A043307 a(n) = A033001(n)/4.
1, 2, 9, 11, 18, 19, 82, 83, 99, 100, 163, 164, 171, 173, 738, 740, 747, 748, 892, 893, 900, 902, 1467, 1469, 1476, 1477, 1540, 1541, 1557, 1558, 6643, 6644, 6660, 6661, 6724, 6725, 6732, 6734, 8028, 8030, 8037, 8038, 8101, 8102, 8118, 8119
Offset: 1
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
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Maple
A[1]:= [1,2]: for d from 2 to 6 do A[d]:= map(t -> seq(9*t+j,j=subs(t mod 9 = NULL, [0,1,2])), A[d-1]) od: seq(op(A[d]),d=1..6); # Robert Israel, Jan 29 2017
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Mathematica
Table[FromDigits[#,9]&/@Select[Tuples[{0,1,2},n],Min[Abs[Differences[#]]]>0&],{n,2,5}]// Flatten// Union (* Harvey P. Dale, May 27 2023 *) -
PARI
is_A043307(n)=(n=[n])&&!until(!n[1],((n=divrem(n[1],9))[2]<3 && n[1]%3!=n[2])||return) \\ M. F. Hasler, Feb 03 2014
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PARI
a(n) = my(v=binary(n+1)); v[1]=0; for(i=2,#v, v[i]+=(v[i]>=v[i-1])); fromdigits(v,9); \\ Kevin Ryde, Mar 13 2021
Formula
From Robert Israel, Jan 29 2017: (Start)
If a(n) == 0 (mod 3) then a(2*n+1) = 9*a(n) + 1 else a(2*n+1) = 9*a(n).
If a(n) == 2 (mod 3) then a(2*n+2) = 9*a(n) + 1 else a(2*n+1) = 9*a(n)+2.
a(4k+5) = 9*a(2k+2).
(End)
Comments