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

1, 331, 159391, 76825981, 37029963301, 17848365484951, 8602875133782931, 4146567966117887641, 1998637156793688059881, 963338963006591526974851, 464327381532020322313818151, 223804834559470788763733373781

Offset: 1

Author

Richard Choulet, Sep 21 2007

Keywords

• nonn
• easy

Comments

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

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);

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);

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);

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).

Links

• Colin Barker, Table of n, a(n) for n = 1..373
• Index entries for linear recurrences with constant coefficients, signature (483,-483,1).

Crossrefs

Cf. A003215, A005891, A254782, A133285.

Programs

Mathematica

LinearRecurrence[{483,-483,1},{1,331,159391},20] (* Paolo Xausa, Jan 07 2024 *)

PARI

Vec(-x*(x^2-152*x+1)/((x-1)*(x^2-482*x+1)) + O(x^100)) \\ Colin Barker, Feb 07 2015

Formula

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

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

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

Extensions

More terms from Paolo P. Lava, Sep 26 2008