A133141 - OEIS

%I #23 Jan 07 2024 15:11:01

%S 1,331,159391,76825981,37029963301,17848365484951,8602875133782931,

%T 4146567966117887641,1998637156793688059881,963338963006591526974851,

%U 464327381532020322313818151,223804834559470788763733373781

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

%C 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 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 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 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 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 Colin Barker, <a href="/A133141/b133141.txt">Table of n, a(n) for n = 1..373</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (483,-483,1).

%F a(n+2) = 482*a(n+1) - a(n) - 150.

%F a(n+1) = 241*a(n) - 75 + 11*sqrt(480*a(n)^2 - 300*a(n) + 45).

%F G.f.: z*(1-152*z+z^2)/((1-z)*(1-482*z+z^2)).

%t LinearRecurrence[{483,-483,1},{1,331,159391},20] (* _Paolo Xausa_, Jan 07 2024 *)

%o (PARI) Vec(-x*(x^2-152*x+1)/((x-1)*(x^2-482*x+1)) + O(x^100)) \\ _Colin Barker_, Feb 07 2015

%Y Cf. A003215, A005891, A254782, A133285.

%K nonn,easy

%O 1,2

%A _Richard Choulet_, Sep 21 2007

%E More terms from _Paolo P. Lava_, Sep 26 2008