A255935 Triangle read by rows: a(n) = Pascal's triangle A007318(n) + A197870(n+1).

0, 1, 2, 1, 2, 0, 1, 3, 3, 2, 1, 4, 6, 4, 0, 1, 5, 10, 10, 5, 2, 1, 6, 15, 20, 15, 6, 0, 1, 7, 21, 35, 35, 21, 7, 2, 1, 8, 28, 56, 70, 56, 28, 8, 0, 1, 9, 36, 84, 126, 126, 84, 36, 9, 2, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 0

Offset: 0

Author

Paul Curtz, Mar 11 2015

Keywords

• nonn
• tabl

Comments

Consider the difference table of a sequence with A000004(n)=0's as main diagonal. (Example: A000045(n).) We call this sequence an autosequence of the first kind.

Based on Pascal's triangle, a(n) =

0, T1

1, 2,

1, 2, 0,

1, 3, 3, 2,

etc.

transforms every sequence s(n) in an autosequence of the first kind via the multiplication by the triangle

s0, T2

s0, s1,

s0, s1, s2,

s0, s1, s2, s3,

etc.

Examples.

1) s(n) = A198631(n)/A006519(n+1), the second fractional Euler numbers (see A209308). This yields 0*1, 1*1+2*1/2=2, 1*1+2*1/2+0*0=2, 1*1+3*1/2++3*0+2*(-1/4)=2, ... .

The autosequence is 0 followed by 2's or 2*(0,1,1,1,1,1,1,1,... = b(n)).

b(n), the basic autosequence of the first kind, is not in the OEIS (see A140575 and A054977).

2) s(n) = A164555(n)/A027642(n), the second Bernoulli numbers, yields 0,2,2,3,4,5,6,7,... = A254667(n).

Row sums of T1: A062510(n) = 3*A001045(n).

Antidiagonal sums of T1: A111573(n).

With 0's instead of the spaces, every column, i.e.,

0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...

0, 2, 2, 3, 4, 5, 6, 7, 8, 9, 10, ... = A001477(n) with 0 instead of 1 = A254667(n)

0, 0, 0, 3, 6, 10, 15, 21, 28, 36, 45, ... = A161680(n) with 0 instead of 1

0, 0, 0, 2, 4, 10, 20, 35, 56, 84, 120, ...

etc., is an autosequence of the first kind.

With T(0,0) = 1, it is (1, 0, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (2, -2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, May 24 2015

Examples

			Triangle starts:
0;
1, 2;
1, 2, 0;
1, 3, 3, 2;
1, 4, 6, 4, 0;
1, 5, 10, 10, 5, 2;
1, 6, 15, 20, 15, 6, 0;
...
		

Crossrefs

Cf. A001045, A001477, A010673, A011973, A054977, A062510, A007318, A197870, A006519, A198631, A140575, A209308, A164555, A027642, A111573, A161680, A254667.

Programs

Mathematica

a[n_, k_] := If[k == n, 2*Mod[n, 2], Binomial[n, k]]; Table[a[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 23 2015 *)

Formula

a(n) = Pascal's triangle A007318(n) with main diagonal A010673(n) (= period 2: repeat 0, 2) instead of 1's=A000012(n).

a(n) = reversal abs(A140575(n)).

a(n) = A007318(n) + A197870(n+1).

T(n,k) = T(n-1,k) + T(n-2,k-1) + T(n-2,k-2), T(0,0) = 0, T(1,0) = 1, T(1,1) = 2, T(n,k) = 0 if k>n or if k<0 . - Philippe Deléham, May 24 2015

G.f.: (-1-2*x*y+x^2*y+x^2*y^2)/((x*y+1)*(x*y+x-1)) - 1. - R. J. Mathar, Aug 12 2015