

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.


7



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,4


COMMENTS

For n > 1, a(n) is the largest proper divisor d of n 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 polynomial division is done over field GF(2). See the example.


LINKS



FORMULA



EXAMPLE

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.


PROG

(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)))));
(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)


CROSSREFS

Cf. A325559 (positions of ones, after the initial 1).


KEYWORD

nonn


AUTHOR



STATUS

approved



