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%I #16 May 11 2014 04:41:37
%S 1,1,1,1,1,1,2,1,1,2,1,3,1,5,1,1,2,4,1,5,8,1,9,1,14,1,1,2,4,8,1,9,14,
%T 22,1,23,32,1,33,1,47,1,1,2,4,8,16,1,17,26,40,62,1,63,86,118,1,119,
%U 152,1,153,1,200,1
%N Triangle, read by rows, where terms in row n equal the partial sums of row n-1 with 1's inserted at positions [0,n,2n-1,3n-3,4n-6,5n-10,...,n(n+1)/2-1] for n>0, with T(0,0)=1.
%H Paul D. Hanna, <a href="/A182961/b182961.txt">Rows n = 0..30, flattened.</a>
%F Row sums equal A129867;
%F n-th row sum = 1 + Sum_{k=1..n} k*(n-k+1)!.
%F T(n,n(n+1)/2) = A129867(n) for n>0, with T(0,0) = 1.
%e This triangle T(n,k), where k=0..n(n+1)/2 in row n>=0, begins:
%e 1;
%e (1),1;
%e (1),1,(1),2;
%e (1),1,2,(1),3,(1),5;
%e (1),1,2,4,(1),5,8,(1),9,(1),14;
%e (1),1,2,4,8,(1),9,14,22,(1),23,32,(1),33,(1),47;
%e (1),1,2,4,8,16,(1),17,26,40,62,(1),63,86,118,(1),119,152,(1),153,(1),200;
%e (1),1,2,4,8,16,32,(1),33,50,76,116,178,(1),179,242,328,446,(1),447,566,718,(1),719,872,(1),873,(1),1073;
%e ...
%e where row n is equal to the partial sums of terms in row n-1, with 1's inserted at positions [0,n,2n-1,3n-3,4n-6,5n-10,...,n(n+1)/2-1].
%e The row sums and rightmost border form sequence A129867, which equals the row sums of triangle A130469.
%e Triangle A130469 begins:
%e 1;
%e 1, 1;
%e 2, 2, 1;
%e 6, 4, 3, 1;
%e 24, 12, 6, 4, 1;
%e 120, 48, 18, 8, 5, 1;
%e 720, 240, 72, 24, 10, 6, 1; ...
%e which has the same row sums as this triangle.
%o (PARI) {T(n,k)=local(A=[1],B); for(m=0,n, t=0;B=[];
%o for(j=0,#A-1, if(j==t*m-t*(t+1)/2, t+=1;B=concat(B,1)); B=concat(B,A[j+1]));
%o A=Vec( Ser(B)/(1-x+O(x^#B)) ) ); if(k+1>#A, 0, B[k+1])}
%o for(n=0,12,for(k=0,n*(n+1)/2,print1(T(n, k), ", ")); print(""))
%Y Cf. A129867, A130469; variant: A131338.
%K nonn,tabf
%O 0,7
%A _Paul D. Hanna_, Dec 31 2010
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