OFFSET
0,3
COMMENTS
Number of compositions of n+1 containing exactly one 1. - Emeric Deutsch, Mar 08 2002
Number of permutations with one fixed point avoiding 231 and 321.
A singleton is a run of length 1. - Michael Somos, Nov 29 2014
Second column of A105422. - Michael Somos, Nov 29 2014
Number of weak compositions of n with one 0 and no 1's. Example: Combine one 0 with the compositions of 5 without 1 to get a(5) = 8 weak compositions: 0,5; 5,0; 0,2,3; 0,3,2; 2,0,3; 3,0,2; 2,3,0; 3,2,0. - Gregory L. Simay, Mar 21 2018
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Ricardo Gómez Aíza, Symbolic dynamical scales: modes, orbitals, and transversals, arXiv:2009.02669 [math.DS], 2020.
Jia Huang, Partially Palindromic Compositions, J. Int. Seq. (2023) Vol. 26, Art. 23.4.1. See pp. 4, 11.
Mengmeng Liu and Andrew Yezhou Wang, The Number of Designated Parts in Compositions with Restricted Parts, J. Int. Seq., Vol. 23 (2020), Article 20.1.8.
J. J. Madden, A generating function for the distribution of runs in binary words, arXiv:1707.04351 [math.CO], 2017, Theorem 1.1, r=k=1.
T. Mansour and A. Robertson, Refined restricted permutations avoiding subsets of patterns of length three, arXiv:math/0204005 [math.CO], 2002.
Index entries for linear recurrences with constant coefficients, signature (2,1,-2,-1).
FORMULA
a(n) = a(n-1) + a(n-2) + Fibonacci(n-3).
G.f.: (1-x)^2/(1-x-x^2)^2. - Emeric Deutsch, Mar 08 2002
Convolution square of A212804. - Michael Somos, Nov 29 2014
a(n) = -(-1)^n * A004798(-1-n) for all n in Z. - Michael Somos, Nov 29 2014
0 = a(n)*(-2*a(n) - 7*a(n+1) + 2*a(n+2) + a(n+3)) + a(n+1)*(-4*a(n+1) + 10*a(n+2) - 2*a(n+3)) + a(n+2)*(+4*a(n+2) - 7*a(n+3)) + a(n+3)*(+2*a(n+3)) for all n in Z. - Michael Somos, Nov 29 2014
a(n) = (n*Lucas(n-2) + Fibonacci(n))/5 + Fibonacci(n-1). - Ehren Metcalfe, Jul 29 2017
EXAMPLE
a(4) = 5 because among the 2^4 compositions of 5 only 4+1,1+4,2+2+1,2+1+2,1+2+2 contain exactly one 1.
a(4) = 5 because the binary vectors of length 4+1 beginning with 0 and with exactly one singleton are: 00001, 00100, 00110, 01100, 01111. - Michael Somos, Nov 29 2014
G.f. = 1 + 2*x^2 + 2*x^3 + 5*x^4 + 8*x^5 + 15*x^6 + 26*x^7 + 46*x^8 + ...
MATHEMATICA
nn=36; CoefficientList[Series[1/(1 -x/(1-x) +x)^2, {x, 0, nn}], x] (* Geoffrey Critzer, Feb 18 2014 *)
a[n_]:= If[ n<0, SeriesCoefficient[((1-x)/(1+x-x^2))^2, {x, 0, -2-n}], SeriesCoefficient[((1-x)/(1-x-x^2))^2, {x, 0, n}]]; (* Michael Somos, Nov 29 2014 *)
PROG
(Magma) I:=[1, 0]; [n le 2 select I[n] else Self(n-1)+Self(n-2)+Fibonacci(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 20 2014
(PARI) Vec( (1-x)^2/(1-x-x^2)^2 + O(x^66) ) \\ Joerg Arndt, Feb 20 2014
(PARI) {a(n) = if( n<0, n = -2-n; polcoeff( (1 - x)^2 / (1 + x - x^2)^2 + x * O(x^n), n), polcoeff( (1 - x)^2 / (1 - x - x^2)^2 + x * O(x^n), n))}; /* Michael Somos, Nov 29 2014 */
(Python)
from sympy import fibonacci
from sympy.core.cache import cacheit
@cacheit
def a(n): return 1 if n==0 else 0 if n==1 else a(n - 1) + a(n - 2) + fibonacci(n - 3)
print([a(n) for n in range(51)]) # Indranil Ghosh, Jul 20 2017
(SageMath)
def A006367(n): return (1/5)*(n*lucas_number2(n-2, 1, -1) + fibonacci(n+1) + 4*fibonacci(n-1))
[A006367(n) for n in (0..40)] # G. C. Greubel, Apr 06 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
David M. Bloom
STATUS
approved