A261116 Pairs of integers (a,b) such a^2 + (a+1)^2 = b^2.
0, 1, 3, 5, 20, 29, 119, 169, 696, 985, 4059, 5741, 23660, 33461, 137903, 195025, 803760, 1136689, 4684659, 6625109, 27304196, 38613965, 159140519, 225058681, 927538920, 1311738121, 5406093003, 7645370045, 31509019100, 44560482149, 183648021599, 259717522849
Offset: 1
Examples
a(5) = 20 and a(6) = 29, because 20^2 + 21^2 = 29^2.
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- V. Pletser, Finding all squared integers expressible as the sum of consecutive squared integers using generalized Pell equation solutions with Chebyshev polynomials, arXiv preprint arXiv:1409.7972 [math.NT], 2014. See Table 3 p. 8.
- Index entries for linear recurrences with constant coefficients, signature (0,7,0,-7,0,1).
Programs
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Mathematica
LinearRecurrence[{0, 7, 0, -7, 0, 1}, {0, 1, 3, 5, 20, 29}, 50] (* Paolo Xausa, Jan 31 2024 *)
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PARI
concat(0, Vec(-x^2*(x^4-x^3-2*x^2+3*x+1) / ((x-1)*(x+1)*(x^2-2*x-1)*(x^2+2*x-1)) + O(x^50))) \\ Colin Barker, Aug 12 2015
Formula
From Colin Barker, Aug 09 2015: (Start)
a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6) for n>6.
G.f.: -x^2*(x^4-x^3-2*x^2+3*x+1) / ((x-1)*(x+1)*(x^2-2*x-1)*(x^2+2*x-1)).
(End)
Extensions
Edited by M. F. Hasler, Aug 11 2015
Comments