A176895 -id:A176895 - 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

A188134 a(4n) = n, a(1+2n) = 4+8n, a(2+4n) = 2+4*n.

Original entry on oeis.org

0, 4, 2, 12, 1, 20, 6, 28, 2, 36, 10, 44, 3, 52, 14, 60, 4, 68, 18, 76, 5, 84, 22, 92, 6, 100, 26, 108, 7, 116, 30, 124, 8, 132, 34, 140, 9, 148, 38, 156, 10, 164, 42, 172, 11, 180, 46, 188, 12, 196, 50, 204, 13, 212, 54, 220, 14, 228, 58, 236, 15, 244, 62

Offset: 0

Paul Curtz, Mar 21 2011

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

Cf. A016825, A017113, A060819, A061037, A064680, A146160, A176895, A177499.

[(64-3*(1+(-1)^n)*(9+(-1)^(n div 2)))*n/16 : n in [0..80]]; // Wesley Ivan Hurt, Jul 06 2016

A188134:=n->8*n/(11 + 9*cos(Pi*n) + 12*cos(n*Pi/2)): seq(A188134(n), n=0..100); # Wesley Ivan Hurt, Jul 06 2016

Table[8 n/(11 + 9 Cos[Pi*n] + 12 Cos[n*Pi/2]), {n, 0, 80}] (* Wesley Ivan Hurt, Jul 06 2016 *)
CoefficientList[Series[x*(4+2*x+12*x^2+x^3+12*x^4+2*x^5+4*x^6)/(1-x^4)^2, {x, 0, 50}], x] (* G. C. Greubel, Sep 20 2018 *)
LinearRecurrence[{0,0,0,2,0,0,0,-1},{0,4,2,12,1,20,6,28},70] (* Harvey P. Dale, Aug 14 2019 *)

x='x+O('x^50); concat([0], Vec(x*(4+2*x+12*x^2+x^3+12*x^4+ 2*x^5 +4*x^6)/(1-x^4)^2)) \\ G. C. Greubel, Sep 20 2018

a(n) = 2*a(n-4) - a(n-8) for n>7.
a(n) = A176895(n) * A060819(n).
a(n) = (4*A061037(n+2))/(n+4).
a(n) = 4*n / A146160(n).
a(2*n) = A064680(n).
a(1+2*n) = A017113(n).
a(4*n) = a(-4+4*n) +  1.
a(1+4*n) = a(-3+4*n) + 16.
a(2+4*n) = a(-2+4*n) +  4.
a(3+4*n) = a(-1+4*n) + 16.  See A177499.
From Bruno Berselli, Mar 22 2011: (Start)
G.f.: x*(4+2*x+12*x^2+x^3+12*x^4+2*x^5+4*x^6)/(1-x^4)^2.
a(n) = (64-3*(1+(-1)^n)*(9+i^n))*n/16 with i=sqrt(-1).
a(n)/a(n-4) = n/(n-4) for n>4. (End)
a(n) = 8*n/(11 + 9*cos(Pi*n) + 12*cos(n*Pi/2)). - Wesley Ivan Hurt, Jul 06 2016
a(n) = lcm(4,n)/gcd(4,n). - R. J. Mathar, Feb 12 2019
Sum_{k=1..n} a(k) ~ (37/32)*n^2. - Amiram Eldar, Oct 07 2023

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

A177499 Period 4: repeat [1, 16, 4, 16].

Original entry on oeis.org

1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16, 1, 16, 4, 16

Offset: 0

Paul Curtz, May 10 2010

From Klaus Brockhaus, May 14 2010: (Start)
Interleaving of A000012, A010855, A010709 and A010855.
Continued fraction expansion of (44+sqrt(2442))/88. (End)

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

Cf. A010674, A177838. [Klaus Brockhaus, May 14 2010]

Cf. A000012, A010709, A010855, A176895.

&cat[[1, 16, 4, 16]^^26]; // Klaus Brockhaus, May 14 2010

seq(op([1, 16, 4, 16]), n=0..50); # Wesley Ivan Hurt, Jul 09 2016

Table[{1, 16, 4, 16}, {21}] // Flatten (* Jean-François Alcover, May 24 2013 *)

From Klaus Brockhaus, May 14 2010: (Start)
a(n+2) - a(n) = A010674(n).
a(n) = a(n-4) for n > 3.
G.f.: (1+16*x+4*x^2+16*x^3)/(1-x^4). (End)
a(n) = A176895(n)^2. - Paul Curtz, Mar 21 2011
a(n) = (37 - 6*cos(n*Pi/2) - 27*cos(n*Pi) - 27*I*sin(n*Pi))/4. - Wesley Ivan Hurt, Jul 09 2016

A213268 Denominators of the Inverse semi-binomial transform of A001477(n) read downwards antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 4, 4, 1, 2, 2, 8, 2, 1, 1, 4, 1, 16, 16, 1, 2, 1, 8, 8, 32, 16, 1, 1, 4, 4, 16, 8, 64, 64, 1, 2, 2, 8, 4, 32, 32, 128, 16, 1, 1, 4, 2, 16, 16, 64, 8, 256, 256, 1, 2, 1, 8, 8, 32, 4, 128, 128, 512, 256, 1, 1, 4, 4, 16, 2, 64, 64, 256, 128, 1024, 1024

Offset: 0

Paul Curtz, Jun 08 2012

Starting from any sequence a(k) in the first row, define the array T(n,k) of the inverse semi-binomial transform  by T(0,k) = a(k), T(n,k) = T(n-1,k+1) -T(n-1,k)/2, n>=1.
Here, where the first row is the nonnegative integers, the array is
0        1      2      3    4      5      6      7     8   =A001477(n)
1       3/2     2    5/2    3    7/2      4    9/2     5   =A026741(n+2)/A000034(n)
1       5/4   3/2    7/4    2    9/4    5/2   11/4     3   =A060819(n+4)/A176895(n)
3/4     7/8     1    9/8  5/4   11/8    3/2   13/8   7/4   =A106609(n+6)/A205383(n+6)
1/2    9/16   5/8  11/16  3/4  13/16    7/8  15/16     1   =A106617(n+8)/TBD
5/16  11/32   3/8  13/32 7/16  15/32    1/2  17/32  9/16
3/16  13/64  7/32  15/64  1/4  17/64   9/32  19/64  5/16
7/64 15/128   1/8 17/128 9/64 19/128   5/32 21/128 11/64
1/16 17/256 9/128 19/256 5/64 21/256 11/128 23/256  3/32.
The first column contains 0, followed by fractions A000265/A084623, that is Oresme numbers n/2^n multiplied by 2 (see A209308).

			The array of denominators starts:
  1   1   1   1   1   1   1   1   1   1   1 ...
  1   2   1   2   1   2   1   2   1   2   1 ...
  1   4   2   4   1   4   2   4   1   4   2 ...
  4   8   1   8   4   8   2   8   4   8   1 ...
  2  16   8  16   4  16   8  16   1  16   8 ...
16  32   8  32  16  32   2  32  16  32   8 ...
16  64  32  64   4  64  32  64  16  64  32 ...
64 128   8 128  64 128  32 128  64 128  16 ...
16 256 128 256  64 256 128 256  32 256 128 ...
256 512 128 512 256 512  64 512 256 512 128 ...
All entries are powers of 2.

A213268frac := proc(n,k)
        if n = 0 then
                return k ;
        else
                return procname(n-1,k+1)-procname(n-1,k)/2 ;
        end if;
end proc:
A213268 := proc(n,k)
        denom(A213268frac(n,k)) ;
end proc: # R. J. Mathar, Jun 30 2012

T[0, k_] := k; T[n_, k_] := T[n, k] = T[n-1, k+1] - T[n-1, k]/2; Table[T[n-k, k] // Denominator, {n, 0, 11}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Sep 12 2014 *)

A185138 a(4n) = n(4n-1); a(2n+1) = n(n+1)/2; a(4n+2) = (2n+1)(4*n+1).

Original entry on oeis.org

0, 0, 1, 1, 3, 3, 15, 6, 14, 10, 45, 15, 33, 21, 91, 28, 60, 36, 153, 45, 95, 55, 231, 66, 138, 78, 325, 91, 189, 105, 435, 120, 248, 136, 561, 153, 315, 171, 703, 190, 390, 210, 861, 231, 473, 253, 1035, 276, 564, 300, 1225, 325

Offset: 0

Paul Curtz, Mar 12 2012

a(n) is divisible by the n-th term of the sequence 3, 3, 1, 1, 3, 3 (periodically repeated with period 6).
a(n) is divisible by b(floor((n-1)/3)), where b(n) = 1, 3, 2, 3, 7, 3, 5, 3, 13, 3, 8, 3, 19, 3,... , n>=0, is defined by inserting a 3 after each entry of A165355.
(n+1)*(n+2)*(n+3)/2=3*A000292(n+1) is divisible by a(n+2), so there is an integer sequence c(n)= 3*A000292(n+1)/a(n+2) = 3, 12, 10, 20, 7, 28, 18,... with c(2*n)=A123167(n+1) and c(n)/A109613(n+2)=A176895(n).
The sequence of denominators of a(n+2)/n has period length 8: 1, 2, 1, 4, 1, 1, 1, 4.
A table T(k,c) = a(1+c*(1+2k)) of (2*k+1)-sections starts as follows:
0  1  1   3   3   15...
0  3  6  45  21   60...
0 15 15  60  55  325...
0 14 28 231 105  315...
0 45 45 189 171 1035...
The table of T'(k,c) = T(k,c)/(2k+1), columns c>=0, looks as follows, construction similar to A165943:
0 1 1  3  3  15  6  14   k=0
0 1 2 15  7  20 15  77   k=1
0 3 3 12 11  65 24  63   k=2
0 2 4 33 15  45 33 175   k=3
0 5 5 21 19 115 42 112   k=4
0 3 6 51 23  70 51 273   k=5
The entries T'(k,c) are divisible by A060819(c).
Differences are T'(2,c)-T'(0,c) = T'(4,c)-T'(2,c) = 0, 2, 2, 9, 8, 50, 18, 49, 32, ... which is A168077(c) multiplied by the c-th term of the period-4 sequence 2, 2, 2, 1.
Differences are T'(3,c)- T'(1,c) = T'(5,c)-T'(3,c) = 0, 1, 2, 18, 8, 25, 18, 98, 32,... which is  A168077(c) multiplied by the period-4 sequence 2, 1, 2, 2.
The reduced fractions T'(0,c)/T'(1,c) = 1, 1/2, 1/5, 3/7, 3/4, 2/5, 2/11, 5/13, 5/7, 3/8, 3/17, 7/19, .., c>=1, have a numerator sequence A026741(floor(c/2)+1). The denominator sequence is f(c) = 1, 2, 5, 7, 4, 5,.. = A001651(c+1)/A130658(c+1), with f(2*c+1) +f(2*c+2) = 3, 12, 9, 24 .. =3*A022998(c).

G. C. Greubel, 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. A000217, A014634, A026741, A033991, A064038, A060819, A165355, A208950.

A185138 := proc(n)
        if n mod 4 = 0 then
                return n/4*(n-1) ;
        elif n mod 2 = 1 then
                return (n-1)*(n+1)/8 ;
        else
                return (n-1)*n/2 ;
        end if;
end proc: # R. J. Mathar, Apr 05 2012

Clear[b];b[1] = 0; b[2] = 0; b[3] = 1; b[4] = 1; b[5] = 3; b[6] = 3; b[7] = 15;b[8] = 6;b[n_Integer] := b[n] = ((-2 + n) (-4 (-4 + n) (-3 + n) (-2 + n) (8 + n (-9 + 2 n)) b[-3 + n] + (-5 + n) ((-3 +n) ((-4 + n) (211 + 2 n (-215 + n (147 + n (-41 + 4 n)))) - 4 (-1 + n) (19 + n (-13 + 2 n)) b[-2 + n]) - 4 (-4 + n)^2 (8 + n (-9 + 2 n)) b[-1 + n])))/(4 (-5 + n) (-4 + n) (-3 + n)^2 (19 + n (-13 + 2 n)))
a = Table[b[n], {n, 1, 52}] (* Roger L. Bagula, Mar 14 2012 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{0,0,1,1,3,3,15,6,14,10,45,15},60] (* Harvey P. Dale, Nov 23 2015 *)

x='x+O('x^50); concat([0,0], Vec(-x^2*(3*x^8+x^7+5*x^6+3*x^5+12*x^4+3*x^3+3*x^2+x+1)/ ((x-1)^3*(x+1)^3*(x^2+1)^3))) \\ G. C. Greubel, Jun 23 2017

a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(2*n) = A064038(2*n), a(2*n+1) = A000217(n).
a(n) = 3*A208950(n)/A109613(n).
a(n+1) = A060819(n) * A026741(n+2)(floor(n/2)).
G.f.: -x^2*(3*x^8+x^7+5*x^6+3*x^5+12*x^4+3*x^3+3*x^2+x+1)/ ((x-1)^3*(x+1)^3*(x^2+1)^3). - R. J. Mathar, Mar 22 2012
a(n) = (4*n^2-3*n-1+(2*n^2-3*n+1)*(-1)^n + n*(n-1)*(1+(-1)^n)*(-1)^((2*n-3-(-1)^n)/4))/16. - Luce ETIENNE, May 13 2016
Sum_{n>=2} 1/a(n) = 2 - Pi/4 + 7*log(2)/2. - Amiram Eldar, Aug 12 2022

A246416 A permutation of essentially the duplicate nonnegative numbers: a(4n) = n + 1/2 - (-1)^n/2, a(2n+1) = a(4n+2) = 2n+1.

Original entry on oeis.org

0, 1, 1, 3, 2, 5, 3, 7, 2, 9, 5, 11, 4, 13, 7, 15, 4, 17, 9, 19, 6, 21, 11, 23, 6, 25, 13, 27, 8, 29, 15, 31, 8, 33, 17, 35, 10, 37, 19, 39, 10, 41, 21, 43, 12, 45, 23, 47, 12, 49, 25, 51, 14, 53, 27, 55, 14, 57, 29, 59, 16, 61, 31, 63, 16

Offset: 0

Paul Curtz, Sep 14 2014

A permutation of A004526 (n > 0).
0 is at its own place. Distance between the two (2*k+1)'s: 2*k+1 terms. 0 is in position 0, the first 1 in position 1, the second 1 in position 2, the first 2 in position 4, the second 2 in position 8. Hence, r(n) = 0, 1, 2, 4, 8, 3, 6, 12, 16, 5, 10, 20, 24, ..., a permutation of A001477. See A225055. The recurrence r(n) = r(n-4) + r(n-8) - r(n-12) is the same as for a(n).
A061037(n+3) is divisible by a(n+5) (= 5, 3, 7, 1, 9, 5, 11, 3, 13, 7, 15, 3, ...). Hence a link, via A212831 and A214282, between the Catalan numbers A000108 and the Balmer series.

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

Cf. A000108, A004526, A052928, A061037, A176895, A212831, A214282, A226279, A225055, A247617.

I:=[0,1,1,3,2,5,3,7,2,9,5,11,4,13,7,15,4,17,9,19,6,21,11,23]; [n le 24 select I[n] else 3*Self(n-8)-3*Self(n-16)+Self(n-24): n in [1..80]]; // Vincenzo Librandi, Oct 15 2014

A246416:=n->n*(1+floor((2-n)/4)+floor((n-2)/4))/2+n*(1+floor((1-n)/2)+floor((n-1)/2))+(-n-2+2*(-1)^(n/4))*(ceil(n/4)-floor(n/4)-1)/4: seq(A246416(n), n=0..50); # Wesley Ivan Hurt, Sep 14 2014

Table[n (1 + Floor[(2 - n)/4] + Floor[(n - 2)/4])/2 + n (1 + Floor[(1 - n)/2] + Floor[(n - 1)/2]) + (-n - 2 + 2 (-1)^(n/4)) (Ceiling[n/4] - Floor[n/4] - 1)/4, {n, 0, 50}] (* Wesley Ivan Hurt, Sep 14 2014 *)
a[n_] := Switch[Mod[n, 4], 0, n/4-(-1)^(n/4)/2+1/2, 1|3, n, 2, n/2]; Table[a[n], {n, 0, 64}] (* Jean-François Alcover, Oct 09 2014 *)
LinearRecurrence[{0,0,0,1,0,0,0,1,0,0,0,-1},{0,1,1,3,2,5,3,7,2,9,5,11},70] (* Harvey P. Dale, Mar 23 2015 *)

a(n)=if(n%4,n/(2-n%2),if(n%8,1,0)+n/4) \\ Charles R Greathouse IV, Sep 14 2014

a(n) = 3*a(n-8) - 3*a(n-16) + a(n-24).
a(n+4) = a(n) + period 8: repeat [2, 4, 2, 4, 0, 4, 2, 4].
a(n+8) = a(n) + period 4: repeat [2, 8, 4, 8] (= 2 * A176895).
a(2n) = A212831(n).
a(n) = n*(1+floor((2-n)/4)+floor((n-2)/4))/2+n*(1+floor((1-n)/2)+floor((n-1)/2))+(-n-2+2*(-1)^(n/4))*(ceiling(n/4)-floor(n/4)-1)/4. - Wesley Ivan Hurt, Sep 14 2014
a(n) = a(n-4) + a(n-8) - a(n-12). - Charles R Greathouse IV, Sep 14 2014
G.f.: x*(x^10+x^9+3*x^8+4*x^6+2*x^5+4*x^4+2*x^3+3*x^2+x+1) / ((x-1)^2*(x+1)^2*(x^2+1)^2*(x^4+1)). - Colin Barker, Sep 15 2014

A247617 a(4n) = n + 1/2 - (-1)^n/2 + (-1)^n, a(2n+1) = 2n + 5, a(4n+2) = 2n + 3.

Original entry on oeis.org

1, 5, 3, 7, 1, 9, 5, 11, 3, 13, 7, 15, 3, 17, 9, 19, 5, 21, 11, 23, 5, 25, 13, 27, 7, 29, 15, 31, 7, 33, 17, 35, 9, 37, 19, 39, 9, 41, 21, 43, 11, 45, 23, 47, 11, 49, 25, 51, 13, 53, 27, 55, 13, 57, 29, 59, 15, 61, 31, 63, 15, 65, 33, 67

Offset: 0

Paul Curtz, Sep 21 2014

Essentially a permutation of A129756 (odd numbers repeated four times).
a(-1) = 3, a(-2) = a(-3) = 1.
Distance between the first two (2*k+1)'s: 2*k+1 terms. Distance between the last two (2*n+1)'s: 4 terms. Essentially same distances as in -a(-n) = -1, -3, -1, -1, 1, 1, 1, 3, 1, 5, 3, 7, 3, 9, 5, 11, 3, 13, 7, 15, 5, 17, 9, 19, 5, 21, 11, 23, 7, 25, 13, 27, 7, ... .

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

Cf. A010709, A061037, A121262, A129756, A176895, A246416.

I:=[1,5,3,7,1,9,5,11,3,13,7,15]; [n le 12 select I[n] else Self(n-4)+Self(n-8)-Self(n-12): n in [1..80]]; // Vincenzo Librandi, Oct 15 2014

A247617:=n->(n+4)*(1-ceil((2-n)/4)-ceil((n-2)/4))/2+(n+4)*(1+floor((1-n)/2)+floor((n-1)/2))-(n+2+2*(-1)^(n/4))*(ceil(n/4)-floor(n/4)-1)/4: seq(A247617(n), n=0..50); # Wesley Ivan Hurt, Sep 21 2014

Table[(n + 4) (1 - Ceiling[(2 - n)/4] - Ceiling[(n - 2)/4])/2 + (n + 4) (1 + Floor[(1 - n)/2] + Floor[(n - 1)/2]) - (n + 2 + 2 (-1)^(n/4)) (Ceiling[n/4] - Floor[n/4] - 1)/4, {n, 0, 50}] (* Wesley Ivan Hurt, Sep 21 2014 *)

Vec(-(3*x^11+x^10+x^9-x^8-4*x^7-2*x^6-4*x^5-7*x^3-3*x^2-5*x-1)/((x-1)^2*(x+1)^2*(x^2+1)^2*(x^4+1)) + O(x^100)) \\ Colin Barker, Sep 21 2014

a(n) = a(n-4) + a(n-8) - a(n-12).
a(n) * A246416(n) = A061037(n+2).
A246416(n+4) - a(n) = sequence of period 4: [1, 0, 0, 0].
a(n+4) - a(n) = sequence of period 8: [0, 4, 2, 4, 2, 4, 2, 4].
G.f.: -(3*x^11+x^10+x^9-x^8-4*x^7-2*x^6-4*x^5-7*x^3-3*x^2-5*x-1) / ((x-1)^2*(x+1)^2*(x^2+1)^2*(x^4+1)). - Colin Barker, Sep 21 2014
a(n) = a(n-8) + sequence of period 4: [2, 8, 4, 8] (= 2*A176895(n)).
a(-n) * A246416(-n) = A061037(n-2).
a(n) = (n+4)*(1-ceiling((2-n)/4)-ceiling((n-2)/4))/2+(n+4)*(1+floor((1-n)/2)+floor((n-1)/2))-(n+2+2(-1)^(n/4))*(ceiling(n/4)-floor(n/4)-1)/4. - Wesley Ivan Hurt, Sep 21 2014

A225058 a(4n) = n-1. a(2n+1) = a(4n+2) = 2n+1.

Original entry on oeis.org

-1, 1, 1, 3, 0, 5, 3, 7, 1, 9, 5, 11, 2, 13, 7, 15, 3, 17, 9, 19, 4, 21, 11, 23, 5, 25, 13, 27, 6, 29, 15, 31, 7, 33, 17, 35, 8, 37, 19, 39, 9, 41, 21, 43, 10, 45, 23, 47, 11, 49, 25, 51, 12, 53, 27, 55, 13, 57, 29, 59, 14, 61, 31, 63, 15, 65, 33, 67, 16, 69

Offset: 0

Paul Curtz, Apr 26 2013

Consider the family of sequences with recurrence a(n) = 2*a(n-4)-a(n-8) where a(0) and a(4) move up in steps of 1. This here is characterized by a(0)=-1, a(4)=0:
-2, 1, 1, 3, -1, 5, 3, 7, 0, 9, 5, 11,...
-1, 1, 1, 3,  0, 5, 3, 7, 1, 9, 5, 11,...  = a(n)
0,  1, 1, 3,  1, 5, 3, 7, 2, 9, 5, 11,...  = A060819
1,  1, 1, 3,  2, 5, 3, 7, 3, 9, 5, 11,...  = b(n)
2,  1, 1, 3,  3, 5, 3, 7, 4, 9, 5, 11,... .
a(n+4)+b(n) = A145979(n).
a(n+4)*b(n) = A061037(n+2).
a(n+4)-b(n) = repeat -1, 4, 2, 4 with period of length 4.

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

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

a[n_] := 1/16*(11*n-(-1)^n*(5*n+4)-2*(n+4)*Re[I^n]-4); Table[a[n], {n, 0, 47}] (* Jean-François Alcover, Apr 30 2013 *)
LinearRecurrence[{0,0,0,2,0,0,0,-1},{-1,1,1,3,0,5,3,7},80] (* Harvey P. Dale, Jul 14 2019 *)

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

a(n) = 2*a(n-4) - a(n-8).
a(n+4) - a(n) = A176895(n).
G.f.: (-1+x+x^2+3*x^3+3*x^5+x^6+x^7+2*x^4)/((-1+x)^2*(1+x)^2*(x^2+1)^2). - R. J. Mathar, Apr 28 2013

A251091 a(n) = n^2 / gcd(n+2, 4).

Original entry on oeis.org

0, 1, 1, 9, 8, 25, 9, 49, 32, 81, 25, 121, 72, 169, 49, 225, 128, 289, 81, 361, 200, 441, 121, 529, 288, 625, 169, 729, 392, 841, 225, 961, 512, 1089, 289, 1225, 648, 1369, 361, 1521, 800, 1681, 441, 1849, 968, 2025, 529, 2209, 1152, 2401, 625, 2601, 1352

Offset: 0

Paul Curtz, May 08 2015

A061038(n), which appears in 4*a(n) formula, is a permutation of n^2.
Origin. In December 2010, I wrote in my 192-page Exercise Book no. 5, page 41, the array (difference table of the first row):
   1     0,   1/3,     1,   9/5,    8/3,   25/7,    9/2,   49/9, ...
  -1,  1/3,   2/3,   4/5, 13/15,  19/21,  13/14,  17/18,  43/45, ...
Numerators are listed in A176126, denominators are in A064038, and denominator - numerator = 2, 2, 1, 1,... (A014695).
  4/3, 1/3,  2/15,  1/15, 4/105,   1/42,   1/63,   1/90,  4/495, ...
  -1, -1/5, -1/15, -1/35, -1/70, -1/126, -1/210, -1/330, -1/495, ...
where the denominators of the second row are listed in A000332.
Also for those of the inverse binomial transform
  1, -1, 4/3, -1, 4/5, -2/3, 4/7, -1/2, 4/9, -2/5, 4/11, -1/3, ... ?
a(n) is the (n+1)-th term of the numerators of the first row.

			a(0) = 0/2, a(1) = 1/1, a(2) = 4/4, a(3) = 9/1.

Colin Barker, 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. A000290, A000332, A016754, A026741, A060819, A061038, A109008, A139098, A168077, A176895, A181900.

[(1-(1/16)*(1+(-1)^n)*(5-(-1)^(n div 2)) )*n^2: n in [0..60]]; // Vincenzo Librandi, Jun 12 2015

seq(seq((4*i+j-1)^2/[2,1,4,1][j],j=1..4),i=0..30); # Robert Israel, May 14 2015

f[n_] := Switch[ Mod[n, 4], 0, n^2/2, 1, n^2, 2, n^2/4, 3, n^2]; Array[f, 50, 0] (* or *) Table[(4 i + j - 1)^2/{2, 1, 4, 1}[[j]], {i, 0, 12}, {j, 4}] // Flatten (* after Robert Israel *) (* or *) LinearRecurrence[{0, 0, 0, 3, 0, 0, 0, -3, 0, 0, 0, 1}, {0, 1, 1, 9, 8, 25, 9, 49, 32, 81, 25, 121}, 53] (* or *) CoefficientList[ Series[-((x (1 + x (1 + x (9 + x (8 + x (22 + x (6 + x (22 + x (8 + x (9 + x + x^2))))))))))/(-1 + x^4)^3), {x, 0, 52}], x] (* Robert G. Wilson v, May 19 2015 *)

concat(0, Vec(-x*(x^10 + x^9 + 9*x^8 + 8*x^7 + 22*x^6 + 6*x^5 + 22*x^4 + 8*x^3 + 9*x^2 + x + 1) / ((x-1)^3*(x+1)^3*(x^2+1)^3) + O(x^100))) \\ Colin Barker, May 14 2015

a(n) = n^2/(period 4: repeat 2, 1, 4, 1).
a(4n) = 8*n^2, a(2n+1) = a(4n+2) = (2*n+1)^2.
a(n+4) = a(n) + 8*A060819(n).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12), n>11.
4*a(n) = (period 4: repeat 2, 1, 4, 1) * A061038(n).
G.f.: -x*(x^10+x^9+9*x^8+8*x^7+22*x^6+6*x^5+22*x^4+8*x^3+9*x^2+x+1) / ((x-1)^3*(x+1)^3*(x^2+1)^3). - Colin Barker, May 14 2015
a(2n) = A181900(n), a(2n+1) = A016754(n). [Bruno Berselli, May 14 2015]
a(n) = ( 1 - (1/16)*(1+(-1)^n)*(5-(-1)^(n/2)) )*n^2. - Bruno Berselli, May 14 2015
Sum_{n>=1} 1/a(n) = 13*Pi^2/48. - Amiram Eldar, Aug 12 2022

Missing term (1521) inserted in the sequence by Colin Barker, May 14 2015
Definition uses a formula by Jean-François Alcover, Jul 01 2015
Keyword:mult added by Andrew Howroyd, Aug 06 2018

cp's OEIS Frontend

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

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A188134 a(4*n) = n, a(1+2*n) = 4+8*n, a(2+4*n) = 2+4*n.

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

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A177499 Period 4: repeat [1, 16, 4, 16].

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A213268 Denominators of the Inverse semi-binomial transform of A001477(n) read downwards antidiagonals.

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A185138 a(4*n) = n*(4*n-1); a(2*n+1) = n*(n+1)/2; a(4*n+2) = (2*n+1)*(4*n+1).

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A246416 A permutation of essentially the duplicate nonnegative numbers: a(4n) = n + 1/2 - (-1)^n/2, a(2n+1) = a(4n+2) = 2n+1.

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A188134 a(4n) = n, a(1+2n) = 4+8n, a(2+4n) = 2+4*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.

A185138 a(4n) = n(4n-1); a(2n+1) = n(n+1)/2; a(4n+2) = (2n+1)(4*n+1).

A247617 a(4n) = n + 1/2 - (-1)^n/2 + (-1)^n, a(2n+1) = 2n + 5, a(4n+2) = 2n + 3.

A225058 a(4n) = n-1. a(2n+1) = a(4n+2) = 2n+1.