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 A060961 Number of compositions (ordered partitions) of n into 1's, 3's and 5's. 6
 1, 1, 1, 2, 3, 5, 8, 12, 19, 30, 47, 74, 116, 182, 286, 449, 705, 1107, 1738, 2729, 4285, 6728, 10564, 16587, 26044, 40893, 64208, 100816, 158296, 248548, 390257, 612761, 962125, 1510678, 2371987, 3724369, 5847808, 9181920, 14416967, 22636762, 35543051 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS Lim_{n->infinity} a(n)/a(n-1) = 1.57014... = A293506. This is the largest absolute value of a root of the characteristic polynomial of the recursion (x^5 - x^4 - x^2 - 1), same as the inverse of smallest absolute value of a root of the reciprocal (here -x^5 - x^3 - x + 1, the denominator of the g.f.) of the characteristic polynomial. - Bob Selcoe, Jun 09 2013 From Bob Selcoe, May 01 2014: (Start) Since a(n) is a recurrence of the form: a(n) = Sum_(a(n-Fi)), i=1..z; where F(i)-F(i-1) is constant (C), and seed values are a(0)=1 and a(<0)=0 exclusively; then apply the following definitions: I.  For T-nomial triangles T(m,k), let T be defined as the number of terms in the recurrence equaling a(n).  That is, T => z => ((Fz-F1)/C)+1.  In this sequence, F1=1, C=2 and T => z => ((5-1)/2)+1 = 3.  Therefore, the applicable triangle is trinomial for this sequence. II.  Let m' be defined as the maxval of m and k' the minval of k such that n = m'*F1+k'*C.  For example, in this sequence:  n=7: m'=7 and k'=0 because 7*1+0*2=7. (Note that m' always equals n and k' always equals 0 when F1=1) III.  THEN:  a(n) = Sum_T((m'-C*j/G),(k'+F1*j/G)), j=0..q; where (m'-C*q)) is the floor and (k'+F1*q) the ceiling for the T-nomial triangle, and G is the greatest common factor of all Fi.  In general, T, F1, C and G are invariant across n; while m', k' and q vary (the exception being k' always equaling 0 when F1=1). In this sequence, T=3, F1=1, C=2, G=1 and k'=0;  m' and q vary with a(n). Example 1.  a(11):  T=3, F1=1, C=2, G=1, k'=0 (invariant);  m'=11, q=4.  a(11) = 74 => T(11,0) + T(9,1) + T(7,2) + T(5,3) + T(3,4) for T==trinomial triangle.  T(11,0)=1, T(9,1)=9, T(7,2)=28, T(5,3)=30 and T(3,4)=6.  1+9+28+30+6 = 74  (Note that T(3,4) is the final term because the ostensible next term [T(1,5)] is not contained in the trinomial triangle.  Therefore q=4.) Example 2.  a(14):  m'=14, q=5.  a(14) = 286 => T(14,0) + T(12,1) + T(10,2) + T(8,3) + T(6,4) + T(4,5) => 1+12+55+112+90+16 = 286. (End) LINKS Harry J. Smith, Table of n, a(n) for n = 0..500 Index entries for linear recurrences with constant coefficients, signature (1,0,1,0,1). FORMULA a(n) = a(n-1) + a(n-3) + a(n-5). G.f.: 1 / (1-(x+x^3+x^5)). a(n) = Sum_{m=floor((n+1)/2)..n} (Sum_{j=0..2*m-n} binomial(j,3*n-5*m+2*j)*binomial(2*m-n,j))), n>0, a(0)=1. - Vladimir Kruchinin, Mar 11 2013 MATHEMATICA CoefficientList[Series[ 1 /(1 - z - z^3 - z^5), {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jun 10 2011 *) PROG (PARI) N=66; x='x+O('x^N); Vec(1/(1-x-x^3-x^5)) \\ Joerg Arndt, Oct 21 2012 (Maxima) a(n):=sum((sum(binomial(j, 3*n-5*m+2*j)*binomial(2*m-n, j), j, 0, 2*m-n)), m, floor((n+1)/2), n); \\ Vladimir Kruchinin, Mar 11 2013 CROSSREFS Cf. A293506 (growth power). Cf. A060945 (compositions into 1's, 2's, and 4's). Cf. A027907 (trinomial coefficients triangle). Sequence in context: A023436 A024567 A303668 * A225393 A243850 A179018 Adjacent sequences:  A060958 A060959 A060960 * A060962 A060963 A060964 KEYWORD nonn,easy AUTHOR Len Smiley, May 08 2001 EXTENSIONS a(0)=1 prepended by Joerg Arndt, Oct 21 2012 STATUS approved

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Last modified June 14 20:21 EDT 2021. Contains 345038 sequences. (Running on oeis4.)