A135877 Triangle, read by rows of n(n+1)+1 terms, where row n+1 is generated from row n by inserting zeros at positions [(m+2)^2/4 - 1] for m=1..2n+2 and then taking partial sums from right to left, starting with a single 1 in row 0.

1, 1, 1, 1, 3, 3, 3, 2, 2, 1, 1, 15, 15, 15, 12, 12, 9, 9, 6, 4, 4, 2, 1, 1, 105, 105, 105, 90, 90, 75, 75, 60, 48, 48, 36, 27, 27, 18, 12, 8, 8, 4, 2, 1, 1, 945, 945, 945, 840, 840, 735, 735, 630, 540, 540, 450, 375, 375, 300, 240, 192, 192, 144, 108, 81, 81, 54, 36, 24, 16, 16, 8

Offset: 0

Author

Paul D. Hanna, Dec 14 2007

Keywords

• nonn
• tabl

Comments

Compare to square array A135876 which is generated by a complementary process. Compare to triangle A127452 which generates the factorials in column 0. A very interesting variant is triangle A135879.

Examples

			Triangle begins:
1;
1, 1, 1;
3, 3, 3, 2, 2, 1, 1;
15, 15, 15, 12, 12, 9, 9, 6, 4, 4, 2, 1, 1;
105, 105, 105, 90, 90, 75, 75, 60, 48, 48, 36, 27, 27, 18, 12, 8, 8, 4, 2, 1, 1;
945, 945, 945, 840, 840, 735, 735, 630, 540, 540, 450, 375, 375, 300, 240, 192, 192, 144, 108, 81, 81, 54, 36, 24, 16, 16, 8, 4, 2, 1, 1; ...
To generate the triangle, start with a single 1 in row 0,
and then obtain row n+1 from row n by inserting zeros
at positions [(m+2)^2/4 - 1] for m=1..2n+2 and then
taking reverse partial sums (i.e., summing from right to left).
Start with row 0, insert 2 zeros in front of the '1':
[0,0,1];
take reverse partial sums to get row 1:
[1,1,1];
insert zeros at positions [0,1,3,5]:
[0,0,1,0,1,0,1];
take reverse partial sums to get row 2:
[3,3,3,2,2,1,1];
insert zeros at positions [0,1,3,5,8,11]:
[0,0,3,0,3,0,3,2,0,2,1,0,1];
take reverse partial sums to get row 3:
[15,15,15,12,12,9,9,6,4,4,2,1,1];
insert zeros at positions [0,1,3,5,8,11,15,19]:
[0,0,15,0,15,0,15,12,0,12,9,0,9,6,4,0,4,2,1,0,1];
take reverse partial sums to get row 4:
[105,105,105,90,90,75,75,60,48,48,36,27,27,18,12,8,8,4,2,1,1].
		

Crossrefs

Cf. A001147; A135876; variants: A135879, A127452, A125781.

Programs

PARI

{T(n,k)=local(A=[1],B);if(n>0,for(i=1,n,m=1;B=[0]; for(j=1,#A,if(j+m-1==floor((m+2)^2/4)-1,m+=1;B=concat(B,0));B=concat(B,A[j])); A=Vec(Polrev(Vec(Pol(B)/(1-x+O(x^#B)))))));if(k+1>#A,0,A[k+1])} /* for(n=0,8,for(k=0,n*(n+1),print1(T(n,k),","));print("")) */

Formula

Column 0 equals the double factorials A001147(n) = (2n)!/(n!*2^n).