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Search: id:A130280
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| A130280 |
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a(n) = smallest integer k>1 such that n(k^2-1)+1 is a perfect square, or 0 if no such number exists. |
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+0
7
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| 2, 5, 3, 0, 2, 3, 5, 2, 0, 3, 7, 5, 4, 11, 3, 2, 4, 13, 9, 7, 2, 5, 19, 4, 0, 5, 21, 3, 11, 9, 11, 14, 2, 29, 5, 3, 6, 31, 21, 2, 13, 11, 13, 169, 3, 7, 41, 6, 0, 7, 5, 11, 22, 419, 3, 2, 5, 23, 461, 27, 8, 55, 7, 4, 2, 3, 49, 29
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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A084702(n) = a(n)^2-1, resp. a(n) = sqrt(A084702(n)+1). See A130283 for values where A130280(n)=0.
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LINKS
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M. F. Hasler, Table of n, a(n) for n = 1..1000
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FORMULA
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If n=(2k)^2, then A130280(n) <= k, since (2k)^2(k^2-1)+1 = (2k^2-1)^2. See A130281 for the cases where equality does not hold. If n=k^2-1, then A130280(n) <= k-1 since (k^2-1)((k-1)^2-1)+1 = (k^2-k-1)^2. See A130282 for the cases where equality does not hold.
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EXAMPLE
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a( (2k)^2 ) <= k since (2k)^2(k^2-1)+1 = (2k^2-1)^2 (but k=1 is excluded since with k^2-1=0 this would be a trivial solution for any n).
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MAPLE
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A130280:=proc(n) local x, y, z; if n=1 then return 2 fi; isolve(n*(x^2-1)+1=y^2, z); select(has, `union`(%), x); map(rhs, %); simplify(eval(%, z=1) union eval(%, z=0)) minus {-1, 1}; if %={} then 0 else (min@op@map)(abs, %) fi end;
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PROGRAM
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(PARI) {A130280(n, L=10^15)=if(issquare(n), L=2+sqrtint(n>>2)); for( k=2, L, if( issquare(n*(k^2-1)+1), return(k)))}
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CROSSREFS
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Cf. A084702, A094357, A130281, A130282, A130283, A130284.
Adjacent sequences: A130277 A130278 A130279 this_sequence A130281 A130282 A130283
Sequence in context: A019295 A055385 A108429 this_sequence A011035 A102892 A132898
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KEYWORD
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easy,nonn
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AUTHOR
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M. F. Hasler (Maximilian.Hasler(AT)gmail.com), May 20 2007, May 25 2007
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