A165355 -id:A165355 - cp's OEIS frontend

A165520 Antidiagonal writing from three rows trio A165351,A165355,A165367 (first,second and third trisections of A026741).

Original entry on oeis.org

0, 1, 3, 1, 2, 3, 5, 7, 9, 4, 5, 6, 11, 13, 15, 7, 8, 9, 17, 19, 21, 10, 11, 12, 23, 25, 27, 13, 14, 15, 29, 31, 33

Offset: 1

Paul Curtz, Sep 21 2009

(6n+3)-th term is 6n+3=A016945. 6n-th term is 3n=3*n=A008585.

Mix (A004273(3n),A004273(3n+1),A004273(3n+2)), (A000027(3n),A000027(3n+1),A000027(3n+2)).

A254051 Square array A by downward antidiagonals: A(n,k) = (3 + 3^n(2floor(3*k/2) - 1))/6, n,k >= 1; read as A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), ...

Original entry on oeis.org

1, 3, 2, 4, 8, 5, 6, 11, 23, 14, 7, 17, 32, 68, 41, 9, 20, 50, 95, 203, 122, 10, 26, 59, 149, 284, 608, 365, 12, 29, 77, 176, 446, 851, 1823, 1094, 13, 35, 86, 230, 527, 1337, 2552, 5468, 3281, 15, 38, 104, 257, 689, 1580, 4010, 7655, 16403, 9842, 16, 44, 113, 311, 770, 2066, 4739, 12029, 22964, 49208, 29525, 18, 47

Offset: 1

L. Edson Jeffery & Antti Karttunen, Jan 24 2015

This is transposed dispersion of (3n-1), starting from its complement A032766 as the first row of square array A(row,col). Please see the transposed array A191450 for references and background discussion about dispersions.

For any odd number x = A135765(row,col), the result after one combined Collatz step (3x+1)/2 -> x (A165355) is found in this array at A(row+1,col).

			The top left corner of the array:
   1,   3,   4,   6,   7,   9,  10,  12,   13,   15,   16,   18,   19,   21
   2,   8,  11,  17,  20,  26,  29,  35,   38,   44,   47,   53,   56,   62
   5,  23,  32,  50,  59,  77,  86, 104,  113,  131,  140,  158,  167,  185
  14,  68,  95, 149, 176, 230, 257, 311,  338,  392,  419,  473,  500,  554
  41, 203, 284, 446, 527, 689, 770, 932, 1013, 1175, 1256, 1418, 1499, 1661
...

Inverse: A254052.
Transpose: A191450.
Row 1: A032766.
Cf. A007051, A057198, A199109, A199113 (columns 1-4).
Cf. A254046 (row index of n in this array, see also A253786), A253887 (column index).
Array A135765(n,k) = 2*A(n,k) - 1.
Other related arrays: A254055, A254101, A254102.
Cf. also A000079, A000244, A003961, A007310, A032766, A002260, A004736, A038722, A165355, A254050.
Related permutations: A048673, A254053, A183209, A249745, A254103, A254104.

In A(n,k)-formulas below, n is the row, and k the column index, both starting from 1:
A(n,k) = (3 + ( A000244(n) * (2*A032766(k) - 1) )) / 6. - Antti Karttunen after L. Edson Jeffery's direct formula for A191450, Jan 24 2015
A(n,k) = A048673(A254053(n,k)). [Alternative formula.]
A(n,k) = (1/2) * (1 + A003961((2^(n-1)) * A254050(k))). [The above expands to this.]
A(n,k) = (1/2) * (1 + (A000244(n-1) * A007310(k))). [Which further reduces to this, equivalent to L. Edson Jeffery's original formula above.]
A(1,k) = A032766(k) and for n > 1: A(n,k) = (3 * A254051(n-1,k)) - 1. [The definition of transposed dispersion of (3n-1).]
A(n,k) = (1+A135765(n,k))/2, or when expressed one-dimensionally, a(n) = (1+A135765(n))/2.
A(n+1,k) = A165355(A135765(n,k)).
As a composition of related permutations. All sequences interpreted as one-dimensional:
a(n) = A048673(A254053(n)). [Proved above.]
a(n) = A191450(A038722(n)). [Transpose of array A191450.]

A067745 Numerator of ((3n - 2)/(n^(2n - 1)(2n - 1)*4^(n - 1))).

Original entry on oeis.org

1, 1, 7, 5, 13, 1, 19, 11, 25, 7, 31, 17, 37, 5, 43, 23, 49, 13, 55, 29, 61, 1, 67, 35, 73, 19, 79, 41, 85, 11, 91, 47, 97, 25, 103, 53, 109, 7, 115, 59, 121, 31, 127, 65, 133, 17, 139, 71, 145, 37, 151, 77, 157, 5, 163, 83, 169, 43, 175, 89, 181, 23, 187, 95, 193, 49, 199

Offset: 1

Marc LeBrun, Jan 29 2002

Conjecture: Odd part of 3n-2. - Ralf Stephan, Nov 18 2010
Conjecture is true.  Note that gcd(3n-2,2n-1)=1 (because 2(3n-2)-3(2n-1) = -1) and gcd(3n-2,n) = 1 or 2.  If 2^k | (3n-2), then k <= log_2(3n-2) < (n-1)/2 for n >= 11.  So only the cases n <= 10 need to be checked individually. - Robert Israel, May 16 2017
This sequence is equivalent to A165355 where each element is reduced by the highest possible power of two. - Joe Slater, Nov 30 2016
Selecting each odd term gives b(n) = 6n+1 (A016921). A075677 is the even bisection of this sequence, while this sequence is the odd bisection of A075677. - Cory Kalm, Apr 29 2021
Numerator of n/2^n + (n-1)/2^(n-1), two Oresme numbers. - Paul Curtz, Dec 07 2021

Robert Israel, Table of n, a(n) for n = 1..10000
D. S. Mitrinovic and Slavko Simic, Special Functions from Long Ago: 5626, Amer. Math. Monthly 109, (2002), p. 83-84.

Cf. A067746, A165355, A000265, A075677, A016921, A007310 (range).

Cf. A075101.

[Numerator(((3*n - 2)/(n^(2*n - 1)*(2*n - 1)*4^(n - 1)))): n in [1..80]]; // Vincenzo Librandi, Feb 16 2015

f:= n -> (3*n-2)/2^padic:-ordp(3*n-2,2):
map(f, [$1..100]); # Robert Israel, May 16 2017

(* Assuming the above conjecture: *)
a067745[n_] := (3*n - 2)/2^IntegerExponent[3*n - 2, 2]; Table[a067745[n], {n, 67}] (* L. Edson Jeffery, Feb 15 2015 *)

vector(80, n, numerator(((3*n - 2)/(n^(2*n - 1)*(2*n - 1)*4^(n - 1))))) \\ Michel Marcus, Feb 16 2015

Assuming the above conjecture, a(n) = a((8+(3*n-2)*4^k)/12), for all k >= 1. - L. Edson Jeffery, Feb 15 2015
a(n) = A000265(A165355(n-1)). - Joe Slater, Nov 30 2016
a(n) = A000265(3*n-2). - R. J. Mathar, Aug 23 2020
a(n) = A075677(2*n-1). a(2*n) = A075677(n); a(2*n-1) = A016921(n). - Cory Kalm, May 03 2021
Sum_{k=1..n} a(k) ~ n^2. - Amiram Eldar, Aug 26 2024
G.f.: Sum_{k>=1} ((3 + 2*(-1)^(k + 1))*x^(3*2^(k - 1) - (2*(-2)^(k - 1))/3 - 1/3) + (3 - 2*(-1)^(k + 1))*x^(2^(k - 1)*(3 + 2*(-1)^k)/3 - 1/3))/(x^(2^(k + 1)) - 2*x^(2^k) + 1). - Miles Wilson, Jan 12 2025

A135765 Distribute the odd numbers in columns based on the occurrence of "3" in each prime factorization; square array A(row, col) = 3^(row-1) * A007310(col), read by antidiagonals A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), ...

Original entry on oeis.org

1, 5, 3, 7, 15, 9, 11, 21, 45, 27, 13, 33, 63, 135, 81, 17, 39, 99, 189, 405, 243, 19, 51, 117, 297, 567, 1215, 729, 23, 57, 153, 351, 891, 1701, 3645, 2187, 25, 69, 171, 459, 1053, 2673, 5103, 10935, 6561, 29, 75, 207, 513, 1377, 3159, 8019, 15309, 32805

Offset: 1

Alford Arnold, Nov 28 2007

The Table can be constructed by multiplying sequence A000244 by A007310.
From Antti Karttunen, Jan 26 2015: (Start)
A permutation of odd numbers. Adding one to each term and then dividing by two gives a related table A254051, which for any odd number, located in this array as x = A(row,col), gives the result at A254051(row+1,col) after one combined Collatz step (3x+1)/2 -> x (A165355) has been applied.
Each odd number n occurs here in position A(A007949(n), A126760(n)).
Compare also to A135764.
(End)

			The top left corner of the array:
    1,    5,    7,   11,   13,   17,   19,   23,   25,   29,   31,   35, ...
    3,   15,   21,   33,   39,   51,   57,   69,   75,   87,   93,  105, ...
    9,   45,   63,   99,  117,  153,  171,  207,  225,  261,  279,  315, ...
   27,  135,  189,  297,  351,  459,  513,  621,  675,  783,  837,  945, ...
   81,  405,  567,  891, 1053, 1377, 1539, 1863, 2025, 2349, 2511, 2835, ...
  243, 1215, 1701, 2673, 3159, 4131, 4617, 5589, 6075, 7047, 7533, 8505, ...
etc.
For n = 6, we have [A002260(6), A004736(6)] = [3, 1] (that is 6 corresponds to location 3,1 (row,col) in above table) and A(3,1) = A000244(3-1) * A007310(1) = 3^2 * 1 = 9.
For n = 9, we have [A002260(9), A004736(9)] = [3, 2] (9 corresponds to location 3,2) and A(3,2) = A000244(3-1) * A007310(2) = 3^2 * 5 = 9*5 = 45.
For n = 13, we have [A002260(13), A004736(13)] = [3, 3] (13 corresponds to location 3,3) and A(3,3) = A000244(3-1) * A007310(3) = 3^2 * 7 = 9*7 = 63.
For n = 23, we have [A002260(23), A004736(23)] = [2, 6] (23 corresponds to location 2,6) and A(2,6) = A000244(2-1) * A007310(6) = 3^1 * 17 = 51.

Row 1: A007310.
Column 1: A000244.
Cf. A007949 (row index), A126760 (column index).
Cf. also A000265, A002260, A003961, A004736, A005408, A165355, A249745.
Related arrays: A135764, A254051, A254055, A254101, A254102.

N:= 20:
B:= [seq(op([6*n+1,6*n+5]),n=0..floor((N-1)/2))]:
[seq(seq(3^j*B[i-j],j=0..i-1),i=1..N)]; # Robert Israel, Jan 26 2015

From Antti Karttunen, Jan 26 2015: (Start)
With both row and col starting from 1:
A(row, col) = A000244(row-1) * A007310(col) = 3^(row-1) * A007310(col).
a(n) = (2*A254051(n))-1.
a(n) = A003961(A254053(n)).
Above in array form:
A(row,col) = A003961(A254053(row,col)) = A003961(A135764(row,A249745(col))).
(End)

Name amended and examples edited by Antti Karttunen, Jan 26 2015

A165351 Numerator of 3*n/2.

Original entry on oeis.org

0, 3, 3, 9, 6, 15, 9, 21, 12, 27, 15, 33, 18, 39, 21, 45, 24, 51, 27, 57, 30, 63, 33, 69, 36, 75, 39, 81, 42, 87, 45, 93, 48, 99, 51, 105, 54, 111, 57, 117, 60, 123, 63, 129, 66, 135, 69, 141, 72, 147, 75, 153, 78, 159, 81, 165, 84, 171, 87, 177, 90, 183, 93, 189, 96, 195

Offset: 0

Paul Curtz, Sep 16 2009

First trisection of A026741. The other trisections are A165355 and A165367.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A000034 (denominator).

Cf. A008585, A016945, A026741, A098832, A165355, A165367.

[Numerator(3*n/2): n in [0..100]]; // Vincenzo Librandi, Mar 03 2014

A165351:=n->numer(3*n/2); seq(A165351(k), k=0..100); # Wesley Ivan Hurt, Oct 11 2013

Table[Numerator[3n/2], {n, 0, 100}] (* Wesley Ivan Hurt, Oct 11 2013 *)
CoefficientList[Series[3*x*(1+x+x^2)/(1-x^2)^2, {x, 0, 70}], x] (* Vincenzo Librandi, Mar 03 2014 *)
LinearRecurrence[{0,2,0,-1},{0,3,3,9},70] (* Harvey P. Dale, Jun 20 2021 *)

[3*n*(3-(-1)^n)/4 for n in (0..100)] # G. C. Greubel, Jul 31 2022

a(n) = A026741(3*n) = 3*A026741(n).
a(2n) = A008585(n).
a(2n+1) = A016945(n).
G.f.: 3*x*(1+x+x^2)/((1-x)^2 * (1+x)^2).
a(n) = numerator(3n/2). - Wesley Ivan Hurt, Oct 11 2013
a(n) = 3*n / (1 + ((n+1) mod 2)). - Wesley Ivan Hurt, Feb 25 2014
From G. C. Greubel, Jul 31 2022: (Start)
a(n) = 3*n*(3 - (-1)^n)/4.
E.g.f.: (3*x/2)*( 2*cosh(x) + sinh(x) ). (End)

Edited and extended by R. J. Mathar, Sep 26 2009

New name from Wesley Ivan Hurt, Oct 13 2013

A165367 Trisection a(n) = A026741(3n + 2).

Original entry on oeis.org

1, 5, 4, 11, 7, 17, 10, 23, 13, 29, 16, 35, 19, 41, 22, 47, 25, 53, 28, 59, 31, 65, 34, 71, 37, 77, 40, 83, 43, 89, 46, 95, 49, 101, 52, 107, 55, 113, 58, 119, 61, 125, 64, 131, 67, 137, 70, 143, 73, 149, 76, 155, 79, 161, 82, 167, 85, 173, 88, 179, 91, 185, 94, 191, 97, 197

Offset: 0

Paul Curtz, Sep 17 2009

The other trisections are A165351 and A165355.

John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1)

Cf. A000326, A016777, A016969, A022998, A026741, A045944, A165351, A165355.

A026741 := proc(n) if type(n,'odd') then n; else n/2 ; fi; end:
A165367 := proc(n) A026741(3*n+2) ; end: seq(A165367(n),n=0..100) ; # R. J. Mathar, Nov 22 2009

LinearRecurrence[{0, 2, 0, -1}, {1, 5, 4, 11}, 66] (* Jean-François Alcover, Nov 15 2017 *)

a(n) = (3*n+2)>>!(n%2); \\ Ruud H.G. van Tol, Oct 09 2023

a(n)*A022998(n) = A045944(n).
a(n)*A026741(n+1) = A000326(n+1).
a(2n) = A016777(n); a(2n+1) = A016969(n).
From R. J. Mathar Nov 22 2009: (Start)
a(n) = 2*a(n-2) - a(n-4).
G.f.: (1 + 5*x + 2*x^2 + x^3)/((1-x)^2*(1+x)^2). (End)

All comments rewritten as formulas by R. J. Mathar, Nov 22 2009

A165988 First trisection of A022998.

Original entry on oeis.org

0, 3, 12, 9, 24, 15, 36, 21, 48, 27, 60, 33, 72, 39, 84, 45, 96, 51, 108, 57, 120, 63, 132, 69, 144, 75, 156, 81, 168, 87, 180, 93, 192, 99, 204, 105, 216, 111, 228, 117, 240, 123, 252, 129, 264, 135, 276, 141, 288, 147, 300, 153, 312, 159, 324, 165, 336, 171, 348, 177

Offset: 0

Paul Curtz, Oct 03 2009

Read modulo 10, this yields a sequence with a period of length 10 containing all 10 digits: 0, 3, 2, 9, 4, 5, 6, 1, 8, 7.
The other two trisections start 1, 8, 7, 20, 13, 32, 19, 44.... and 4, 5, 16, 11, 28, 17, 40, 23....
The Pisano period lengths for reading the sequence modulo m>=1 are 1,  2,  1,  4, 10,  2, 14,  8,  6, 10, 22,  4, 26, 14, 10, 16, 34,  6, 38, 20, 14, 22, 46,  8, 50, 26, 18, 28, 58... - R. J. Mathar, Oct 08 2011

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A165351, A165355, A165367.

LinearRecurrence[{0, 2, 0, -1}, {0, 3, 12, 9}, 50] (* G. C. Greubel, Apr 20 2016 *)

a(n) = my(n=3*n); if (n % 2, n, 2*n); \\ Michel Marcus, Apr 21 2016

a(n) = A022998(3n) = 3*A022998(n) = 3*n*(3 +(-1)^n)/2 .
a(n) = 2*a(n-2) - a(n-4).
G.f.: 3*x*(1+4*x+x^2)/((x-1)^2 *(1+x)^2).
E.g.f.: 3*x*(-1 + 3*exp(2*x))*exp(-x)/2. - Ilya Gutkovskiy, Apr 21 2016

A166304 Third trisection of A022998.

Original entry on oeis.org

4, 5, 16, 11, 28, 17, 40, 23, 52, 29, 64, 35, 76, 41, 88, 47, 100, 53, 112, 59, 124, 65, 136, 71, 148, 77, 160, 83, 172, 89, 184, 95, 196, 101, 208, 107, 220, 113, 232, 119, 244, 125, 256, 131, 268, 137, 280, 143, 292, 149, 304, 155, 316, 161, 328, 167, 340, 173, 352, 179

Offset: 0

Paul Curtz, Oct 11 2009

The sequence read modulo 9 is the periodic sequence 4, 5, 7, 2, 1, 8 (repeat..)
The same set of numbers in a period of length 6 is in A153130,
A165355 read modulo 9, A165367 read modulo 9, and A166138 read modulo 9.

G. C. Greubel, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (0, 2, 0, -1).

Cf. A165988 (first trisection), A166138 (2nd trisection).

LinearRecurrence[{0, 2, 0, -1}, {4, 5, 16, 11}, 100] (* G. C. Greubel, May 09 2016 *)

a(n) = A022998(3*n+2).
a(n) = 2*a(n-2)-a(n-4).
G.f.: (4+5*x+8*x^2+x^3)/((x-1)^2 *(1+x)^2 ).
a(2*n) = A017569(n). a(2n+1) = A016969(n) .

Edited and extended by R. J. Mathar, Oct 14 2009

A111329 Number of partitions of T where T = (3n + 1) if n is even and T=(3n + 1)/2 if n is odd.

Original entry on oeis.org

2, 15, 7, 101, 22, 490, 56, 1958, 135, 6842, 297, 21637, 627, 63261, 1255, 173525, 2436, 451276, 4565, 1121505, 8349, 2679689, 14883, 6185689, 26015, 13848650, 44583, 30167357, 75175, 64112359, 124754, 133230930, 204226, 271248950, 329931

Offset: 1

Parthasarathy Nambi, Nov 04 2005

nonn

			If n=1 then T = 2 and a(1) = 2.

Jeffrey C. Lagarias The 3x+1 problem: An annotated bibliography arXiv:math/0309224 [math.NT], 2003-2011.
Jeffrey C. Lagarias, "The Problem and Its Generalizations." Amer. Math. Monthly 92, 3-23, 1985.
Eric Weisstein's World of Mathematics, Collatz Problem

Cf. A000546, A070165, A006577.

f[n_] := If[EvenQ[n], PartitionsP[3n + 1], PartitionsP[(3n + 1)/2]]; Table[ f[n], {n, 35}] (* Robert G. Wilson v, Nov 07 2005 *)

a(n) = A000041(A165355(n-1)). [Reinhard Zumkeller, Nov 19 2009]

A140777 a(n) = 2*prime(n) - 4.

Original entry on oeis.org

0, 2, 6, 10, 18, 22, 30, 34, 42, 54, 58, 70, 78, 82, 90, 102, 114, 118, 130, 138, 142, 154, 162, 174, 190, 198, 202, 210, 214, 222, 250, 258, 270, 274, 294, 298, 310, 322, 330, 342, 354, 358, 378, 382, 390, 394, 418, 442, 450, 454, 462, 474, 478, 498, 510, 522

Offset: 1

Leroy Quet, May 29 2008, May 31 2008

A number n is included if (p + n/p) is prime, where p is the smallest prime that divides n. Since all terms of this sequence are even (or otherwise p + n/p would be even and not a prime), p is always 2. So this sequence is the set of all even numbers n where (2 + n/2) is prime.
The entries are also encountered via the bilinear transform approximation to the natural log (unit circle). Specifically, evaluating 2(x-1)/(x+1) at x = 2, 3, 4, ..., the terms of this sequence are seen ahead of each new prime encountered. Additionally, the position of those same primes will occur at the entry positions. For clarity, the evaluation output is 2, 3, 1, 1, 6, 5, 4, 3, 10, 7, 3, 2, 14, 9, 8, 5, 18, 11, ..., where the entries ahead of each new prime are 2, 6, 10, 18, ... . As an aside, the same mechanism links this sequence to A165355. - Bill McEachen, Jan 08 2015
As a follow-up to previous comment, it appears that the numerators and denominators of 2(x-1)/(x+1) are respectively given by A145979 and A060819, but with different offsets. - Michel Marcus, Jan 14 2015
Subset of the union of A017641 & A017593. - Michel Marcus, Sep 01 2020

			The smallest prime dividing 42 is 2. Since 2 + 42/2 = 23 is prime, 42 is included in this sequence.

Cf. A000040, A020639, A040976, A111234, A140775, A140776.
Cf. A060819, A145979, A165355.
Cf. A017641, A017593.

[2*NthPrime(n)-4: n in [1..80]]; // Vincenzo Librandi, Feb 19 2015

A020639 := proc(n) local dvs,p ; dvs := sort(convert(numtheory[divisors](n),list)) ; for p in dvs do if isprime(p) then RETURN(p) ; fi ; od: error("%d",n) ; end: A111234 := proc(n) local p ; p := A020639(n) ; p+n/p ; end: isA140777 := proc(n) RETURN(isprime(A111234(n))) ; end: for n from 2 to 1200 do if isA140777(n) then printf("%d,",n) ; fi ; od: # R. J. Mathar, May 31 2008
seq(2*ithprime(i)-4, i=1..1000); # Robert Israel, Jan 09 2015

fQ[n_] := Block[{p = First@ First@ Transpose@ FactorInteger@ n}, PrimeQ[p + n/p] == True]; Select[ Range[2, 533], fQ@# &] (* Robert G. Wilson v, May 30 2008 *)
Table[2 Prime[n] - 4, {n, 60}] (* Vincenzo Librandi, Feb 19 2015 *)

vector(100, n, 2*prime(n) - 4) \\ Michel Marcus, Jan 09 2015

a(n) = 2*A040976(n). - Michel Marcus, Jan 09 2015

More terms from Robert G. Wilson v and R. J. Mathar, May 30 2008

cp's OEIS Frontend

A165520 Antidiagonal writing from three rows trio A165351,A165355,A165367 (first,second and third trisections of A026741).

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A254051 Square array A by downward antidiagonals: A(n,k) = (3 + 3^n*(2*floor(3*k/2) - 1))/6, n,k >= 1; read as A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), ...

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A067745 Numerator of ((3*n - 2)/(n^(2*n - 1)*(2*n - 1)*4^(n - 1))).

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A135765 Distribute the odd numbers in columns based on the occurrence of "3" in each prime factorization; square array A(row, col) = 3^(row-1) * A007310(col), read by antidiagonals A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), ...

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A165351 Numerator of 3*n/2.

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A165367 Trisection a(n) = A026741(3n + 2).

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A165988 First trisection of A022998.

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A166304 Third trisection of A022998.

A254051 Square array A by downward antidiagonals: A(n,k) = (3 + 3^n(2floor(3*k/2) - 1))/6, n,k >= 1; read as A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), ...

A067745 Numerator of ((3n - 2)/(n^(2n - 1)(2n - 1)*4^(n - 1))).