A003948 Expansion of (1+x)/(1-5*x).

1, 6, 30, 150, 750, 3750, 18750, 93750, 468750, 2343750, 11718750, 58593750, 292968750, 1464843750, 7324218750, 36621093750, 183105468750, 915527343750, 4577636718750, 22888183593750, 114440917968750, 572204589843750, 2861022949218750, 14305114746093750

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice
• walk

Comments

Coordination sequence for infinite tree with valency 6.

The n-th term of the coordination sequence of the infinite tree with valency 2m is the same as the number of reduced words of size n in the free group on m generators. In the five sequences A003946, A003948, A003950, A003952, A003954, m is 2, 3, 4, 5, 6. - Avi Peretz (njk(AT)netvision.net.il), Feb 23 2001 and Ola Veshta (olaveshta(AT)my-deja.com), Mar 30 2001

Hamiltonian path in S_4 X P_2n.

For n>=1, a(n+1) is equal to the number of functions f:{1,2,...,n+1}->{1,2,3,4,5,6} such that for fixed, different x_1, x_2,...,x_n in {1,2,...,n+1} and fixed y_1, y_2,...,y_n in {1,2,3,4,5,6} we have f(x_i)<>y_i, (i=1..n). - Milan Janjic, May 10 2007

For n>=1, a(n) equals the numbers of words of length n over the alphabet {0..5} with no two adjacent letters identical. - Milan Janjic, Jan 31 2015 [Corrected by David Nacin, May 30 2017]

a(n) equals the numbers of sequences of length n on {0,...,5} where no two adjacent terms differ by three. - David Nacin, May 30 2017

It appears that these are the only n>1 for which alpha(n)=2n, where alpha(n) is the entry point of n in the Fibonacci sequence, see A001177. - Philippe Schnoebelen, Apr 11 2024

Links

• T. D. Noe, Table of n, a(n) for n = 0..200
• F. Faase, Counting Hamiltonian cycles in product graphs
• INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 307
• Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
• A. M. Nemirovsky et al., Marriage of exact enumeration and 1/d expansion methods: lattice model of dilute polymers, J. Statist. Phys., 67 (1992), 1083-1108.
• Index to divisibility sequences
• Index entries for linear recurrences with constant coefficients, signature (5).
• Index entries for sequences related to trees

Crossrefs

Cf. A003946, A003947, A003949, A003950, A003952, A003954, A029653.

Programs

GAP

Concatenation([1], List([1..30], n-> 6*5^(n-1) )); # G. C. Greubel, Sep 24 2019

Magma

[1] cat [6*5^(n-1): n in [1..30]]; // G. C. Greubel, Sep 24 2019

Maple

k := 6; if n = 0 then 1 else k*(k-1)^(n-1); fi;

Mathematica

q = 6; Join[{a = 1}, Table[If[n != 0, a = q*a - a, a = q*a], {n, 0, 25}]] (* and *) Join[{1}, 6*5^Range[0, 25]] (* Vladimir Joseph Stephan Orlovsky, Jul 11 2011 *)
Join[{1},NestList[5#&,6,30]] (* Harvey P. Dale, Dec 31 2013 *)
CoefficientList[Series[(1+x)/(1-5x), {x,0,30}], x] (* Michael De Vlieger, Dec 10 2016 *)

PARI

Vec((1+x)/(1-5*x)+O(x^30)) \\ Charles R Greathouse IV, Nov 20 2012

Sage

[1]+[6*5^(n-1) for n in (1..30)] # G. C. Greubel, Sep 24 2019

Formula

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

a(n) = Sum_{k=0..n} A029653(n, k)*x^k for x = 4. - Philippe Deléham, Jul 10 2005

The Hankel transform of this sequence is [1,-6,0,0,0,0,0,0,0,0,...]. - Philippe Deléham, Nov 21 2007

a(n) = 6*5^(n-1) for n>0, a(0)=1. - Vincenzo Librandi, Nov 18 2010

G.f.: 2/x - 5 - 8/(x*U(0)) where U(k)= 1 + 2/(3^k - 3^k/(2 + 1 - 12*x*3^k/(6*x*3^k + 1/U(k+1)))) ; (continued fraction, 4-step). - Sergei N. Gladkovskii, Oct 30 2012

E.g.f.: (6*exp(5*x) - 1)/5. - Ilya Gutkovskiy, Dec 10 2016

Sum_{n>=0} 1/a(n) = 29/24. - Bernard Schott, Oct 25 2021

Extensions

Definition corrected by Frans J. Faase, Feb 07 2009

Edited by N. J. A. Sloane, Dec 04 2009