cp's OEIS Frontend

A133141 Numbers which are both centered pentagonal (A005891) and centered hexagonal numbers (A003215).

%I A133141 #23 Jan 07 2024 15:11:01
%S A133141 1,331,159391,76825981,37029963301,17848365484951,8602875133782931,
%T A133141 4146567966117887641,1998637156793688059881,963338963006591526974851,
%U A133141 464327381532020322313818151,223804834559470788763733373781
%N A133141 Numbers which are both centered pentagonal (A005891) and centered hexagonal numbers (A003215).
%C A133141 The problem is to find p and r such that 6*(2*p-1)^2 = 5*(2*r+1)^2 + 1 equivalent to 3*p^2 - 3*p + 1 = (5*r^2 + 5*r + 2)/2. The Diophantine equation (6*X)^2 = 30*Y^2 + 6 is such that
%C A133141 X is given by 1, 21, 461, 10121, ... with a(n+2) = 22*a(n+1) - a(n) and also a(n+1) = 11*a(n) + sqrt(120*a(n)^2 - 20);
%C A133141 Y is given by 1, 23, 805, 11087, ... with a(n+2) = 22*a(n+1) - a(n) and also a(n+1) = 11*a(n) + sqrt(120*a(n)^2+24);
%C A133141 r is given by 0, 11, 252, 5543, 121704, ... with a(n+2) = 22*a(n+1) - a(n) + 10 and also a(n+1) = 11*a(n) + 5 + sqrt(120*a(n)^2 + 120*a(n) + 36);
%C A133141 p is given by 1, 11, 231, 5061, ... with a(n+2) = 22*a(n+1) - a(n) - 10 and also a(n+1) = 11*a(n) - 5 + sqrt(120*a(n)^2 - 120*a(n) + 25).
%H A133141 Colin Barker, <a href="/A133141/b133141.txt">Table of n, a(n) for n = 1..373</a>
%H A133141 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (483,-483,1).
%F A133141 a(n+2) = 482*a(n+1) - a(n) - 150.
%F A133141 a(n+1) = 241*a(n) - 75 + 11*sqrt(480*a(n)^2 - 300*a(n) + 45).
%F A133141 G.f.: z*(1-152*z+z^2)/((1-z)*(1-482*z+z^2)).
%t A133141 LinearRecurrence[{483,-483,1},{1,331,159391},20] (* _Paolo Xausa_, Jan 07 2024 *)
%o A133141 (PARI) Vec(-x*(x^2-152*x+1)/((x-1)*(x^2-482*x+1)) + O(x^100)) \\ _Colin Barker_, Feb 07 2015
%Y A133141 Cf. A003215, A005891, A254782, A133285.
%K A133141 nonn,easy
%O A133141 1,2
%A A133141 _Richard Choulet_, Sep 21 2007
%E A133141 More terms from _Paolo P. Lava_, Sep 26 2008