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A281166
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a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3) for n>2, a(0)=a(1)=1, a(2)=3.
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2
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1, 1, 3, 8, 17, 33, 64, 127, 255, 512, 1025, 2049, 4096, 8191, 16383, 32768, 65537, 131073, 262144, 524287, 1048575, 2097152, 4194305, 8388609, 16777216, 33554431, 67108863, 134217728, 268435457, 536870913, 1073741824, 2147483647, 4294967295, 8589934592
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OFFSET
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0,3
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COMMENTS
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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].
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LINKS
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FORMULA
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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].
G.f.: (1 - 2*x + 3*x^2)/((1 - 2*x)*(1 - x + x^2)). - Colin Barker, Jan 16 2017
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MATHEMATICA
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PROG
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(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
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CROSSREFS
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Cf. A000079, A001045, A005010, A007283, A014551 (a diagonal), A057079, A062510 (a diagonal), A128834, A130750, A130752, A130755, A153234, A153237, A259713.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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