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%I #19 Sep 08 2022 08:46:12
%S 3,29,75,432,998,3624,8310,44717,102443,370269,848195,4561352,
%T 10448838,37764464,86508230,465213837,1065679683,3851605709,
%U 8822991915,47447250672,108688879478,392826018504,899858667750,4839154355357,11085200027723,40064402282349
%N Numbers n such that T(n) + T(n+1) + ... + T(n+12) is a square, where T = A000217 (triangular numbers).
%C It is well known that T(n)+T(n+1) is always a square. T(n)+T(n+1)+T(n+2) is a square for n in A165517. T(n)+T(n+1)+T(n+2)+T(n+3) is a square for n in A202391. There is no sequence of 5, 6, 7, 8, 9 or 10 consecutive T(i)'s which sum to a square, cf. A176541. The next possible length is 11, see A116476. Then comes this sequence, corresponding to length 13.
%C Positive integers y in the solutions to 2*x^2-13*y^2-169*y-728 = 0. - _Colin Barker_, May 04 2015
%H Colin Barker, <a href="/A257293/b257293.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,102,-102,0,0,-1,1).
%F G.f.: x*(3*x^8+7*x^7+6*x^6+26*x^5-260*x^4-357*x^3-46*x^2-26*x-3) / ((x-1)*(x^4-10*x^2-1)*(x^4+10*x^2-1)). - _Colin Barker_, May 04 2015
%t Select[Range[10^5],IntegerQ[Sqrt[(#^2+13*#+56)*13/2]]&] (* _Ivan N. Ianakiev_, May 04 2015 *)
%t LinearRecurrence[{1, 0, 0, 102, -102, 0, 0, -1, 1}, {3, 29, 75, 432, 998, 3624, 8310, 44717, 102443}, 50] (* _Vincenzo Librandi_, May 05 2015 *)
%o (PARI) for(n=0,10^8,issquare(binomial(n+14,3)-binomial(n+1,3))&&print1(n","))
%o (PARI) Vec(x*(3*x^8+7*x^7+6*x^6+26*x^5-260*x^4-357*x^3-46*x^2-26*x-3) / ((x-1)*(x^4-10*x^2-1)*(x^4+10*x^2-1)) + O(x^100)) \\ _Colin Barker_, May 04 2015
%o (Magma) I:=[3,29,75,432,998,3624,8310,44717,102443]; [n le 9 select I[n] else Self(n-1)+102*Self(n-4)-102*Self(n-5)-Self(n-8)+Self(n-9): n in [1..40]]; // _Vincenzo Librandi_, May 05 2015
%Y Cf. A176541, A176542, A000217, A000292, A001110, A077415.
%Y Cf. A116476 (length 11).
%K nonn,easy
%O 1,1
%A _M. F. Hasler_, May 04 2015
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