

A005708


a(n) = a(n1) + a(n6), with a(i) = 1 for i = 0..5.
(Formerly M0496)


25



1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 9, 12, 16, 21, 27, 34, 43, 55, 71, 92, 119, 153, 196, 251, 322, 414, 533, 686, 882, 1133, 1455, 1869, 2402, 3088, 3970, 5103, 6558, 8427, 10829, 13917, 17887, 22990, 29548, 37975, 48804, 62721, 80608, 103598, 133146, 171121, 219925, 282646
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OFFSET

0,7


COMMENTS

This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n1) + a(nm), with a(n) = 1 for n = 0...m1. The generating function is 1/(1xx^m). Also a(n) = sum_{i=0..n/m} binomial(n(m1)*i, i). This family of binomial summations or recurrences gives the number of ways to cover (without overlapping) a linear lattice of n sites with molecules that are m sites wide. Special case: m=1: A000079; m=4: A003269; m=5: A003520; m=6: A005708; m=7: A005709; m=8: A005710.
For n>=6, a(n6) = number of compositions of n in which each part is >=6.  Milan Janjic, Jun 28 2010
Number of compositions of n into parts 1 and 6.  Joerg Arndt, Jun 24 2011
The compositions of n in which each natural number is colored by one of p different colors are called pcolored compositions of n. For n>=6, 2*a(n6) equals the number of 2colored compositions of n with all parts >=6, such that no adjacent parts have the same color.  Milan Janjic, Nov 27 2011
a(n+5) equals the number of binary words of length n having at least 5 zeros between every two successive ones.  Milan Janjic, Feb 07 2015


REFERENCES

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS

T. D. Noe, Table of n, a(n) for n=0..500
E. Di Cera and Y. Kong, Theory of multivalent binding in one and twodimensional lattices, Biophysical Chemistry, Vol. 61 (1996), pp. 107124.
V. C. Harris, C. C. Styles, A generalization of Fibonacci numbers, Fib. Quart. 2 (1964) 277289, sequence u(n,5,1).
D. Kleitman, Solution to Problem E3274, Amer. Math. Monthly, 98 (1991), 958959.
D. Newman, Problem E3274, Amer. Math. Monthly, 95 (1988), 555.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.
Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 379
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,1).


FORMULA

G.f.: 1/(1xx^6).  Simon Plouffe in his 1992 dissertation
a(n) = term (1,1) in the 6 X 6 matrix [1,1,0,0,0,0; 0,0,1,0,0,0; 0,0,0,1,0,0; 0,0,0,0,1,0; 0,0,0,0,0,1]; 1,0,0,0,0,0]^n.  Alois P. Heinz, Jul 27 2008
For positive integers n and k such that k <= n <= 6*k and 5 divides nk, define c(n,k) = binomial(k,(nk)/5), and c(n,k)=0, otherwise. Then, for n>= 1, a(n) = sum_{k=1..n} c(n,k).  Milan Janjic, Dec 09 2011
Apparently a(n) = hypergeometric([1/6n/6, 1/3n/6, 1/2n/6, 2/3n/6, 5/6n/6, n/6], [1/5n/5, 2/5n/5, 3/5 n/5, 4/5n/5, n/5], 6^6/5^5) for n>=25.  Peter Luschny, Sep 19 2014


MAPLE

with(combstruct): SeqSetU := [S, {S=Sequence(U), U=Set(Z, card > 5)}, unlabeled]: seq(count(SeqSetU, size=j), j=6..59); # Zerinvary Lajos, Oct 10 2006
ZL:=[S, {a = Atom, b = Atom, S = Prod(X, Sequence(Prod(X, b))), X = Sequence(b, card >= 5)}, unlabelled]: seq(combstruct[count](ZL, size=n), n=5..58); # Zerinvary Lajos, Mar 26 2008
M := Matrix(6, (i, j)> if j=1 and member(i, [1, 6]) then 1 elif (i=j1) then 1 else 0 fi); a:= n> (M^(n))[1, 1]; seq(a(n), n=0..60); # Alois P. Heinz, Jul 27 2008


MATHEMATICA

LinearRecurrence[{1, 0, 0, 0, 0, 1}, {1, 1, 1, 1, 1, 1}, 80] (* Vladimir Joseph Stephan Orlovsky, Feb 16 2012 *)


PROG

(PARI) x='x+O('x^66); Vec(x/(1(x+x^6))) /* Joerg Arndt, Jun 25 2011 */


CROSSREFS

Cf. A000045, A000079, A000930, A003269, A003520, A005709, A005710, A005711.
Sequence in context: A193286 A098132 A017900 * A085793 A281809 A143286
Adjacent sequences: A005705 A005706 A005707 * A005709 A005710 A005711


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane


EXTENSIONS

Additional comments from Yong Kong (ykong(AT)curagen.com), Dec 16 2000


STATUS

approved



