A181318 -id:A181318 - cp's OEIS frontend

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

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

A168077 a(2n) = A129194(2n)/2; a(2n+1) = A129194(2n+1).

Original entry on oeis.org

0, 1, 1, 9, 4, 25, 9, 49, 16, 81, 25, 121, 36, 169, 49, 225, 64, 289, 81, 361, 100, 441, 121, 529, 144, 625, 169, 729, 196, 841, 225, 961, 256, 1089, 289, 1225, 324, 1369, 361, 1521, 400, 1681, 441, 1849, 484, 2025, 529, 2209, 576, 2401, 625, 2601

Offset: 0

Paul Curtz, Nov 18 2009

From Paul Curtz, Mar 26 2011: (Start)
Successive A026741(n) * A026741(n+p):
  p=0:  0, 1,  1,  9,  4, 25,  9,     a(n),
  p=1:  0, 1,  3,  6, 10, 15, 21,  A000217,
  p=2:  0, 3,  2, 15,  6, 35, 12,  A142705,
  p=3:  0, 2,  5,  9, 14, 20, 27,  A000096,
  p=4:  0, 5,  3, 21,  8, 45, 15,  A171621,
  p=5:  0, 3,  7, 12, 18, 25, 33,  A055998,
  p=6:  0, 7,  4, 27, 10, 55, 18,
  p=7:  0, 4,  9, 15, 22, 30, 39,  A055999,
  p=8:  0, 9,  5, 33, 12, 65, 21,  (see A061041),
  p=9:  0, 5, 11, 18, 26, 35, 45,  A056000. (End)
The moment generating function of p(x, m=2, n=1, mu=2) = 4*x*E(x, 2, 1), see A163931 and A274181, is given by M(a) = (-4 * log(1-a) - 4 * polylog(2, a))/a^2. The series expansion of M(a) leads to the sequence given above. - Johannes W. Meijer, Jul 03 2016
Multiplicative because both A129194 and A040001 are. - Andrew Howroyd, Jul 26 2018

G. C. Greubel, Table of n, a(n) for n = 0..1000
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 (0,3,0,-3,0,1).

Cf. A040001, A168068, A181318.

I:=[0,1,1,9,4,25]; [n le 6 select I[n] else 3*Self(n-2)-3*Self(n-4)+Self(n-6): n in [1..60]]; // Vincenzo Librandi, Jul 10 2016

a := proc(n): n^2*(5-3*(-1)^n)/8 end: seq(a(n), n=0..46); # Johannes W. Meijer, Jul 03 2016

LinearRecurrence[{0,3,0,-3,0,1},{0,1,1,9,4,25},60] (* Harvey P. Dale, May 14 2011 *)
f[n_] := Numerator[(n/2)^2]; Array[f, 60, 0] (* Robert G. Wilson v, Dec 18 2012 *)
CoefficientList[Series[x(1+x+6x^2+x^3+x^4)/((1-x)^3(1+x)^3), {x,0,60}], x] (* Vincenzo Librandi, Jul 10 2016 *)

concat(0, Vec(x*(1+x+6*x^2+x^3+x^4)/((1-x)^3*(1+x)^3) + O(x^60))) \\ Altug Alkan, Jul 04 2016

a(n) = lcm(4, n^2)/4; \\ Andrew Howroyd, Jul 26 2018

(x*(1+x+6*x^2+x^3+x^4)/(1-x^2)^3).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Feb 20 2019

From R. J. Mathar, Jan 22 2011: (Start)
G.f.: x*(1 + x + 6*x^2 + x^3 + x^4) / ((1-x)^3*(1+x)^3).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
a(n) = n^2*(5 - 3*(-1)^n)/8. (End)
a(n) = A026741(n)^2.
a(2*n) = A000290(n); a(2*n+1) = A016754(n).
a(n) - a(n-4) = 4*A064680(n+2). - Paul Curtz, Mar 27 2011
4*a(n) = A061038(n) * A010121(n+2) = A109043(n)^2, n >= 2. - Paul Curtz, Apr 07 2011
a(n) = A129194(n) / A040001(n). - Andrew Howroyd, Jul 26 2018
From Peter Bala, Feb 19 2019: (Start)
a(n) = numerator(n^2/(n^2 + 4)) = n^2/(gcd(n^2,4)) = (n/gcd(n,2))^2.
a(n) = n^2/b(n), where b(n) = [1, 4, 1, 4, ...] is a purely periodic sequence of period 2. Thus a(n) is a quasi-polynomial in n.
O.g.f.: x*(1 + x)/(1 - x)^3 - 3*x^2*(1 + x^2)/(1 - x^2)^3.
Cf. A181318. (End)
From Werner Schulte, Aug 30 2020: (Start)
Multiplicative with a(2^e) = 2^(2*e-2) for e > 0, and a(p^e) = p^(2*e) for prime p > 2.
Dirichlet g.f.: zeta(s-2) * (1 - 3/2^s).
Dirichlet convolution with A259346 equals A000290.
Sum_{n>0} 1/a(n) = Pi^2 * 7 / 24. (End)
Sum_{k=1..n} a(k) ~ (5/24) * n^3. - Amiram Eldar, Nov 28 2022

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

A165943 a(n) = A061037(7*n+2).

Original entry on oeis.org

0, 77, 63, 525, 56, 1365, 483, 2597, 210, 4221, 1295, 6237, 462, 8645, 2499, 11445, 812, 14637, 4095, 18221, 1260, 22197, 6083, 26565, 1806, 31325, 8463, 36477, 2450, 42021, 11235, 47957, 3192, 54285, 14399, 61005, 4032, 68117, 17955, 75621, 4970

Offset: 0

Paul Curtz, Oct 01 2009

The (2k+1)-sections A061037((2*k+1)*n+2) are multiples of 2k+1:
0,...21,...15,..117,...12,..285,...99,..525,...42,..837,..255, k=1, A142590
0,...45,...35,..285,...30,..725,..255,.1365,..110,.2205,..675, k=2, A165248
0,...77,...63,..525,...56,.1365,..483,.2597,..210,.4221,.1295, k=3, here
0,..117,...99,..837,...90,.2205,..783,.4221,..342,.6885,.2115, k=4,
0,..165,..143,.1221,..132,.3245,.1155,.6237,..506,10197,.3135, k=5
0,..221,..195,.1677,..182,.4485,.1599,.8645,..702,14157,.4355, k=6
After division by 2k+1 these define a table T'(k,c) :
0,....7,....5,...39,....4,...95,...33,..175,...14,..279,...85, k=1, A142883
0,....9,....7,...57,....6,..145,...51,..273,...22,..441,..135, k=2
0,...11,....9,...75,....8,..195,...69,..371,...30,..603,..185, k=3
0,...13,...11,...93,...10,..245,...87,..469,...38,..765,..235, k=4
0,...15,...13,..111,...12,..295,..105,..567,...46,..927,..285, k=5
0,...17,...15,..129,...14,..345,..123,..665,...54,.1089,..335, k=6
Differences downwards each second column in this second table are 2 = 7-5 = 9-7..; 18 = 57-39 = 75-57..; 50 = 145-95 = 195-145... = A077591(n+1) = 2*A016754.
The difference T'(k+1,c)-T'(k,c) is 0, 2, 2, 18, 2, 50, 18, 98, 8 ... = 2*A181318(c) =A061037(c-2)+A061037(c+2). - Paul Curtz, Mar 12 2012
Let b(n)= a(n) mod 11. The sequence b(n) has the property b(n+44) = b(n) with the first 43 values being {0, 0 , 8, 1, 1, 10, 1, 1, 8, 8, 0, 0, 10, 5, 5, 9, 7, 1, 5, 6, 10, 7, 0, 2, 1, 1, 8, 1, 5, 8, 2, 0, 8, 10, 6, 5, 10, 7, 9, 5, 0, 10, 0}. - G. C. Greubel, Apr 18 2016

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

m:=25; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3))); // G. C. Greubel, Sep 19 2018

seq(numer(1/4 - 1/(7*n+2)^2), n=0..50); # Robert Israel, Apr 20 2016

Table[Numerator[1/4 - 1/(7 n + 2)^2], {n, 0, 40}] (* Michael De Vlieger, Apr 19 2016 *)
CoefficientList[Series[7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3), {x, 0, 50}], x] (* G. C. Greubel, Sep 19 2018 *)

a(n) = numerator(1/4 - 1/(7*n+2)^2); \\  Altug Alkan, Apr 18 2016

x='x+O('x^50); concat([0], Vec(7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3))) \\ G. C. Greubel, Sep 19 2018

a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12), n>12. - Conjectured by R. J. Mathar, Mar 02 2010, proved by Robert Israel, Apr 20 2016
From Ilya Gutkovskiy, Apr 19 2016: (Start)
G.f.: 7*x*(11 + 9*x + 75*x^2 + 8*x^3 + 162*x^4 + 42*x^5 + 146*x^6 + 6*x^7 + 51*x^8 + 5*x^9 + 3*x^10)/((1 - x)^3*(1 + x)^3*(1 + x^2)^3).
a(n) = -7*n*(7*n + 4)*(27*(-1)^n + 6*cos((Pi*n)/2) - 37)/64. (End)

Partially edited and extended by R. J. Mathar, Mar 02 2010

Removed division by 7 in definition and formula - R. J. Mathar, Mar 23 2010

A198148 a(n) = n(n+2)(9 - 7*(-1)^n)/16.

Original entry on oeis.org

0, 3, 1, 15, 3, 35, 6, 63, 10, 99, 15, 143, 21, 195, 28, 255, 36, 323, 45, 399, 55, 483, 66, 575, 78, 675, 91, 783, 105, 899, 120, 1023, 136, 1155, 153, 1295, 171, 1443, 190, 1599, 210, 1763, 231, 1935, 253, 2115, 276, 2303, 300, 2499, 325

Offset: 0

Paul Curtz, Oct 21 2011

See, in A181318(n), A060819(n)*A060819(n+p): A060819(n)^2, A064038(n), a(n), A160050(n), A061037(n), A178242(n). The second differences a(n+2)-2*a(n+1)+a(n) = -5, 16, -26, 44, -61, 86, -110, 142, -173, 212, -250, 296, -341, 394, -446, 506, taken modulo 9 are periodic with the palindromic period 4, 7, 1, 8, 2, 5, 7, 7, 7, 5, 2, 8, 1, 7, 4.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. quadrisections A014105, A001539, A000384, A141759.

Cf. A060819, A000217, A000466, A142705, A000034, A005563, A010689, A028552, A050534.

List([0..60], n -> n*(n+2)*(9-7*(-1)^n)/16); # G. C. Greubel, Feb 21 2019

[n*(n+2)*(9-7*(-1)^n)/16: n in [0..60]]; // Vincenzo Librandi, Nov 25 2011

A198148:=n->n*(n+2)*(9-7*(-1)^n)/16; seq(A198148(k), k=0..100); # Wesley Ivan Hurt, Oct 16 2013

LinearRecurrence[{0,3,0,-3,0,1},{0,3,1,15,3,35},60] (* Vincenzo Librandi, Nov 25 2011 *)

a(n)=n*(n+2)*(9-7*(-1)^n)/16 \\ Charles R Greathouse IV, Oct 16 2015

[n*(n+2)*(9-7*(-1)^n)/16 for n in (0..60)] # G. C. Greubel, Feb 21 2019

a(n) = A060819(n)*A060819(n+2).
a(2n) = n*(n+1)/2 = A000217(n).
a(2n+1) = (2*n+1)*(2*n+3) = A000466(n+1).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6), n>5.
a(n+1) - a(n) = (7*(-1)^n *(2*n^2+6*n+3) +18*n +27)/16.
a(n) = A142705(n) / A000034(n+1).
a(n) = A005563(n) / A010689(n+1). - Franklin T. Adams-Watters, Oct 21 2011
G.f. x*(3 +x +6*x^2 -x^4)/(1-x^2)^3. - R. J. Mathar, Oct 25 2011
a(n)*a(n+1) = a(A028552(n)) = A050534(n+2). - Bruno Berselli, Oct 26 2011
a(n) = numerator( binomial((n+2)/2,2) ). - Wesley Ivan Hurt, Oct 16 2013
E.g.f.: x*((24+x)*cosh(x) + (3+8*x)*sinh(x))/8. - G. C. Greubel, Sep 20 2018
Sum_{n>=1} 1/a(n) = 5/2. - Amiram Eldar, Aug 12 2022

A206771 0 followed by the numerators of the reduced (A001803(n) + A001790(n)) / (2*A046161(n)).

Original entry on oeis.org

0, 1, 1, 9, 5, 175, 189, 1617, 429, 57915, 60775, 508079, 264537, 8788507, 9100525, 75218625, 9694845, 5109183315, 5250613995, 43106892675, 22090789875, 723694276305, 740104577355, 6049284520695, 1543768261425, 201547523019375

Offset: 0

Paul Curtz, Jan 10 2013

We write the fractions a(n)/b(n) and higher order differences as a matrix:
0,         1,      1,    9/8,    5/4,...
1,         0,    1/8,    1/8, 15/128,... = (A001790(n)/A046161(n) +Lorbeta(n)) /2
-1,      1/8,      0, -1/128, -1/128,... = (Lorbeta(n+1) +A161200(n+1)/A046161(n+1)) / 2
9/8,    -1/8, -1/128,      0, 1/1024,...
-5/4, 15/128,  1/128, 1/1024,      0,...
Here, Lorbeta(0)=1 and Lorbeta(n) = -A098597(n-1)/A046161(n) for n>0 is the inverse of the Lorentz factor.
The first line with numerators a(n) and denominators b(n) is 0, 1, 1, 9/8, 5/4, 175/128, 189/128, 1617/1024, 429/256, 57915/32768, 60775/32768,... It is an autosequence: Its inverse binomial transform is the signed sequence.
a(n+1)/(2*n-1)= 1, 3, 1, 25, 21, 147, 33, 3861, 3575, 26741,... .
a(n+1)/A146535(n) = 9, 5, 35, 27, 539, 39, 4455,... .
A001790(n)/A046161(n) yields the coefficients of the Lorentz factor (or Lorentz gamma factor). With b for beta and g for gamma:
g = (1-b^2)^-1 = 1 + (b^2)/2 + 3*(b^4)/8 + 5*(b^6)/16 + ... .
b = (1-g^-2)^-1 = 1 - (g^-2)/2 - (g^-4)/8 - (g^-6)/16  - ... .
Are the denominators of the first subdiagonal 1, 1/8, -1/128, 1/1024,...  A061549(n) ?
a(n+1)/(A000108(n)*b(n)) = 1, 1, 9/16, 1/4, 25/256, 9/256, 49/4096, 1/256, 81/65536, 25/65536, 121/1048576,... = A191871(n+1)/ A084623(n+1)^2 ?

			From the first formula: a(1)=1*1, a(2)=1*1, a(3)=3*3, a(4)=1*5, a(5)=5*35, a(6)=3*63.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
OEIS Wiki, Autosequence
Wikipedia, Lorentz Factor.

Cf. A187791, A000108.

/* By definition: */ m:=25; R:=PowerSeriesRing(Rationals(), m); p:=Coefficients(R!(1/(1-x)^(1/2))); q:=Coefficients(R!((1-x)^(-3/2))); A001790:=[Numerator(p[i]): i in [1..m]]; A001803:=[Numerator(q[i]): i in [1..m]]; A046161:=[Denominator(Binomial(2*n,n)/4^n): n in [0..m-1]]; [0] cat [Numerator((A001803[n]+A001790[n])/(2*A046161[n])): n in [1..m]]; // Bruno Berselli, Mar 11 2013

A206771 := proc(n)
        A001790(n)+A001803(n) ;
        %/2/A046161(n) ;
        numer(%) ;
end proc: # R. J. Mathar, Jan 18 2013

max = 25; A001803 = CoefficientList[Series[(1 - x)^(-3/2), {x, 0, max}], x] // Numerator; A001790 = CoefficientList[Series[1/Sqrt[(1 - x)], {x, 0, max}], x] // Numerator; A046161 = Table[Binomial[2n, n]/4^n, {n, 0, max}] // Denominator; a[n_] := (A001803[[n]] + A001790[[n]])/(2*A046161[[n]]) // Numerator; a[0] = 0; Table[a[n], {n, 0, max}]
(* or (from 1st formula) : *) Table[ n*Numerator[4^(1-n)*Binomial[2n-2, n-1]]/2^IntegerExponent[n, 2], {n, 0, max}]
(* or (from 2nd formula) : *) Table[ Numerator[ CatalanNumber[n-1]/2^(2n-1)]*Numerator[n^2/2^n], {n, 0, max}] (* Jean-François Alcover, Jan 31 2013 *)

a(n) = A000265(n) * A001790(n-1).
a(n) = A098597(n-1) * A191871(n). See also A181318
a(n) = numerator of n*binomial(2n-2,n-1)/4^(n-1). - Nathaniel Johnston, Dec 16 2022

a(11)-a(25) from Jean-François Alcover, Jan 13 2013

A181407 a(n) = (n-4)(n-3)2^(n-2).

Original entry on oeis.org

3, 3, 2, 0, 0, 16, 96, 384, 1280, 3840, 10752, 28672, 73728, 184320, 450560, 1081344, 2555904, 5963776, 13762560, 31457280, 71303168, 160432128, 358612992, 796917760, 1761607680, 3875536896, 8489271296, 18522046464, 40265318400, 87241523200, 188441690112

Offset: 0

Paul Curtz, Jan 28 2011

Binomial transform of (3, 0, -1, followed by A005563).
The sequence and its successive differences are:
   3,  3,  2,   0,   0,   16,   96,  384,      a(n),
   0, -1, -2,   0,  16,   80,  288,  896,      A178987,
  -1, -1,  2,  16,  64,  208,  608, 2688,     -A127276,
   0,  3, 14,  48, 144,  400, 1056, 2688,      A176027,
   3, 11, 34,  96, 256,  656, 1632, 3968,      A084266(n+1)
   8, 23, 62, 160, 400,  976, 2336, 5504,
  15, 39, 98, 240, 576, 1360, 3168, 7296.
Division of the k-th column by abs(A174882(k)) gives
   3,  3,  1,  0,  0,  1,  3,  3,  5,  15,  21, 14,   A064038(n-3),
   0, -1, -1,  0,  1,  5,  9,  7, 10,  27,  35, 22,   A160050(n-3),
  -1, -1,  1,  2,  4, 13, 19, 13, 17,  43,  53, 32,   A176126,
   0,  3,  7,  6,  9, 25, 33, 21, 26,  63,  75, 44,   A178242,
   3, 11, 17, 12, 16, 41, 51, 31, 37,  87, 101, 58,
   8  23, 31, 20, 25, 61, 73, 43, 50, 115, 131, 74,
  15, 39, 49, 30, 36, 85, 99, 57, 65, 147, 165, 92.
Columns are (or from) A005563, A142463, A056220, A002378, A000290, A001844, A058331, A002061, A002522, A097080, A093328, A014206.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-12,8).

Cf. A176027, A181318.

List([0..40], n-> (n-4)*(n-3)*2^(n-2)); # G. C. Greubel, Feb 21 2019

[(n-4)*(n-3)*2^(n-2): n in [0..40] ]; // Vincenzo Librandi, Feb 01 2011

Table[(n-4)*(n-3)*2^(n-2), {n,0,40}] (* G. C. Greubel, Feb 21 2019 *)

vector(40, n, n--; (n-4)*(n-3)*2^(n-2)) \\ G. C. Greubel, Feb 21 2019

[(n-4)*(n-3)*2^(n-2) for n in (0..40)] # G. C. Greubel, Feb 21 2019

a(n) = 16*A001788(n-4).
a(n+1) - a(n) = A178987(n).
G.f.: (3 - 15*x + 20*x^2) / (1-2*x)^3. - R. J. Mathar, Jan 30 2011
E.g.f.: (x^2 - 3*x + 3)*exp(2*x). - G. C. Greubel, Feb 21 2019

A191567 Four interlaced 2nd order polynomials: a(4k) = k(1+2k); a(1+2k) = 2(1+2k)(3+2k); a(2+4k) = 4(1+k)(1+2k).

Original entry on oeis.org

0, 6, 4, 30, 3, 70, 24, 126, 10, 198, 60, 286, 21, 390, 112, 510, 36, 646, 180, 798, 55, 966, 264, 1150, 78, 1350, 364, 1566, 105, 1798, 480, 2046, 136, 2310, 612, 2590, 171, 2886, 760, 3198, 210, 3526, 924, 3870, 253, 4230, 1104, 4606, 300, 4998, 1300, 5406, 351

Offset: 0

Paul Curtz, Jun 12 2011

a(n) = T(0,n) and differences T(n,k) = T(n-1,k+1) - T(n-1,k) define the array
0,   6,  4,  30,    3,  70,   24,  126,   10,  198,   60,  286,   21,  390,  ..
6,  -2, 26, -27,   67, -46,  102, -116,  188, -138,  226, -265,  369, -278, ..
-8, 28 -53,  94, -113, 148, -218,  304, -326,  364, -491,  634, -647,  676, ...
T(3,n) mod 9 is the sequence 1, 1, 1, 4, 4, 4, 7, 7, 7, 4, 4, 4 (and periodically repeated with period 12).
A064680(2+n) divides a(n), where b(n) = a(n)/A064680(2+n) = 0, 1, 2, 3, 1, 5, 6, 7, 2,... for n>=0, obeys b(4*k) = k and has recurrence b(n) = 2*b(n-4) - b(n-8).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A014105, A000466, A000384, A177427.

a:=[0,6,4,30,3,70,24,126,10,198,60,286];; for n in [13..60] do a[n]:= 3*a[n-4]-3*a[n-8]+a[n-12]; od; a; # G. C. Greubel, Feb 26 2019

I:=[0,6,4,30,3,70,24,126,10,198,60,286]; [n le 12 select I[n] else 3*Self(n-4)-3*Self(n-8)+Self(n-12): n in [1..60]]; // Vincenzo Librandi, Apr 23 2017

Table[Which[OddQ@ n, 2 (1 + 2 #) (3 + 2 #) &[(n - 1)/2], Mod[n, 4] == 0, # (1 + 2 #) &[n/4], True, 4 (1 + #) (1 + 2 #) &[(n - 2)/4]], {n, 0, 60}] (* or *)
CoefficientList[Series[x(6 +4x +30x^2 +3x^3 +52x^4 +12x^5 +36x^6 +x^7 +6x^8 -2x^10)/((1-x)^3*(1+x)^3*(1+x^2)^3), {x, 0, 60}], x] (* Michael De Vlieger, Apr 22 2017 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1}, {0,6,4,30,3,70,24,126,10,198,60, 286}, 80] (* Vincenzo Librandi, Apr 23 2017 *)

m=60; v=concat([0,6,4,30,3,70,24,126,10,198,60,286], vector(m-12)); for(n=13, m, v[n]=3*v[n-4]-3*v[n-8]+v[n-12]); v \\ G. C. Greubel, Feb 26 2019

(x*(6+4*x+30*x^2+3*x^3+52*x^4+12*x^5+36*x^6+x^7+6*x^8-2*x^10)/((1-x)^3 *(1+x)^3*(1+x^2)^3 )).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Feb 26 2019

a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(n) = A061037(n+2) + A181318(n). - Paul Curtz, Jul 19 2011
a(n) = A060819(n) * A145979(n). - Paul Curtz, Sep 06 2011
G.f.: x*(6+4*x+30*x^2+3*x^3+52*x^4+12*x^5+36*x^6+x^7+6*x^8-2*x^10) /( (1-x)^3 *(1+x)^3 *(1+x^2)^3 ). - R. J. Mathar, Jun 17 2011
Let BEB(n) = a(n)/A061038(n+2) = A060819(n)/A145979(n). Then (BEB(n))^2 = A181318(n)/A061038(n+2) = BEB(n) - A061037(n+2)/A061038(n+2). - Paul Curtz, Jul 19 2011, index corrected by R. J. Mathar, Sep 09 2011
From Luce ETIENNE, Apr 18 2017: (Start)
a(n) = n*(n + 2)*(37 - 27*(-1)^n - 3*((-1)^((2*n + 1 - (-1)^n)/4) + (-1)^((2*n - 1 + (-1)^n)/4)))/32.
a(n) = n*(n+2)*(37-27*cos(n*Pi) - 6*cos(n*Pi/2))/32.
a(n) = n*(n + 2)*(37 - 27*(-1)^n - 3*(i^n + (-i)^n))/32, where i=sqrt(-1). (End)

A217366 a(n) = ((n+6) / gcd(n+6,4)) * (n / gcd(n,4)).

Original entry on oeis.org

0, 7, 2, 27, 5, 55, 9, 91, 14, 135, 20, 187, 27, 247, 35, 315, 44, 391, 54, 475, 65, 567, 77, 667, 90, 775, 104, 891, 119, 1015, 135, 1147, 152, 1287, 170, 1435, 189, 1591, 209, 1755, 230, 1927, 252, 2107, 275, 2295, 299, 2491, 324, 2695, 350, 2907, 377

Offset: 0

Jean-François Alcover, Oct 01 2012

The 6th sequence (p=6) of the family A060819(n)*A060819(n+p).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A060819, A181318.

[(9-7*(-1)^n)/16*n*(n+6): n in [0..50]]; // G. C. Greubel, Sep 21 2018

a[n_] := 8^(Mod[n, 2] - 1)*n*(n + 6); Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Oct 01 2012 *)
CoefficientList[Series[x*(7 + 2*x + 6*x^2 - x^3 - 5*x^4)/(1 - x^2)^3, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 15 2012 *)
LinearRecurrence[{0,3,0,-3,0,1},{0,7,2,27,5,55},60] (* Harvey P. Dale, Sep 14 2022 *)

concat(0, Vec(x*(7+2*x+6*x^2-x^3-5*x^4)/((1-x)^3*(1+x)^3) + O(x^100))) \\ Colin Barker, Jan 27 2016

vector(50, n, n--; (9-7*(-1)^n)/16*n*(n+6)) \\ G. C. Greubel, Sep 21 2018

a(n) = 8^(n mod 2 - 1)*n*(n + 6).
G.f.: x*(7 + 2*x + 6*x^2 - x^3 - 5*x^4)/(1 - x^2)^3. - Bruno Berselli, Oct 01 2012
From Colin Barker, Jan 27 2016: (Start)
a(n) = (9 - 7*(-1)^n)*n*(n + 6)/16.
a(n) = (n^2 + 6*n)/8 for n even.
a(n) = n^2 + 6*n for n odd. (End)
Sum_{n>=1} 1/a(n) = 133/90. - Amiram Eldar, Aug 12 2022

A217367 a(n) = ((n+7) / gcd(n+7,4)) * (n / gcd(n,4)).

Original entry on oeis.org

0, 2, 9, 15, 11, 15, 39, 49, 30, 36, 85, 99, 57, 65, 147, 165, 92, 102, 225, 247, 135, 147, 319, 345, 186, 200, 429, 459, 245, 261, 555, 589, 312, 330, 697, 735, 387, 407, 855, 897, 470, 492, 1029, 1075, 561, 585, 1219, 1269, 660, 686, 1425, 1479, 767, 795, 1647

Offset: 0

Jean-François Alcover, Oct 01 2012

The 7th sequence (p=7) of the family A060819(n)*A060819(n+p).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wikipedia, Quasi-polynomial.
Index entries for linear recurrences with constant coefficients, signature (3,-6,10,-12,12,-10,6,-3,1).

Cf. A060819, A181318.

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(2+3*x+6*x^4-6*x^5+4*x^6-3*x^7)/(1-x+x^2-x^3)^3)); // G. C. Greubel, Sep 20 2018

a[n_] := n*(n+7)/(2*Mod[1 + Floor[n/2], 2] + 2); Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Oct 01 2012 *)
CoefficientList[Series[x (2 + 3 x + 6 x^4 - 6 x^5 + 4 x^6 - 3 x^7) / (1 - x + x^2 - x^3)^3, {x, 0, 33}], x] (* Vincenzo Librandi, Jul 17 2013 *)

my(x='x+O('x^50)); Vec(x*(2+3*x+6*x^4-6*x^5+4*x^6-3*x^7)/(1-x+x^2-x^3)^3) \\ G. C. Greubel, Sep 20 2018

a(n) = n*(n+7)/(2*mod(1 + floor(n/2), 2) + 2).
G.f.: x*(2 + 3*x + 6*x^4 - 6*x^5 + 4*x^6 - 3*x^7)/(1 - x + x^2 - x^3)^3.
From Peter Bala, Aug 07 2022: (Start)
a(n) = numerator of n*(n+7)/4.
a(n) is quasi-polynomial in n: if p(n) = n*(n+7)/4 then a(4*n) = p(4*n), a(4*n+1) = p(4*n+1), a(4*n+2) = 2*p(4*n+2) and a(4*n+3) = 2*p(4*n+3) = A303295(n+1) for n >= 1. (End)
Sum_{n>=1} 1/a(n) = 697/735 + Pi/14. - Amiram Eldar, Aug 16 2022

cp's OEIS Frontend

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

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A168077 a(2n) = A129194(2n)/2; a(2n+1) = A129194(2n+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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A165943 a(n) = A061037(7*n+2).

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A198148 a(n) = n*(n+2)*(9 - 7*(-1)^n)/16.

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A206771 0 followed by the numerators of the reduced (A001803(n) + A001790(n)) / (2*A046161(n)).

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A198148 a(n) = n(n+2)(9 - 7*(-1)^n)/16.

A181407 a(n) = (n-4)(n-3)2^(n-2).

A191567 Four interlaced 2nd order polynomials: a(4k) = k(1+2k); a(1+2k) = 2(1+2k)(3+2k); a(2+4k) = 4(1+k)(1+2k).