A145979 -id:A145979 - cp's OEIS frontend

A173630 Denominator of A002445(n)/A145979(n).

Original entry on oeis.org

1, 1, 2, 5, 1, 7, 4, 3, 1, 11, 2, 13, 1, 5, 8, 17, 3, 19, 2, 7, 1, 23, 4, 25, 1, 9, 14, 29, 1, 31

Offset: 0

Paul Curtz, Nov 23 2010

Fractions: 1,1,15/2,21/5,10,33/7,1365/4,1/3,102,...

Second bisection is A146535(n+2).

A060789.

A060819 a(n) = n / gcd(n,4).

Original entry on oeis.org

1, 1, 3, 1, 5, 3, 7, 2, 9, 5, 11, 3, 13, 7, 15, 4, 17, 9, 19, 5, 21, 11, 23, 6, 25, 13, 27, 7, 29, 15, 31, 8, 33, 17, 35, 9, 37, 19, 39, 10, 41, 21, 43, 11, 45, 23, 47, 12, 49, 25, 51, 13, 53, 27, 55, 14, 57, 29, 59, 15, 61, 31, 63, 16, 65, 33, 67, 17, 69, 35, 71, 18, 73, 37, 75, 19

Offset: 1

Len Smiley, Apr 30 2001

From Peter Bala, Feb 19 2019: (Start)
We make some general remarks about the sequence a(n) = numerator(n/(n + k)) = n/gcd(n,k) for k a fixed positive integer. The present sequence is the case k = 4. Several other cases are listed in the Crossrefs. In addition to being multiplicative these sequences are also strong divisibility sequences, that is, gcd(a(n),a(m)) = a(gcd(n, m)) for n, m >= 1. In particular, it follows that a(n) is a divisibility sequence: if n divides m then a(n) divides a(m).
By the multiplicativeness and strong divisibility property of the sequence a(n) it follows that if gcd(n, m) = 1 then a(a(n)*a(m) ) = a(a(n)) * a(a(m)), a(a(a(n))*a(a(m)) ) = a(a(a(n))) * a(a(a(m))) and so on.
The sequence a(n) has the rational generating function Sum_{d divides k} f(d)*x^d/(1 - x^d)^2, where f(n) is the Dirichlet inverse of the Euler totient function A000010. f(n) is a multiplicative function defined on prime powers p^k by f(p^k) = 1 - p. See A023900. Cf. A181318. (End)

			From _Peter Bala_, Feb 21 2019: (Start)
Sum_{n >= 1} n*a(n)*x^n = G(x) - 2*G(x^2) - 4*G(x^4), where G(x) = x*(1 + x)/(1 - x)^3.
Sum_{n >= 1} (1/n)*a(n)*x^n = H(x) - (1/2)*H(x^2) - (1/4)*H(x^4), where H(x) = x/(1 - x).
Sum_{n >= 1} (1/n^2)*a(n)*x^n = L(x) - (1/2^2)*L(x^2) - (1/4^2)*L(x^4), where L(x) = Log(1/(1 - x)).
Sum_{n >= 1} (1/a(n))*x^n = L(x) + (1/2)*L(x^2) + (1/2)*L(x^4). (End)

Harry J. Smith, Table of n, a(n) for n = 1..1000
Peter Bala, A note on the sequence of numerators of a rational function, 2019.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).

Cf. A026741, A051176, A060791, A060789. Cf. Other sequences given by the formula numerator(n/(n + k)): A106608 thru A106612 (k = 7 thru 11), A051724 (k = 12), A106614 thru A106621 (k = 13 thru 20).

Cf. A000010, A023900, A061037, A061038, A109008, A145979, A167192, A176895, A181318, A220466.

List([1..80],n->n/Gcd(n,4)); # Muniru A Asiru, Feb 20 2019

a060819 n = n `div` a109008 n  -- Reinhard Zumkeller, Nov 25 2013

[n/GCD(n,4): n in [1..80]]; // G. C. Greubel, Sep 19 2018

A060819 := n -> numer(1/2-1/(n+2)): seq(A060819(n),n=1..75); # Gary Detlefs, Sep 16 2011

Mathematica
```
f[n_]:= n/GCD[n, 4]; Array[f, 80]
```

a(n) = { n / gcd(n, 4) } \\ Harry J. Smith, Jul 12 2009

[lcm(n,4)/4 for n in (1..80)] # Zerinvary Lajos, Jun 07 2009

G.f.: x*(1 +x +3*x^2 +x^3 +3*x^4 +x^5 +x^6)/(1 - x^4)^2.
a(n) = 2*a(n-4) - a(n-8).
a(n) = (n/16)*(11 - 5*(-1)^n - i^n - (-i)^n). - Ralf Stephan, Mar 15 2003
a(2*n+1) = a(4*n+2) = 2*n+1, a(4*n+4) = n+1. - Ralf Stephan, Jun 10 2005
Multiplicative with a(2^e) = 2^max(0, e-2), a(p^e) = p^e, p >= 3. - Mitch Harris, Jun 29 2005
a(n) = A167192(n+4,4). - Reinhard Zumkeller, Oct 30 2009
From R. J. Mathar, Apr 18 2011: (Start)
a(n) = A109045(n)/4.
Dirichlet g.f.: zeta(s-1)*(1-1/2^s-1/2^(2s)). (End)
a(n+4) - a(n) = A176895(n).  - Paul Curtz, Apr 05 2011
a(n) = numerator(Sum_{k=1..n} 1/((k+1)*(k+2))). This summation has a closed form of 1/2 - 1/(n+2) and denominator of A145979(n). - Gary Detlefs, Sep 16 2011
a((2*n-1)*2^p) = ceiling(2^(p-2))*(2*n-1), p >= 0 and n >= 1. - Johannes W. Meijer, Feb 06 2013
a(n) = n / A109008(n). - Reinhard Zumkeller, Nov 25 2013
a(n) = denominator((2n-4)/n). - Wesley Ivan Hurt, Dec 22 2016
From Peter Bala, Feb 21 2019: (Start)
O.g.f.: Sum_{n >= 0} a(n)*x^n = F(x) - F(x^2) - F(x^4), where F(x) = x/(1 - x)^2.
More generally, Sum_{n >= 0} (a(n)^m)*x^n = F(m,x) + (1 - 2^m)*( F(m,x^2) + F(m,x^4) ), where F(m,x) = A(m,x)/(1 - x)^(m+1) with A(m,x) the m_th Eulerian polynomial: A(1,x) = x, A(2,x) = x*(1 + x), A(3,x) = x*(1 + 4*x + x^2) - see A008292.
Repeatedly applying the Euler operator x*d/dx or its inverse operator to the o.g.f. for the sequence a(n) produces generating functions for the sequences ((n^m)*a(n))n>=1 for m in Z. Some examples are given below.
(End)
Sum_{k=1..n} a(k) ~ (11/32) * n^2. - Amiram Eldar, Nov 25 2022
E.g.f.: x*(8*cosh(x) + sin(x) + 3*sinh(x))/8. - Stefano Spezia, Dec 02 2023

A061038 Denominator of 1/4 - 1/n^2.

Original entry on oeis.org

1, 36, 16, 100, 9, 196, 64, 324, 25, 484, 144, 676, 49, 900, 256, 1156, 81, 1444, 400, 1764, 121, 2116, 576, 2500, 169, 2916, 784, 3364, 225, 3844, 1024, 4356, 289, 4900, 1296, 5476, 361, 6084, 1600, 6724, 441, 7396, 1936, 8100, 529, 8836

Offset: 2

N. J. A. Sloane, May 26 2001

Harry J. Smith, Table of n, a(n) for n = 2..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

See A061037 for comments, references, links.

Cf. A145979. - Bruno Berselli, Mar 21 2011

import Data.Ratio ((%), denominator)
a061038 n = denominator (1%4 - 1%n^2)  -- Reinhard Zumkeller, Jan 22 2012

[ Denominator(1/4-1/n^2): n in [2..50] ]; // Vincenzo Librandi, Feb 10 2011

Table[Denominator[1/4 - 1/n^2], {n, 2, 60}] (* Stefan Steinerberger, Apr 08 2006 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{1,36,16,100,9,196,64,324, 25,484,144,676},50] (* Harvey P. Dale, Aug 05 2018 *)

a(n) = { denominator(1/4 - 1/n^2) } \\ Harry J. Smith, Jul 17 2009

def A061038(n): return denominator(1/4 - 1/n^2)
[A061038(n) for n in range(2,51)] # G. C. Greubel, Apr 18 2023

a(4n+2) = (2n+1)^2, a(2n+3) = (4n+6)^2, a(4n+4) = (4n+4)^2. - Ralf Stephan, Jun 10 2005
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12). - Paul Curtz, Feb 25 2011
From Bruno Berselli, Mar 21 2011: (Start)
G.f.: x^2*(1 +36*x +16*x^2 +100*x^3 +6*x^4 +88*x^5 +16*x^6 +24*x^7 +x^8 +4*x^9 +4*x^11)/(1-x^4)^3.
a(n) = (1/64)*( n*(16 - (1+(-1)^n)*(5-i^n)) )^2 with i=sqrt(-1).
a(n) = (n/(n-4))^2 * a(n-4) for n>5. (End)
a(n) = 4*n^2 / gcd(4*n^2, (n^2-4)). - Colin Barker, Jan 13 2014
Sum_{n>=2} 1/a(n) = Pi^2/6 - 1/4. - Amiram Eldar, Aug 12 2022

More terms from Stefan Steinerberger, Apr 08 2006

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

A214297 a(0)=-1, a(1)=0, a(2)=-3; thereafter a(n+2) - 2*a(n+1) + a(n) has period 4: repeat -4, 8, -4, 2.

Original entry on oeis.org

-1, 0, -3, 2, 3, 6, 5, 12, 15, 20, 21, 30, 35, 42, 45, 56, 63, 72, 77, 90, 99, 110, 117, 132, 143, 156, 165, 182, 195, 210, 221, 240, 255, 272, 285, 306, 323, 342, 357, 380, 399, 420, 437, 462, 483, 506, 525, 552, 575, 600, 621, 650, 675, 702, 725, 756, 783, 812, 837, 870, 899, 930, 957, 992, 1023, 1056, 1085, 1122, 1155, 1190

Offset: 0

Paul Curtz, Jul 11 2012

Let a(n)/A000290(n) = [-1/0, 0/1, -3/4, 2/9, 3/16, 6/25, 5/36, 12/49, 15/64, 20/81, 21/100, 30/121, ...] = a(n)/b(n) (say).
Then b(n)-4*a(n)=4, 1, 16, 1 (period of length 4).
Permutation from a(n) to A061037(n): 1, 3, 2, 7, 5, 11, 4, 15, 9, 19, 6, ... = shifted A145979 + 1.
A061037(n) - a(n) = 0, 3, -3, -3, 0, -15, 3, -33, 0 -57, 15, -87, 0, -123, ...
First 3 rows:
-1 0 -3 2  3  6  5 12 15 20 21 30 35
1 -3  5 1  3 -1  7  3  5  1  9  5  7
-4 8 -4 2 -4  8 -4  2 -4  8 -4  2 -4.
Note that the terms of a(n) increase from 12. Compare to increasing terms permutation of A061037(n): -3,-1,0,2,3,5,6,12,15, .... and A129647.
c(n) =  0, -1, 0, -1, 2, 1, 2, 1, 4, 3, 4, 3, 6, 5, 6, 5, ... (cf. A134967)
d(n) = -1,  1, 1,  3, 1, 3, 3, 5, 3, 5, 5, 7, 5, 7, 7, 9, ..., hence:
a(n) = c(n+1) * d(n+1).

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

[(2*n^2-11-9*(-1)^n+6*((-1)^((2*n+1-(-1)^n)/4)+(-1)^((2*n-1+(-1)^n)/4)))/8: n in [0..100]]; // G. C. Greubel, Sep 19 2018

A214297 := proc(n)
    option remember;
    if n <=5 then
        op(n+1,[-1,0,-3,2,3,6]) ;
    else
        2*procname(n-1)-procname(n-2)+procname(n-4)-2*procname(n-5)+procname(n-6) ;
    end if;
end proc: # R. J. Mathar, Jun 28 2013

Table[(2 n^2 - 11 - 9 (-1)^n + 6 ((-1)^((2 n + 1 - (-1)^n)/4) + (-1)^((2 n - 1 + (-1)^n)/4)))/8, {n, 0, 69}] (* or *)
CoefficientList[Series[-(1 - 2 x + 4 x^2 - 8 x^3 + 3 x^4)/((1 - x)^2*(1 - x^4)), {x, 0, 69}], x] (* Michael De Vlieger, Mar 24 2017 *)

vector(100, n, n--; (2*n^2-11-9*(-1)^n+6*((-1)^((2*n+1-(-1)^n)/4)+(-1)^((2*n-1+(-1)^n)/4)))/8) \\ G. C. Greubel, Sep 19 2018

a(k+4)  - a(k)    =  2*k + 4.
a(k+2)  - a(k-2)  =  2*k.
a(k+6)  - a(k-6)  =  6*k.
a(k+10) - a(k-10) = 10*k.
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(2*k)   = -1, -3, followed by 3, 5, 15, 21, 35, 45, ... (A142717);
a(2*k+1) = k*(k+1) (see A002378).
A198442(n) = -1,0,0,2,3,6,8,12,  minus 3 at A198442(4*n+2).
G.f. -( 1-2*x+4*x^2-8*x^3+3*x^4 )/( (1-x)^2*(1-x^4) ). - R. J. Mathar, Jul 17 2012; edited by N. J. A. Sloane, Jul 22 2012
From R. J. Mathar, Jun 28 2013: (Start)
a(4*k)   = A000466(k);
a(4*k+1) = A002943(k);
a(4*k+2) = A078371(k-1) for k>0;
a(4*k+3) = A002939(k+1). (End)
a(n) = (2*n^2-11-9*(-1)^n+6*((-1)^((2*n+1-(-1)^n)/4)+(-1)^((2*n-1+(-1)^n)/4)))/8. - Luce ETIENNE, Oct 27 2016

Edited by N. J. A. Sloane, Jul 22 2012

A221918 Triangle of denominators of sum of two unit fractions: 1/n + 1/m, n >= m >= 1.

Original entry on oeis.org

1, 2, 1, 3, 6, 3, 4, 4, 12, 2, 5, 10, 15, 20, 5, 6, 3, 2, 12, 30, 3, 7, 14, 21, 28, 35, 42, 7, 8, 8, 24, 8, 40, 24, 56, 4, 9, 18, 9, 36, 45, 18, 63, 72, 9, 10, 5, 30, 20, 10, 15, 70, 40, 90, 5, 11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 11, 12, 12, 12, 3, 60, 4, 84, 24, 36, 60, 132, 6

Offset: 1

Wolfdieter Lang, Feb 21 2013

The corresponding triangle of numerators is A221919.
The law for the electrical resistance in a parallel circuit with two resistors R1 and R2 is 1/R = 1/R1 + 1/R2. Here we take 1/R(n,m) =  1/n + 1/m, with n >= m> =1, and R(n,m) = a(n,m)/A221919(n,m).
The reduced mass mu in a two body problem with masses m1 and m2 is given by 1/mu = 1/m1 + 1/m2.
The radius R of the twin circles of Archimedes' arbelos with the radii of the two small half-circles r1 and r2 is given by 1/R = 1/r1 +1/r2. The large half-circle has radius r =  r1 + r2. See, e.g., the Bankoff reference (according to which one should speak of a triple of such radius R circles). There are much more such radius R circles. See the Arbelos references given by Schoch, especially reference [3].
The columns give A000027, A145979(n-2), A221920, A221921, A222463 for m = 1, 2, ..., 5.
This and the companion entry resulted from a remark on the twin circles in Archimedes' arbelos in the Strick reference, p. 13, and the obvious question about their radii and centers. See the MathWorld link, also for more references.
The rationals R(n,m) = a(n,m)/A221919(n,m) (in lowest terms) equal H(n,m)/2, where H(n,m) = A227041(n,m)/A227042(n,m) is the harmonic mean of m and n. - Wolfdieter Lang, Jul 02 2013

			The triangle a(n,m) begins:
n\m    1    2    3    4    5    6   7   8   9   10   11  12 ...
1:     1
2:     2    1
3:     3    6    3
4:     4    4   12    2
5:     5   10   15   20    5
6:     6    3    2   12   30    3
7:     7   14   21   28   35   42   7
8:     8    8   24    8   40   24  56   4
9:     9   18    9   36   45   18  63  72   9
10:   10    5   30   20   10   15  70  40  90    5
11:   11   22   33   44   55   66  77  88  99  110   11
12:   12   12   12    3   60    4  84  24  36   60  132   6
...
a(n,1) = n because 1/R(n,1) =  1/n +1/1 = (n+1)/n, hence a(n,1) = denominator(1/n +1/1/) = n =  numerator(R(n,1)).
a(5,3) = denominator(1/5 + 1/3) = denominator(8/15 ) = 15.
a(6,3) = denominator(1/6 + 1/3) = denominator(9/18 ) = denominator(1/2) = 2.
The triangle of rationals R(n,m) = n*m/(n+m) = a(n,m)/A221919(n,m) given by 1/R(n,m) = 1/n + 1/m starts:
n\m    1    2      3     4      5     6     7     8     9   10
1:    1/2
2:    2/3    1
3:    3/4   6/5   3/2
4:    4/5   4/3  12/7    2
5:    5/6  10/7  15/8   20/9   5/2
6:    6/7   3/2    2    12/5  30/11   3
7:    7/8  14/9  21/10  28/11 35/12 42/13  7/2
8:    8/9   8/5  24/11   8/3  40/13 24/7  56/15   4
9:   9/10  18/11  9/4   36/13 45/14 18/5  63/16 72/17  9/2
10:  10/11  5/3  30/13  20/7  10/3  15/4  70/17 40/9  90/19  5
...

L. Bankoff, Are the Twin Circles of Archimedes Really Twins?, Mathematics Mag. 47,4 (1974) 214-218.
H. K. Strick, Geschichten aus der Mathematik, Spektrum der Wissenschaft - Spezial 2/2009.

T. Schoch, Arbelos References.
Eric W. Weisstein, Arbelos (MathWorld).

Cf. A221919 (companion).

a[n_, m_] := Denominator[1/n + 1/m]; Table[a[n, m], {n, 1, 12}, {m, 1, n}] // Flatten  (* Jean-François Alcover, Feb 25 2013 *)

a(n,m) = denominator(1/n +1/m) = numerator(n*m/(n+m)), n >= m >= 1 and 0 otherwise.

a(n,m)/A221919(n,m) = R(n,m) = n*m/(n+m). 1/R(n,m) = 1/n + 1/m.

A177427 Numerators of the Inverse Akiyama-Tanigawa transform of the aerated even-indexed Bernoulli numbers 1, 0, 1/6, 0, -1/30, 0, 1/42, ...

Original entry on oeis.org

1, 1, 13, 7, 149, 157, 383, 199, 7409, 7633, 86231, 88331, 1173713, 1197473, 1219781, 620401, 42862943, 43503583, 279379879, 283055551, 57313183, 19328341, 449489867, 1362695813, 34409471059, 34738962067, 315510823603, 45467560829, 9307359944587, 9382319148907, 293103346860157, 147643434162641, 594812856101039, 54448301591149

Offset: 0

Paul Curtz, May 07 2010

These are the numerators of the first row of a Table T(n,k) which contains the even-indexed Bernoulli numbers in the first column: T(2n,0) = A000367(n)/A002445(n), T(2n+1,0)=0, and which generates rows with the Akiyama-Tanigawa transform. (Because the first column is given, the algorithm is an inverse Akiyama-Tanigawa transform.)

These are the absolute values of the numerators of the Taylor expansion of sinh(log(x+1))*log(x+1)at x=0. - Gary Detlefs, Aug 31 2011

			The table T(n,k) of fractions generated by the Akiyama-Tanigawa transform, with the column T(n,0) equal to Bernoulli(n) for even n and equal to 0 for odd n, starts in row n=0 as:
1, 1, 13/12, 7/6, 149/120, 157/120, 383/280, 199/140, ...
0, -1/6, -1/4, -3/10, -1/3, -5/14, -3/8, -7/18, -2/5, -9/22, ...
1/6, 1/6, 3/20, 2/15, 5/42, 3/28, 7/72, 4/45, 9/110, 5/66, ...
0, 1/30, 1/20, 2/35, 5/84, 5/84, 7/120, 28/495, 3/55, 15/286, ...
-1/30, -1/30, -3/140, -1/105, 0, 1/140, 49/3960, 8/495, ...
0, -1/42, -1/28, -4/105, -1/28, -29/924, -7/264, -28/1287, -87/5005, ...
1/42, 1/42, 1/140, -1/105, -5/231, -9/308, -343/10296, -1576/45045, ...

L. A. Medina, V. H. Moll, E. S. Rowland, Iterated primitives of logarithmic powers, arXiv:0911.1325, arXiv:0911.1325 [math.NT], 2009-2010.
D. Merlini, R. Sprugnoli, M. C. Verri, The Akiyama-Tanigawa Transformation, Integers, 5 (1) (2005) #A05.

Cf. A177690 (denominators).

t[n_, 0] := BernoulliB[n]; t[1, 0]=0; t[n_, k_] := t[n, k] = (t[n, k-1] + (k-1)*t[n, k-1] - t[n+1, k-1])/k; Table[t[0, k], {k, 0, 33}] // Numerator (* Jean-François Alcover, Aug 09 2012 *)

From Groux Roland, Jan 07 2011: (Start)
T(0,k) = H(k)/2 + 1/(k+1) with H(k) harmonic number of order k.
T(0,k)= -(1/2)*(k+1)*Integral_{x=0..1} x^n*log(x*(1-x)) dx.
G.f.: Sum_{k>=0} T(0,k) x^k = (x-2)*(log(1-x))/(2*x*(1-x)). (End)
T(1,n) = -A191567(n)/A061038(n+2) = -A060819(n)/A145979(n). - Paul Curtz, Jul 19 2011
(T(1,n))^2 = A181318(n)/A061038(n+2). - Paul Curtz, Jul 19 2011, index corrected by R. J. Mathar, Sep 09 2011

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

Original entry on oeis.org

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

A208950 a(4n) = n(16n^2-1)/3, a(2n+1) = n(n+1)(2n+1)/6, a(4n+2) = (4n+1)(4n+2)(4*n+3)/6.

Original entry on oeis.org

0, 0, 1, 1, 5, 5, 35, 14, 42, 30, 165, 55, 143, 91, 455, 140, 340, 204, 969, 285, 665, 385, 1771, 506, 1150, 650, 2925, 819, 1827, 1015, 4495, 1240, 2728, 1496, 6545, 1785, 3885, 2109, 9139, 2470, 5330, 2870, 12341, 3311, 7095, 3795, 16215, 4324

Offset: 0

Paul Curtz, Mar 03 2012

a(n+2) is divisible by A060819(floor(n/3)).
a(n) is divisible by A176672(floor(n/3)).
Denominator of a(n)/n is of period 24: 1,1,3,4,1,6,1,4,3,1,1,12,1,2,3,4,1,3,1,4,3,2,1,12 (two successive palindromes).
This is the fifth column of the triangle A107711, hence the formula involving gcd(n+2,4) given below follows. - Wolfdieter Lang, Feb 24 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,4,0,0,0,-6,0,0,0,4,0,0,0,-1).

Cf. A107711 (fifth column).

Cf. A000034, A000292, A002415, A051724, A060819, A061037, A107711, A138190, A145979, A176672, A176895.

[Binomial(n+1,3)*GCD(n+2,4)/4: n in [0..50]]; // G. C. Greubel, Sep 20 2018

CoefficientList[Series[(x^2 + x^3 + 5 x^4 + 5 x^5 + 31 x^6 + 10 x^7 + 22 x^8 + 10 x^9 + 31 x^10 + 5 x^11 + 5 x^12 + x^13 + x^14)/((1 - x)^4 (1 + x)^4 (1 + 4 x^2 + 6 x^4 + 4 x^6 + x^8)), {x, 0, 47}], x] (* Bruno Berselli, Mar 11 2012 *)

A208950(n) := block(
        [a,npr] ,
        if equal(mod(n,4), 0) then (
                a : n/12*(n^2-1)
        ) else if equal(mod(n,2),0) then (
                a : (n-1)*n*(n+1)/6
        ) else (
                npr : (n-1)/2,
                a : npr*(npr+1)*n/6
        ) ,
        return(a)
)$ /* R. J. Mathar, Mar 10 2012 */

vector(50, n, n--; binomial(n+1,3)*gcd(n+2,4)/4) \\ G. C. Greubel, Sep 20 2018

a(n) = 4*a(n-4) - 6*a(n-8) + 4*a(n-12) - a(n-16).
a(n+1) = A002415(n+1)/A145979(n-1).
a(n) = A051724(n-1) * A051724(n) * A051724(n+1).
a(n) = A060819(n-1) * A060819(n) * A060819(n+1) / 3.
a(n) * a(n+4) = A061037(n+1) * A061037(n+2) * A061037(n+3) / 9.
a(n) = A138190(n)/A000034(n) for n > 0.
a(n) = A000292(n-1)/A176895(n+2) for n > 0.
a(n)/a(n+4) = n*(n^2-1)/((n+3)*(n+4)*(n+5)).
a(n)/a(n+12) = (n-1)*n*(n+1)/((n+11)*(n+12)*(n+13)).
G.f.: (x^2 + x^3 + 5*x^4 + 5*x^5 + 31*x^6 + 10*x^7 + 22*x^8 + 10*x^9 + 31*x^10 + 5*x^11 + 5*x^12 + x^13 + x^14) / ((1-x)^4*(1+x)^4*(1 + 4*x^2 + 6*x^4 + 4*x^6 + x^8)). - R. J. Mathar, Mar 10 2012
From Wolfdieter Lang, Feb 24 2014: (Start)
G.f.: (1 + x^12 + x*(1+x^10) + 5*x^2*(1+x^8) + 5*x^3*(1+x^7) + 31*x^4*(1+x^4) + 10*x^5*(1+x^2) + 22*x^6)/(1-x^4)^4. This is the preceding g.f. rewritten.
a(n) = binomial(n+1,3)*gcd(n+2,4)/4, n >= 0. From the g.f., see a comment above on A107711. (End)
a(n) = (n*(n-1)*((n+1)*(4+2*(-1)^n + (1+(-1)^n)*(-1)^((2*n+3+(-1)^n)/4))))/48. - Luce ETIENNE, Jan 01 2015
Sum_{n>=2} 1/a(n) = 12 - 27*log(2)/2. - Amiram Eldar, Aug 12 2022

A225975 Square root of A226008(n).

Original entry on oeis.org

0, 2, 2, 6, 1, 10, 6, 14, 4, 18, 10, 22, 3, 26, 14, 30, 8, 34, 18, 38, 5, 42, 22, 46, 12, 50, 26, 54, 7, 58, 30, 62, 16, 66, 34, 70, 9, 74, 38, 78, 20, 82, 42, 86, 11, 90, 46, 94, 24, 98, 50, 102, 13, 106, 54, 110, 28, 114, 58

Offset: 0

Paul Curtz, May 22 2013

Repeated terms of A016825 are in the positions 1,2,3,6,5,10,... (A043547).
From Wolfdieter Lang, Dec 04 2013: (Start)
This sequence a(n), n>=1, appears in the formula 2*sin(2*Pi/n) = R(p(n), x) modulo C(a(n), x), with x = rho(a(n)) = 2*cos(Pi/a(n)), the R-polynomials given in A127672 and the minimal C-polynomials of rho given in A187360. This follows from the identity 2*sin(2*Pi/n) = 2*cos(Pi*p(n)/a(n)) with gcd(p(n), a(n)) = 1. For p(n) see a comment on A106609,
Because R is an integer polynomial it shows that 2*sin(2*Pi/n) is an integer in the algebraic number field Q(rho(a(n))) of degree delta(a(n)) (the degree of C(a(n), x)), with delta(k) = A055034(k). This degree is given in A093819. For the coefficients of 2*sin(2*Pi/n) in the power basis of Q(rho(a(n))) see A231189 . (End)

			For the first formula: a(0)=-1+1=0, a(1)=-3+5=2, a(2)=-1+3=2, a(3)=-1+7=6, a(4)=0+1=1.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,-1).

Cf. A016825, A017089, A017137, A022998, A106609, A145979, A226008.

a[0]=0; a[n_] := Sqrt[Denominator[1/4 - 4/n^2]]; Table[a[n], {n, 0, 58}] (* Jean-François Alcover, May 30 2013 *)
LinearRecurrence[{0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,-1},{0,2,2,6,1,10,6,14,4,18,10,22,3,26,14,30},60] (* Harvey P. Dale, Nov 21 2019 *)

a(n) = A106609(n-4) + A106609(n+4) with A106609(-4)=-1, A106609(-3)=-3, A106609(-2)=-1, A106609(-1)=-1.
a(n) = 2*a(n-8) -a(n-16).
a(2n+1) = A016825(n), a(2n) = A145979(n-2) for n>1, a(0)=0, a(2)=2.
a(4n)   = A022998(n).
a(4n+1) = A017089(n).
a(4n+2) = A016825(n).
a(4n+3) = A017137(n).
G.f.: x*(2 +2*x +6*x^2 +x^3 +10*x^4 +6*x^5 +14*x^6 +4*x^7 +14*x^8 +6*x^9 +10*x^10 +x^11 +6*x^12 +2*x^13 +2*x^14)/((1-x)^2*(1+x)^2*(1+x^2)^2*(1+x^4)^2). [Bruno Berselli, May 23 2013]
From Wolfdieter Lang, Dec 04 2013: (Start)
a(n) = 2*n if n is odd; if n is even then a(n) is n if n/2 == 1, 3, 5, 7 (mod 8), it is n/2 if n/2 == 0, 4 (mod 8) and it is n/4 if n/2 == 2, 6 (mod 8). This leads to the given G.f..
With c(n) = A178182(n), n>=1, a(n) = c(n)/2 if c(n) is even and c(n) if c(n) is odd. This leads to the preceding formula. (End)

Edited by Bruno Berselli, May 24 2013

cp's OEIS Frontend

A173630 Denominator of A002445(n)/A145979(n).

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A060819 a(n) = n / gcd(n,4).

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A061038 Denominator of 1/4 - 1/n^2.

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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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A214297 a(0)=-1, a(1)=0, a(2)=-3; thereafter a(n+2) - 2*a(n+1) + a(n) has period 4: repeat -4, 8, -4, 2.

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A221918 Triangle of denominators of sum of two unit fractions: 1/n + 1/m, n >= m >= 1.

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A177427 Numerators of the Inverse Akiyama-Tanigawa transform of the aerated even-indexed Bernoulli numbers 1, 0, 1/6, 0, -1/30, 0, 1/42, ...

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

A208950 a(4n) = n(16n^2-1)/3, a(2n+1) = n(n+1)(2n+1)/6, a(4n+2) = (4n+1)(4n+2)(4*n+3)/6.