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A106230 least k > 0 for n > 0 such that (n^2 + 1)*(k^2) + (n^2 + 1)*k + 1 = j^2 j sequence = A106229. +0
3
3, 8, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168, 195, 224, 255, 288, 323, 360, 399, 440 (list; graph; listen)
OFFSET

1,1

COMMENT

For (n^2 + 1)*(k^2) + (n^2 +1)*k + 1 = j^2 there is a sequence k(i,n) with a recurrence for n=1 k(1,1) = 0, k(2,1) = 3, k(i,1) = 6*k(i-1,1) + 2 - k(i-2,1) for n=2 k(1,2) = 1, k(2,2) = 19, k(i,2) = 18*k(i-1,2) + 8 -k(i-2,2) for n>2 k(1,n) = 0, k(2,n) = n^2 - 2*n, k(3,n) = 2*n^2 -k(2), k(4,n) = (4*n^2 + 2)*k(2,n) + 2*n^2 then k(i,n) = (4*n^2 + 2)*k(i-2,n) + 2*n^2 - k(i-4,n) As i increases the ratio j(i,n)/k(i,n) tends to sqrt(n^2 + 1)

FORMULA

for n > 2 k(n) = n^2 - 2*n

CROSSREFS

Cf. A003777, A005563, A005899, A106229.

Sequence in context: A125025 A019919 A135608 this_sequence A016623 A046543 A035292

Adjacent sequences: A106227 A106228 A106229 this_sequence A106231 A106232 A106233

KEYWORD

nonn

AUTHOR

Pierre CAMI (pierrecami(AT)tele2.fr), Apr 26 2005

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Last modified August 28 22:44 EDT 2008. Contains 143251 sequences.


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