A048328 Numbers that are repdigits in base 3.

0, 1, 2, 4, 8, 13, 26, 40, 80, 121, 242, 364, 728, 1093, 2186, 3280, 6560, 9841, 19682, 29524, 59048, 88573, 177146, 265720, 531440, 797161, 1594322, 2391484, 4782968, 7174453, 14348906, 21523360, 43046720, 64570081, 129140162, 193710244, 387420488, 581130733

Offset: 0

Author

Patrick De Geest, Feb 15 1999

Keywords

• nonn
• base
• easy

Comments

Case for base 2 see A000225: 2^n - 1.

If the sequence b(n) represents the number of paths of length n, n >= 1, starting at node 1 and ending at nodes 1, 2, 3 and 4 on the path graph P_5 then a(n-1) = b(n) - 1. - Johannes W. Meijer, May 29 2010

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000
• Eric Weisstein's World of Mathematics, Repdigit.
• Index entries for 3-automatic sequences.
• Index entries for linear recurrences with constant coefficients, signature (0,4,0,-3).

Crossrefs

Bisections: A024023 and A003462.

Cf. A000225, A010785, A028987, A028988, A033016, A038754, A068911, A124302, A214369.

Programs

Maple

nmax := 35; a(0) := 0: for n from 1 to nmax do a(2*n) := a(2*n-2) + 2*3^(n-1); od: a(1) := 1: for n from 1 to nmax do a(2*n+1) := 1*a(2*n-1) + 3^n; od: seq(a(n), n=0..nmax);
# End program 1
with(GraphTheory): G := PathGraph(5): A:= AdjacencyMatrix(G): nmax := nmax; for n from 1 to nmax+1 do B(n) := A^n; b(n) := add(B(n)[1, k], k=1..4); a1(n-1) := b(n)-1; od: seq(a1(n), n=0..nmax);
# End program 2
# From Johannes W. Meijer, May 29 2010, revised Sep 23 2012
# third Maple program:
a:= n->(<<0|1>, <-3|4>>^iquo(n, 2, 'r').`if`(r=0, <<0, 2>>, <<1, 4>>))[1, 1]:
seq (a(n), n=0..60);  # Alois P. Heinz, Sep 23 2012

Mathematica

Rest[FromDigits[#, 3]&/@Flatten[Table[{PadRight[{1}, n, 1], PadRight[{2}, n, 2]}, {n, 0, 20}], 1]] (* Harvey P. Dale, Feb 03 2011 *)

PARI

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -3,0,4,0]^n*[0;1;2;4])[1,1] \\ Charles R Greathouse IV, Oct 07 2015

Formula

G.f.: (2*x^2+x)/(1-4*x^2+3*x^4). - Alois P. Heinz, Sep 23 2012

Sum_{n>=1} 1/a(n) = 3 * A214369 = 2.04646050781571420028... - Amiram Eldar, Jan 21 2022

a(n) = (3^(n/2)*(sqrt(3) + 2 - (-1)^n*(sqrt(3) - 2)) - 3 - (-1)^n)/4. - Stefano Spezia, Feb 18 2022