A336014 Irregular triangle read by rows: T(n,1) = T(n,2) = T(n,3*n-2) = T(n,3*n-1) = n for n >= 1 and T(n,k) = T(n-1,k-2) + T(n-1,k-1) for n > 1, 3 <= k <= 3*(n-1).

1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 3, 3, 4, 4, 6, 7, 8, 8, 8, 7, 6, 4, 4, 5, 5, 8, 10, 13, 15, 16, 16, 15, 13, 10, 8, 5, 5, 6, 6, 10, 13, 18, 23, 28, 31, 32, 31, 28, 23, 18, 13, 10, 6, 6, 7, 7, 12, 16, 23, 31, 41, 51, 59, 63, 63, 59, 51, 41, 31, 23, 16, 12, 7, 7

Offset: 1

Author

Lechoslaw Ratajczak, Jul 04 2020

Keywords

• nonn
• tabf

Comments

The number of terms in row n is 3*n-1 = A016789(n-1).

The sum of row n is equal to 2*A094002(n-1) = 2*A188589(n).

Fibonacci(n) = T(n+k,n) - T(n+k-1,n) for n >= 1, k = 1,2,3,...

The elements b(k) of the main diagonal, superdiagonal 1 and all subdiagonals have the recursive formula: b(k) = 2*b(k-1) + b(k-2) - 2*b(k-3) - b(k-4) for k > 4.

Examples

			Triangle begins:
n\k 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20...
1   1  1
2   2  2  2  2  2
3   3  3  4  4  4  4  3  3
4   4  4  6  7  8  8  8  7  6  4  4
5   5  5  8 10 13 15 16 16 15 13 10  8  5  5
6   6  6 10 13 18 23 28 31 32 31 28 23 18 13 10  6  6
7   7  7 12 16 23 31 41 51 59 63 63 59 51 41 31 23 16 12  7  7
...
		

Crossrefs

Superdiagonal 1 is A029907 for n >= 1.

The main diagonal is A208354 for n >= 1.

Subdiagonal 1 is A102702(n-1) for n >= 1.

Subdiagonal 2 is A206268(n+2) for n >= 1 (conjectured).

Subdiagonal 3 is A191830(n+3) for n >= 1.

Cf. A007318, A051597, A228196.

Formula

T(n,k) = T(n,3*k-n) for 1 <= k <= 3*n-1.

T(n,k) = Sum_{u=2*(n-k)+3..2*n-k+1} ceiling(u/2)*A065941(k-2,u-2*(n-k)-3) for n >= 3, 3 <= k <= n.

T(n,k) = Sum_{m1=1..k-n} A208354(m1)*binomial(n-m1-1, k-n-m1) + Sum_{m2=1..2*n-k} A208354(m2)*binomial(n-m2-1, 2*n-k-m2) for n >= 2, n+1 <= k <= 2*n-1.

T(n,k) = Sum_{u=2*(k-2*n)+3..k-n+1} ceiling(u/2)*A065941(3*n-k-2,u-2*(k-2*n)-3) for n>= 3, 2*n <= k <= 3*(n-1).

T(n,k) = A208354(k) + (n-k)*Fibonacci(k) for n >= 3, 3 <= k <= n.

T(n,k) = A029907(k-1) + (n-k+1)*Fibonacci(k) for n >= 2, 3 <= k <= n+1.