A396457 Triangle T(n, k) of convolved numbers of k-sections of the Fibonacci sequence at the first convolution order, for n >= 1 and 1 <= k <= n, read by rows.

1, 2, 1, 5, 6, 1, 10, 25, 8, 1, 20, 90, 50, 14, 1, 38, 300, 280, 145, 22, 1, 71, 954, 1475, 1330, 365, 36, 1, 130, 2939, 7472, 11420, 5390, 970, 58, 1, 235, 8850, 36836, 94066, 74660, 23220, 2525, 94, 1, 420, 26195, 178000, 752979, 993058, 520995, 97730, 6625, 152, 1

Offset

1,2

Comments

This triangle forms an inter-sectional cross-slice of the family of matrices of convolved k-sections of the Fibonacci sequence. The row index n represents the element position, while the column index k represents the Fibonacci section parameter.

The table is constructed by gathering the terms where the convolution order is fixed at s = 1. The columns of this triangle aggregate properties across different k-sections studied in our previous records. The first column (k=1) represents the standard convolutions of the base Fibonacci sequence (A001629). The second column (k=2) represents the convolution of A001906(n) for n >= 1 (even-indexed Fibonacci numbers) with itself (A001871). The third column (k=3) corresponds directly to A395431.

Links

Table of n, a(n) for n=1..55.

D. Dmytryshyn, D. Gray, V. Khamitov, and Alex Stokolos, Convolved numbers of k-section of the Fibonacci sequence: properties, consequences, arXiv:2603.08636 [math.CA], 2026; see also alternate link, Herald of Advanced Information Technology, 9, pp. 129-139 (2026).

Formula

T(n, k) = A(n-k+1, k, 1) for n >= 1 and 1 <= k <= n, where for odd k: A(n, k, s) = (-i)^(n-1) * GegenbauerC(n-1, 1+s, i * L(k) / 2), for even k: A(n, k, s) = GegenbauerC(n-1, 1+s, L(k) / 2), where i is the imaginary unit, s=1 and L(k) is the k-th Lucas number.

T(n, k) = A(n-k+1, k, 1) for n >= 1 and 1 <= k <= n, where A(n, k, s) is defined as follows:

A(n, k, 1) = Sum_{j=0..floor((n-1)/2)} (-1)^(j*(k-1)) * (n-j) * binomial(n-1-j, j) * A000032(k)^(n-1-2j), or

A(n, k, 1) = A000032(k) * n/(n-1) * A(n-1, k, 1) + (-1)^(k+1) * (n+1)/(n-1) * A(n-2, k, 1) for n >= 3, with A(1, k, 1) = 1 and A(2, k, 1) = 2*A000032(k).

Example

The triangle starts:
       k=1   k=2   k=3   k=4   k=5   k=6
  n=1:  1;
  n=2:  2,    1;
  n=3:  5,    6,    1;
  n=4: 10,   25,    8,    1;
  n=5: 20,   90,   50,   14,    1;
  n=6: 38,  300,  280,  145,   22,    1;
T(4, 2) = A(3, 2, 1) = GegenbauerC(2, 2, L_2/2) = GegenbauerC(2, 2, 3/2) = 25.
We can also compute this term in two ways using A000032(2) = 3:
1) Using the explicit sum (with n=3, k=2):
j=0: (-1)^(0*(2-1)) * (3-0) * binomial(3-1-0, 0) * 3^(3-1-2*0) = 1 * 3 * 1 * 9 = 27,
j=1: (-1)^(1*(2-1)) * (3-1) * binomial(3-1-1, 1) * 3^(3-1-2*1) = -1 * 2 * 1 * 1 = -2.
Total: 27 + (-2) = 25.
2) Using the non-stationary recurrence (with n=3, k=2):
x_0 = 0, x_1 = 1, x_2 = 3 * 2/(2-1) * x_1 - (2+1)/(2-1) * x_0 = 6 - 0 = 6, x_3 = 3 * 3/(3-1) * x_2 - (3+1)/(3-1) * x_1 = 3 * (3/2) * 6 - 2 * 1 = 27 - 2 = 25.

Maple

A_sum := proc(n, k) local Lk, j; Lk := round((1/2 + 1/2*sqrt(5))^k + (1/2 - 1/2*sqrt(5))^k); sum((-1)^(j*(k - 1))*(n - j)*binomial(n - 1 - j, j)*Lk^(n - 1 - 2*j), j = 0 .. floor(1/2*n - 1/2)); end proc;
for n to 10 do
seq(A_sum(n - k + 1, k), k = 1 .. n);
end do;
A_rec := proc(n, k) local Lk, x, j; if n = 1 then return 1; end if; Lk := round((1/2 + 1/2*sqrt(5))^k + (1/2 - 1/2*sqrt(5))^k); if n = 2 then return 2*Lk; end if; x := array(1 .. n); x[1] := 1; x[2] := 2*Lk; for j from 3 to n do x[j] := j*Lk*x[j - 1]/(j - 1) + (-1)^(k + 1)*(j + 1)*x[j - 2]/(j - 1); end do; return x[n]; end proc;
for n to 10 do
    seq(A_rec(n - k + 1, k), k = 1 .. n);
end do;

Crossrefs

Cf. A001629, A001871, A395431, A395902, A393329, A396358.

Sequence in context: A193635 A241168 A145324 * A260613 A375044 A375042

Adjacent sequences: A396454 A396455 A396456 * A396458 A396459 A396460

Keyword

nonn,tabl

Author

Alex Stokolos, May 26 2026

Status

approved