A325563 a(1) = 1; for n > 1, a(n) is the largest proper divisor d of n such that A048720(d,k) = n for some k.
1, 1, 1, 2, 1, 3, 1, 4, 3, 5, 1, 6, 1, 7, 5, 8, 1, 9, 1, 10, 7, 11, 1, 12, 1, 13, 9, 14, 1, 15, 1, 16, 3, 17, 7, 18, 1, 19, 3, 20, 1, 21, 1, 22, 15, 23, 1, 24, 7, 25, 17, 26, 1, 27, 1, 28, 3, 29, 1, 30, 1, 31, 21, 32, 5, 33, 1, 34, 1, 35, 1, 36, 1, 37, 15, 38, 1, 39, 1, 40, 1, 41, 1, 42, 17, 43, 1, 44, 1, 45, 1, 46, 31, 47, 19, 48, 1, 49, 33, 50
Offset: 1
Keywords
Examples
For n = 39 = 3*13, A032742(39) = 13, but 13 is not the answer because X^3 + X^2 + 1 does not divide X^5 + X^2 + X + 1 (39 is "100111" in binary) over GF(2). However, the next smaller divisor 3 works because X^5 + X^2 + X + 1 = (X^1 + 1)(X^4 + X^3 + X^2 + 1) when multiplication is done over GF(2). Note that 39 = A048720(3,29), where 29 is "11101" in binary. Thus a(39) = 3.
Links
Crossrefs
Programs
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PARI
A325563(n) = if(1==n,n, my(p = Pol(binary(n))*Mod(1, 2)); fordiv(n,d,if((d>1),my(q = Pol(binary(n/d))*Mod(1, 2)); if(0==(p%q), return(n/d)))));
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PARI
A048720(b,c) = fromdigits(Vec(Pol(binary(b))*Pol(binary(c)))%2, 2); A325563(n) = if(1==n,n,fordiv(n,d,if((d>1),for(t=1,n,if(A048720(n/d,t)==n,return(n/d)))))); \\ (Slow)
Formula
For all n, a(n) <= A032742(n).
Comments