A105438 Triangle, row sums = (Fibonacci numbers - 2).

1, 2, 1, 3, 2, 1, 4, 4, 2, 1, 5, 6, 5, 2, 1, 6, 9, 8, 6, 2, 1, 7, 12, 14, 10, 7, 2, 1, 8, 16, 20, 20, 12, 8, 2, 1, 9, 20, 30, 30, 27, 14, 9, 2, 1, 10, 25, 40, 50, 42, 35, 16, 10, 2, 1, 11, 30, 55, 70, 77, 56, 44, 18, 11, 2, 1

Offset: 0

Author

Gary W. Adamson, Apr 09 2005

Keywords

• nonn
• tabl
• easy

Comments

Row sums = 1, 3, 6, 11, 19, 32, 53...(Fibonacci numbers - 2; starting with F(4)) The first few rows of the triangle are:

Row sums = (Fibonacci numbers - 2; starting 1, 3, 6...).

Column 1 = A002620; Column 2 = A006918; Column 3 = A096338.

Inverse array is A105522. - Paul Barry, Apr 11 2005

Diagonal sums are A027383(n). - Philippe Deléham, Jan 16 2014

Examples

			Column 2: 1, 2, 5, 8, 14, 20, 30...is generated by using the partial sum operator on 1, 1, 3, 3, 6, 6, 10, 10...
The first few rows of the triangle are:
  1;
  2, 1;
  3, 2, 1;
  4, 4, 2, 1;
  5, 6, 5, 2, 1;
  6, 9, 8, 6, 2, 1;
  7, 12, 14, 10, 7, 2, 1;
  8, 16, 20, 20, 12, 8, 2, 1;
  9, 20, 30, 30, 27, 14, 9, 2, 1;
  10, 25, 40, 50, 42, 35, 16, 10, 2, 1;
  ...
		

Links

• T. Mansour and A. O. Munagi, Alternating subsets modulo m, Rocky Mt. J. Math. 42, No. 4, 1313-1325 (2012), eq. (2)

Crossrefs

Cf. A001911, A002620, A006918, A096338, A105108.

Formula

By columns (k = 0, 1, 2...); use partial sum operator on (bin(n, k) numbers repeated).

T(n,k) = Sum_{j=0..n-k} C((j+2k)/2, k)*(1+(-1)^j)+C((j-1+2k)/2, k)*(1-(-1)^j)/2; Riordan array (1/(1-x)^2, x/(1-x^2)). - Paul Barry, Apr 11 2005

T(n,k) = T(n-1,k)+T(n-1,k-1)+T(n-2,k)-T(n-2,k-1)-T(n-3,k), T(0,0)=1, T(1,0)=2, T(1,1)= 1, T(n,k)= 0 if k<0 or if k>n. - Philippe Deléham, Jan 16 2014