A165367 -id:A165367 - cp's OEIS frontend

A176672 a(2n) = 1 + 6n, a(2*n+1) = A165367(n).

1, 1, 7, 5, 13, 4, 19, 11, 25, 7, 31, 17, 37, 10, 43, 23, 49, 13, 55, 29, 61, 16, 67, 35, 73, 19, 79, 41, 85, 22, 91, 47, 97, 25, 103, 53, 109, 28, 115, 59, 121, 31, 127, 65, 133, 34, 139, 71, 145, 37, 151, 77, 157, 40, 163, 83, 169, 43, 175, 89, 181, 46, 187, 95, 193

Offset: 0

Paul Curtz, Apr 23 2010

Motivation: Start an array from a left column of fractions 0, 1/6, 0, -1/30, 0, ... = A176327(.)/A176592(.), which is zero followed by the Bernoulli numbers from B_2 onwards.
Construct more columns of the array by iteration of the Akiyama-Tanigawa algorithm working backwards through the rows of the table. In our case, the array starts with column indices k>=0:
0, -1/6, -1/4, -3/10, -1/3, -5/14, -3/8, -7/18, ...
1/6, 1/6, 3/20, 2/15, 5/42, 3/28, 7/72, 4/45, 9/110, ...
0, 1/30, 1/20, 2/35, 5/84, 5/84, 7/120, 28/495, ...
-1/30, -1/30, -3/140, -1/105, 0, 1/140, 49/3960, ...
0, -1/42, -1/28, -4/105, -1/28, -29/924, ...
1/42, 1/42, 1/140, -1/105, -5/231, -9/308, -343/10296, ...
The fractions of the top row are -A060819(n)/A145979(n). The current sequence contains essentially the difference between numerator and denominator of each fraction, a(2)=6+1, a(3)=4+1, a(4)=10+3, ... The sum of numerator and denominator is essentially A060819.
Also, numerators of (3*n + 1)/12. - Bruno Berselli, Apr 13 2018
Also, numerators of (3*n + 1)/4. - Altug Alkan, Apr 17 2018

M. Kaneko, The Akiyama-Tanigawa algorithm for Bernoulli numbers, J. Integer Sequences, 3 (2000), #00.2.9.
D. Merlini, R. Sprugnoli, M. C. Verri, The Akiyama-Tanigawa Transformation, Integers, 5 (1) (2005) #A05
Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).

Cf. A061038, A145979.

From R. J. Mathar, Jan 06 2011: (Start)
a(n) = 2*a(n-4) - a(n-8).
G.f.: (1 + x + 7*x^2 + 5*x^3 + 11*x^4 + 2*x^5 + 5*x^6 + x^7) / ((1 - x)^2*(1 + x)^2*(1 + x^2)^2). (End)
a(n) = (2*(3*n + 1)*(11 + 5*(-1)^n) + (6*n + 5 + 3*(-1)^n)*(1 - (-1)^n)*(-1)^((2*n + 3 + (-1)^n)/4))/32. - Luce ETIENNE, Jan 28 2015

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)).

A165355 a(n) = 3n + 1 if n is even, or a(n) = (3n + 1)/2 if n is odd.

Original entry on oeis.org

1, 2, 7, 5, 13, 8, 19, 11, 25, 14, 31, 17, 37, 20, 43, 23, 49, 26, 55, 29, 61, 32, 67, 35, 73, 38, 79, 41, 85, 44, 91, 47, 97, 50, 103, 53, 109, 56, 115, 59, 121, 62, 127, 65, 133, 68, 139, 71, 145, 74, 151, 77, 157, 80, 163, 83, 169, 86, 175, 89, 181, 92, 187, 95, 193, 98

Offset: 0

Paul Curtz, Sep 16 2009

Second trisection of A026741.
A111329(n+1) = A000041(a(n)). - Reinhard Zumkeller, Nov 19 2009
We observe that this sequence is a particular case of sequence for which there exists q: a(n+3) = (a(n+2)*a(n+1)+q)/a(n) for every n >= n0. Here q=-9 and n0=0. - Richard Choulet, Mar 01 2010
The entries are also encountered via the bilinear transform approximation to the natural log (unit circle). Specifically, evaluating 2(z-1)/(z+1) at z = 2, 3, 4, ..., A165355 entries stem from the pair (sums) seen 2 ahead of each new successive prime. 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 (1+1), (4+3), (3+2), (8+5), ... generate the A165355 entries (after the first). As an aside, the same mechanism links A165355 to A140777. - Bill McEachen, Jan 08 2015
As a follow-up to the previous comment, it appears that the numerators and denominators of 2(z-1)/(z+1) are respectively given by A145979 and A060819, but with different offsets. - Michel Marcus, Jan 14 2015
Odd parts of the terms give A067745. E.g.: 1, 2/2, 7, 5, 13, 8/8 .... - Joe Slater, Nov 30 2016

G. C. Greubel, Table of n, a(n) for n = 0..1000
D. H. Lehmer, Continued fractions containing arithmetic progressions, Scripta Mathematica, 29 (1973): 17-24. [Annotated copy of offprint]
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A165351, A165367, A067745, A007814.

f[n_] := If[ OddQ@ n, (3n +1)/2, (3n +1)]; Array[f, 66, 0] (* Robert G. Wilson v, Jan 26 2015 *)
f[n_] := (3 (-1)^(2n) + (-1)^(1 + n)) (-2 + 3n)/4; Array[f, 66] (* or *)
CoefficientList[ Series[(x^3 + 5x^2 + 2x + 1)/(x^2 - 1)^2, {x, 0, 65}], x] (* or *)
LinearRecurrence[{0, 2, 0, -1}, {1, 2, 7, 5}, 66] (* Robert G. Wilson v, Apr 13 2017 *)

a(n)=n+=2*n+1; if(n%2,n,n/2) \\ Charles R Greathouse IV, Jan 13 2015

a(n) = A026741(3*n+1).
a(n)*A026741(n) = A005449(n).
a(n)*A022998(n+1) = A000567(n+1).
a(n) = A026741(n+1) + A022998(n).
a(2n) = A016921(n). a(2n+1) = A016789(n).
a(2n+1)*A026741(2n) = A045944(n).
G.f.: (1+2*x+5*x^2+x^3)/((x-1)^2 * (1+x)^2). - R. J. Mathar, Sep 26 2009
a(n) = (3+9*n)/4 + (-1)^n*(1+3*n)/4. - R. J. Mathar, Sep 26 2009
a(n) = 2*(3n+1)/(4-((2n+2) mod 4)). - Bill McEachen, Jan 09 2015
If a(2n-1) = x then a(2n) = 2x+3. - Robert G. Wilson v, Jan 26 2015
Let the reduced Collatz procedure be defined as Cr(n) = (3*n+1)/2. For odd n, a(n) = Cr(n). For even n, a(n) = Cr(4*n+1)/2. - Joe Slater, Nov 29 2016
a(n) = A067745(n+1) * 2^A007814((3n+1)/2). - Joe Slater, Nov 30 2016
a(n) = 2*a(n-2) - a(n-4). - G. C. Greubel, Apr 13 2017

All comments changed to formulas by R. J. Mathar, Sep 26 2009
New name from Charles R Greathouse IV, Jan 13 2015
Name corrected by Joe Slater, Nov 29 2016

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

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

cp's OEIS Frontend

A176672 a(2*n) = 1 + 6*n, a(2*n+1) = A165367(n).

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

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A165355 a(n) = 3n + 1 if n is even, or a(n) = (3n + 1)/2 if n is odd.

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A165351 Numerator of 3*n/2.

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A165988 First trisection of A022998.

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A166304 Third trisection of A022998.

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A176672 a(2n) = 1 + 6n, a(2*n+1) = A165367(n).