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A304357
Antidiagonal sums of the first quadrant of array A(k,m) = F_k(m), F_k(m) being the k-th Fibonacci polynomial evaluated at m.
5
0, 1, 1, 3, 5, 13, 32, 94, 297, 1036, 3911, 15918, 69350, 321779, 1582745, 8220349, 44925187, 257563819, 1544896976, 9671289892, 63051738167, 427254561854, 3003872526303, 21876513464296, 164790822258172, 1282198404741305, 10292007232817249, 85126350266370355
OFFSET
0,4
COMMENTS
Equivalently, antidiagonal sums of the third quadrant of array A(k,m).
It seems that: a(n+1) is the sum of the n-th antidiagonal of triangle A101494; a(n)-(n mod 2) is the sum of the n-th antidiagonal of array A172236; and a(n+1)+(n mod 2) is the sum of row n of triangle A157103. - Mathew Englander, Feb 28 2021
FORMULA
a(n) = Sum_{j=0..n} F_j(n-j).
a(n+1) = Sum_{j = 0..n} Sum_{i = j..floor((n+j)/2)} binomial(i,j)*(n+j-2*i)^j (empirically). - Mathew Englander, Feb 28 2021
MAPLE
F:= (n, k)-> (<<0|1>, <1|k>>^n)[1, 2]:
a:= n-> add(F(j, n-j), j=0..n):
seq(a(n), n=0..30);
# second Maple program:
F:= proc(n, k) option remember;
`if`(n<2, n, k*F(n-1, k)+F(n-2, k))
end:
a:= n-> add(F(j, n-j), j=0..n):
seq(a(n), n=0..30);
# third Maple program:
a:= n-> add(combinat[fibonacci](j, n-j), j=0..n):
seq(a(n), n=0..30);
MATHEMATICA
a[n_] := Sum[Fibonacci[j, n - j], {j, 0, n}];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jun 02 2018, from 3rd Maple program *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Alois P. Heinz, May 11 2018
STATUS
approved