OFFSET
0,3
COMMENTS
a(n) is the first sequence on three (with its first and second differences):
1, 1, 3, 8, 17, 33, 64, 127, ...;
0, 2, 5, 9, 16, 31, 63, 128, ..., that is 0 followed by A130752;
2, 3, 4, 7, 15, 32, 65, 129, ..., that is 2 followed by A130755;
1, 1, 3, 8, 17, 33, 64, 127, ..., this sequence.
The main diagonal is 2^n.
The sum of the first three lines is 3*2^n.
Alternated sum and subtraction of a(n) and its inverse binomial transform (period 3: repeat [1, 0, 2]) gives the autosequence of the first kind b(n):
0, 1, 1, 9, 17, 35, 63, 127, ...
1, 0, 8, 8, 18, 28, 64, 126, ...
-1, 8, 0, 10, 10, 36, 62, 134, ...
9, -8, 10, 0, 26, 26, 72, 118, ... .
The main diagonal is 0's. The first two upper diagonals are A259713.
The sum of the first three lines gives 9*A001045.
a(n) mod 9 gives a periodic sequence of length 6: repeat [1, 1, 3, 8, 8, 6].
a(n) = A130750(n-1) for n > 2. - Georg Fischer, Oct 23 2018
LINKS
Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,2).
FORMULA
Binomial transform of the sequence of length 3: repeat [1, 0, 2].
a(n+3) = -a(n) + 9*2^n.
a(n) = 2^n - periodic 6: repeat [0, 1, 1, 0, -1, -1, 0].
a(n+6) = a(n) + 63*2^n.
a(n+1) = 2*a(n) - period 6: repeat [1, -1, -2, -1, 1, 2].
a(n) = 2^n - 2*sin(Pi*n/3)/sqrt(3). - Jean-François Alcover and Colin Barker, Jan 16 2017
G.f.: (1 - 2*x + 3*x^2)/((1 - 2*x)*(1 - x + x^2)). - Colin Barker, Jan 16 2017
MATHEMATICA
LinearRecurrence[{3, -3, 2}, {1, 1, 3}, 30] (* Jean-François Alcover, Jan 16 2017 *)
PROG
(PARI) Vec((1 - 2*x + 3*x^2) / ((1 - 2*x)*(1 - x + x^2)) + O(x^40)) \\ Colin Barker, Jan 16 2017
(Magma) I:=[1, 1, 3]; [n le 3 select I[n] else 3*Self(n-1) - 3*Self(n-2) + 2*Self(n-3): n in [1..30]]; // G. C. Greubel, Jan 15 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul Curtz, Jan 16 2017
STATUS
approved