|
|
A039966
|
|
a(0) = 1; thereafter a(3n+2) = 0, a(3n) = a(3n+1) = a(n).
|
|
32
|
|
|
1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,1
|
|
COMMENTS
|
Number of partitions of n into distinct powers of 3.
Trajectory of 1 under the morphism: 1 -> 110, 0 -> 000. Thus 1 -> 110 ->110110000 -> 110110000110110000000000000 -> ... - Philippe Deléham, Jul 09 2005
Also, an example of a d-perfect sequence.
This is a composite of two earlier sequences contributed at different times by N. J. A. Sloane and by Reinhard Zumkeller, Mar 05 2005. Christian G. Bower extended them and found that they agreed for at least 512 terms. The proof that they were identical was found by Ralf Stephan, Jun 13 2005, based on the fact that they were both 3-regular sequences.
|
|
LINKS
|
D. Kohel, S. Ling and C. Xing, Explicit Sequence Expansions, in Sequences and their Applications, C. Ding, T. Helleseth, and H. Niederreiter, eds., Proceedings of SETA'98 (Singapore, 1998), 308-317, 1999.
|
|
FORMULA
|
a(0) = 1, a(1) = 0, a(n) = b(n-2), where b is the sequence defined by b(0) = 1, b(3n+2) = 0, b(3n) = b(3n+1) = b(n). - Ralf Stephan
Euler transform of sequence b(n) where b(3^k) = 1, b(2*3^k) = -1 and zero otherwise. - Michael Somos, Jul 15 2005
G.f. A(x) satisfies A(x) = (1+x)*A(x^3). - Michael Somos, Jul 15 2005
G.f.: Product{k>=0} 1+x^(3^k). Exponents give A005836.
|
|
EXAMPLE
|
The triples of elements (a(3k), a(3k+1), a(3k+2)) are (1,1,0) if a(k) = 1 and (0,0,0) if a(k) = 0. So since a(2) = 0, a(6) = a(7) = a(8) = 0, and since a(3) = 1, a(9) = a(10) = 1 and a(11) = 0. - Michael B. Porter, Jul 11 2016
|
|
MAPLE
|
a := proc(n) option remember; if n <= 1 then RETURN(1) end if; if n = 2 then RETURN(0) end if; if n mod 3 = 2 then RETURN(0) end if; if n mod 3 = 0 then RETURN(a(1/3*n)) end if; if n mod 3 = 1 then RETURN(a(1/3*n - 1/3)) end if end proc; # Ralf Stephan, Jun 13 2005
|
|
MATHEMATICA
|
(* first do *) Needs["DiscreteMath`Combinatorica`"] (* then *) s = Rest[ Sort[ Plus @@@ Table[UnrankSubset[n, Table[3^i, {i, 0, 4}]], {n, 32}]]]; Table[ If[ Position[s, n] == {}, 0, 1], {n, 105}] (* Robert G. Wilson v, Jun 14 2005 *)
CoefficientList[Series[Product[(1 + x^(3^k)), {k, 0, 5}], {x, 0, 111}], x] (* or *)
Nest[ Flatten[ # /. {0 -> {0, 0, 0}, 1 -> {1, 1, 0}}] &, {1}, 5] (* Robert G. Wilson v, Mar 29 2006 *)
|
|
PROG
|
(PARI) {a(n)=local(A, m); if(n<0, 0, m=1; A=1+O(x); while(m<=n, m*=3; A=(1+x)*subst(A, x, x^3)); polcoeff(A, n))} /* Michael Somos, Jul 15 2005 */
(PARI) A039966(n)=vecmax(digits(n+!n, 3))<2;
(Haskell)
a039966 n = fromEnum (n < 2 || m < 2 && a039966 n' == 1)
where (n', m) = divMod n 3
(Python)
while n > 2:
n, r = divmod(n, 3)
if r==2: return 0
|
|
CROSSREFS
|
For generating functions Product_{k>=0} (1+a*x^(b^k)) for the following values of (a,b) see: (1,2) A000012 and A000027, (1,3) A039966 and A005836, (1,4) A151666 and A000695, (1,5) A151667 and A033042, (2,2) A001316, (2,3) A151668, (2,4) A151669, (2,5) A151670, (3,2) A048883, (3,3) A117940, (3,4) A151665, (3,5) A151671, (4,2) A102376, (4,3) A151672, (4,4) A151673, (4,5) A151674.
Characteristic function of A005836 (and apart from offset of A003278).
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
Entry revised Jun 30 2005
|
|
STATUS
|
approved
|
|
|
|