A133217 Indices of decagonal numbers (A001107) that are also triangular (A000217).

0, 1, 2, 20, 55, 667, 1856, 22646, 63037, 769285, 2141390, 26133032, 72744211, 887753791, 2471161772, 30157495850, 83946756025, 1024467105097, 2851718543066, 34801724077436, 96874483708207, 1182234151527715, 3290880727535960, 40161159427864862

Offset

1,3

Comments

For n>0, a(n) = (A055979(n) - A056161(n))/2, with those two sequences related through the Diophantine equation 2x^2 + 3x + 2 = r^2. - Richard R. Forberg, Nov 24 2013

Links

Table of n, a(n) for n=1..24.

Index entries for linear recurrences with constant coefficients, signature (1, 34, -34, -1, 1).

Formula

For n>5, a(n) = 34*a(n-2) - a(n-4) - 12.

For n>6, a(n) = a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5).

For n>1, a(n) = 1/16 * ((2*sqrt(2) + (-1)^n)*(1 + sqrt(2))^(2*n - 3) - (2*sqrt(2) - (-1)^n)*(1 - sqrt(2))^(2*n - 3) + 6).

For n>1, a(n) = ceiling (1/16*(2*sqrt(2) + (-1)^n)*(1 + sqrt(2))^(2*n - 3)).

G.f.: ( 1 - 33*x^2 + 18*x^3 + 2*x^4 ) / ((1 - x ) * (1 - 6*x + x^2 ) * (1 + 6*x + x^2)).

lim (n -> Infinity, a(2n+1)/a(2n)) = 1/7*(43 + 30*sqrt(2)).

lim (n -> Infinity, a(2n)/a(2n-1)) = 1/7*(11 + 6*sqrt(2)).

Example

The third number which is both decagonal (A001107) and triangular (A000217) is A133216(3)=10. As this is the second decagonal number, we have a(3) = 2.

Mathematica

LinearRecurrence[{1, 34, -34, -1, 1} , {0, 1, 2, 20, 55, 667}, 24] (* first term 0 corrected by Georg Fischer, Apr 02 2019 *)

Crossrefs

Cf. A000217, A001107, A077443, A077442, A133216, A133218.

Sequence in context: A059211 A331090 A139271 * A001504 A192351 A136905

Adjacent sequences: A133214 A133215 A133216 * A133218 A133219 A133220

Keyword

nonn

Author

Richard Choulet, Oct 11 2007; Ant King, Nov 04 2011

Extensions

Entry revised by Max Alekseyev, Nov 06 2011

Status

approved