A380851 Riordan array ((1-x)^(m-1), x/(1-x)) with factor r^(2*n) on row n, for m = 3/2, r = 2.

1, -2, 4, -2, 8, 16, -4, 24, 96, 64, -10, 80, 480, 640, 256, -28, 280, 2240, 4480, 3584, 1024, -84, 1008, 10080, 26880, 32256, 18432, 4096, -264, 3696, 44352, 147840, 236544, 202752, 90112, 16384, -858, 13728, 192192, 768768, 1537536, 1757184, 1171456, 425984, 65536

Offset: 0

Author

Igor Victorovich Statsenko, Feb 06 2025

Keywords

• sign
• tabl

Examples

			Triangle starts:
       k = 0      1       2        3        4        5       6
  n=0:     1;
  n=1:    -2,     4;
  n=2:    -2,     8,     16;
  n=3:    -4,    24,     96,      64;
  n=4:   -10,    80,    480,     640,     256;
  n=5:   -28,   280,   2240,    4480,    3584,    1024;
  n=6:   -84,  1008,  10080,   26880,   32256,   18432,   4096;
		

Links

• Igor Victorovich Statsenko, Identification of Riordan generalizations of binomial coefficients, Innovation science No 02-1, State Ufa, Aeterna Publishing House, 2025, pp. 11-18, see table 10. In Russian.

Crossrefs

Columns: A002420 (k=0); A240530 (k=1).

Triangle for m=-3, r=1: A104713; for m=-2, r=1: A104712; for m=-1, r=1: A135278; for m=0, r=1: A007318; for m=1, r=1: A097805; for m=2, r=1: A159854.

Programs

Maple

T:=(m,r,n,k)->add(binomial(i+m,m)*binomial(n+1,n-k-i)*r^(2*n)*(-1)^(i),i=0..n-k): m:=3/2: r:=2: seq(print(seq(T(m,r,n,k), k=0..n)), n=0..10);

Mathematica

T[n_, k_] := 4^n Binomial[n, k] Hypergeometric2F1[3/2, k - n, k + 1, 1];
Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten  (* Peter Luschny, Feb 07 2025 *)

SageMath

# Using function riordan_array from A256893.
RA = riordan_array((1 - x)^(3/2 - 1), x/(1-x), 7)
for n in range(7): print(4^n * RA.row(n)[:n+1])  # Peter Luschny, Feb 28 2025

Formula

T(n,k) = Sum_{i=0..n-k} binomial(i+m, m)*binomial(n+1, n-k-i)*r^(2*n)*(-1)^(i), for m = 3/2 and r = 2.

From Peter Luschny, Feb 07 2025: (Start)

T(n,k) = r^(2*n)*JacobiP(n - k, 1 + k, m - 1 - n, -1).

T(n,k) = 4^n*binomial(n, k)*hypergeom([3/2, k - n], [k + 1], 1). (End)