A135876 Square array, read by antidiagonals, where row n+1 is generated from row n by first removing terms at positions [(m+2)^2/4 - 1] for m>=0 and then taking partial sums, starting with all 1's in row 0.

1, 1, 1, 3, 2, 1, 15, 8, 3, 1, 105, 48, 15, 4, 1, 945, 384, 105, 23, 5, 1, 10395, 3840, 945, 176, 33, 6, 1, 135135, 46080, 10395, 1689, 279, 44, 7, 1, 2027025, 645120, 135135, 19524, 2895, 400, 57, 8, 1, 34459425, 10321920, 2027025, 264207, 35685, 4384, 561

Offset: 0

Author

Paul D. Hanna, Dec 14 2007

Keywords

• nonn
• tabl

Comments

This is the double factorial analog of Moessner's factorial array (A125714). Compare to triangle A135877 which is generated by a complementary process. A very interesting variant is square array A135878.

Examples

			Square array begins:
(1),(1),1,(1),1,(1),1,1,(1),1,1,(1),1,1,1,(1),1,1,1,(1),1,1,1,1,(1),...;
(1),(2),3,(4),5,(6),7,8,(9),10,11,(12),13,14,15,(16),17,18,19,(20),...;
(3),(8),15,(23),33,(44),57,71,(86),103,121,(140),161,183,206,(230),..;
(15),(48),105,(176),279,(400),561,744,(950),1206,1489,(1800),2171,..;
(105),(384),945,(1689),2895,(4384),6555,9129,(12139),16161,20763,..;
(945),(3840),10395,(19524),35685,(56448),89055,129072,(177331),245778,...;
(10395),(46080),135135,(264207),509985,(836352),1381905,2071215,(2924172),.;
(135135),(645120),2027025,(4098240),8294895,(14026752),24137505,...; ...
where terms in parenthesis are removed before taking partial sums.
For example, to generate row 2 from row 1, remove terms at positions
{[(m+2)^2/4-1], m>=0} = [0,1,3,5,8,11,15,19,24,29,35,...] to obtain:
[3, 5, 7,8, 10,11, 13,14,15, 17,18,19, 21,22,23,24, 25,26,27,28, ...]
then take partial sums to get row 2:
[3, 8, 15,23, 33,44, 57,71,86, 103,121,140, 161,183,206,230, ...].
Repeating this process will generate all the rows of the triangle,
where column 0 will be the odd double factorials (A001147)
and column 1 will be the even double factorials (A000165).
		

Crossrefs

Cf. columns: A001147, A000165, A004041, A129890; variants: A135878, A125714.

Programs

PARI

{T(n, k)=local(A=0, b=0, c=0, d=0); if(n==0, A=1, until(d>k, if(c==floor((b+2)^2/4)-1, b+=1, A+=T(n-1, c); d+=1); c+=1)); A}

Formula

T(n,0) = (2n)!/n!/2^n; T(n,1) = 2^n*n!; T(n,2) = (2n+1)!/n!/2^n; T(n,3) = A004041(n) = (2n+1)!/n!/2^n * Sum_{k=0..n} 1/(2k+1). T(n,4) = A129890(n) = 2^(n+1)*(n+1)! - (2n+1)!/n!/2^n = T(n+1,1)-T(n+1,0).