0,2

Comments from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 22 2007 (Start): Sequences which equal the sequence of their d-th differences obey linear recurrences with constant binomial coefficients of the form sum_{i=0..d} binomial(d,d-i)*(-1)^i*a(n-i)=a(n-d).

If d is even, this simplifies to sum_{i=0..d-1} binomial(d,d-i)*(-1)^i*a(n-i)=0.

This binding of d (d odd) or d-1 (d even) consecutive terms by the recurrences leaves d or d-1, respectively, free parameters to choose a(0),a(1),...a(d) or a(0),a(1),...a(d-1), respectively, which ultimately define the individual sequence.

The generating functions are

d=2: a(0)/(1-2*x).

d=3: 1/3*(-a(0)+a(1)-a(2))/(-1+2*x)+1/3*(-4*a(0)*x-x*a(2)+4*a(1)*x-a(2)+2*a(0)+a(1))/(x^2-x+1).

d=4: 1/2*(-2*a(0)+2*a(1)-a(2))/(-1+2*x)+1/2*(2*a(1)*x-4*a(0)*x-a(2)+2*a(1))/(1-2*x+2*x^2) .

In the present sequence we have d=3 and g.f. = (x-1)/(x^2-x+1)-2/(-1+2*x) . (End)

Also binomial transform of A130784. a(n)=2^(n+1) + A010892(n+4).

Recurrence in shorter form: a(n)=2a(n) + periodicly extended 2 1 -1 -2 -1 1.

See A130750, A130752, A130755 for other examples of d=3 sequences, A130781 for an example of d=4.

a(n)=-(1/2)*{1/2-(1/2)*I*sqrt(3)}^n-(1/2)*{1/2+(1/2)*I*sqrt(3)}^n+2*2^n+(1/6)*I*{1/2-(1/2) *I*sqrt(3)}^n*sqrt(3)-(1/6)*I*{1/2+(1/2)*I*sqrt(3)}^n*sqrt(3), with n>=0 and I=sqrt(-1) - Paolo P. Lava (ppl(AT)spl.at), Jun 09 2008

Triangle of sequence and 1st, 2nd, 3rd differences:

1..4..9..17..32..63..127..256..513

.3..5..8..15..31..64..129..257

...2..3..7..16..33..65..128

....1..4..9...17..32..63 ... equal to 1st row

Sequence in context: A008093 A027374 A009922 this_sequence A008236 A088365 A139468

Adjacent sequences: A130782 A130783 A130784 this_sequence A130786 A130787 A130788

nonn

Paul Curtz (bpcrtz(AT)free.fr), Jul 15 2007

Edited and extended by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 22 2007