A133853 a(n) = (64^n - 1)/63.

0, 1, 65, 4161, 266305, 17043521, 1090785345, 69810262081, 4467856773185, 285942833483841, 18300341342965825, 1171221845949812801, 74958198140788019265, 4797324681010433232961, 307028779584667726909505, 19649841893418734522208321, 1257589881178799009421332545

Offset: 0

Author

Paul Curtz, Jan 07 2008

Keywords

• nonn
• easy

Comments

Partial sums of powers of 64 (A089357), a.k.a. q-numbers for q=64.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..500
• Quynh Nguyen, Jean Pedersen, and Hien T. Vu, New Integer Sequences Arising From 3-Period Folding Numbers, Vol. 19 (2016), Article 16.3.1. See Table 1.
• Index entries related to partial sums
• Index entries related to q-numbers
• Index entries for linear recurrences with constant coefficients, signature (65,-64).

Crossrefs

Cf. A000364.

Cf. similar sequences of the form (k^n-1)/(k-1) listed in A269025.

Programs

Magma

[(64^n-1)/63: n in [0..20]]; // Vincenzo Librandi, Aug 10 2011

Mathematica

LinearRecurrence[{65,-64},{0,1},20] (* Harvey P. Dale, Aug 20 2017 *)

Maxima

makelist((64^n-1)/63, n, 0, 20); /* Martin Ettl, Nov 12 2012 */

PARI

A133853(n)=64^n\63

Formula

From Wolfdieter Lang, Apr 08 2022: (Start)

a(n) = Sum_{j=0..n-1} 2^(6*j). See the comment.

G.f.: x/((1 - 64*x)*(1 - x)).

E.g.f.: exp(x)*(exp(63*x) - 1)/63. (End)

Extensions

a(6)-a(15) from Vincenzo Librandi, Aug 10 2011