%I #9 Feb 03 2020 15:25:46

%S 0,10,101,100,110,1001,1000,1010,1101,1100,1110,-1,-1,-1,-1,10001,

%T 10000,10010,10101,10100,10110,11001,11000,11010,11101,11100,11110,-1,

%U -1,-1,-1,100001,100000,100010,100101,100100,100110,101001,101000,101010,101101,101100,101110,-1,-1,-1,-1,110001

%N Binary string which equals n when 1's, 2's, 4's and 8's bits have weights -1, 1, 3, 6 respectively, while the other bits have their usual weights. -1 if no such string exists.

%C After a(10), the pattern seems to be sequences of sixteen a(n), four of which without solution, then 12 formed by placing a member of the binary sequence 1,10,11,11,100,101 etc. in front of re-occurring list of the same 12 4-digit numbers. The description does not lead to a unique sequence: a(0)=0 and a(0)=11 are both valid. a(3)=111 and a(3)=100 are both valid. - _R. J. Mathar_, Mar 14 2006

%D John M, Yarbough, Digital Logic Applications and Design, West Publishing, 1997. p. 25

%o (PARI) dig(n,digno,base) = { local(nshif) ; nshif=n ; for(shifr=0,digno-1, nshif = floor(nshif/base) ) ; nshif % base ; } binrep(n) = { local(nshif,resul) ; nshif=n; resul = Str(dig(nshif,0,2)) ; nshif=floor(nshif/2) ; while (nshif != 0, resul = concat(Str(dig(nshif,0,2)),resul) ; nshif=floor(nshif/2) ; ) ; return(resul) ; } modN(n) = { local(resul) ; resul = 16*floor(n/16) ; resul += -1*dig(n,0,2) ; resul += 1*dig(n,1,2) ; resul += 3*dig(n,2,2) ; resul += 6*dig(n,3,2) ; return(resul) ; } { for (n = 0, 60, for(an =0, 1000, if( modN(an) == n, anS = binrep(an) ; print1(anS,",") ; break ; ) ; if( an==1000, print("-1,") ); ) ; ) } - _R. J. Mathar_, Mar 14 2006

%Y Cf. A066335.

%K easy,sign

%O 0,2

%A _George E. Antoniou_, Dec 15 2001

%E More terms from _R. J. Mathar_, Mar 14 2006