A134511 abs(A049310) * A128174 provided both arrays are read with offset (n,k) = (0,0).

1, 0, 1, 2, 0, 1, 0, 3, 0, 1, 5, 0, 4, 0, 1, 0, 8, 0, 5, 0, 1, 13, 0, 12, 0, 6, 0, 1, 0, 21, 0, 17, 0, 7, 0, 1, 34, 0, 33, 0, 23, 0, 8, 0, 1, 0, 55, 0, 50, 0, 30, 0, 9, 0, 1, 89, 0, 88, 0, 73, 0, 38, 0, 10, 0, 1, 0, 144, 0, 138, 0, 103, 0, 47, 0, 11, 0, 1, 233, 0, 232, 0, 211, 0, 141, 0, 57, 0, 12, 0, 1

Offset: 0

Author

Gary W. Adamson, Oct 28 2007

Keywords

• nonn
• tabl

Comments

A112552(unsigned) = A128174 * A049310.

Row sums = A134512: (1, 1, 3, 4, 10, 14, 32, 46, 99, 145, ...).

From Petros Hadjicostas, Sep 03 2019: (Start)

To prove Alois P. Heinz's claim (see the Formula section and his Maple program below) we note that, for n >= 0 and 0 <= k <= n, T(n, n-k) = Sum_{r = 0 .. infinity} abs(A049310(n,r)) * A128174(r,n-k) = Sum_{r = n-k..n} abs(A049310(n,r)) * A128174(r,n-k). But A049310(n,r) = 0 when n + r is odd and A128174(r,n-k) = 1 iff r + n - k is even. Thus, when k is odd, T(n, n-k) = 0.

Assume now k is even. Then T(n, n-k) = Sum_{r = n-k..n and n+r even} abs(A049310(n,r)) = Sum_{r = n-k..n and n+r even} binomial((n+r)/2, r). Letting m = n-r (which is even), we see that the summation ranges from m = 0 to k over even numbers. Thus, let s = m/2, and so T(n, n-k) = Sum_{s = 0 .. k/2} binomial(n-s, n-2*s) = Sum_{s = 0 .. k/2} binomial(n-s, s) = F(n+1, k/2), where F(.,.) is the incomplete Fibonacci number from the references (see also the Formula section below).

(End)

Examples

			First few rows of the triangle T(n,k):
   1;
   0,  1;
   2,  0,  1;
   0,  3,  0,  1;
   5,  0,  4,  0,  1;
   0,  8,  0,  5,  0,  1;
  13,  0, 12,  0,  6,  0,  1;
   0, 21,  0, 17,  0,  7,  0,  1;
  34,  0, 33,  0, 23,  0,  8,  0,  1;
   0, 55,  0, 50,  0, 30,  0,  9,  0,  1;
  ...
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10010 (rows 0 to 141, flattened)
• H. Belbachir and A. Belkhir, Combinatorial Expressions Involving Fibonacci, Hyperfibonacci, and Incomplete Fibonacci Numbers, Journal of Integer Sequences, Vol. 17 (2014), Article 14.4.3.
• A. Dil and I. Mezo, A symmetric algorithm for hyperharmonic and Fibonacci numbers, Appl. Math. Comp. 206 (2008), 942-951; in Eqs. (11), see the incomplete Fibonacci numbers.
• Piero Filipponi, Incomplete Fibonacci and Lucas numbers, P. Rend. Circ. Mat. Palermo (Serie II) 45(1) (1996), 37-56; see Table 1 (p. 39) that contains the incomplete Fibonacci numbers.
• A. Pintér and H.M. Srivastava, Generating functions of the incomplete Fibonacci and Lucas numbers, Rend. Circ. Mat. Palermo (Serie II) 48(3) (1999), 591-596.

Crossrefs

Cf. A038730, A038792, A049310, A128174, A134512.

A(4n,2n) gives: A038736.

Programs

Maple

N:= 20: # for the first N rows
T128174:= Matrix(N,N,(i,j) -> `if`(j<=i, (i-j+1) mod 2, 0)):
T049310:= Matrix(N,N):
for i from 1 to N do
     P:= orthopoly[U](i-1,x/2);
     for j from 1 to i do
       T049310[i,j]:= abs(coeff(P,x,j-1))
     od
od:
A:= T049310 . T128174:
for i from 1 to N do
convert(A[i,1..i],list)
od;  # Robert Israel, Mar 02 2018
# second Maple program:
T:= (n, k)-> `if`((n+k)::odd, 0, add(binomial(n-s, s), s=0..(n-k)/2)):
seq(seq(T(n, k), k=0..n), n=0..12); # Alois P. Heinz, Sep 02 2019

Mathematica

T[n_, k_] := If[OddQ[n+k], 0, Sum[Binomial[n-s, s], {s, 0, (n-k)/2}]];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 12}] // Flatten (* Jean-François Alcover, Dec 31 2021, after Alois P. Heinz *)

Formula

abs(A049310) * A128174 as infinite lower triangular matrices assuming both of them have offset (n,k) = (0,0).

From Petros Hadjicostas, Sep 03 2019: (Start)

Let F(m,r) = Sum_{j = 0..r} binomial(m-1-j, j) be the incomplete Fibonacci numbers from the references (defined for m >= 1 and 0 <= r <= floor((m-1)/2)).

As Alois P. Heinz observed, for n >= 0 and 0 <= k <= n, T(n, n-k) = F(n+1, k/2) when k is even, and = 0 otherwise (see his Maple program below).

(End)

Extensions

Edited by Robert Israel, Mar 02 2018