login
Triangle, read by rows, where T(n,k) = A049020([n/2],k)*A049020([(n+1)/2],k).
4

%I #14 Nov 02 2020 09:09:31

%S 1,1,1,1,2,3,4,9,1,10,30,6,25,100,36,1,75,370,186,10,225,1369,961,100,

%T 1,780,5587,4960,750,15,2704,22801,25600,5625,225,1,10556,101774,

%U 136960,39000,2325,21,41209,454276,732736,270400,24025,441,1,178031,2199262,4110512,1849120,217000,6027,28,769129,10647169,23059204,12645136,1960000,82369,784,1,3630780,55493841,136074274,87570056,16787400,944230,13720,36

%N Triangle, read by rows, where T(n,k) = A049020([n/2],k)*A049020([(n+1)/2],k).

%C Row n has 1+floor(n/2) terms.

%C T(n,0) = A124419(n).

%C A124418(n,k) = k!*T(n,k) (conjecture).

%C A000110(n) = Sum_{k=0..[n/2]} k!*T(n,k), where A000110 is the Bell numbers.

%C Inspired by triangle A124418 and the work of _Emeric Deutsch_.

%F T(n,k) = A049020([n/2],k) * A049020([(n+1)/2],k), where A049020(n,k) = Sum_{i=0..n} S2(n,i) * C(i,k) and S2(n,k) = (1/k!)*Sum_{j=0..k} (-1)^(k-j)*C(k,j)*j^n (the Stirling numbers of 2nd kind).

%e Triangle begins:

%e 1;

%e 1;

%e 1, 1;

%e 2, 3;

%e 4, 9, 1;

%e 10, 30, 6;

%e 25, 100, 36, 1;

%e 75, 370, 186, 10;

%e 225, 1369, 961, 100, 1;

%e 780, 5587, 4960, 750, 15;

%e 2704, 22801, 25600, 5625, 225, 1;

%e 10556, 101774, 136960, 39000, 2325, 21;

%e 41209, 454276, 732736, 270400, 24025, 441, 1;

%e 178031, 2199262, 4110512, 1849120, 217000, 6027, 28;

%e 769129, 10647169, 23059204, 12645136, 1960000, 82369, 784, 1;

%e 3630780, 55493841, 136074274, 87570056, 16787400, 944230, 13720, 36; ...

%t S[n_, k_] = Sum[StirlingS2[n, i] Binomial[i, k], {i, 0, n}];

%t T[n_, k_] := S[Floor[n/2], k] S[Floor[(n+1)/2], k];

%t Table[T[n, k], {n, 0, 15}, {k, 0, Floor[n/2]}] // Flatten (* _Jean-François Alcover_, Nov 02 2020 *)

%o (PARI) {T(n,k) = (n\2)!*((n+1)\2)!*polcoeff(polcoeff(exp((1+y)*(exp(x+x*O(x^n))-1)),n\2),k) *polcoeff(polcoeff(exp((1+y)*(exp(x+x*O(x^n))-1)),(n+1)\2),k)}

%o for(n=0,15, for(k=0,n\2, print1(T(n,k),", "));print(""))

%Y Cf. A124418, A124419, A049020; A124527, A124528, A124529.

%K nonn,tabf

%O 0,5

%A _Paul D. Hanna_, Nov 08 2006