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%I #33 Jan 31 2024 08:07:15
%S 0,1,3,5,20,29,119,169,696,985,4059,5741,23660,33461,137903,195025,
%T 803760,1136689,4684659,6625109,27304196,38613965,159140519,225058681,
%U 927538920,1311738121,5406093003,7645370045,31509019100,44560482149,183648021599,259717522849
%N Pairs of integers (a,b) such a^2 + (a+1)^2 = b^2.
%C Can also be seen as a table with two columns, read by rows: T[n,1] = a(2n-1) = A001652(n), T[n,2] = a(2n) = A001653(n).
%C The conjectured recurrence formula and g.f. are proved by the formulas for A001652. - _M. F. Hasler_, Aug 11 2015
%H Colin Barker, <a href="/A261116/b261116.txt">Table of n, a(n) for n = 1..1000</a>
%H V. Pletser, <a href="http://arxiv.org/abs/1409.7972">Finding all squared integers expressible as the sum of consecutive squared integers using generalized Pell equation solutions with Chebyshev polynomials</a>, arXiv preprint arXiv:1409.7972 [math.NT], 2014. See Table 3 p. 8.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,7,0,-7,0,1).
%F From _Colin Barker_, Aug 09 2015: (Start)
%F a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6) for n>6.
%F 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)).
%F (End)
%F a(2n-1)=A001652(n), a(2n)=A001653(n). - _M. F. Hasler_, Aug 11 2015
%e a(5) = 20 and a(6) = 29, because 20^2 + 21^2 = 29^2.
%t LinearRecurrence[{0, 7, 0, -7, 0, 1}, {0, 1, 3, 5, 20, 29}, 50] (* _Paolo Xausa_, Jan 31 2024 *)
%o (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
%Y Cf. A001652, A001653, A260819.
%K nonn,tabf,easy
%O 1,3
%A _Marco Ripà_, Aug 08 2015
%E Edited by _M. F. Hasler_, Aug 11 2015
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