A198862 Sum of the n-th antidiagonal in the triangle A192011.

-1, 2, 2, 2, 3, 1, -1, -3, -6, -7, -6, -3, 3, 10, 16, 19, 16, 6, -10, -29, -45, -51, -41, -12, 33, 84, 125, 137, 104, 20, -105, -242, -346, -366, -261, -19, 327, 693, 954, 973, 646, -47, -1001, -1974, -2620, -2573, -1572, 402, 3022

Offset: 0

Author

Paul Curtz, Oct 30 2011

Keywords

• sign
• easy

Comments

The current sequence and its successive differences are:

-1, 2, 2, 2, 3, 1, -1 ,-3, -6, -7, ...

3, 0, 0, 1, -2, -2, -2, -3, -1, 1, 3, 6, 7, 6, 3, -3, -10, -16, ...

-3, 0, 1, -3, 0, 0, -1, 2, 2, 2, 3, 1, -1, -3, -6, -7, -6, ...

3, 1, -4, 3, 0, -1, 3, 0, 0, 1, -2, -2, -2, -3, -1, 1, 3, 6, 7, ...

-2, -5, 7, -3, -1, 4, -3, 0, 1, -3, 0, 0, -1, 2, 2, 2, 3, 1, ...

-3, 12, -10, 2, 5, -7, 3, 1, -4, 3, 0, -1, 3, 0, 0, 1, -2, ...

15, -22, 12, 3, -12, 10, -2, -5, 7, -3, -1, 4, -3, 0, ...

-37, 34, -9, -15, 22, -12, -3, 12, -10, 2, 5, -7, 3, 1, -4, ...

Each row obeys the same linear recurrence and is a version of the row 4 lines farther up in the same array shifted right by 12 places.

Links

• Index entries for linear recurrences with constant coefficients, signature (1,0,0,-1).

Programs

Maple

A198862 := proc(n)
        add( A192011(n-k,k),k=0..floor(n/2)) ;
end proc:
seq(A198862(n),n=0..80) ; # R. J. Mathar, Nov 03 2011

Formula

a(n) = Sum_{k=0..floor(n/2)} A192011(n-k,k).

a(n) = a(n-1) - a(n-4), n > 3.

From R. J. Mathar, Nov 02 2011: (Start)

G.f.: (-1 + 3*x) / (1 - x + x^4).

a(n) = 3*A099530(n-1) - A099530(n). (End)