OFFSET
1,3
COMMENTS
It is much easier to find the terms by backtracking (see PARI code). It seems that the sequence is infinite and log(a(n))/(n*log(3))->1 as n goes to infinity. - Robert Gerbicz, Jun 04 2010
LINKS
Robert Gerbicz, Table of n, a(n) for n = 1..208
EXAMPLE
2217 = 100010101001 (base 2)
2217 = 10001010.... (base 3)
531795 = 10000001110101010011 (base 2)
531795 = 1000000111010....... (base 3)
50218140 = 10111111100100010010011100 (base 2) = 10111111100100010 (base 3)
63858541 = 11110011100110011101101101 (base 2) = 11110011100110011 (base 3)
MATHEMATICA
Do[i3=IntegerDigits[n, 2]; l3=Length[i3]; i=FromDigits[i3, 3]; i2=IntegerDigits[i, 2]; If[i3==Take[i2, l3], Print[i]]; , {n, 0, 10^7}]; (* Ray Chandler, Jun 03 2010 *)
PROG
(PARI)
ct=1; print("1 0"); for(L=1, 200, pos=1; a=vector(L); a[1]=0; \
while(pos>0, backtrack=0; a[pos]++; \
if(a[pos]>1, backtrack=1, \
n1=sum(i=1, pos, a[i]*3^(L-i)); n2=n1+3^(L-pos); n=sum(i=1, L, a[i]*2^(L-i)); \
if(pos==L, L1=length(binary(n1)); \
if(shift(n1, L-L1)==n, ct++; print(ct" "n1)), \
b1=binary(n1); b2=binary(n2); L1=length(b1); L2=length(b2); ext=1; \
if(L1==L2, s=0; while(s+1<=L1&&s+1<=pos&&b1[s+1]==b2[s+1], s++); \
if(sum(i=1, s, abs(b1[i]-a[i]))!=0, ext=0)); if(ext, pos++; a[pos]=-1))); \
if(backtrack, pos--)))
\\ Robert Gerbicz, Jun 04 2010
CROSSREFS
KEYWORD
nonn
AUTHOR
Zak Seidov and Ray Chandler, Jun 03 2010
STATUS
approved