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Search: id:A136325
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| A136325 |
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Integers x such that Sqrt[3(5x^2+3)] is a perfect square. |
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+0
1
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| 0, 3, 24, 189, 1488, 11715, 92232, 726141, 5716896, 45009027, 354355320, 2789833533, 21964312944, 172924670019, 1361433047208, 10718539707645, 84386884613952, 664376537203971, 5230625413017816, 41180626766938557
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OFFSET
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0,2
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COMMENT
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The difference equation solution is a[n]=(+or -)Sqrt[15]((4 + Sqrt[15])^n - (4 - Sqrt[15])^n)/30, where the sign indicates the positive or negative series choice. The series is then generated from 3a[n]=+or -)Sqrt[15]((4 + Sqrt[15])^n - (4 - Sqrt[15])^n)/10.
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FORMULA
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The recursion that generates the sequence divided by three is a[n]=8a[n-1]-a[n-2] with a[0]=0 and a[1]=+1 for the positive sequence and a[1]=-1 for the negative sequence. The series is then 3*a[n]. The sub series (dividing the series by 3) would be {0, 1, 8, 63, 496, 3905, 30744, 242047, ...} which is the sequence A001090.
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MATHEMATICA
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Do[If[IntegerQ[Sqrt[3 (3 + 5 x^2)]], Print[{x, Sqrt[3 (3 + 5 x^2)]}]], {x, 0, 2000000}]
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PROGRAM
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(PARI) {a(n) = subst(poltchebi(n+1) - 4 * poltchebi(n), x, 4) / 5} /* Michael Somos Apr 05 2008 */
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CROSSREFS
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3*A001090(n) = a(n).
Adjacent sequences: A136322 A136323 A136324 this_sequence A136326 A136327 A136328
Sequence in context: A110347 A027324 A122741 this_sequence A103333 A037762 A037650
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KEYWORD
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nonn
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AUTHOR
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Lorenz H. Menke, Jr. (lnz2004(AT)mindspring.com), Mar 26 2008
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