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A337442
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Number of output sequences from the linear feedback shift register whose feedback polynomial coefficients (excluding the constant term) correspond to the binary representation of n.
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0
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1, 2, 3, 2, 4, 2, 2, 4, 6, 2, 4, 4, 2, 6, 4, 4, 8, 4, 2, 6, 2, 4, 8, 2, 4, 8, 4, 2, 6, 2, 2, 8, 14, 2, 6, 4, 8, 8, 4, 6, 6, 8, 12, 4, 4, 2, 8, 6, 2, 12, 8, 2, 8, 8, 2, 4, 4, 2, 4, 12, 6, 4, 6, 10, 20, 2, 4, 8, 2, 12, 6, 2, 2, 6, 4, 8, 16, 8, 2, 8, 4, 4, 16, 2
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OFFSET
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0,2
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COMMENTS
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a(n) > 1 for n > 0.
It appears that every term after a(2) is even.
It appears that a(2^n) is greater than each preceding term and is greater than or equal to each term up to a(2^(n+1)).
If a(n) = 2, then the nonzero shift register sequence is an m-sequence.
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LINKS
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EXAMPLE
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For n = 3 = 11 in binary, the polynomial is 1+x+x^2 and the 2 shift register sequences are {00..., 01101...}.
For n = 4 = 100 in binary, the polynomial is 1+x^3 and the 4 shift register sequences are {000..., 001001..., 011011..., 111...}.
For n = 6 = 110 in binary, the polynomial is 1+x^2+x^3 and the 2 shift register sequences are {000..., 0010111001...}.
For n = 10 = 1010 in binary, the polynomial is 1+x^2+x^4 and the 4 shift register sequences are {0000..., 0001010001..., 0011110011..., 0110110...}.
For n = 11 = 1011 in binary, the polynomial in 1+x+x^2+x^4. Using a Fibonacci LSFR, if the current state of the register is 0001, the next input bit is 0+0+1=1, and the next state is 0011. If the current state is 0100, the next input bit is 0+0+0=0, and the next state is 1000. The 4 shift register sequences are {0000..., 00011010001..., 00101110010..., 1111...}.
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CROSSREFS
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A011260 counts how many 2's are in the interval [2^(n-1),(2^n)-1].
a(n) = 2 if and only if 2n+1 is in A091250.
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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