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A121548
Triangle read by rows: T(n,k) is the number of compositions of n into k Fibonacci numbers (1 <= k <= n; only one 1 is considered as a Fibonacci number).
15
1, 1, 1, 1, 2, 1, 0, 3, 3, 1, 1, 2, 6, 4, 1, 0, 3, 7, 10, 5, 1, 0, 2, 9, 16, 15, 6, 1, 1, 2, 9, 23, 30, 21, 7, 1, 0, 2, 10, 28, 50, 50, 28, 8, 1, 0, 3, 9, 34, 71, 96, 77, 36, 9, 1, 0, 2, 12, 36, 95, 156, 168, 112, 45, 10, 1, 0, 0, 12, 43, 115, 231, 308, 274, 156, 55, 11, 1, 1, 2, 9, 48, 140, 312, 504, 560, 423, 210, 66, 12, 1
OFFSET
1,5
LINKS
FORMULA
G.f.: G(t,z) = 1 / (1 - t*Sum_{i>=2} z^Fibonacci(i)) - 1.
Sum of terms in row n = A076739(n).
T(n,1) = A010056(n) (the characteristic function of the Fibonacci numbers);
T(n,2) = A121549(n);
T(n,3) = A121550(n);
Sum_{k=1..n} k*T(n,k) = A121551(n).
EXAMPLE
T(5,3)=6 because we have [1,2,2], [2,1,2], [2,2,1], [1,1,3], [1,3,1] and [3,1,1].
Triangle starts:
1;
1, 1;
1, 2, 1;
0, 3, 3, 1;
1, 2, 6, 4, 1;
0, 3, 7, 10, 5, 1;
0, 2, 9, 16, 15, 6, 1;
...
MAPLE
with(combinat): G:=1/(1-t*sum(z^fibonacci(i), i=2..40))-1: Gser:=simplify(series(G, z=0, 25)): for n from 1 to 23 do P[n]:=sort(coeff(Gser, z, n)) od: for n from 1 to 15 do seq(coeff(P[n], t, j), j=1..n) od; # yields sequence in triangular form
# second Maple program:
g:= proc(n) g(n):= (t-> issqr(t+4) or issqr(t-4))(5*n^2) end:
T:= proc(n, t) option remember;
`if`(n=0, `if`(t=0, 1, 0), `if`(t<1, 0, add(
`if`(g(j), T(n-j, t-1), 0), j=1..n)))
end:
seq(seq(T(n, k), k=1..n), n=1..14); # Alois P. Heinz, Oct 10 2022
MATHEMATICA
nmax = 14;
T = Rest@CoefficientList[#, t]& /@ Rest@(1/(1 - t*Sum[z^Fibonacci[i],
{i, 2, nmax}]) - 1 + O[z]^(nmax+1) // CoefficientList[#, z]&);
Table[T[[n, k]], {n, 1, nmax}, {k, 1, n}] // Flatten (* Jean-François Alcover, May 02 2022 *)
CROSSREFS
T(2n,n) gives A341072.
Sequence in context: A293109 A233530 A079123 * A180179 A346038 A215062
KEYWORD
nonn,tabl
AUTHOR
Emeric Deutsch, Aug 07 2006
STATUS
approved