A074939 -id:A074939 - cp's OEIS frontend

A055246 At step number k >= 1 the 2^(k-1) open intervals that are erased from [0,1] in the Cantor middle-third set construction are I(k,n) = (a(n)/3^k, (1+a(n))/3^k), n=1..2^(k-1).

1, 7, 19, 25, 55, 61, 73, 79, 163, 169, 181, 187, 217, 223, 235, 241, 487, 493, 505, 511, 541, 547, 559, 565, 649, 655, 667, 673, 703, 709, 721, 727, 1459, 1465, 1477, 1483, 1513, 1519, 1531, 1537, 1621, 1627, 1639, 1645, 1675, 1681, 1693, 1699

Offset: 1

Wolfdieter Lang, May 23 2000

Related to A005836. Gives boundaries of open intervals that have to be erased in the Cantor middle-third set construction.
Let g(n) = Sum_{i=0..n} (i*binomial(n+i,i)^3*binomial(n,i)^2) = A112035(n). Let b = {m>0 : g(m) != 0 (mod 3)}. Then b(n) = a(n). - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 08 2004
Conjecture: Similarly to A191107, this increasing sequence is generated by the rules: a(1) = 1, and if x is in the sequence, then 3*x-2 and 3*x+4 are also in the sequence. - L. Edson Jeffery, Nov 17 2015

			k=1: (1/3, 2/3);
k=2: (1/9, 2/9), (7/9, 8/9);
k=3: (1/27, 2/27), (7/27, 8/27), (19/27, 20/27), (25/27, 26/27); ...

Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Index entries for 3-automatic sequences.

Cf. A003278, A005836, A005823, A055247, A112035.

Cf. A191107.

(* (Conjectured) Choose rows large enough to guarantee that all terms < max are generated. *)
rows = 1000; max = 10^4; a[1] = {1}; i = 1; Do[a[n_] = {}; Do[If[1 < 3*a[n - 1][[k]] - 2 < max, AppendTo[a[n], 3*a[n - 1][[k]] - 2], Break]; If[3*a[n - 1][[k]] + 4 < max, AppendTo[a[n], 3*a[n - 1][[k]] + 4], Break], {k, Length[a[n - 1]]}]; If[a[n] == {}, Break, i++], {n, 2, 1000}]; a055246 = Take[Flatten[Table[a[n], {n, i}]], 48] (* L. Edson Jeffery, Nov 17 2015 *)
Join[{1}, 1 + 6 Accumulate[Table[(3^IntegerExponent[n, 2] + 1)/2, {n, 60}]]] (* Vincenzo Librandi, Nov 26 2015 *)

g(n)=sum(i=0,n,i*binomial(n+i,i)^3*binomial(n,i)^2);
for (i=1,2000,if(Mod(g(i),3)<>0,print1(i,",")))

a(n) = fromdigits(binary(n-1),3)*6 + 1; \\ Kevin Ryde, Apr 23 2021

def A055246(n): return int(bin(n-1)[2:],3)*6|1 # Chai Wah Wu, Jun 26 2025

a(n) = 1+6*A005836(n), n >= 1.
a(n) = 1+3*A005823(n), n >= 1.
a(n+1) = A074938(n) + A074939(n); A074938: odd numbers in A005836, A074939: even numbers in A005836. - Philippe Deléham, Jul 10 2005
Conjecture: a(n) = 2*A191107(n) - 1 = 6*A003278(n) - 5 = (a((2*n-1)*2^(k-1))+2)/3^k, k>0. - L. Edson Jeffery, Nov 25 2015

Edited by N. J. A. Sloane, Nov 20 2015: used first comment to give more precise definition, and edited a comment at the suggestion of L. Edson Jeffery.

A006996 a(n) = C(2n,n) mod 3.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, James Propp

Removing 0's from the sequence gives Thue-Morse sequence A001285 : 1,2,0,2,1,0,0,0,0,2,1,0,1,2,..->1,2,2,1,2,1,1,2,... - Benoit Cloitre, Jan 04 2004
a(n) = 0 if n in A074940, a(n) = 1 if n in A074939, a(n) = 2 if n in A074938.
Central terms of the triangle in A083093. - Reinhard Zumkeller, Jul 11 2013

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Table of n, a(n) for n = 0..2187=3^7
Michael Gilleland, Some Self-Similar Integer Sequences
Index entries for sequences that are fixed points of mappings

Cf. A001285.
Cf. A074938, A074939, A074940.
Cf. A083093.
Cf. A000984, A005704.

a006996 n = a083093 (2 * n) n  -- Reinhard Zumkeller, Jul 11 2013

Table[ Mod[ Binomial[2n, n], 3], {n, 0, 104}] (* Or *)
Nest[ Function[ l, {Flatten[(l /. {0 -> {0, 0, 0}, 1 -> {1, 2, 0}, 2 -> {2, 1, 0}})]}], {1}, 7] (* Robert G. Wilson v, Mar 28 2005 *)

a(n)=if(n==0, return(1)); if(vecmax(Set(digits(n,3)))>1, 0, 1 + n%2) \\ Charles R Greathouse IV, May 09 2016

from gmpy2 import digits
def A006996(n): return 0 if '2' in digits(n,3) else 1+(n&1) # Chai Wah Wu, Jun 26 2025

a(n) = A000984(n) mod 3.
a(n) = A005704(n) mod 3. - Benoit Cloitre, Jan 04 2004
A fixed point of the morphism : 1 -> 120, 2 -> 210, 0 -> 000. - Philippe Deléham, Jan 08 2004

A074938 Odd numbers such that base 3 representation contains no 2.

Original entry on oeis.org

1, 3, 9, 13, 27, 31, 37, 39, 81, 85, 91, 93, 109, 111, 117, 121, 243, 247, 253, 255, 271, 273, 279, 283, 325, 327, 333, 337, 351, 355, 361, 363, 729, 733, 739, 741, 757, 759, 765, 769, 811, 813, 819, 823, 837, 841, 847, 849, 973, 975, 981, 985, 999, 1003, 1009

Offset: 0

Benoit Cloitre, Oct 04 2002; Nov 15 2003

Odd numbers in A005836.
Numbers m such that coefficient of x^m equals -1 in Product_{k>=0} 1-x^(3^k).
Numbers k such that binomial(2k, k) == 2 (mod 3).
Sum of an odd number of distinct powers of 3. - Emeric Deutsch, Dec 03 2003

Amiram Eldar, Table of n, a(n) for n = 0..10000
Emeric Deutsch and B. E. Sagan, Congruences for Catalan and Motzkin numbers and related sequences, arXiv:math/0407326 [math.CO], 2004; J. Num. Theory 117 (2006), 191-215.

Intersection of A005408 and A005836.

Cf. A006996, A074939.

Select[Range[1,1111,2],Count[IntegerDigits[#,3],2]==0&] (* Harvey P. Dale, Dec 19 2010 *)

def A074938(n): return int(bin((n<<1)+(n.bit_count()&1^1))[2:],3) # Chai Wah Wu, Jun 26 2025

a(n) (mod 3) = A010059(n).

((a(n)-1)/2) (mod 3) = A010060(n) = (1/2)*{binomial(2*a(n)+1, a(n)) (mod 3)}.

A083095 a(n) = A083094(n)/4.

Original entry on oeis.org

0, 2, 5, 6, 14, 15, 18, 20, 41, 42, 45, 47, 54, 56, 59, 60, 122, 123, 126, 128, 135, 137, 140, 141, 162, 164, 167, 168, 176, 177, 180, 182, 365, 366, 369, 371, 378, 380, 383, 384, 405, 407, 410, 411, 419, 420, 423, 425, 486, 488, 491, 492, 500

Offset: 1

Benoit Cloitre, Apr 22 2003

Is this the same as A083097? - Andrew S. Plewe, May 30 2007

Hugo Pfoertner, Table of n, a(n) for n = 1..16384 (terms 1..106 from Paul F. Marrero Romero)

Cf. A010060, A051638, A083093, A083094, A083097, A074939.

Select[Range[0,2000], OddQ[Sum[Mod[Binomial[#,j],3],{j,0,#}]]&]/4 (* Paul F. Marrero Romero, Dec 28 2024 *)

def A083095(n): return int(bin(((m:=n-1).bit_count()&1)+(m<<1))[2:],3)>>1 # Chai Wah Wu, Jun 26 2025

Apparently (2*a(n)) mod 3 = A010060(n-1), the Thue-Morse sequence.
Numbers k such that C(4*k, 2*k) == 1 (mod 3). - Benoit Cloitre, Jul 30 2003
Numbers k such that the base-3 digits of 2k contains no 2's, i.e. a(n) = A074939(n-1)/2. - Chai Wah Wu, Jun 26 2025

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 29 2003

A083094 Numbers k such that Sum_{j=0..k} (binomial(k,j) mod 3) is odd.

Original entry on oeis.org

0, 8, 20, 24, 56, 60, 72, 80, 164, 168, 180, 188, 216, 224, 236, 240, 488, 492, 504, 512, 540, 548, 560, 564, 648, 656, 668, 672, 704, 708, 720, 728, 1460, 1464, 1476, 1484, 1512, 1520, 1532, 1536, 1620, 1628, 1640, 1644, 1676, 1680, 1692, 1700, 1944, 1952

Offset: 1

Benoit Cloitre, Apr 22 2003

Apparently a(n)/2 (mod 3) = A010060(n), the Thue-Morse sequence.

Cf. A010060, A051638, A074939, A083093, A083095 (gives a b-file of 16384 terms).

Select[Range[0, 2000],OddQ[Sum[Mod[Binomial[#, j], 3], {j, 0, #}]] &] (* Paul F. Marrero Romero, Dec 28 2024 *)

isok(n) = sum(k=0, n, binomial(n,k) % 3) % 2; \\ Michel Marcus, Dec 05 2013

def A083094(n): return int(bin(((m:=n-1).bit_count()&1)+(m<<1))[2:],3)<<1 # Chai Wah Wu, Jun 26 2025

a(n) = 4*A083095(n). - Hugo Pfoertner, Jan 12 2025

Numbers that are multiples of 4 and such that base-3 digits contain no 1's, or equivalently, numbers such that base-3 digits contains an even number of 2's and no 1's, i.e. a(n) = 2*A074939(n-1). This characterization can be derived from the formula in A051638. - Chai Wah Wu, Jun 26 2025

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 29 2003

A246510 G.f.: Sum_{n>=0} x^n / (1-4x)^(2n+1) * [Sum_{k=0..n} C(n,k)^2 * 3^k * x^k] * [Sum_{k=0..n} C(n,k)^2 * 4^k * x^k].

Original entry on oeis.org

1, 5, 36, 305, 2821, 27690, 282699, 2967285, 31785786, 345815975, 3808549531, 42360017130, 474990254821, 5362633500755, 60897115958286, 695012481567465, 7966829676299139, 91674042449673960, 1058486539560201051, 12258669983923625475, 142359286920427682046, 1657287004720545992505

Offset: 0

Paul D. Hanna, Aug 27 2014

nonn

a(n) == 1 (mod 3) iff n = A074939(k) for k>=0, where A074939 gives even numbers such that base 3 representation contains no 2.

a(n) == 2 (mod 3) iff n = A074938(k) for k>=0, where A074938 gives odd numbers such that base 3 representation contains no 2.

			G.f.: A(x) = 1 + 5*x + 36*x^2 + 305*x^3 + 2821*x^4 + 27690*x^5 +...

Cf. A246509, A246056, A074938, A074939.

Table[Sum[3^k * Sum[Binomial[n-k, k+j]^2 * Binomial[k+j, j]^2 * 4^j,{j,0,n-2*k}],{k,0,Floor[n/2]}],{n,0,20}] (* Vaclav Kotesovec, Oct 04 2014 *)

/* By definition: */
{a(n)=local(A=1); A=sum(m=0, n, x^m/(1-4*x)^(2*m+1) * sum(k=0, m, binomial(m, k)^2 * 3^k * x^k) * sum(k=0, m, binomial(m, k)^2 * 4^k * x^k) +x*O(x^n)); polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))

/* By a binomial identity: */
{a(n)=local(A=1); A=sum(m=0, n, x^m/(1-x)^(2*m+1) * sum(k=0, m, binomial(m, k)^2 * 4^(m-k) * 3^k * x^k) * sum(k=0, m, binomial(m, k)^2 * x^k) +x*O(x^n)); polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))

/* By a binomial identity: */
{a(n)=polcoeff(sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^2 * sum(j=0, k, binomial(k, j)^2 * 4^(k-j) * 3^j * x^j)+x*O(x^n))), n)}
for(n=0, 25, print1(a(n), ", "))

/* By a binomial identity: */
{a(n)=polcoeff(sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^2 * 4^(m-k) * sum(j=0, k, binomial(k, j)^2 * 3^j * x^j)+x*O(x^n))), n)}
for(n=0, 25, print1(a(n), ", "))

/* Formula for a(n): */
{a(n)=sum(k=0, n\2, sum(j=0, n-2*k, 3^k * binomial(n-k, k+j)^2 * binomial(k+j, j)^2 * 4^j))}
for(n=0, 25, print1(a(n), ", "))

G.f.: Sum_{n>=0} x^n / (1-x)^(2*n+1) * [Sum_{k=0..n} C(n,k)^2 * x^k] * [Sum_{k=0..n} C(n,k)^2 * 4^(n-k) * 3^k * x^k].
G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k)^2 * Sum_{j=0..k} C(k,j)^2 * 4^(k-j) * 3^j * x^j.
G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k)^2 * 4^(n-k) * Sum_{j=0..k} C(k,j)^2 * 3^j * x^j.
a(n) = Sum_{k=0..[n/2]} 3^k * Sum_{j=0..n-2*k} C(n-k, k+j)^2 * C(k+j, j)^2 * 4^j.
a(n) ~ sqrt(36 + 29*sqrt(3) + 3*sqrt(423 + 232*sqrt(3))) *  (9/2 + sqrt(3) + 3/2*sqrt(9 + 4*sqrt(3)))^n / (8*Pi*n). - Vaclav Kotesovec, Oct 04 2014

A107757 Numbers k such that Sum_{j=1..k} Catalan(j) == 2 (mod 3).

Original entry on oeis.org

3, 9, 11, 27, 29, 35, 39, 81, 83, 89, 93, 107, 111, 117, 119, 243, 245, 251, 255, 269, 273, 279, 281, 323, 327, 333, 335, 351, 353, 359, 363, 729, 731, 737, 741, 755, 759, 765, 767, 809, 813, 819, 821, 837, 839, 845, 849, 971, 975, 981, 983, 999, 1001, 1007, 1011

Offset: 1

N. J. A. Sloane, Jun 11 2005

Y. More, Problem 11165, Amer. Math. Monthly, 112 (2005), 568.

Cf. A000108, A107755, A107756.

Equals A074939 - 1.

c:=n->binomial(2*n,n)/(n+1): s:=0: for n from 1 to 1500 do s:=s+c(n): a[n]:=s mod 3: od: A:=[seq(a[n],n=1..1500)]: p:=proc(n) if A[n]=2 then n else fi end: seq(p(n),n=1..1500); # Emeric Deutsch, Jun 12 2005

s0 = s2 = {}; s = 0; Do[s = Mod[s + (2 n)!/n!/(n + 1)!, 3]; Switch[ Mod[s, 3], 0, AppendTo[s0, n], 2, AppendTo[s2, n]], {n, 1055}]; s2 (* Robert G. Wilson v, Jun 14 2005 *)

More terms from Emeric Deutsch, Jun 12 2005

cp's OEIS Frontend

A055246 At step number k >= 1 the 2^(k-1) open intervals that are erased from [0,1] in the Cantor middle-third set construction are I(k,n) = (a(n)/3^k, (1+a(n))/3^k), n=1..2^(k-1).

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A006996 a(n) = C(2n,n) mod 3.

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A074938 Odd numbers such that base 3 representation contains no 2.

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A083095 a(n) = A083094(n)/4.

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A083094 Numbers k such that Sum_{j=0..k} (binomial(k,j) mod 3) is odd.

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A246510 G.f.: Sum_{n>=0} x^n / (1-4*x)^(2*n+1) * [Sum_{k=0..n} C(n,k)^2 * 3^k * x^k] * [Sum_{k=0..n} C(n,k)^2 * 4^k * x^k].

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Mathematica

PARI

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Formula

A246510 G.f.: Sum_{n>=0} x^n / (1-4x)^(2n+1) * [Sum_{k=0..n} C(n,k)^2 * 3^k * x^k] * [Sum_{k=0..n} C(n,k)^2 * 4^k * x^k].