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A178679 Numbers n such that the binary expansion of n starts with the base 3 expansion of n. 3
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 (list; graph; refs; listen; history; text; internal format)
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*)

PROG

(PARI code from Robert Gerbicz)

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--)))

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

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Last modified August 19 14:06 EDT 2017. Contains 290808 sequences.