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Triangle built from partial sums of Catalan numbers A000108 multiplied by powers.
12

%I #18 Aug 29 2022 10:31:31

%S 1,1,1,1,2,1,1,4,3,1,1,9,11,4,1,1,23,51,22,5,1,1,65,275,157,37,6,1,1,

%T 197,1619,1291,357,56,7,1,1,626,10067,11497,3941,681,79,8,1,1,2056,

%U 64979,107725,46949,9431,1159,106,9,1,1,6918,431059,1045948,587621,140681,19303,1821,137,10,1

%N Triangle built from partial sums of Catalan numbers A000108 multiplied by powers.

%C The column sequences (without leading zeros) begin with A000012 (powers of 1), A112705 (partial sums Catalan), A112696-A112704, for m=0..10.

%H Wolfdieter Lang, <a href="/A112705/a112705.txt">First 10 rows.</a>

%F a(n, m) = sum(C(k)*m^k, k=0..n-m), n>m>0, with C(n):=A000108(n); a(n, n)=1; a(n, 0)=1; a(n, m)=0 if n<m.

%F G.f. for column m>=0 (without leading zeros): c(m*x)/(1-x), where c(x):=(1-sqrt(1-4*x))/(2*x) is the o.g.f. of Catalan numbers A000108.

%e Triangle starts:

%e 1;

%e 1, 1;

%e 1, 2, 1;

%e 1, 4, 3, 1;

%e 1, 9, 11, 4, 1;

%e 1, 23, 51, 22, 5, 1;

%e 1, 65, 275, 157, 37, 6, 1;

%e ...

%t col[m_] := col[m] = CatalanNumber[#]*m^#& /@ Range[0, 20] // Accumulate;

%t T[n_, m_] := If[m == 0, 1, col[m][[n - m + 1]]];

%t Table[T[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* _Jean-François Alcover_, Aug 29 2022 *)

%o (PARI) t(n, m) = if (m==0, 1, if (n==m, 1, sum(kk=0, n-m, m^kk*binomial(2*kk, kk)/(kk+1))));

%o tabl(nn) = {for (n=0, nn, for (m=0, n, print1(t(n, m), ", ");); print(););} \\ _Michel Marcus_, Nov 25 2015

%Y Row sums give A112706.

%K nonn,easy,tabl

%O 0,5

%A _Wolfdieter Lang_, Oct 31 2005