

A003269


a(n) = a(n1) + a(n4) with a(0) = 0, a(1) = a(2) = a(3) = 1.
(Formerly M0526)


91



0, 1, 1, 1, 1, 2, 3, 4, 5, 7, 10, 14, 19, 26, 36, 50, 69, 95, 131, 181, 250, 345, 476, 657, 907, 1252, 1728, 2385, 3292, 4544, 6272, 8657, 11949, 16493, 22765, 31422, 43371, 59864, 82629, 114051, 157422, 217286, 299915, 413966, 571388, 788674, 1088589, 1502555, 2073943
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OFFSET

0,6


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 this family of sequences, a(n+1) is the number of compositions of n+1 into parts 1 and m. For n>=m, a(nm+1)is the number of compositions of n in which each part is greater than m or equivalently, in which parts 1 through m are excluded.  Gregory L. Simay, Jul 14 2016
For this family of sequences, let a(m,n) = a(n1) + a(nm). Then the number of compositions of n having m as a least summand is a(m, nm)  a(m+1, nm1).  Gregory L. Simay, Jul 14 2016
For n>=3, a(n3) = number of compositions of n in which each part is >=4.  Milan Janjic, Jun 28 2010
For n>=1, number of compositions of n into parts == 1 (mod 4). Example: a(8)=5 because there are 5 compositions of 8 into parts 1 or 5: (1,1,1,1,1,1,1,1), (1,1,1,5), (1,1,5,1), (1,5,1,1), (5,1,1,1).  Adi Dani, Jun 16 2011
a(n+1) is the number of compositions of n into parts 1 and 4.  Joerg Arndt, Jun 25 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>=4, 2*a(n3) equals the number of 2colored compositions of n with all parts >= 4, such that no adjacent parts have the same color.  Milan Janjic, Nov 27 2011
Number of permutations satisfying k<=p(i)i<=r and p(i)i not in I, i=1..n, with k=1, r=3, I={1,2}.  Vladimir Baltic, Mar 07 2012
a(n+4) equals the number of binary words of length n having at least 3 zeros between every two successive ones.  Milan Janjic, Feb 07 2015
Let T* be the infinite tree with root 0 generated by these rules: if p is in T*, then p+1 is in T* and x*p is in T*.
Let g(n) be the set of nodes in the nth generation, so that g(0) = {0}, g(1) = {1}, g(2) = {2,x}, g(3) = {3, 2*x, x+1, x^2}, etc.
Let T(r) be the tree obtained by substituting r for x.
If N is a positive integer such that r = N^(1/4) is not an integer, then the number of (not necessarily distinct) integers in g(n) is A003269(n), for n > = 1. See A274142. (End)


REFERENCES

A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 120.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS



FORMULA

G.f.: x/(1xx^4).
G.f.: 1 + 1/(1Sum_{k>=0} x^(4*k+1)).
a(n) = a(n3) + a(n4) + a(n5) + a(n6) for n>4.
a(n) = floor(d*c^n + 1/2) where c is the positive real root of x^4+x^3+1 and d is the positive real root of 283*x^418*x^28*x1 (c=1.38027756909761411... and d=0.3966506381592033124...).  Benoit Cloitre, Nov 30 2002
Equivalently, a(n) = floor(c^(n+3)/(c^4+3) + 1/2) with c as defined above (see A086106).  Greg Dresden and Shuer Jiang, Aug 31 2019
a(n) = term (1,2) in the 4 X 4 matrix [1,1,0,0; 0,0,1,0; 0,0,0,1; 1,0,0,0]^n.  Alois P. Heinz, Jul 27 2008
a(n+1) = Sum_{k=0..n} C((n+3*k)/4,k)*((1+(1)^(nk))/2 + cos(Pi*n/2))/2;
a(n+1) = Sum_{k=0..n} C(k,floor((nk)/3))(2*cos(2*Pi*(nk)/3)+1)/3. (End)
G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1  x*(2*k+1 + x^3)/( x*(2*k+2 + x^3) + 1/Q(k+1) )); (continued fraction).  Sergei N. Gladkovskii, Aug 29 2013
Appears a(n) = hypergeometric([1/4n/4,1/2n/4,3/4n/4,1n/4], [1/3n/3,2/3n/3,1n/3], 4^4/3^3) for n>=10.  Peter Luschny, Sep 18 2014


EXAMPLE

G.f.: x + x^2 + x^3 + x^4 + 2*x^5 + 3*x^6 + 4*x^7 + 5*x^8 + 7*x^9 + 10*x^10 + ...
The number of compositions of 12 having 4 as a least summand is a(4, 12 4 + 1)  a(5, 12  5 + 1) = A003269(9)  A003520(8) = 74 = 3. The compositions are (84), (48) and (444).  Gregory L. Simay, Jul 14 2016


MAPLE

with(combstruct): SeqSetU := [S, {S=Sequence(U), U=Set(Z, card > 3)}, unlabeled]: seq(count(SeqSetU, size=j), j=4..51);
seq(add(binomial(n3*k, k), k=0..floor(n/3)), n=0..47); # Zerinvary Lajos, Apr 03 2007
ZL:=[S, {a = Atom, b = Atom, S = Prod(X, Sequence(Prod(X, b))), X = Sequence(b, card >= 3)}, unlabelled]: seq(combstruct[count](ZL, size=n), n=3..50); # Zerinvary Lajos, Mar 26 2008
M:= Matrix(4, (i, j)> if j=1 then [1, 0, 0, 1][i] elif (i=j1) then 1 else 0 fi); a:= n> (M^(n))[1, 2]; seq(a(n), n=0..48); # Alois P. Heinz, Jul 27 2008


MATHEMATICA

a[0]= 0; a[1]= a[2]= a[3]= 1; a[n_]:= a[n]= a[n1] + a[n4]; Table[a[n], {n, 0, 50}]
CoefficientList[Series[x/(1xx^4), {x, 0, 50}], x] (* Zerinvary Lajos, Mar 29 2007 *)
Table[Sum[Binomial[n3*i1, i], {i, 0, (n1)/3}], {n, 0, 50}]


PROG

(PARI) {a(n) = polcoeff( if( n<0, (1 + x^3) / (1 + x^3  x^4), 1 / (1  x  x^4)) + x * O(x^abs(n)), abs(n))} /* Michael Somos, Jul 12 2003 */
(Haskell)
a003269 n = a003269_list !! n
a003269_list = 0 : 1 : 1 : 1 : zipWith (+) a003269_list
(drop 3 a003269_list)
(Magma) I:=[0, 1, 1, 1]; [n le 4 select I[n] else Self(n1) + Self(n4) :n in [1..50]]; // Marius A. Burtea, Sep 13 2019
(SageMath)
@CachedFunction
def a(n): return ((n+2)//3) if (n<4) else a(n1) + a(n4) # a = A003269


CROSSREFS

See A017898 for an essentially identical sequence.


KEYWORD

nonn,easy


AUTHOR



EXTENSIONS

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


STATUS

approved



