This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A076708 #28 Nov 28 2015 17:37:00
%S A076708 0,5,34,203,1188,6929,40390,235415,1372104,7997213,46611178,271669859,
%T A076708 1583407980,9228778025,53789260174,313506783023,1827251437968,
%U A076708 10650001844789,62072759630770,361786555939835,2108646576008244,12290092900109633,71631910824649558
%N A076708 Numbers n such that triangular numbers T(n) and T(n+1) sum to another triangular number (that is also a perfect square).
%C A076708 From T(k)+T(k+1) = (k*(k+1)+(k+1)*(k+2))/2 = (k+1)^2 any two consecutive triangular numbers sum to a square, the above sequence gives the sums that are also triangular. The units digit cycles through 0, 5, 4, 3, 8, 9, 0, 5, ...
%C A076708 Let P(b,e) be the polynomial 1+4*b+4*b^2+4*e+4*e^2. It appears that sequences A076708 and A076049 are special cases of the sequence of integers b such that P(b,b+n) is a perfect square. A076708 and A076049 for example are respectively the sequences of b's such that P(b,b+1) and P(b,b+2) are perfect squares. In fact it appears to be true that the sequence of integers b such that P(b,b+n) is a perfect square has the property that t(b)+t(b+n) is a triangular number. I have not had time to prove this but I do have evidence produced by Mathematica to support the assertion. - Robert Phillips (bobanne(AT)bellsouth.net), Sep 04 2009; corrected Sep 08 2009
%H A076708 Colin Barker, <a href="/A076708/b076708.txt">Table of n, a(n) for n = 1..1000</a>
%H A076708 B. Polster, M. Ross, <a href="http://arxiv.org/abs/1503.04658">Marching in squares</a>, arXiv preprint arXiv:1503.04658, 2015
%H A076708 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-7,1).
%F A076708 Recursion: a(n+2) = 6*a(n+1)-a(n)+4, with a(0)=0 and a(1)=5.
%F A076708 G.f.: (5*x^2-x^3)/((1-x)*(1-6*x+x^2)).
%F A076708 Closed form: a(n)= ( sqrt(2)*( (3+2*sqrt(2))^(n+1) - (3-2*sqrt(2))^(n+1) )-8 )/8.
%F A076708 Also, if the entries in A001109 are denoted by b(n) then a(n) = b(n+1)-1.
%F A076708 a(n) = sqrt(A001110(n)) - 1. - _Ivan N. Ianakiev_, May 03 2014
%e A076708 a(1) = (sqrt(2)*((3+2*sqrt(2))^2-(3-2*sqrt(2))^2)-8)/8 = (sqrt(2)*(9+12*sqrt(2)+8-9+12*sqrt(2)-8)-8)/8 = (sqrt(2)*24*sqrt(2)-8)/8 = (48-8)/8 = 40/8 = 5.
%e A076708 T(5) + T(6) = 15 + 21 = 36 = T(8).
%t A076708 Table[((3 + 2 Sqrt[2])^n - (3 - 2 Sqrt[2])^n)/(4 Sqrt[2]) - 1, {n, 1, 20}] (* _Zerinvary Lajos_, Jul 14 2009 *)
%o A076708 (PARI) concat(0, Vec(x^2*(x-5)/((x-1)*(x^2-6*x+1)) + O(x^100))) \\ _Colin Barker_, May 15 2015
%Y A076708 Cf. A001108, A001109, A001110.
%K A076708 nonn,easy
%O A076708 1,2
%A A076708 Bruce Corrigan (scentman(AT)myfamily.com), Oct 26 2002