A018903 Define the sequence S(a(0),a(1)) by a(n+2) is the least integer such that a(n+2)/a(n+1) > a(n+1)/a(n) for n >= 0. This is S(1,5).

1, 5, 26, 136, 712, 3728, 19520, 102208, 535168, 2802176, 14672384, 76825600, 402264064, 2106281984, 11028635648, 57746685952, 302365573120, 1583206694912, 8289777876992, 43405840482304, 227275931385856, 1190032226385920, 6231089632772096, 32626408891088896

Offset

0,2

Comments

a(n) is the number of compositions of n when there are 5 types of ones. - Milan Janjic, Aug 13 2010

a(n)/a(n-1) tends to (6 + sqrt(20))/2 = 5.236067.... - Gary W. Adamson, Jul 30 2013

Number of words of length n over {0,1,...,6} in which each 1 is followed by at least one zero. - Milan Janjic, Jan 24 2017

Links

Colin Barker, Table of n, a(n) for n = 0..1000

D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory (Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.

Pamela Fleischmann, Jonas Höfer, Annika Huch, and Dirk Nowotka, alpha-beta-Factorization and the Binary Case of Simon's Congruence, arXiv:2306.14192 [math.CO], 2023.

H. D. Nguyen and D. Taggart, Mining the OEIS: Ten Experimental Conjectures, 2013. Mentions this sequence. - From N. J. A. Sloane, Mar 16 2014

INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 879

Index entries for linear recurrences with constant coefficients, signature (6,-4).

Index entries for Pisot sequences

Formula

a(n) = (a(1)+1)*a(n-1) - (a(1)-1)*a(n-2) = 6*a(n-1) - 4*a(n-2).

G.f.: (1 - x)/(1 - 6*x + 4*x^2). - Colin Barker, Feb 04 2012

a(n) = 2^(n-1)*A000045(2*n+3). - Sergio Falcon, Jan 25 2017

a(n) = ((3-sqrt(5))^n*(-2+sqrt(5)) + (2+sqrt(5))*(3+sqrt(5))^n) / (2*sqrt(5)). - Colin Barker, Jan 20 2017

Equivalent to the first formula: a(n)=a(1)+a(2)+....+a(n-2)+5*a(n-1). - Arie Bos, May 05 2017

a(n) = Sum_{k>=1} binomial(k+n-1,n) * A000045(k) / 2^(k+1). - Diego Rattaggi, Jun 25 2020

Mathematica

LinearRecurrence[{6, -4}, {1, 5}, 24] (* or *)
CoefficientList[Series[(1 - x)/(1 - 6 x + 4 x^2), {x, 0, 23}], x] (* or *)
a = {1, 5}; Do[AppendTo[a, 6 a[[n - 1]] - 4 a[[n - 2]]], {n, 3, 24}];
a (* or *)
S[a_, b_, n_] := Block[{s = {a, b}, k}, Do[k = Last@ s + 1; While[k/s[[i - 1]] <= s[[i - 1]]/s[[i - 2]], k++]; AppendTo[s, k], {i, 3, n}]; s]; S[1, 5, 10] (* Michael De Vlieger, Feb 15 2016 *)

Prog

(PARI) S(a0, a1, maxn) = a=vector(maxn); a[1]=a0; a[2]=a1; for(n=3, maxn, a[n]=a[n-1]^2\a[n-2]+1); a
S(1, 5, 40) \\ Colin Barker, Feb 15 2016
(PARI) Vec((1-x)/(1-6*x+4*x^2) + O(x^40)) \\ Colin Barker, Feb 15 2016

Crossrefs

Sequence in context: A052918 A255633 A255815 * A355361 A083331 A076025

Adjacent sequences: A018900 A018901 A018902 * A018904 A018905 A018906

Keyword

nonn,easy

Author

R. K. Guy

Status

approved