A183202 - OEIS

%I #7 Mar 30 2012 18:37:23

%S 1,1,1,2,1,2,3,3,3,5,4,6,10,9,14,5,10,22,34,29,43,6,15,40,84,122,100,

%T 143,7,21,65,169,334,463,367,510,8,28,98,300,738,1390,1851,1426,1936,

%U 9,36,140,489,1426,3345,6043,7767,5839,7775,10,45,192,749,2510,6990,15735

%N Triangle, read by rows, where T(n,k) equals the sum of (n-k) terms in row n of triangle A131338 starting at position nk - k(k-1)/2, with the main diagonal formed from the row sums.

%e Triangle begins:

%e 1;

%e 1,1;

%e 2,1,2;

%e 3,3,3,5;

%e 4,6,10,9,14;

%e 5,10,22,34,29,43;

%e 6,15,40,84,122,100,143;

%e 7,21,65,169,334,463,367,510;

%e 8,28,98,300,738,1390,1851,1426,1936;

%e 9,36,140,489,1426,3345,6043,7767,5839,7775;

%e 10,45,192,749,2510,6990,15735,27374,34097,25094,32869; ...

%e The rows are derived from triangle A131338 by summing terms in the following manner:

%e (1);

%e (1),(1);

%e (1+1),(1),(2);

%e (1+1+1),(1+2),(3),(5);

%e (1+1+1+1),(1+2+3),(4+6),(9),(14);

%e (1+1+1+1+1),(1+2+3+4),(5+7+10),(14+20),(29),(43);

%e (1+1+1+1+1+1),(1+2+3+4+5),(6+8+11+15),(20+27+37),(51+71),(100),(143); ...

%e where row n of triangle A131338 consists of n '1's followed by the partial sums of the prior row.

%o (PARI) {A131338(n, k)=if(k>n*(n+1)/2|k<0,0,if(k<=n,1,sum(i=0, k-n,A131338(n-1,i))))}

%o {T(n,k)=if(n==k,A131338(n,n*(n+1)/2),sum(j=n*k-k*(k-1)/2,n*k-k*(k-1)/2+n-k-1,A131338(n,j)))}

%Y Cf. A131338, A098568, A098569 (row sums), A183203 (antidiagonal sums).

%K tabl,nonn

%O 0,4

%A _Paul D. Hanna_, Dec 30 2010