A380113 - OEIS

%I #29 Jan 13 2025 08:05:22

%S 1,1,1,3,4,1,10,15,6,1,35,56,28,8,1,126,210,120,45,10,1,462,792,495,

%T 220,66,12,1,1716,3003,2002,1001,364,91,14,1,6435,11440,8008,4368,

%U 1820,560,120,16,1,24310,43758,31824,18564,8568,3060,816,153,18,1

%N Triangle read by rows: The inverse matrix of the central factorials A370707, row n normalized by (-1)^(n - k)*A370707(n, n).

%C The inverse matrix of A370707 is ​​a rational matrix and the normalization serves to make it a matrix over the integers. Note that the normalization factor A370707(n, n) = FallingFactorial(n, n) * RisingFactorial(n, n) extends A002674 to n = 0.

%H Paolo Xausa, <a href="/A380113/b380113.txt">Table of n, a(n) for n = 0..11475</a> (rows 0..150 of triangle, flattened).

%F T(n, k) = (-1)^(n - k) * ff(n, n) * rf(n, n) * M^(-1)(ff(n, k) * rf(n, k)) where ff denotes the falling factorial, rf the rising factorial and M^(-1)(t(n, k)) the matrix inverse to the matrix with entries t(n, k).

%F T(n, k) = binomial(2*n, n - k) for 0 < k < n. T(n, n) = 1; T(n, 0) = (-1)^n*binomial(-n, n).

%F Sum_{k=0..n} T(n, k)*cos(k*x) = 2^(n-1)*(cos(x)+1)^n. (After _Philippe Deléham_ in A008311).

%e Triangle starts:

%e   [0] [    1]

%e   [1] [    1,     1]

%e   [2] [    3,     4,     1]

%e   [3] [   10,    15,     6,     1]

%e   [4] [   35,    56,    28,     8,    1]

%e   [5] [  126,   210,   120,    45,   10,    1]

%e   [6] [  462,   792,   495,   220,   66,   12,   1]

%e   [7] [ 1716,  3003,  2002,  1001,  364,   91,  14,   1]

%e   [8] [ 6435, 11440,  8008,  4368, 1820,  560, 120,  16,  1]

%e   [9] [24310, 43758, 31824, 18564, 8568, 3060, 816, 153, 18, 1]

%e .

%e Row 3 of the matrix inverse of the central factorials is [-1/36, 1/24, -1/60, 1/360]. Normalized with (-1)^(n-k)*360 gives row 3 of T.

%p T := (n, k) -> if n = k then 1 elif k = 0 then binomial(2*n, n - k)/2 else binomial(2*n, n - k) fi: seq(seq(T(n, k), k = 0..n), n = 0..9);

%t A380113[n_, k_] := Binomial[2*n, n - k]/(Boole[k == 0 && n > 0] + 1);

%t Table[A380113[n, k], {n, 0, 10}, {k, 0, n}] (* _Paolo Xausa_, Jan 13 2025 *)

%o (SageMath)

%o def Trow(n):

%o     def cf(n, k): return falling_factorial(n, k)*rising_factorial(n, k)

%o     def w(n): return factorial(n)*rising_factorial(n, n)

%o     m = matrix(QQ, n + 1, lambda x, y: cf(x, y)).inverse()

%o     return [(-1)^(n-k)*w(n)*m[n, k] for k in range(n+1)]

%o for n in range(10): print(Trow(n))

%Y Variant: A094527.

%Y Cf. A370707, A002674, A008311, A088218 and A110556 (column 0), A081294 (row sums), A000007 (alternating row sums), A005810 (central terms).

%K nonn,tabl

%O 0,4

%A _Peter Luschny_, Jan 12 2025