then a(n)=0. |
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+0
8
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| 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
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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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
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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)
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EXAMPLE
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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.
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MATHEMATICA
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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}] (from Robert G. Wilson v Nov 27 2004)
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CROSSREFS
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Sequence in context: A051339 A134141 A110788 this_sequence A119935 A027447 A027517
Adjacent sequences: A100251 A100252 A100253 this_sequence A100255 A100256 A100257
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KEYWORD
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nonn
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AUTHOR
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Charlie Marion (charliemath(AT)optonline.net), Nov 21 2004
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EXTENSIONS
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More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Nov 27 2004
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