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A130281
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Integers n>1 such that A130280(4n^2)<n, i.e. there is an m<n, m>1 such that 4n^2(m^2-1)+1 is a square.
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2
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28, 102, 248, 390, 490, 852, 1358, 2032, 2898, 3465, 3980, 5302, 5432, 6888, 8762, 10948, 13470, 15372, 16352, 19618, 23292, 27398, 31960, 37002, 42548, 48015, 48622, 55248, 62450, 70252, 75658, 78678, 87752, 97498
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| If n>4 is an even square, n=4k^2, then A130280(n) <= k since n(k^2-1)+1 = (2k^2-1)^2. This sequence lists those k for which we have strict inequality. Most terms in this sequence belong to the subsequence b(m)=2m*(2m^2-1), m>1, for which A130280(4 b(m)^2) <= m < b(m), since 4 b(m)^2(m^2-1)+1 = (8m^4-8m^2+1)^2. For other terms k of this sequence (e.g. the subsequence 390, 3465, 15372, 48015,...), A130280(4k^2) is even smaller.
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PROG
| (PARI) checkA130281(n)={local(m=4*n^2); for(i=2, sqrt(n), if(issquare(m*(i^2-1)+1), return(i)))} for(n=1, 99999, if(checkA130281(n), print(n", ")))
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CROSSREFS
| Cf. A130280.
Sequence in context: A005971 A189808 A130085 * A168254 A201469 A010016
Adjacent sequences: A130278 A130279 A130280 * A130282 A130283 A130284
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KEYWORD
| nonn
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AUTHOR
| M. F. Hasler (Maximilian.Hasler(AT)gmail.com), May 20 2007
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