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%I #50 Mar 26 2022 04:11:25
%S 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,
%T 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,
%U 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
%N abs(A049310) * A128174 provided both arrays are read with offset (n,k) = (0,0).
%C A112552(unsigned) = A128174 * A049310.
%C Row sums = A134512: (1, 1, 3, 4, 10, 14, 32, 46, 99, 145, ...).
%C From _Petros Hadjicostas_, Sep 03 2019: (Start)
%C 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.
%C 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).
%C (End)
%H Robert Israel, <a href="/A134511/b134511.txt">Table of n, a(n) for n = 0..10010</a> (rows 0 to 141, flattened)
%H H. Belbachir and A. Belkhir, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Belbachir/belb2.html">Combinatorial Expressions Involving Fibonacci, Hyperfibonacci, and Incomplete Fibonacci Numbers</a>, Journal of Integer Sequences, Vol. 17 (2014), Article 14.4.3.
%H A. Dil and I. Mezo, <a href="http://dx.doi.org/10.1016/j.amc.2008.10.013">A symmetric algorithm for hyperharmonic and Fibonacci numbers</a>, Appl. Math. Comp. 206 (2008), 942-951; in Eqs. (11), see the incomplete Fibonacci numbers.
%H Piero Filipponi, <a href="https://doi.org/10.1007/BF02845088">Incomplete Fibonacci and Lucas numbers</a>, P. Rend. Circ. Mat. Palermo (Serie II) 45(1) (1996), 37-56; see Table 1 (p. 39) that contains the incomplete Fibonacci numbers.
%H A. Pintér and H.M. Srivastava, <a href="https://doi.org/10.1007/BF02844348">Generating functions of the incomplete Fibonacci and Lucas numbers</a>, Rend. Circ. Mat. Palermo (Serie II) 48(3) (1999), 591-596.
%F abs(A049310) * A128174 as infinite lower triangular matrices assuming both of them have offset (n,k) = (0,0).
%F From _Petros Hadjicostas_, Sep 03 2019: (Start)
%F 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)).
%F 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).
%F (End)
%e First few rows of the triangle T(n,k):
%e 1;
%e 0, 1;
%e 2, 0, 1;
%e 0, 3, 0, 1;
%e 5, 0, 4, 0, 1;
%e 0, 8, 0, 5, 0, 1;
%e 13, 0, 12, 0, 6, 0, 1;
%e 0, 21, 0, 17, 0, 7, 0, 1;
%e 34, 0, 33, 0, 23, 0, 8, 0, 1;
%e 0, 55, 0, 50, 0, 30, 0, 9, 0, 1;
%e ...
%p N:= 20: # for the first N rows
%p T128174:= Matrix(N,N,(i,j) -> `if`(j<=i, (i-j+1) mod 2, 0)):
%p T049310:= Matrix(N,N):
%p for i from 1 to N do
%p P:= orthopoly[U](i-1,x/2);
%p for j from 1 to i do
%p T049310[i,j]:= abs(coeff(P,x,j-1))
%p od
%p od:
%p A:= T049310 . T128174:
%p for i from 1 to N do
%p convert(A[i,1..i],list)
%p od; # _Robert Israel_, Mar 02 2018
%p # second Maple program:
%p T:= (n, k)-> `if`((n+k)::odd, 0, add(binomial(n-s, s), s=0..(n-k)/2)):
%p seq(seq(T(n, k), k=0..n), n=0..12); # _Alois P. Heinz_, Sep 02 2019
%t T[n_, k_] := If[OddQ[n+k], 0, Sum[Binomial[n-s, s], {s, 0, (n-k)/2}]];
%t Table[Table[T[n, k], {k, 0, n}], {n, 0, 12}] // Flatten (* _Jean-François Alcover_, Dec 31 2021, after _Alois P. Heinz_ *)
%Y Cf. A038730, A038792, A049310, A128174, A134512.
%Y A(4n,2n) gives: A038736.
%K nonn,tabl
%O 0,4
%A _Gary W. Adamson_, Oct 28 2007
%E Edited by _Robert Israel_, Mar 02 2018
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