A339050 Triangle read by rows T(n, m) = F(2*m-1)*(n-m) + F(2*m), for 1 <= m <= n, where F = A000045 (Fibonacci).

1, 2, 3, 3, 5, 8, 4, 7, 13, 21, 5, 9, 18, 34, 55, 6, 11, 23, 47, 89, 144, 7, 13, 28, 60, 123, 233, 377, 8, 15, 33, 73, 157, 322, 610, 987, 9, 17, 38, 86, 191, 411, 843, 1597, 2584, 10, 19, 43, 99, 225, 500, 1076, 2207, 4181, 6765

Offset: 1

Author

Gary W. Adamson and Wolfdieter Lang, Jan 15 2021

Keywords

• nonn
• tabl
• easy

Comments

This is the partial sum triangle of triangle A143929.

The main diagonal is the INVERT transform of the first column (offset 1 in both sequences).

Examples

			The triangle T(n, m) begins:
n\m   1  2  3  4   5   6    7    8    9   10 ...
1:    1
2:    2  3
3:    3  5  8
4:    4  7 13 21
5:    5  9 18 34  55
6:    6 11 23 47  89 144
7:    7 13 28 60 123 233  377
8:    8 15 33 73 157 322  610  987
9:    9 17 38 86 191 411  843 1597 2584
10:  10 19 43 99 225 500 1076 2207 4181 6765
...
		

Crossrefs

Cf. A000045, A001519, A001906, A143929.

The first columns (without leading zeros) are A001477(n), A005408(n+1), A005408(n+1), for n >= 1.

The first (sub)diagonals are A001906(m), A001519(m+1), A005248(m), for m >= 1.

Formula

T(n, m) = Sum_{k=1..m} A143929(n, k), n >=1, m = 1, 2, ..., n, otherwise 0.

T(n, m) = A(m)*n + B(m), with A(m) = A(m-1) + F(2*(m-1)), for m >= 2 and A(1) = 1, and B(m) = B(m-1) + (m-1)*F(2*(m-1)), for m >= 2 and B(1) = 0, where F(2*m) =A001906(m) and F(2*m-1) = A001519(m).

T(n, 1) = n, for n >= 1; T(n, m) = F(2*(m-1))*(n-m+1), if m >= 2 and n >= m, and 0 otherwise.

G.f. of column m: G(m,x) = x^m*(x*F(2*m-1)/(1-x)^2 + F(2*m)/(1-x)), for m >= 1.

G.f. of row polynomials R(n, x) := Sum{m=1..n} T(n, m)*x^m, that is g.f. of the triangle: G(z,x) = (x*z)*(1 - x*z^2)/((1- 3*x*z + (x*z)^2)*(1 - z)^2).

G.f. of (sub)diagonal k: D(k,x) = x*((k-1)*(1-x) + 1)/(1 - 3*x + x^2), for k >= 1.