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A001196
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Double-bitters: only even length runs in binary expansion.
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45
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0, 3, 12, 15, 48, 51, 60, 63, 192, 195, 204, 207, 240, 243, 252, 255, 768, 771, 780, 783, 816, 819, 828, 831, 960, 963, 972, 975, 1008, 1011, 1020, 1023, 3072, 3075, 3084, 3087, 3120, 3123, 3132, 3135, 3264, 3267, 3276, 3279, 3312, 3315, 3324, 3327, 3840, 3843
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OFFSET
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0,2
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COMMENTS
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Numbers whose set of base 4 digits is {0,3}. - Ray Chandler, Aug 03 2004
n such that there exists a permutation p_1, ..., p_n of 1, ..., n such that i + p_i is a power of 4 for every i. - Ray Chandler, Aug 03 2004
The first 2^n terms of the sequence could be obtained using the Cantor-like process for the segment [0, 4^n-1]. E.g., for n=1 we have [0, {1, 2}, 3] such that numbers outside of braces are the first 2 terms of the sequence; for n=2 we have [0, {1, 2}, 3, {4, 5, 6, 7, 8, 9, 10, 11}, 12, {13, 14}, 15] such that the numbers outside of braces are the first 4 terms of the sequence, etc. - Vladimir Shevelev, Dec 17 2012
Also, the indices of the compositions having only even parts. For the definition of the index of a composition, see A298644. For example, 195 is in the sequence since its binary form is 11000011 and the composition [2,4,2] has only even parts. 132 is not in the sequence since its binary form is 10000100 and the composition [1,4,1,2] also has odd parts.
The command c(n) from the Maple program yields the composition having index n. (End)
After the k-th step of generating the Koch snowflake curve, label the edges of the curve consecutively 0..3*4^k-1 starting from a vertex of the originating triangle. a(0), a(1), ... a(2^k-1) are the labels of the edges contained in one edge of the originating triangle. Add 4^k to each label to get the labels for the next edge of the triangle. Compare with A191108 in respect of the Sierpinski arrowhead curve. - Peter Munn, Aug 18 2019
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LINKS
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FORMULA
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a(2n) = 4*a(n), a(2n+1) = 4*a(n) + 3.
Sum_{n>=1} 1/a(n) = 0.628725478158702414849086504025451177643560169366348272891020450593453403709... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 12 2022
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MAPLE
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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 := {}: for n to 3350 do if type(product(1+c(n)[j], j = 1 .. nops(c(n))), odd) = true then A := `union`(A, {n}) else end if end do: A; # most of the Maple program is due to W. Edwin Clark. - Emeric Deutsch, Jan 26 2018
# second Maple program:
a:= proc(n) option remember;
`if`(n<2, 3*n, 4*a(iquo(n, 2, 'r'))+3*r)
end:
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MATHEMATICA
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fQ[n_] := Union@ Mod[Length@# & /@ Split@ IntegerDigits[n, 2], 2] == {0}; Select[ Range@ 10000, fQ] (* Or *)
fQ[n_] := Union@ Join[IntegerDigits[n, 4], {0, 3}] == {0, 3}; Select[ Range@ 10000, fQ] (* Robert G. Wilson v, Dec 24 2012 *)
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PROG
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(Haskell)
a001196 n = if n == 0 then 0 else 4 * a001196 n' + 3 * b
where (n', b) = divMod n 2
(Python)
def inA001196(n):
while n != 0:
if n%4 == 1 or n%4 == 2:
return 0
n = n//4
return 1
n, a = 0, 0
while n < 20:
if inA001196(a):
print(n, a)
n = n+1
(Python)
from itertools import groupby
def ok2lb(n):
if n == 0: return True # by convention
return all(len(list(g))%2 == 0 for k, g in groupby(bin(n)[2:]))
(Python)
(PARI) a(n) = 3*fromdigits(binary(n), 4); \\ Kevin Ryde, Nov 07 2020
(C) int a_next(int a_n) { int t = a_n << 1; return a_n ^ t ^ (t + 3); } /* Falk Hüffner, Jan 24 2022 */
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CROSSREFS
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3 times the Moser-de Bruijn sequence A000695.
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KEYWORD
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nonn,base,easy
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AUTHOR
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N. J. A. Sloane, based on an email from Bart la Bastide (bart(AT)xs4all.nl)
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STATUS
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approved
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