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A093010 Triangle, read by rows, such that the convolution of the n-th row with the natural numbers forms the n-th diagonal, for n>=0, where each row begins with 1.

%I A093010 #13 Mar 19 2023 21:55:30
%S A093010 1,1,2,1,4,3,1,6,7,4,1,8,14,10,5,1,10,22,22,13,6,1,12,33,40,30,16,7,1,
%T A093010 14,45,66,58,38,19,8,1,16,60,100,104,76,46,22,9,1,18,76,146,168,142,
%U A093010 94,54,25,10,1,20,95,202,262,242,180,112,62,28,11,1,22,115,272,386,394,316
%N A093010 Triangle, read by rows, such that the convolution of the n-th row with the natural numbers forms the n-th diagonal, for n>=0, where each row begins with 1.
%C A093010 Row sums form A000713, the number of partitions of n into parts of 3 kinds. Antidiagonal sums form A000990, the number of 2-line partitions of n.
%F A093010 T(n, k) = sum_{j=0..k} (k-j+1)*T(n-k, j), with T(0, n) = 1 for all n>=0.
%F A093010 A000713(n) = sum_{k=0..n} T(n, k) (row sums).
%F A093010 A000990(n) = sum_{k=0..floor(n/2)} T(n-k, k) (antidiagonal sums).
%e A093010 T(7,3) = 66 = 1*4+8*3+14*2+10*1 = T(4,0)*4+T(4,1)*3+T(4,2)*2+T(4,3)*1; this is also the third term of the 4th-diagonal.
%e A093010 The 6th antidiagonal is {1,10,14,4}, which has a sum of 29 = A000990(6) = number of 2-line partitions of 6.
%e A093010 Rows begin:
%e A093010 {1},
%e A093010 {1,2},
%e A093010 {1,4,3},
%e A093010 {1,6,7,4},
%e A093010 {1,8,14,10,5},
%e A093010 {1,10,22,22,13,6},
%e A093010 {1,12,33,40,30,16,7},
%e A093010 {1,14,45,66,58,38,19,8},
%e A093010 {1,16,60,100,104,76,46,22,9},
%e A093010 {1,18,76,146,168,142,94,54,25,10},
%e A093010 {1,20,95,202,262,242,180,112,62,28,11},
%e A093010 {1,22,115,272,386,394,316,218,130,70,31,12},...
%o A093010 (PARI) T(n,k)=if(n<k || k<0,0,if(k==0,1,sum(j=0,min(k,n-k),(k-j+1)*T(n-k,j))))
%Y A093010 Cf. A000713, A000990, A092905.
%K A093010 nonn,tabl
%O A093010 0,3
%A A093010 _Paul D. Hanna_, Mar 14 2004