OFFSET
0,6
COMMENTS
For n > 0, a(n) is the number of ways to split [n] into an unspecified number of intervals and then choose 4 blocks (i.e., subintervals) from each interval. For example, for n=12, a(12)=1211 since the number of ways to split [12] into intervals and then select 4 blocks from each interval is C(12,4) + C(8,4)*C(4,4) + C(7,4)*C(5,4) + C(6,4)*C(6,4) + C(5,4)*C(7,4) + C(4,4)*C(8,4) + C(4,4)*C(4,4)*C(4,4) for a total of 1211 ways.
For n > 0, a(n) is also the number of compositions of n using parts of size at least 4 where there are binomial(i,4) types of i, i >= 4 (see example).
Number of compositions of 5*n-4 into parts 4 and 5. - Seiichi Manyama, Feb 01 2024
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-4,1).
FORMULA
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 4*a(n-4) + a(n-5), n>=6; a(0)=1, a(1)=a(2)=a(3)=0, a(4)=1, a(5)=5.
G.f.: 1/(1-Sum_{k>=4} binomial(k,4)*x^k).
G.f.: 1/p(S), where p(S) = 1 - S^4 - S^5 and S = x/(1-x).
First differences of A099131. - R. J. Mathar, Jan 29 2024
a(n) = A017827(5*n-4) = Sum_{k=0..floor((5*n-4)/4)} binomial(k,5*n-4-4*k) for n > 0. - Seiichi Manyama, Feb 01 2024
a(n) = Sum_{k=0..floor(n/4)} binomial(n-1+k,n-4*k). - Seiichi Manyama, Feb 02 2024
EXAMPLE
Since there are C(4,4) = 1 type of 4, C(5,4) = 5 types of 5, C(6,4) = 15 types of 6, C(7,4) = 35 types of 7, C(8,4) = 70 types of 8, and (12,4) = 495 types of 12, we can write 12 in the following ways:
12: 495 ways;
8+4: 70 ways;
7+5: 175 ways;
6+6: 225 ways;
5+7: 175 ways;
4+8: 70 ways;
4+4+4: 1 way, for a total of 1211 ways.
MATHEMATICA
CoefficientList[Series[(1 - x)^5/((1 - x)^5 - x^4), {x, 0, 50}], x] (* Wesley Ivan Hurt, Dec 26 2023 *)
PROG
(PARI) Vec((1-x)^5/((1-x)^5 - x^4) + O(x^40)) \\ Michel Marcus, Dec 27 2023
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Enrique Navarrete, Dec 26 2023
STATUS
approved