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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

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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--)))