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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).
0
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
1,2
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).
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.
EXAMPLE
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
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.
Sequence in context: A153643 A053218 A198335 * A296335 A296635 A295051
KEYWORD
nonn,tabl,easy
AUTHOR
STATUS
approved