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Irregular triangle read by rows: coefficients of polynomials related to a family of convolutions of certain central binomial sequences.
4

%I #36 Jun 28 2023 10:49:59

%S 1,1,1,3,3,5,-2,1,30,35,-10,5,70,63,8,-2,-75,35,315,231,56,-14,-245,

%T 105,693,429,-272,36,2268,-525,-5880,2310,12012,6435,-2448,324,9660,

%U -2037,-16632,6006,25740,12155,3968,-304,-31260,3840,73395,-14091,-90090,30030,109395,46189,43648,-3344

%N Irregular triangle read by rows: coefficients of polynomials related to a family of convolutions of certain central binomial sequences.

%C The row length sequence {r(k)} of this irregular triangle a(k, p) is given by r(0) = 1 = r(1) and r(k) = 2*(floor(k/2)) = A052928(k), k >= 2. This is {1,1,2,2,4,4,6,6,8,8,10,10,...}.

%C The array of the k-family of convolutions Sigma(k, n) := Sum_{p=0..n} p^k * binomial(2*p, p) * binomial(2*(n-p), n-p) can be written as Sigma(k, n) = ((4^n)*c(k, n) / A046161(k)) * Sum_{p=0..r(k)-1} a(k, p)*n^p, where c(k, n) = n for even k >= 2 and c(k, n) = n^2 for odd k >= 3, with c(0, n) = 1, c(1, n).

%C The author was led to compute such sums by a question asked by M. Greiter, Jun 27 2008.

%H Wolfdieter Lang, <a href="/A142961/a142961_1.txt">First eleven rows and more</a>

%F a(k, p)= [n^p] P(k, n) where c(k, n)*P(k, n) = A046161(k)*Sigma(k, n)/(4^n), with c(k, n) and the array Sigma(k, n) given above. A046161(k) are the denominators of binomial(2*k, k)/4^k: [1, 2, 8, 16, 128, 256, 1024, 2048, 32768, 65536, 262144,...].

%F Sigma(k, n)/4^n = Sum_{p=0..min(n, k)} binomial(n, p)*(2*p -1)!!*S2(k, p)/2^p, with the double factorials (2*p -1)!!= A001147(p), with (-1)!! := 1, and the Stirling numbers of the second kind S2(k, p):=A048993(k, p). (Proof from the product of the o.g.f.s and the normal ordering (x^d_x)^k = Sum_{p=0..k} (S2(k, p)*x^p*d_x^p), with the derivative operator d_x.)

%e The irregular triangle a(k, p) begins:

%e k\p 0 1 2 3 4 5 ...

%e 0: 1

%e 1: 1

%e 2: 1 3

%e 3: 3 5

%e 4: 2 -1 30 35

%e 5: -10 5 70 63

%e 6: 8 -2 -75 35 315 231

%e ...

%e k=3: Sigma(3, n) = Sum_{p=0..n} p^3 * binomial(2*p, p) * binomial(2*(n-p), n-p) = (4*n/16)*n^2*(3 + 5*n), for n >= 0. This is the sequence {0, 2, 52, 648, 5888, 44800, 304128, 1906688, 11272192, 63700992, ...}.

%Y Cf. A001147, A046161, A048993, A052928, A001147.

%K sign,easy,tabf

%O 0,4

%A _Wolfdieter Lang_, Sep 15 2008

%E Name changed, edited and corrected by _Wolfdieter Lang_, Aug 23 2019