A195156 a(n) = (16^n-1)/3.

0, 5, 85, 1365, 21845, 349525, 5592405, 89478485, 1431655765, 22906492245, 366503875925, 5864062014805, 93824992236885, 1501199875790165, 24019198012642645, 384307168202282325, 6148914691236517205, 98382635059784275285, 1574122160956548404565

Offset: 0

Author

Omar E. Pol, Sep 10 2011

Keywords

• nonn
• easy

Comments

Numbers of A002450 that are multiples of 5. Also sequence found by reading the line from 0, in the direction 0, 5,..., in the square spiral whose edges are the Jacobsthal numbers A001045 and whose vertices are the numbers A000975. This is a semi-diagonal in the spiral.

In binary, these numbers are 101...01 (see A031982). - Alonso del Arte, May 20 2017

0 together with Jacobsthal numbers ending with the decimal digit 5. - Jianing Song, Aug 30 2022

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..800
• Index entries for linear recurrences with constant coefficients, signature (17,-16).

Crossrefs

Bisection of A002450.

Cf. A000975, A001025, A013777, A031982, A131865.

First quadrisection of Jacobsthal numbers A001045; the other quadrisections are A139792 (second), A144864 (third), and A141060 (fourth).

Programs

Magma

[(16^n-1)/3:n in [0..20]]; // Vincenzo Librandi, Sep 20 2011

Maple

A195156:=n->(16^n-1)/3; seq(A195156(k), k=0..50); # Wesley Ivan Hurt, Oct 24 2013

Mathematica

Table[(16^n - 1)/3, {n, 0, 63}] (* Wesley Ivan Hurt, Oct 24 2013 *)

PARI

for(n=0,50, print1((16^n - 1)/3, ", ")) \\ G. C. Greubel, Oct 11 2017

Formula

From Bruno Berselli, Sep 19 2011: (Start)

G.f.: 5*x/((1-x)*(1-16*x)).

a(n) = A002450(2n) = (16^n-1)/3.

a(n) = 5*A131865(n-1) = a(n-1) + 5*A001025(n-1) = 16*a(n-1) + 5 for n > 0. (End)

From Jianing Song, Aug 30 2022: (Start)

a(n) = A001045(4*n).

a(n+1) - a(n) = 10*A013777(n-1) = 80*A001025(n-1) for n >= 1. (End)

E.g.f.: exp(x)*(exp(15*x) - 1)/3. - Stefano Spezia, Dec 17 2022

Extensions

New sequence name suggested by Charles R Greathouse IV using Berselli's formula. - Sep 19 2011