A178679 Numbers n such that the binary expansion of n starts with the base 3 expansion of n.

0, 1, 9, 10, 12, 94, 118, 120, 2217, 22204, 22602, 26608, 27004, 27009, 531795, 5021473, 5321298, 6384861, 50218140, 63858541, 1181639052, 12029699478, 14392731189, 15114438648, 283004031766, 283592574694, 2672433464707, 2835932927661

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

Cf. A169828, A178680.

Sequence in context: A279731 A037408 A178680 * A154766 A116594 A249723

Adjacent sequences: A178676 A178677 A178678 * A178680 A178681 A178682

Keyword

nonn

Author

Zak Seidov and Ray Chandler, Jun 03 2010

Status

approved