A353232 - OEIS

A353232

a(n) is the number of ways to split [n] = {1,2,...,n} into two (possibly empty) complementary intervals {1,2,...,i} and {i+1,i+2,...,n} and then, if both intervals are nonempty, select 2 nonempty blocks/cells (i.e., subintervals) from each of them, or if one of the intervals is empty, select 2 nonempty blocks/cells from the nonempty interval.

0, 2, 6, 13, 26, 51, 98, 182, 324, 552, 902, 1419, 2158, 3185, 4578, 6428, 8840, 11934, 15846, 20729, 26754, 34111, 43010, 53682, 66380, 81380, 98982, 119511, 143318, 170781, 202306, 238328, 279312, 325754, 378182, 437157, 503274, 577163, 659490, 750958

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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

Cf. A095263, A002378, A000389.