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A180926 Numbers k such that 6*k and 10*k are triangular numbers.

%I A180926 #37 Jul 29 2024 10:00:10
%S A180926 0,1,63,3906,242110,15006915,930186621,57656563588,3573776755836,
%T A180926 221516502298245,13730449365735355,851066344173293766,
%U A180926 52752382889378478138,3269796672797292350791,202674641330542747270905
%N A180926 Numbers k such that 6*k and 10*k are triangular numbers.
%C A180926 From _Klaus Purath_, Jul 25 2024: (Start)
%C A180926 Numbers k such that 48k + 1 is a square as well as the sum of two consecutive terms.
%C A180926 a(n) = t(n-1)*t(n)/(8*t(1)^2) where (t) is any recurrence t(k) = 8*t(k-1) - t(k-2) with t(0) = 0 and arbitrary t(1) != 0. (End)
%H A180926 Vincenzo Librandi, <a href="/A180926/b180926.txt">Table of n, a(n) for n = 1..500</a>
%H A180926 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (63,-63,1).
%F A180926 a(n) = (62*a(n-1) + 1 + ((48*a(n-1) + 1)*(80*a(n-1) + 1))^(1/2))/2 with a(1)=0.
%F A180926 G.f.: -x^2 / ((x-1)*(x^2-62*x+1)). - _Colin Barker_, Jun 25 2014
%F A180926 a(n) = (-8+(4+sqrt(15))*(31+8*sqrt(15))^(-n) - (-4+sqrt(15))*(31+8*sqrt(15))^n)/480. - _Colin Barker_, Mar 03 2016
%t A180926 a[1] = 0; a[n_] := a[n] = (62 a[n - 1] + 1 + Sqrt[(48 a[n - 1] + 1)*(80 a[n - 1] + 1)])/2; Array[a, 14] (* _Robert G. Wilson v_, Sep 27 2010 *)
%t A180926 Rest[CoefficientList[Series[-x^2/((x - 1) (x^2 - 62 x + 1)), {x, 0, 30}], x]] (* _Vincenzo Librandi_, Jun 26 2014 *)
%t A180926 LinearRecurrence[{63,-63,1},{0,1,63},20] (* _Harvey P. Dale_, Dec 25 2019 *)
%o A180926 (PARI) isok(n) = ispolygonal(6*n, 3) && ispolygonal(10*n, 3); \\ _Michel Marcus_, Jun 25 2014
%Y A180926 Subsequence of A154293.
%Y A180926 Cf. A069498, A154293.
%K A180926 easy,nonn
%O A180926 1,3
%A A180926 _Vladimir Pletser_, Sep 25 2010
%E A180926 a(8) onwards from _Robert G. Wilson v_, Sep 27 2010