OFFSET
1,2
COMMENTS
See A095263 for the number of ways to split [n] into an unspecified number of intervals and then choose 2 blocks (i.e., subintervals) from each interval.
LINKS
Paolo Xausa, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).
FORMULA
a(n) = 2*C(n,2) + C(n+1,5).
G.f.: x^2*(2 - 6*x + 7*x^2 - 2*x^3)/(1 - x)^6. - Stefano Spezia, May 02 2022
a(n) = n*(n-1)*(n^3 - 4*n^2 + n + 126)/120. - R. J. Mathar, Jul 05 2022
EXAMPLE
a(1)=0 since we can't choose 2 nonempty blocks/cells (i.e., subintervals) from an interval of one block.
a(2)=2 since we have 2 cases: first interval is empty, so we choose both blocks (i.e., subintervals) from the second interval in C(2,2) ways, and similarly for the case of the second interval being empty (note we can't consider the case where [2] splits into 2 intervals of one block each since we can't choose 2 nonempty blocks from a single block; i.e., C(1,2)*C(1,2)=0).
a(6)=51 since the following are the number of ways to split [6] into 2 intervals with k and (n-k) blocks (subintervals) each (written as k|(n-k) below) and to choose the blocks/cells:
6|0 (second interval empty): C(6,2) = 15 from the first interval;
0|6 (first interval empty): C(6,2) = 15 from the second interval;
2|4: C(2,2)*C(4,2) = 6;
3|3: C(3,2)*C(3,2) = 9;
4|2: C(4,2)*C(2,2) = 6.
MATHEMATICA
A353232[n_] := 2*Binomial[n, 2] + Binomial[n + 1, 5];
Array[A353232, 50] (* Paolo Xausa, May 27 2024 *)
PROG
(PARI) a(n) = 2*binomial(n, 2) + binomial(n+1, 5); \\ Michel Marcus, Jul 06 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Enrique Navarrete, May 01 2022
STATUS
approved