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A063037 Numbers without 3 consecutive equal binary digits. 11
0, 1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 13, 18, 19, 20, 21, 22, 25, 26, 27, 36, 37, 38, 41, 42, 43, 44, 45, 50, 51, 52, 53, 54, 73, 74, 75, 76, 77, 82, 83, 84, 85, 86, 89, 90, 91, 100, 101, 102, 105, 106, 107, 108, 109, 146, 147, 148, 149, 150, 153, 154, 155, 164, 165, 166 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
Complement of A136037; intersection of A003796 and A003726. - Reinhard Zumkeller, Dec 11 2007
From Emeric Deutsch, Jan 27 2018: (Start)
Also 0 together with the indices of the compositions that have no parts larger than 2. For the definition of the index of a composition see A298644.
For example, 105 is in the sequence since its binary form is 1101001 and the composition [2,1,1,2,1] has no parts larger than 2.
On the other hand, 132 is not in the sequence since its binary form is 10000100 and the composition [1,4,1,2] has a part larger than 2.
The command c(n) from the Maple program yields the composition having index n. (End)
The sequence contains A000045(n+1) positive terms with binary length n. - Rémy Sigrist, Sep 30 2022
LINKS
FORMULA
It appears (but has not yet been proved) that the terms of {a(n)} can be computed recursively as follows. Let {c(i)} be defined for i >= 4 by c(i) = 2c(i-1) + 1, if i is a multiple of 3, else c(i) = 2c(i-1) - 1, with c(4) = 1. I.e., {c(i)} = {1,1,3,5,9,19,37,73,147,...}, for i=4,5,6,... . Let a(1)=1, a(2)=2, a(3)=3. For n > 3, choose k so that F(k)-2 < n <= F(k+1)-2, where F(k) denotes the k-th Fibonacci number (A000045). Then a(n) = c(k) + 2a(F(k)-2) - a(2F(k)-n-3). This has been verified for n up to 1100. - John W. Layman, May 26 2009
EXAMPLE
The binary representation of 9 (1001) has no 3 consecutive equal digits.
MAPLE
isA063037 := proc(n)
local bdgs, rep, d, i ;
if n = 0 then
return true;
end if;
bdgs := convert(n, base, 2) ;
rep := 1;
d := op(1, bdgs) ;
for i from 2 to nops(bdgs) do
if op(i, bdgs) = op(i-1, bdgs) then
rep := rep+1 ;
else
rep :=1 ;
end if ;
if rep > 2 then
return false;
end if;
end do:
return true ;
end proc:
for n from 0 to 50 do
if isA063037(n) then
printf("%d, ", n) ;
end if;
end do: # R. J. Mathar, Dec 18 2013
# Second Maple program:
Runs := proc (L) local j, r, i, k: j := 1: r[j] := L[1]: for i from 2 to nops(L) do if L[i] = L[i-1] then r[j] := r[j], L[i] else j := j+1: r[j] := L[i] end if end do: [seq([r[k]], k = 1 .. j)] end proc: RunLengths := proc (L) map(nops, Runs(L)) end proc: c := proc (n) ListTools:-Reverse(convert(n, base, 2)): RunLengths(%) end proc: A := {0}: for n to 250 do if `subset`(convert(c(n), set), {1, 2}) then A := `union`(A, {n}) else end if end do: A; # most of this Maple program is due to W. Edwin Clark. - Emeric Deutsch, Jan 27 2018
MATHEMATICA
Select[Range[0, 168], AllTrue[Length /@ Split@ IntegerDigits[#, 2], # < 3 &] &] (* Michael De Vlieger, Aug 20 2017 *)
PROG
(PARI) { n=0; for (m=0, 10^9, x=m; t=1; b=2; while (x>0, d=x-2*(x\2); x\=2; if (d==b, c++; if (c==3, t=x=0), b=d; c=1)); if (t, write("b063037.txt", n++, " ", m); if (n==1000, break)) ) } \\ Harry J. Smith, Aug 16 2009
(PARI) a(n) = { n--; if (n<=1, return (n), my (s=1); for (i=1, oo, my (f=fibonacci(i+1)); if (n<s+f, return (2^i-1-a(2*s-n)), s+=f))) } \\ Rémy Sigrist, Sep 30 2022
(Python)
from itertools import count, islice
def A063037_gen(startvalue=0): # generator of terms >= startvalue
return filter(lambda n: not ('000' in (s:=bin(n)[2:]) or '111' in s), count(max(0, startvalue)))
A063037_list = list(islice(A063037_gen(), 20)) # Chai Wah Wu, Oct 04 2022
CROSSREFS
Sequence in context: A032878 A032845 A023776 * A286262 A330029 A368841
KEYWORD
easy,nonn
AUTHOR
Lior Manor, Jul 05 2001
EXTENSIONS
Missing "less than" sign supplied in the conjectured recurrence (thanks to Franklin T. Adams-Watters for pointing this out) by John W. Layman, Nov 09 2009
STATUS
approved

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Last modified March 19 03:21 EDT 2024. Contains 370952 sequences. (Running on oeis4.)