A004642 -id:A004642 - cp's OEIS frontend

A039966 a(0) = 1; thereafter a(3n+2) = 0, a(3n) = a(3n+1) = a(n).

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

Offset: 0

N. J. A. Sloane, Dec 11 1999

nonn

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.

			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

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.]
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.
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
Index entries for sequences that are fixed points of mappings
Index entries for characteristic functions

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.
Cf. A062051, A000009, A000244, A004642.
Characteristic function of A005836 (and apart from offset of A003278).

a039966 n = fromEnum (n < 2 || m < 2 && a039966 n' == 1)
   where (n',m) = divMod n 3
-- Reinhard Zumkeller, Sep 29 2011

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

(* 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 *)
Nest[ Join[#, #, 0 #] &, {1}, 5] (* Robert G. Wilson v, Jul 27 2014 *)

{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 */

A039966(n)=vecmax(digits(n+!n,3))<2;
apply(A039966, [0..99]) \\ M. F. Hasler, Feb 15 2023

def A039966(n):
    while n > 2:
        n,r = divmod(n,3)
        if r==2: return 0
    return int(n!=2) # M. F. Hasler, Feb 15 2023

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
a(n) = A005043(n-1) mod 3. - Christian G. Bower, Jun 12 2005
a(n) = A002426(n) mod 3. - John M. Campbell, Aug 24 2011
a(n) = A000275(n) mod 3. - John M. Campbell, Jul 08 2016
Properties: 0 <= a(n) <= 1, a(A074940(n)) = 0, a(A005836(n)) = 1; A104406(n) = Sum(a(k), 1 <= k <= n). - Reinhard Zumkeller, Mar 05 2005
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.

Entry revised Jun 30 2005

Offset corrected by John M. Campbell, Aug 24 2011

A102483 Numbers k such that 2^k contains no zeros in base 3.

Original entry on oeis.org

0, 1, 2, 3, 4, 15

Offset: 1

N. J. A. Sloane, Feb 25 2005

I conjectured in 1973 that there are no further terms. This question is still open.
A104320(a(n)) = 0. - Reinhard Zumkeller, Mar 01 2005
No other terms less than 200000. - Robert G. Wilson v, Dec 06 2005
a(7) > 10^7. - Martin Ehrenstein, Jul 27 2021
If it exists, a(7) > 10^21. - Robert Saye, Mar 23 2022

Robert I. Saye, On two conjectures concerning the ternary digits of powers of two, J. Integer Seq. 25 (2022) Article 22.3.4.
N. J. A. Sloane, The persistence of a number, J. Recreational Math., 6 (1973), 97-98.
Eric Weisstein's World of Mathematics, Ternary

Cf. A007377, A004642, A346497.

Select[ Range@1000, FreeQ[ IntegerDigits[2^#, 3], 0] &] (* Robert G. Wilson v, Dec 06 2005 *)

for (n=0, 100, if (vecmin(digits(2^n, 3)), print1(n, ", "))) \\ Michel Marcus, Mar 25 2015

A021823 Decimal expansion of 1/819.

Original entry on oeis.org

0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1

Offset: 0

N. J. A. Sloane

Partial sums of A010892. - Paul Barry, Jun 06 2003
Expansion in any base b >= 3 of 1/((b-1)*(b^2-b+1)) = 1/(b^3-2b^2+2b-1). E.g., 1/14 in base 3, 1/39 in base 4, 1/84 in base 5, etc. - Franklin T. Adams-Watters, Nov 07 2006
a(n) is the second least significant digit in the ternary representation of 2^n (cf. A004642). - Alexandre Herrera, Oct 09 2023

			0.0012210012210012210...

Index entries for linear recurrences with constant coefficients, signature (2,-2,1).

Cf. A077859, A027444.

Cf. A004642, A153130 (2^n mod 9).

Join[{0,0},RealDigits[1/819,10,120][[1]]] (* or *) PadRight[{},120,{0,0,1,2,2,1}] (* or *) LinearRecurrence[{2,-2,1},{0,0,1},120] (* Harvey P. Dale, Aug 19 2012 *)

a(n)=1/819. \\ Charles R Greathouse IV, Sep 24 2015

a(n) = a(n-1)-a(n-2)+1 = 2-a(n-3) = a(n-6). - Henry Bottomley, Apr 12 2000
a(n) = Sum_{k=1..floor(n/2)} (-1)^(k+1)*binomial(n-k, k) = 1-((-1)^floor(n/3)+(-1)^(floor((n+1)/3)))/2. - Vladeta Jovovic, Feb 10 2003
G.f.: x^2/(1-2x+2x^2-x^3)=x^2/((1-x)(x^2-x+1)). - Paul Barry, Jun 06 2003
a(n+2) = sum{k=0..n, binomial(n-2k, n-k)}. - Paul Barry, Jan 15 2005
a(0)=0, a(1)=0, a(2)=1, a(n)=2*a(n-1)-2*a(n-2)+a(n-3). - Harvey P. Dale, Aug 19 2012

A104320 Number of zeros in ternary representation of 2^n.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 1, 2, 2, 1, 1, 1, 4, 1, 0, 4, 2, 3, 3, 3, 3, 3, 7, 7, 9, 5, 6, 6, 4, 4, 3, 5, 6, 7, 9, 9, 10, 6, 6, 9, 9, 8, 9, 8, 7, 13, 12, 13, 9, 5, 9, 8, 6, 16, 13, 9, 10, 11, 11, 7, 14, 13, 13, 9, 12, 14, 15, 15, 11, 11, 17, 15, 19, 14, 19, 12, 18, 15, 11, 10, 16, 15, 14, 14, 13, 17, 14

Offset: 0

Reinhard Zumkeller, Mar 01 2005

Conjecture from N. J. A. Sloane: a(n) > 0 for n > 15, see A102483.

			n=13: 2^13=8192 -> '102020102', a(13) = 4.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A004642, A007089.

[Multiplicity(Intseq(2^n,3),0):n in [0..90]]; // Marius A. Burtea, Nov 17 2019

f:= n -> numboccur(0, convert(2^n,base,3)):
map(f, [$0..100]); # Robert Israel, Nov 17 2019

Table[DigitCount[2^n,3,0],{n,0,90}] (* Harvey P. Dale, May 06 2014 *)

a(n) = my(d=vecsort(digits(2^n, 3))); #setintersect(d, vector(#d)) \\ Felix Fröhlich, Nov 17 2019

a(n) = #select(d->!d, digits(2^n, 3)); \\ Ruud H.G. van Tol, May 09 2024

a(n) = A077267(A000079(n)).

a(A104321(n))=n and a(m)<>n for m < A104321(n).

A000866 2^n written in base 5.

Original entry on oeis.org

1, 2, 4, 13, 31, 112, 224, 1003, 2011, 4022, 13044, 31143, 112341, 230232, 1011014, 2022033, 4044121, 13143242, 31342034, 113234123, 232023301, 1014102102, 2033204204, 4121413413, 13243332331, 32042220212, 114134440424, 233324431403, 1022204413311

Offset: 0

N. J. A. Sloane, Jacques Haubrich (jhaubrich(AT)freeler.nl)

T. D. Noe, Table of n, a(n) for n = 0..300

Cf. A000079, A004642, ..., A004655: powers of 2 written in base 10, 3, 4, ..., 16

Cf. A000244, A004656, A004658, A004659, ... : powers of 3 written in base 10, 2, 4, 5, ...

Table[FromDigits[IntegerDigits[2^n, 5]], {n, 0, 30}] (* T. D. Noe, Jun 20 2012 *)

a(n)=fromdigits(digits(2^n,5)) \\ M. F. Hasler, Jun 23 2018

More terms from Erich Friedman.

A037096 Periodic vertical binary vectors computed for powers of 3: a(n) = Sum_{k=0 .. (2^n)-1} (floor((3^k)/(2^n)) mod 2) * 2^k.

Original entry on oeis.org

1, 2, 0, 204, 30840, 3743473440, 400814250895866480, 192435610587299441243182587501623263200, 2911899996313975217187797869354128351340558818020188112521784134070351919360

Offset: 0

Antti Karttunen, Jan 29 1999

This sequence can be also computed with a recurrence that does not explicitly refer to 3^n. See the C program.

Conjecture: For n >= 3, each term a(n), when considered as a GF(2)[X] polynomial, is divisible by the GF(2)[X] polynomial (x + 1) ^ A055010(n-1). If this holds, then for n >= 3, a(n) = A048720(A136386(n), A048723(3,A055010(n-1))).

			When powers of 3 are written in binary (see A004656), under each other as:
  000000000001 (1)
  000000000011 (3)
  000000001001 (9)
  000000011011 (27)
  000001010001 (81)
  000011110011 (243)
  001011011001 (729)
  100010001011 (2187)
it can be seen that the bits in the n-th column from the right can be arranged in periods of 2^n: 1, 2, 4, 8, ... This sequence is formed from those bits: 1, is binary for 1, thus a(0) = 1. 01, reversed is 10, which is binary for 2, thus a(1) = 2, 0000 is binary for 0, thus a(2)=0, 000110011, reversed is 11001100 = A007088(204), thus a(3) = 204.

S. Wolfram, A New Kind of Science, Wolfram Media Inc., (2002), p. 119.

Antti Karttunen, Table of n, a(n) for n = 0..11
Antti Karttunen, C program for computing this sequence.
S. Wolfram, A New Kind of Science, Wolfram Media Inc., (2002), p. 119.

Cf. A036284, A037095, A037097, A136386 for related sequences.
Cf. A000079, A000215, A000225, A000244, A004656, A007088, A048720, A048723, A055010.
Cf. also A004642, A265209, A265210 (for 2^n written in base 3).

a(n) := sum( 'bit_n(3^i, n)*(2^i)', 'i'=0..(2^(n))-1);
bit_n := (x, n) -> `mod`(floor(x/(2^n)), 2);

a(n) = Sum_{k=0 .. A000225(n)} (floor(A000244(k)/(2^n)) mod 2) * 2^k.
Other identities and observations:
For n >= 2, a(n) = A000215(n-1)*A037097(n) = A048720(A037097(n), A048723(3, A000079(n-1))).

Entry revised by Antti Karttunen, Dec 29 2007

Name changed and the example corrected by Antti Karttunen, Dec 05 2015

A368866 The smallest positive number such that 2^a(n) when written in base n contains adjacent equal digits.

Original entry on oeis.org

2, 2, 4, 5, 6, 3, 6, 12, 16, 14, 11, 15, 8, 4, 8, 23, 16, 14, 16, 21, 9, 17, 20, 14, 30, 27, 16, 15, 10, 5, 10, 29, 48, 14, 46, 19, 18, 15, 32, 36, 27, 36, 18, 12, 56, 41, 37, 24, 58, 22, 26, 46, 58, 40, 29, 24, 36, 14, 20, 18, 12, 6, 12, 60, 62, 50, 49, 50, 20, 35, 36, 55, 61, 52, 53, 77

Offset: 2

Scott R. Shannon, Jan 08 2024

In the first 10000 terms the largest value is a(9031) = 1924, with a corresponding power of 2 of approximately 1.52*10^579.

			a(2) = 2 as 2^2 = 4 written in base 2 = 100_2 which contains adjacent 0's.
a(6) = 6 as 2^6 = 64 written in base 6 = 144_6 which contains adjacent 4's.
a(10) = 16 as 2^16 = 65536 written in base 10 = 65536_10 which contains adjacent 5's.

Scott R. Shannon, Table of n, a(n) for n = 2..10000

Cf. A000079, A171901, A011557, A004642, A004643, A000866.

f:= proc(n) local k,L;
  for k from 1 do
    L:= convert(2^k,base,n);
    if member(0, L[2..-1]-L[1..-2]) then return k fi
  od
end proc:
map(f, [$2..100]); # Robert Israel, Jan 09 2024

from itertools import count
from sympy.ntheory.factor_ import digits
def A368866(n):
    k = 1
    for m in count(1):
        k <<= 1
        s = digits(k,n)[1:]
        if any(s[i]==s[i+1] for i in range(len(s)-1)):
            return m # Chai Wah Wu, Jan 08 2024

A004663 Powers of 3 written in base 9.

Original entry on oeis.org

1, 3, 10, 30, 100, 300, 1000, 3000, 10000, 30000, 100000, 300000, 1000000, 3000000, 10000000, 30000000, 100000000, 300000000, 1000000000, 3000000000, 10000000000, 30000000000, 100000000000, 300000000000, 1000000000000, 3000000000000, 10000000000000, 30000000000000

Offset: 0

N. J. A. Sloane

Robert Israel, Table of n, a(n) for n = 0..1997
Index entries for linear recurrences with constant coefficients, signature (0,10).

Cf. A026383, A016116.
Cf. A000244, A004656, A004658, A004659, ... : powers of 3 in base 10, 2, 4, 5, ...
Cf. A000079, A004642, ..., A004655: powers of 2 written in base 10, 2, 3, ..., 16.

seq(op([10^i,3*10^i]),i=0..100); # Robert Israel, Jun 25 2018

Table[FromDigits[IntegerDigits[3^n, 9]], {n, 0, 100}] (* G. C. Greubel, Oct 12 2018 *)

a(n)=3^bittest(n,0)*10^(n\2) \\ M. F. Hasler, Jun 25 2018

From Paul Barry, Jul 14 2004: (Start)
G.f.: (1 + 3*x)/(1 - 10*x^2);
a(n) = 2*a(n-1) + 3*a(n-2) + 10^floor((n-2)/2);
a(n) = Sum_{k=0..floor(n/2)} binomial(floor(n/2), k)*3^(n-2*k). (End)
a(n) = 3*a(n-1) + ((1 + (-1)^n)/2)*a(n-2) with a(0)=1, a(1)=3. - Taras Goy, Mar 20 2019
E.g.f.: cosh(sqrt(10)*x) + 3*sinh(sqrt(10)*x)/sqrt(10). - Stefano Spezia, Mar 31 2023

A259667 Catalan numbers mod 6.

Original entry on oeis.org

1, 1, 2, 5, 2, 0, 0, 3, 2, 2, 2, 4, 4, 4, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 4, 4, 1, 0, 0, 0, 4, 4, 4, 2, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 4, 4, 4, 0, 0, 0, 4, 4, 4, 2, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 4, 4, 2, 2, 2, 0, 0, 0, 2, 2, 2, 4

Offset: 0

M. F. Hasler, Nov 08 2015

nonn

The only odd terms are those with indices n = 2^k - 1 (k = 0, 1, 2, 3, ...); see also A038003.
It is conjectured that the only k which yield a(2^k-1) = 1 are k = 0, 1 and 5. Are there other k than 2 and 8 that yield a(2^k-1) = 5? Otherwise said, is a(2^k-1) = 3 for all k > 8?
The question is equivalent to: does 2^k - 1 always contain a digit 2 when converted into base 3 for all k > 8? Similar conjecture has been proposed for 2^k, see A004642. - Jianing Song, Sep 04 2018

M. Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems, sect. III: Binomial coefficients modulo integers, binomod.gp (v.1.4, 11/2015).
V. Reshetnikov, A000108(n) ≡ 1 (mod 6), SeqFan list, Nov. 8, 2015.

Cf. A000108, A004642, A038003.

Mod[CatalanNumber[Range[0,120]],6] (* Harvey P. Dale, Oct 24 2020 *)

PARI
```
a(n)=binomial(2*n,n)/(n+1)%6
```

A259667(n)=lift(if(n%3!=1,binomod(2*n+1,n,6)/(2*n+1), if(bittest(n,0),binomod(2*n,n-1,6)/n,binomod(2*n,n,6)/(n+1)))) \\ using binomod.gp by M. Alekseyev, cf. Links.

a(n) = A000108(n) mod 6.

A104406 Number of numbers <= n having no 2 in ternary representation.

Original entry on oeis.org

1, 1, 2, 3, 3, 3, 3, 3, 4, 5, 5, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 9, 9, 10, 11, 11, 11, 11, 11, 12, 13, 13, 14, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15

Offset: 1

Reinhard Zumkeller, Mar 05 2005

nonn

Partial sums of A039966: a(n) = Sum(A039966(k): 1<=k<=n).

N. J. A. Sloane, Table of n, a(n) for n = 1..10000

Cf. A005836, A004642, A039966.

This is a lower bound for A003002.

cp's OEIS Frontend

A039966 a(0) = 1; thereafter a(3n+2) = 0, a(3n) = a(3n+1) = a(n).

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A102483 Numbers k such that 2^k contains no zeros in base 3.

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A021823 Decimal expansion of 1/819.

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A104320 Number of zeros in ternary representation of 2^n.

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A000866 2^n written in base 5.

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A037096 Periodic vertical binary vectors computed for powers of 3: a(n) = Sum_{k=0 .. (2^n)-1} (floor((3^k)/(2^n)) mod 2) * 2^k.

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A368866 The smallest positive number such that 2^a(n) when written in base n contains adjacent equal digits.

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