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A325642
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a(1) = 1; for n > 1, a(n) = k for the least divisor d > 1 of n such that A048720(d,k) = n for some k.
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4
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1, 1, 1, 2, 1, 3, 1, 4, 7, 5, 1, 6, 1, 7, 5, 8, 1, 9, 1, 10, 7, 11, 1, 12, 1, 13, 9, 14, 1, 15, 1, 16, 31, 17, 13, 18, 1, 19, 29, 20, 1, 21, 1, 22, 27, 23, 1, 24, 11, 25, 17, 26, 1, 27, 1, 28, 23, 29, 1, 30, 1, 31, 21, 32, 21, 33, 1, 34, 1, 35, 1, 36, 1, 37, 57, 38, 1, 39, 1, 40, 1, 41, 1, 42, 17, 43, 1, 44, 1, 45, 1, 46, 7, 47, 19
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OFFSET
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1,4
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COMMENTS
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For n > 1, we first find 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). a(n) is then the quotient polynomial converted back to decimal via its binary encoding. See the example.
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LINKS
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FORMULA
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EXAMPLE
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For n = 9, its least nontrivial divisor is 3, and we find that 3 (in binary "11") corresponds to polynomial X + 1, which in this case is a factor of polynomial X^3 + 1 (corresponding to 9 as 9 is "1001" in binary) as the latter factorizes as (X + 1)(X^2 + X + 1) over GF(2), that is, 9 = A048720(3,7). Thus a(9) = 7.
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PROG
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(PARI) A325642(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(fromdigits(Vec(lift(p/q)), 2))))));
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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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