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A106232
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Least k > 0 such that (4*n^2 + 2)*(k^2) + (4*n^2 + 2)*k + 1 = j^2.
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1
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4, 4, 12, 24, 40, 60, 84, 112, 144, 180, 220, 264, 312, 364, 420
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| For k there is alway a recurrence for n=1 k(1,1)=0, k(2,1)=4 then k(i,1)=10*k(i-1,1)+4-k(i-3,n) for n>1 k(1,n)=0, k(2,n)=2*n^2-2*n, k(3,n)=2*n^2+2*n k(4,n)=(8*n^2+2)*k(2,n)+4*n^2 then k(i,n)=(8*n^2+2)*k(i-2,n)+4*n^2-k(i-4,n)
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FORMULA
| for n=1 k=4, for n > 1 k(n) = 2*n^2 - 2*n j sequence = A106231
a(n)=A046092(n-1), n>1. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Aug 28 2008]
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CROSSREFS
| Cf. A106231.
Sequence in context: A120033 A097073 A019085 * A038804 A183362 A088838
Adjacent sequences: A106229 A106230 A106231 * A106233 A106234 A106235
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KEYWORD
| nonn
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AUTHOR
| Pierre CAMI (pierre-cami(AT)bbox.fr), Apr 26 2005
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