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%I #33 Dec 18 2023 12:20:26
%S 1,4,9,17,32,63,127,256,513,1025,2048,4095,8191,16384,32769,65537,
%T 131072,262143,524287,1048576,2097153,4194305,8388608,16777215,
%U 33554431,67108864,134217729,268435457,536870912,1073741823,2147483647,4294967296,8589934593
%N 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.
%C From _R. J. Mathar_, Nov 22 2007: (Start)
%C 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).
%C If d is even, this simplifies to Sum_{i=0..d-1} binomial(d,d-i)*(-1)^i*a(n-i) = 0.
%C 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.
%C The generating functions are
%C d=2: a(0)/(1-2*x).
%C 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).
%C 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).
%C In the present sequence we have d=3 and g.f. = (x-1)/(x^2-x+1) - 2/(-1+2*x). (End)
%C Also binomial transform of A130784. a(n) = 2^(n+1) + A010892(n+4).
%C Recurrence in shorter form: a(n) = 2*a(n) + periodically extended [2, 1, -1, -2, -1, 1].
%C See A130750, A130752, A130755 for other examples of d=3 sequences, A130781 for an example of d=4.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,2).
%F a(n) = 2^(n+1) - cos((2*n+1)*Pi/6) * 2/sqrt(3). - _Vladimir Reshetnikov_, Oct 15 2017
%F G.f.: (1+x)/((1-2*x)*(1-x+x^2)). - _Joerg Arndt_, Oct 16 2017
%e Triangle of sequence and 1st, 2nd, 3rd differences:
%e 1 4 9 17 32 63 127 256 513
%e 3 5 8 15 31 64 129 257
%e 2 3 7 16 33 65 128
%e 1 4 9 17 32 63 ... equal to first row
%t 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 *)
%t LinearRecurrence[{3, -3, 2},{1, 4, 9},21] (* _Ray Chandler_, Sep 23 2015 *)
%t Table[2^(n + 1) - Cos[(2 n + 1) Pi/6] 2/Sqrt[3], {n, 0, 32}] (* _Vladimir Reshetnikov_, Oct 15 2017 *)
%K nonn,easy
%O 0,2
%A _Paul Curtz_, Jul 15 2007
%E Edited and extended by _R. J. Mathar_, Nov 22 2007
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