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 A164055 #87 Jan 05 2025 19:51:39
%S A164055 1,10,325,11026,374545,12723490,432224101,14682895930,498786237505,
%T A164055 16944049179226,575598885856165,19553418069930370,664240615491776401,
%U A164055 22564627508650467250,766533094678624110085,26039560591564569275626
%N A164055 Triangular numbers that are one plus a perfect square.
%C A164055 a(n+1) is the second element of the n-th set of three consecutive triangular numbers whose product is a perfect square. - _Arkadiusz Wesolowski_, Apr 27 2012
%C A164055 The triangular numbers of this sequence satisfy the Diophantine equation T(k) = k*(k+1)/2 = m^2 + 1, which is equivalent to (2k+1)^2 - 2*(2m)^2 = 9. Now, with x=2k+1 and y=2m, the Pell-Fermat equation x^2 - 2*y^2 = 9 appears. The solutions x and y of this equation are respectively in A106329 and A075848. The indices k=(x-1)/2 of the triangular numbers of this sequence are in A072221, while the indices m=y/2 of the corresponding square numbers are in A106328. - _Bernard Schott_, Mar 09 2019
%H A164055 Reinhard Zumkeller, <a href="/A164055/b164055.txt">Table of n, a(n) for n = 1..500</a>
%H A164055 K. B. Subramaniam, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/37-3/subramaniam.pdf">Almost Square Triangular Numbers</a>, The Fibonacci Quarterly, Vol. 37, No. 3 (1999), pp. 194-197.
%H A164055 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (35,-35,1).
%F A164055 A000217 INTERSECT A002522.
%F A164055 a(n) = A000217(A072221(n-1)).
%F A164055 From _R. J. Mathar_, Sep 22 2009: (Start)
%F A164055 a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
%F A164055 G.f.: x*(1-25*x+10*x^2)/((1-x)*(x^2-34*x+1)). (End)
%F A164055 A010054(a(n)) * A010052(a(n) - 1) = 1. - _Reinhard Zumkeller_, Apr 29 2012
%F A164055 a(n) = (A106329(n) - 1)*(A106329(n) + 1)/8 = (A106328(n))^2 + 1. - _Bernard Schott_, Mar 10 2019
%F A164055 a(n) = (14 + 9*(17+12*sqrt(2))^(1-n) - 9*(-17+12*sqrt(2))*(17+12*sqrt(2))^n) / 32. - _Colin Barker_, Mar 23 2019
%F A164055 a(n) = 9 * A001110(n) + 1 (Subramaniam, 1999). - _Amiram Eldar_, Jan 13 2022
%e A164055 10 is in this sequence because it is a triangular number A000217(4) and is equal to a square plus 1: 10 = 3^2 + 1.
%t A164055 LinearRecurrence[{35,-35,1}, {1,10,325}, 50] (* _G. C. Greubel_, Sep 09 2017 *)
%o A164055 (Haskell)
%o A164055 a164055 n = a164055_list !! (n-1)
%o A164055 a164055_list = 1 : 10 : 325 : zipWith (+) a164055_list
%o A164055 (map (* 35) $ tail $ zipWith (-) (tail a164055_list) a164055_list)
%o A164055 -- _Reinhard Zumkeller_, Apr 29 2012
%o A164055 (PARI) my(x='x+O('x^50)); Vec(x*(1-25*x+10*x^2)/((1-x)*(x^2-34*x+1))) \\ _G. C. Greubel_, Sep 09 2017
%Y A164055 Cf. A001110, A106329, A075848 (solutions of x^2 - 2 * y^2 = 9).
%Y A164055 Cf. A072221 (indices of the triangular terms of this sequence), A106328 (indices of the corresponding square numbers).
%K A164055 nonn,easy
%O A164055 1,2
%A A164055 _Tanya Khovanova_ & Alexey Radul, Aug 08 2009
%E A164055 Comment molded into formula by _R. J. Mathar_, Sep 22 2009