A039969 -id:A039969 - cp's OEIS frontend

A039967 Duplicate of A039969.

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

Offset: 1

dead

A085296 Runs of zeros in Catalan sequence modulo 3: consecutive occurrences of binomial(2*k,k)/(k+1) == 0 (mod 3).

Original entry on oeis.org

3, 12, 3, 39, 3, 12, 3, 120, 3, 12, 3, 39, 3, 12, 3, 363, 3, 12, 3, 39, 3, 12, 3, 120, 3, 12, 3, 39, 3, 12, 3, 1092, 3, 12, 3, 39, 3, 12, 3, 120, 3, 12, 3, 39, 3, 12, 3, 363, 3, 12, 3, 39, 3, 12, 3, 120, 3, 12, 3, 39, 3, 12, 3, 3279, 3, 12, 3, 39, 3, 12, 3, 120, 3, 12, 3, 39, 3, 12, 3

Offset: 1

Paul D. Hanna, Jun 24 2003

nonn

When we prepend a '1' to the Catalan sequence modulo 3, the only nonzero digit strings are {1,1,1,2,2,2} and {2,2,2,1,1,1}; see A085297 for the occurrences of these digit strings.

Antti Karttunen, Table of n, a(n) for n = 1..16384
R. Stephan, Some divide-and-conquer sequences ...
R. Stephan, Table of generating functions

Cf. A000108, A039969, A085297, A220466.

nmax:=79: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p) := (3^(p+2)-3)/2 od: od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Feb 11 2013

Map[If[First@ # == 0, Length@ #, Nothing] &, SplitBy[Array[Mod[CatalanNumber@ #, 3] &, 10^4], # == 0 &]] (* Michael De Vlieger, Nov 02 2018 *)

A085296(n) = if(n%2,3,3*(1+A085296(n/2))); \\ Antti Karttunen, Nov 01 2018

a(2*n-1) = 3, a(2*n) = 3*(a(n)+1), for n >= 1.
a(n) = (9 * 3^A007814(n) - 1) / 2 - 1. - Ralf Stephan, Oct 10 2003
From Johannes W. Meijer, Feb 11 2013: (Start)
a((2*n-1)*2^p) = (3^(p+2)-3)/2, p >= 0 and n >= 1. Observe that a(2^p) = A029858(p+2).
a(2^(p+3)*n + 2^(p+2) - 1) = a(2^(p+2)*n + 2^(p+1) - 1) for p >= 0 and n >= 1. (End)

A113047 a(n) = C(3n,n)/(2n+1) mod 3.

Original entry on oeis.org

1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 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, 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, 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

Offset: 0

Paul Barry, Oct 11 2005

a(n) differs from 0 only when n=(3^j-1)/2, j>=0. [Conjecture confirmed by Kevin Ryde, Jun 23 2021; see links]

Characteristic function of the ternary repunits, a(n) = 1 iff n is a ternary repunit (A003462). - Kevin Ryde, Jun 23 2021

Antti Karttunen, Table of n, a(n) for n = 0..29524
Kevin Ryde, Proof of characteristic function of the ternary repunits.

Cf. A001764, A003462 (indices of 1's), A010872, A039969.

Table[Mod[Binomial[3 n, n]/(2 n + 1), 3], {n, 0, 72}] (* Michael De Vlieger, Mar 24 2015 *)

A113047(n) = ((binomial(3*n,n)/(n+n+1))%3); \\ Antti Karttunen, Aug 28 2017

a(n) = while(n, my(r);[n,r]=divrem(n,3); if(r!=1,return(0))); 1; \\ Kevin Ryde, Jun 23 2021

G.f.: A(x) satisfies A(x)=1+x*A(x^3). - Vladimir Kruchinin, Mar 24 2015
a(n) = A001764(n) mod 3. - Michel Marcus, Mar 24 2015
a(n) = floor(log_3(2*n + 1)) - floor(log_3(2*n - 1)), for n>=1. - Ridouane Oudra, Aug 24 2021

More terms from Antti Karttunen, Aug 28 2017

A039972 An example of a d-perfect sequence: a(n) = A007317(n) mod 3.

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane

nonn

Odd bisection seems to be A006996. See also A039969. - Antti Karttunen, Aug 15 2017

Antti Karttunen, Table of n, a(n) for n = 1..6561
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. DOI: 10.1007/978-1-4471-0551-0_23

Cf. A006996, A007317, A010872, A039969.

write_to_bfile(start_offset,vec,bfilename) = { for(n=1, length(vec), write(bfilename, (n+start_offset)-1, " ", vec[n])); }
A039972_as_a_vector(size)= { my(A=vector(size)); for(j=1, size, A[j]=1+sum(k=1, j-1, A[k]*A[j-k])); apply(n->(n%3),A); }; \\ After Michael Somos' May 23 2005 code for A007317 and Christian G. Bower's formula for A039972.
write_to_bfile(1,A039972_as_a_vector(6561),"b039972_upto6561.txt"); \\ Antti Karttunen, Aug 15 2017

a(n) = A007317(n) mod 3. - Christian G. Bower, Jun 12 2005

More terms from Christian G. Bower, Jun 12 2005

Formula added to the name by Antti Karttunen, Aug 15 2017

A352409 G.f. A(x) satisfies: [x^(n+1)] (1+x - x^2A(x))^(n(2*n+1)) = 0, for n >= 0.

Original entry on oeis.org

1, 3, 45, 1267, 51597, 2761539, 182885885, 14415019395, 1316237331069, 136512958750979, 15842506286290173, 2033176597680449283, 285833727841312233725, 43677225803116362273795, 7207197437612731825348605, 1277141936892060488486787075

Offset: 0

Paul D. Hanna, Mar 15 2022

nonn

Conjectures:
(1) a(3*n) = A039969(n) (mod 3) = A000108(n) (mod 3) for n >= 0, where A000108 is the Catalan sequence.
(2) a(3*n+1) = 0 (mod 3) for n >= 0.
(3) a(3*n+2) = 0 (mod 3) for n >= 0.
(4) a(n) = 1 (mod 2) for n >= 0.

			G.f.: A(x) = 1 + 3*x + 45*x^2 + 1267*x^3 + 51597*x^4 + 2761539*x^5 + 182885885*x^6 + 14415019395*x^7 + 1316237331069*x^8 + ...
Related table.
The table of coefficients of x^k in (1+x - x^2*A(x))^(n*(2*n+1)) begins:
n=0: [1,  0,    0,     0,       0,        0,         0, ...];
n=1: [1,  3,    0,   -14,    -153,    -4059,   -162214, ...];
n=2: [1, 10,   35,     0,    -825,   -17758,   -642015, ...];
n=3: [1, 21,  189,   847,       0,   -55818,  -1835218, ...];
n=4: [1, 36,  594,  5772,   32715,        0,  -4524660, ...];
n=5: [1, 55, 1430, 23100,  252450,  1762706,         0, ...];
n=6: [1, 78, 2925, 69836, 1179672, 14597856, 122423756, 0, ...];
...
in which a diagonal equals all zeros, illustrating that
[x^(n+1)] (1+x - x^2*A(x))^(n*(2*n+1)) = 0, for n >= 0.
Congruence modulo 3:
(1) The terms of this sequence appear to be divisible by 3 when the index is not divisible by 3:
a(3*n+k) = 0 (mod 3) for n >= 0 and k = 1 or 2.
(2) For the terms a(3*n), the residues modulo 3 begin:
a(3*n) (mod 3) = [1, 1, 2, 2, 2, 0, 0, 0, 2, 2, 2, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 1, 1, 1, 0, ...],
which appears to be congruent to the Catalan sequence A000108 modulo 3; i.e.,
a(3*n) = binomial(2*n,n)/(n+1) (mod 3), for n >= 0.
The above conjectures have been verified for the initial 1201 terms of this sequence.

Paul D. Hanna, Table of n, a(n) for n = 0..300

Cf. A352408, A039969, A000108.

{a(n) = my(A=[1]); for(i=1,n, A=concat(A,0); m=#A;
A[#A] = polcoeff( (1+x - x^2*Ser(A))^(m*(2*m+1)) / (m*(2*m+1)) ,m+1););A[n+1]}
for(n=0,10,print1(a(n),", "))

Conjectures: g.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
(1) A(x) = 1 + x^3*A(x)^2 (mod 3),
(2) A(x) = C(x^3) (mod 3) = C(x)^3 (mod 3), where C(x) = 1 + x*C(x)^2 is the Catalan function (A000108).
a(n) ~ c * 2^(3*n) * (n-1)! / (-LambertW(-2*exp(-2)) * (2 + LambertW(-2*exp(-2))))^n, where c = 0.87591174815917817179... - Vaclav Kotesovec, Mar 19 2022

A039970 An example of a d-perfect sequence: a(2n) = 0, a(2n+1) = Catalan(n) mod 3.

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane

nonn

Antti Karttunen, Table of n, a(n) for n = 1..65537
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. DOI: 10.1007/978-1-4471-0551-0_23

Cf. A039969.

[n mod 2 eq 0 select 0 else Catalan(Floor((n-1)/2)) mod 3: n in [1..100]]; // G. C. Greubel, Feb 13 2019

Table[If[IntegerQ[n/2], 0, Mod[CatalanNumber[(n-1)/2], 3]], {n, 1, 100}] (* G. C. Greubel, Feb 13 2019 *)

A039969(n) = ((binomial(2*n, n)/(n+1))%3);
A039970(n) = if(n%2,A039969((n-1)/2),0); \\ Antti Karttunen, Feb 13 2019

def A039970(n):
    if (mod(n,2)==0):
        return 0
    else:
        return mod(catalan_number((n-1)/2), 3)
[A039970(n) for n in (1..100)] # G. C. Greubel, Feb 13 2019

a(2*n) = 0, a(2*n+1) = A039969(n). - Christian G. Bower, Jun 12 2005, sign edited because of changed offset of A039969. - Antti Karttunen, Feb 13 2019

More terms from Christian G. Bower, Jun 12 2005

Formula added to the name by Antti Karttunen, Feb 13 2019

A039975 An example of a d-perfect sequence: a(n) = A006318(n-1) mod 3.

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane

nonn

Antti Karttunen, Table of n, a(n) for n = 1..11000
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. DOI: 10.1007/978-1-4471-0551-0_23

Cf. A006318.

Cf. also A039969.

A006318(n) = if( n<1, 1, sum( k=0, n, 2^k * binomial( n, k) * binomial( n, k-1)) / n);
A039975(n) = (A006318(n-1) % 3); \\ Antti Karttunen, Feb 13 2019

a(n) = A006318(n-1) mod 3. - Christian G. Bower, Jun 12 2005

More terms from Christian G. Bower, Jun 12 2005

Bower's formula added to the name by Antti Karttunen, Feb 13 2019

A085297 Nonzero residues of Catalan sequence modulo 3; related to the Thue-Morse sequence (A001285).

Original entry on oeis.org

1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2

Offset: 1

Paul D. Hanna, Jun 24 2003

nonn

The runs of zeros in between the digit strings are given in A085296.

Cf. A001285 (Thue-Morse), A000108 (Catalan), A039969 (d-perfect), A085296.

If a leading '1' is added to the Catalan sequence modulo 3, the only nonzero digit strings are {1, 1, 1, 2, 2, 2} and {2, 2, 2, 1, 1, 1}. Replacing these digit strings with their first digit, {1, 1, 1, 2, 2, 2} -> '1' and {2, 2, 2, 1, 1, 1} -> '2', then omitting all zeros, results in the Thue-Morse sequence.

A113048 Binomial(4n,n)/(3n+1) mod 4.

Original entry on oeis.org

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

Offset: 0

Paul Barry, Oct 11 2005

It would appear that the behavior of this sequence is determined by values of the Jacobsthal numbers J(n)=A001045(n): a(n)=1 only if n=J(2k),k=0,1,...; a(n)=2 only if n=J(2k+3)+2J(2j), some k,j>=0, otherwise a(n)=0.

Cf. A039969, A113047, A113049.

Table[Mod[Binomial[4n,n]/(3n+1),4],{n,0,120}] (* Harvey P. Dale, Mar 18 2018 *)

A159979 Lodumo_3 of Catalan numbers .

Original entry on oeis.org

1, 4, 2, 5, 8, 0, 3, 6, 11, 14, 17, 7, 10, 13, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 20, 23, 26, 16, 19, 22, 45, 48, 51, 25, 28, 31, 29, 32, 35, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129

Offset: 0

Philippe Deléham, Apr 28 2009

Permutation of nonnegative integers.

Cf. A000108, A039969

a(n)=lod_3(A000108(n)).

99 inserted by R. J. Mathar, Apr 30 2009

cp's OEIS Frontend

A039967 Duplicate of A039969.

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A085296 Runs of zeros in Catalan sequence modulo 3: consecutive occurrences of binomial(2*k,k)/(k+1) == 0 (mod 3).

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Maple

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A113047 a(n) = C(3n,n)/(2n+1) mod 3.

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A039972 An example of a d-perfect sequence: a(n) = A007317(n) mod 3.

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A352409 G.f. A(x) satisfies: [x^(n+1)] (1+x - x^2*A(x))^(n*(2*n+1)) = 0, for n >= 0.

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A039970 An example of a d-perfect sequence: a(2*n) = 0, a(2*n+1) = Catalan(n) mod 3.

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Magma

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A039975 An example of a d-perfect sequence: a(n) = A006318(n-1) mod 3.

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PARI

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A085297 Nonzero residues of Catalan sequence modulo 3; related to the Thue-Morse sequence (A001285).

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A352409 G.f. A(x) satisfies: [x^(n+1)] (1+x - x^2A(x))^(n(2*n+1)) = 0, for n >= 0.

A039970 An example of a d-perfect sequence: a(2n) = 0, a(2n+1) = Catalan(n) mod 3.