cp's OEIS Frontend

A136218 Triangle, read by rows of A136219(n) terms, where row n+1 is generated from row n by first inserting zeros in row n at positions {[m*(m+7)/6], m>=0} and then taking partial sums, starting with a '1' in row 0.

%I A136218 #2 Mar 30 2012 18:37:08
%S A136218 1,1,1,1,3,3,3,2,2,1,1,15,15,15,12,12,9,9,6,6,4,2,2,1,108,108,108,93,
%T A136218 93,78,78,63,63,51,39,39,30,21,21,15,9,9,5,3,1,1,1036,1036,1036,928,
%U A136218 928,820,820,712,712,619,526,526,448,370,370,307,244,244,193,154,115,115,85
%N A136218 Triangle, read by rows of A136219(n) terms, where row n+1 is generated from row n by first inserting zeros in row n at positions {[m*(m+7)/6], m>=0} and then taking partial sums, starting with a '1' in row 0.
%C A136218 A variant of the triple factorial triangle A136213. Compare to square array A136217, which is generated by a complementary process.
%e A136218 Triangle begins:
%e A136218 1;
%e A136218 1,1,1;
%e A136218 3,3,3,2,2,1,1;
%e A136218 15,15,15,12,12,9,9,6,6,4,2,2,1;
%e A136218 108,108,108,93,93,78,78,63,63,51,39,39,30,21,21,15,9,9,5,3,1,1;
%e A136218 1036,1036,1036,928,928,820,820,712,712,619,526,526,448,370,370,307,244,244,193,154,115,115,85,64,43,43,28,19,10,10,5,2,1;
%e A136218 12569,12569,12569,11533,11533,10497,10497,9461,9461,8533,7605,7605,6785,5965,5965,5253,4541,4541,3922,3396,2870,2870,2422,2052,1682,1682,1375,1131,887,887,694,540,425,310,310,225,161,118,75,75,47,28,18,8,8,3,1;
%e A136218 ...
%e A136218 Number of terms in rows is given by A136219, which starts:
%e A136218 [1,3,7,13,22,33,47,64,84,106,131,159,190,224,261,301,343,388,...].
%e A136218 To generate row 3, start with row 2:
%e A136218 [3,3,3,2,2,1,1];
%e A136218 insert zeros at positions [0,1,3,5,7,10] to get:
%e A136218 [0,0,3,0,3,0,3,0,2,2,0,1,1],
%e A136218 then take reverse partial sums (from right to left) to obtain row 3:
%e A136218 [15,15,15,12,12,9,9,6,6,4,2,2,1].
%e A136218 For row 4, insert zeros in row 3 at positions [0,1,3,5,7,10,13,16,20]:
%e A136218 [0,0,15,0,15,0,15,0,12,12,0,9,9,0,6,6,0,4,2,2,0,1]
%e A136218 then take reverse partial sums to obtain row 4:
%e A136218 [108,108,108,93,93,78,78,63,63,51,39,39,30,21,21,15,9,9,5,3,1,1].
%e A136218 Continuing in this way will generate all the rows of this triangle.
%e A136218 Amazingly, column 0 of this triangle = column 0 of triangle P=A136220:
%e A136218 1;
%e A136218 1, 1;
%e A136218 3, 2, 1;
%e A136218 15, 10, 3, 1;
%e A136218 108, 75, 21, 4, 1;
%e A136218 1036, 753, 208, 36, 5, 1;
%e A136218 12569, 9534, 2637, 442, 55, 6, 1;
%e A136218 185704, 146353, 40731, 6742, 805, 78, 7, 1; ...
%e A136218 where column k of P^3 = column 0 of P^(3k+3) such that
%e A136218 column 0 of P^3 = column 0 of P shift one place left.
%o A136218 (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==(m*(m+7))\6,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=0,6,for(k=0,2*n^2,if(T(n,k)==0,break,print1(T(n,k),",")));print("")) */
%Y A136218 Cf. A136221 (column 0), A136219; A136213, A136220.
%K A136218 nonn,tabl
%O A136218 0,5
%A A136218 _Paul D. Hanna_, Dec 23 2007