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 A156640 #42 Apr 12 2025 12:16:17
%S A156640 29,338,985,1970,3293,4954,6953,9290,11965,14978,18329,22018,26045,
%T A156640 30410,35113,40154,45533,51250,57305,63698,70429,77498,84905,92650,
%U A156640 100733,109154,117913,127010,136445,146218,156329,166778
%N A156640 a(n) = 169*n^2 + 140*n + 29.
%C A156640 The identity (57122*n^2 +47320*n +9801)^2 - (169*n^2 +140*n +29)*(4394*n +1820)^2 = 1 can be written as A156735(n)^2 - a(n)*A156636(n)^2 = 1.
%C A156640 The continued fraction expansion of sqrt(a(n)) is [13n+5; {2, 1, 1, 2, 26n+10}]. - _Magus K. Chu_, Sep 15 2022
%C A156640 From _Klaus Purath_, Apr 06 2025: (Start)
%C A156640 a(n)*13^2-1 is a square, and a(n) is the sum of two squares (see FORMULA). There are no squares in this sequence. The odd prime factors of these terms are always of the form 4*k + 1.
%C A156640 All a(n) = D satisfy the Pell equation (k*x)^2 - D*(13*y)^2 = -1 for any integer n where a(1-n) = A156639(n). The values for k and the solutions x, y can be calculated using the following algorithm: k = sqrt(D*13^2 - 1), x(0) = 1, x(1) = 4*D*13^2 - 1, y(0) = 1, y(1) = 4*D*13^2 - 3. The two recurrences are of the form (4*D*13^2 - 2, -1).
%C A156640 It follows from the above that this sequence and A156639 belong to A031396. (End)
%H A156640 Vincenzo Librandi, <a href="/A156640/b156640.txt">Table of n, a(n) for n = 0..10000</a>
%H A156640 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A156640 a(n) = 3*a(n-1) -3*a(n-2) +a(n-3) for n>2.
%F A156640 G.f.: (29 + 251*x + 58*x^2)/(1-x)^3. - _Vincenzo Librandi_, May 03 2014
%F A156640 E.g.f.: (29 +309*x +169*x^2)*exp(x). - _G. C. Greubel_, Feb 28 2021
%F A156640 From _Klaus Purath_, Apr 06 2025: (Start)
%F A156640 a(n) = (5*n + 2)^2 + (12*n + 5)^2 for any integer n.
%F A156640 169*a(n) - 1 = (169*n + 70)^2 for any integer n. (End)
%p A156640 A156640:= n-> 169*n^2 + 140*n + 29; seq(A156640(n), n=0..50); # _G. C. Greubel_, Feb 28 2021
%t A156640 LinearRecurrence[{3,-3,1},{29,338,985},50]
%t A156640 CoefficientList[Series[(29 +251x +58x^2)/(1-x)^3, {x, 0, 60}], x] (* _Vincenzo Librandi_, May 03 2014 *)
%o A156640 (Magma) I:=[29, 338, 985]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]];
%o A156640 (PARI) a(n)=169*n^2+140*n+29 \\ _Charles R Greathouse IV_, Dec 23 2011
%o A156640 (Sage) [169*n^2 + 140*n + 29 for n in (0..50)] # _G. C. Greubel_, Feb 28 2021
%Y A156640 Cf. A156636, A156718, A156735.
%Y A156640 Cf. A154609 (13n+5).
%Y A156640 Subsequence of A031396.
%K A156640 nonn,easy
%O A156640 0,1
%A A156640 _Vincenzo Librandi_, Feb 15 2009
%E A156640 Edited by _Charles R Greathouse IV_, Jul 25 2010