A130785 Sequence identical to its third differences: a(n+3) = 3a(n+2)-3a(n+1)+2a(n), with a(0)=1, a(1)=4, a(2)=9.

1, 4, 9, 17, 32, 63, 127, 256, 513, 1025, 2048, 4095, 8191, 16384, 32769, 65537, 131072, 262143, 524287, 1048576, 2097153, 4194305, 8388608, 16777215, 33554431, 67108864, 134217729, 268435457, 536870912, 1073741823, 2147483647, 4294967296, 8589934593

Offset: 0

Author

Paul Curtz, Jul 15 2007

Keywords

• nonn
• easy

Comments

From R. J. Mathar, 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) = 2*a(n) + periodically 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.

Examples

			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 first row
		

Links

• Index entries for linear recurrences with constant coefficients, signature (3,-3,2).

Programs

Mathematica

d = 3; nmax = 20; a[n_ /; n < d] := (n+1)^2; seq = Table[a[n], {n, 0, nmax}]; seq /. Solve[ Thread[ Take[seq, nmax - d + 1] == Differences[seq, d]]] // First (* Jean-François Alcover, Nov 07 2013 *)
LinearRecurrence[{3, -3, 2},{1, 4, 9},21] (* Ray Chandler, Sep 23 2015 *)
Table[2^(n + 1) - Cos[(2 n + 1) Pi/6] 2/Sqrt[3], {n, 0, 32}] (* Vladimir Reshetnikov, Oct 15 2017 *)

Formula

a(n) = 2^(n+1) - cos((2*n+1)*Pi/6) * 2/sqrt(3). - Vladimir Reshetnikov, Oct 15 2017

G.f.: (1+x)/((1-2*x)*(1-x+x^2)). - Joerg Arndt, Oct 16 2017

Extensions

Edited and extended by R. J. Mathar, Nov 22 2007