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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 where j sequence = A106229. 3
 3, 8, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168, 195, 224, 255, 288, 323, 360, 399, 440, 483, 528, 575, 624, 675, 728, 783, 840, 899, 960, 1023, 1088, 1155, 1224, 1295, 1368, 1443, 1520, 1599, 1680, 1763, 1848, 1935, 2024, 2115, 2208, 2303, 2400, 2499 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS 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) LINKS G. C. Greubel, Table of n, a(n) for n = 1..5000 FORMULA For n > 2 a(n) = n^2 - 2*n. a(n) = A005563(n-2), n>2. - R. J. Mathar, Aug 28 2008 G.f.: (3 - x - 12*x^2 + 20*x^3 - 8*x^4)/(1 - x)^3. - G. C. Greubel, May 11 2017 MATHEMATICA CoefficientList[Series[(-3 + z + 12*z^2 - 20*z^3 + 8*z^4)/(-1 + z)^3, {z, 0, 60}], z] (* Vladimir Joseph Stephan Orlovsky, Feb 17 2012 *) PROG (PARI) x='x+O('x^50); Vec((3-x-12*x^2+20*x^3-8*x^4)/(1-x)^3) \\ G. C. Greubel, May 11 2017 CROSSREFS Cf. A003777, A005563, A005899, A106229. Sequence in context: A019919 A213277 A135608 * A205126 A016623 A273100 Adjacent sequences:  A106227 A106228 A106229 * A106231 A106232 A106233 KEYWORD nonn AUTHOR Pierre CAMI, Apr 26 2005 STATUS approved

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Last modified January 27 09:09 EST 2020. Contains 331293 sequences. (Running on oeis4.)