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, 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

Offset: 1

Author

Pierre CAMI, Apr 26 2005

Keywords

• nonn
• easy

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) = n^2 + 2*n, 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
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A003777, A005563, A005899, A106229.

Programs

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 *)
LinearRecurrence[{3,-3,1},{3,8,3,8,15},60] (* Harvey P. Dale, Jul 05 2022 *)

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

PARI

a(n)=n*if(n>2, n-2, n+2) \\ Charles R Greathouse IV, Oct 19 2022

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