A107118 Numbers that are both centered triangular numbers (A005448) and centered hexagonal numbers (A003215).

1, 19, 631, 21421, 727669, 24719311, 839728891, 28526062969, 969046412041, 32919051946411, 1118278719765919, 37988557420094821, 1290492673563457981, 43838762343737476519, 1489227427013510743651, 50589893756115627807601, 1718567160280917834714769

Offset: 1

Author

Richard Choulet, Sep 18 2007

Keywords

• nonn
• easy

Comments

The centered hexagonal numbers are given by 3*p^2 - 3*p + 1 while the centered triangular numbers are given by (3*r^2 + 3*r + 2)/2. A natural number is both of the above numbers if and only if there exist numbers p and r such that 2*(2p-1)^2 = (2*r+1)^2+1. The Diophantine equation X^2 = 2*Y^2 - 1 has the following solutions: X is given by 1, 7, 41, 239, ..., i.e., A002315, and Y is given by A001653. The first equation gives r with 0, 3, 20, 119, 6906, i.e., A001652, and p with 1, 3, 15, 85, 493, ..., i.e., A011900.

Links

• Colin Barker, Table of n, a(n) for n = 1..654
• Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Crossrefs

Cf. A003215 (Centered hexagonal numbers), A005448 (Centered triangular numbers).

Cf. A001652, A001653, A002315, A011900.

Programs

Mathematica

a[n_] := 17*n - 7 + Sqrt[288*n^2 - 252*n + 45]; NestList[a, 1, 20] (* Stefan Steinerberger, Sep 18 2007 *)
LinearRecurrence[{35,-35,1},{1,19,631},30] (* Harvey P. Dale, Jan 16 2016 *)

PARI

Vec(-x*(x^2-16*x+1)/((x-1)*(x^2-34*x+1)) + O(x^100)) \\ Colin Barker, Jan 02 2015

Formula

a(n+2) = 34*a(n+1) - a(n) - 14.

a(n+1) = 17*a(n) - 7 + sqrt(288*a(n)^2 - 252*a(n) + 45).

G.f.: h(z)=(z*(1-16*z+z^2))/((1-z)*(1-34*z+z^2)).

a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3). - Colin Barker, Jan 02 2015

a(n) = (14+(9+6*sqrt(2))*(17+12*sqrt(2))^(-n)+(9-6*sqrt(2))*(17+12*sqrt(2))^n)/32. - Colin Barker, Mar 02 2016

Extensions

More terms from Stefan Steinerberger, Sep 18 2007