A155078 - OEIS

%I #24 Jun 13 2017 18:53:45

%S 3,4,5,4,3,3,4,4,5,3,13,3,15,4,8,4,8,5,21,8,4,22,25,6,27,26,5,4,31,3,

%T 4,4,8,8,32,3,39,38,38,8,43,4,45,22,8,23,19,6,9,50,8,26,24,5,46,4,5,

%U 29,18,3,63,4,5,4,7,6,69,8,25,30,47,6,4,74,17,38,79,12,60,8,79,82,85,4,8,43

%N The representation of n=1,2,3,... in binary is a divisor of the same representation in another base. The sequence is the first such base.

%C The pattern of solutions for each binary representation is notable. For 1001= decimal 9, the bases as solutions are 5,8,11,14,... whereas the pattern for 111=decimal 7 is 4,9,11,16,18,....

%C The binary representation of n corresponds to the unique polynomial p_n(x) with coefficients in {0,1} such that p(2) = n. a(n) is the least x >= 3 such that p_n(x) == 0 mod n. Thus 3 <= a(n) <= n + 2. - _Robert Israel_, Dec 15 2014

%C From _Rémy Sigrist_, Mar 15 2017: (Start)

%C If n is even then a(n) <= max(4, n).

%C If n is odd then a(n) <= n + 2.

%C If n is odd then n and a(n) are coprime.

%C If a(n)=4 then n belongs to A062846.

%C (End)

%H Robert Israel, <a href="/A155078/b155078.txt">Table of n, a(n) for n = 1..10000</a>

%F It is simply a matter of converting a binary number to another base to see if the resulting number is a multiple of n. The lowest other base is listed.

%e The n-th term is solved by converting the decimal n to binary then asking to what other base is this representation a multiple of n. For the 5th term, the binary representation is 101; if this is converted to base 3, 101 = 9+0+1 = 10, a multiple of 5. The base 3 is the first base producing a multiple of n: the 5th term is therefore 3.

%p A155078 := proc(n) local bdgs,b ; bdgs := convert(n,base,2) ; for b from 3 do add(op(i,bdgs)*b^(i-1),i=1..nops(bdgs)) ; if mod n = 0 then RETURN(b); fi; od: end: seq(A155078(n),n=1..100) ; # _R. J. Mathar_, Mar 14 2009

%p # second Maple program:

%p f:= proc(n) local b, L,r, sols;

%p      L:= convert(n,base,2);

%p      r:= add(L[i]*b^(i-1),i=1..nops(L));

%p      sols:= subs(0=n,1=n+1,2=n+2,map(t -> rhs(op(t)),{msolve(r,n)})) ;

%p      min(sols);

%p end proc:

%p 3, seq(f(n),n=2..100); # _Robert Israel_, Dec 15 2014

%Y Cf. A062846.

%K base,easy,nonn

%O 1,1

%A _J. M. Bergot_, Jan 19 2009

%E Corrected and extended by _R. J. Mathar_, Mar 14 2009