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A325643
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a(1) = 1; for n > 1, a(n) is the least divisor d > 1 of n such that A048720(d,k) = n for some k.
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6
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1, 2, 3, 2, 5, 2, 7, 2, 3, 2, 11, 2, 13, 2, 3, 2, 17, 2, 19, 2, 7, 2, 23, 2, 25, 2, 3, 2, 29, 2, 31, 2, 3, 2, 7, 2, 37, 2, 3, 2, 41, 2, 43, 2, 3, 2, 47, 2, 7, 2, 3, 2, 53, 2, 55, 2, 3, 2, 59, 2, 61, 2, 3, 2, 5, 2, 67, 2, 69, 2, 71, 2, 73, 2, 3, 2, 77, 2, 79, 2, 81, 2, 83, 2, 5, 2, 87, 2, 89, 2, 91, 2, 31, 2, 5, 2, 97, 2, 3, 2, 101, 2, 103, 2, 3
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OFFSET
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1,2
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COMMENTS
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For n > 1, a(n) is the least divisor d of n that is larger than 1 and for which it holds that when the binary expansion of d is converted to a (0,1)-polynomial (e.g., 13=1101[2] encodes X^3 + X^2 + 1), then that polynomial is a divisor of (0,1)-polynomial similarly converted from n, when the division is done over GF(2). See the example.
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LINKS
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FORMULA
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a(2n) = 2.
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EXAMPLE
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For n = 21 = 3*7, 3 is not the answer because X^1 + 1 does not divide X^4 + X^2 + 1 (21 is "10101" in binary) over GF(2). However, the next larger divisor 7 works because X^4 + X^2 + 1 = (X^2 + X^1 + 1)^2 when multiplication is done over GF(2) (note that A048720(7,7) = 21). Thus a(21) = 7.
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PROG
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(PARI) A325643(n) = if(1==n, n, my(p = Pol(binary(n))*Mod(1, 2)); fordiv(n, d, if((d>1), my(q = Pol(binary(d))*Mod(1, 2)); if(0==(p%q), return(d)))));
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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