A363882 - OEIS

%I #33 Aug 02 2023 14:07:40

%S 0,1,1,0,2,0,1,3,3,1,0,4,4,4,0,1,5,10,10,5,1,0,6,12,20,12,6,0,1,7,21,

%T 35,35,21,7,1,0,8,24,56,64,56,24,8,0,1,9,36,84,126,126,84,36,9,1,0,10,

%U 40,120,200,252,200,120,40,10,0,1,11,55,165,330,462,462,330,165,55,11,1

%N Take 2 copies of Pascal's triangle. One copy has one inch between the terms of each row and the other copy has two inches between the terms of each row. Put one on top of the other so that the 1's at the very top of each copy coincide. Sequence is a triangle giving the differences between the overlapping terms.

%F T(n,k) = binomial(n,k) - [n mod 2 = k mod 2 = 0] * binomial(n/2,k/2).

%e Row n=8 is formed by taking row 8 of Pascal's triangle (1, 8, 28, 56, 70, 56, 28, 8, 1) and subtracting row 4 (1, 4, 6, 4, 1) spaced 2 apart. The numbers that overlap are 1, 28, 70, 28, 1 over 1, 4, 6, 4, 1, from which 1-1=0, 28-4=24, 70-6=64, 28-4=24, and 1-1=0. Thus, row 8 of the present triangle is 0, 8, 24, 56, 64, 56, 24, 8, 0.

%e Subtracting:

%e   1;                1;               0,

%e   1, 1;                              1, 1;

%e   1, 2, 1;       -  1,    1;      =  0, 2, 0;

%e   1, 3, 3, 1;                        1, 3, 3, 1;

%e   1, 4, 6, 4, 1;    1,    2,    1;   0, 4, 4, 4, 0;

%e   ...

%e Resulting triangle begins:

%e       k=0  1  2  3  4

%e   n=0:  0;

%e   n=1:  1, 1;

%e   n=2:  0, 2, 0;

%e   n=3:  1, 3, 3, 1;

%e   n=4:  0, 4, 4, 4, 0;

%e   ...

%Y Cf. A007318.

%K tabl,nonn,easy

%O 0,5

%A _J. Lowell_, Jun 25 2023