A051176 -id:A051176 - cp's OEIS frontend

A106448 Table of (x+y)/gcd(x,y) where (x,y) runs through the pairs (1,1), (1,2), (2,1), (1,3), (2,2), (3,1), ...

2, 3, 3, 4, 2, 4, 5, 5, 5, 5, 6, 3, 2, 3, 6, 7, 7, 7, 7, 7, 7, 8, 4, 8, 2, 8, 4, 8, 9, 9, 3, 9, 9, 3, 9, 9, 10, 5, 10, 5, 2, 5, 10, 5, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 6, 4, 3, 12, 2, 12, 3, 4, 6, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 7, 14, 7, 14, 7, 2, 7, 14, 7, 14, 7, 14

Offset: 1

Antti Karttunen, May 21 2005

Can also be viewed as a triangular table T(n,k) (n>=1, 1<=k<=n) read by rows: T(1,1); T(2,1), T(2,2); T(3,1), T(3,2), T(3,3); T(4,1), T(4,2), T(4,3), T(4,4); ... where T(n,k) gives the least value v>0 such that v*k = 0 modulo n+1, i.e., in other words, T(n,k) = (n+1)/gcd(n+1,k).

			The top left corner of the square array is:
   2  3  4  5  6  7  8  9 10 11 ...
   3  2  5  3  7  4  9  5 11 ...
   4  5  2  7  8  3 10 11 ...
   5  3  7  2  9  5 11 ...
   6  7  8  9  2 11 ...
   7  4  3  5 11 ...
   8  9 10 11 ...
   9  5 11 ...
  10 11 ...
  11 ...

GF(2)[X] analog: A106449. Row 1 is n+1, row 2 is LEFT(LEFT(LEFT(A026741))), row 3 is LEFT^4(A051176). Essentially the same as A054531, but without its right-hand edge of all-1's.

Equals A003057/A003989.

T(n, k) = numerator((n+k)/n) = numerator((n+k)/k). - Michel Marcus, Dec 29 2013

A106610 Numerator of n/(n+9).

Original entry on oeis.org

0, 1, 2, 1, 4, 5, 2, 7, 8, 1, 10, 11, 4, 13, 14, 5, 16, 17, 2, 19, 20, 7, 22, 23, 8, 25, 26, 3, 28, 29, 10, 31, 32, 11, 34, 35, 4, 37, 38, 13, 40, 41, 14, 43, 44, 5, 46, 47, 16, 49, 50, 17, 52, 53, 6, 55, 56, 19, 58, 59, 20, 61, 62, 7, 64, 65, 22, 67, 68, 23, 70, 71, 8, 73, 74, 25, 76, 77

Offset: 0

N. J. A. Sloane, May 15 2005

Apart from 0, also numerator of Sum_{i=1..n} (1/((i+2)*(i+3))) = n/(3n+9). - Bruno Berselli, Nov 07 2012

In addition to being multiplicative, a(n) is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n >= 1, m >= 1. In particular, a(n) is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Feb 21 2019

			For n = 12, n/(n+9) = 12/21 = 4/7. So, a(12) = 4. - _Indranil Ghosh_, Jan 31 2017
From _Peter Bala_, Feb 21 2019: (Start)
Sum_{n >= 1} n*a(n)*x^n = G(x) - (2*3)*G(x^3) - (2*9)*G(x^9) , where G(x) = x*(1 + x)/(1 - x)^3.
Sum_{n >= 1} (1/n)*a(n)*x^n = H(x) - (2/3)*H(x^3) - (2/9)*H(x^9), where H(x) = x/(1 - x).
Sum_{n >= 1} (1/n^2)*a(n)*x^n = L(x) - (2/3^2)*L(x^3) - (2/9^2)*L(x^9), where L(x) = Log(1/(1 - x)).
Sum_{n >= 1} (1/a(n))*x^n = L(x) + (2/3)*L(x^3) + (2/3)*L(x^9). (End)

Raffaello Giusti, editore, Supplemento al Periodico di Matematica (Livorno), Jul 1902, p. 138 (Problem 421, case k=3).

Indranil Ghosh, Table of n, a(n) for n = 0..20000 (terms 0..1000 from Vincenzo Librandi)
Peter Bala, A note on the sequence of numerators of a rational function, 2019.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0, 0,-1).

Cf. A008292, A109050.

Cf. Sequences given by the formula numerator(n/(n + k)): A026741 (k = 2), A051176 (k = 3), A060819 (k = 4), A060791 (k = 5), A060789 (k = 6), A106608 thru A106612 (k = 7 thru 11), A051724 (k = 12), A106614 thru A106621 (k = 13 thru 20).

List([0..80],n->NumeratorRat(n/(n+9))); # Muniru A Asiru, Feb 19 2019

[Numerator(n/(n+9)): n in [0..100]]; // Vincenzo Librandi, Apr 18 2011

seq(numer(n/(n+9)),n=0..80); # Muniru A Asiru, Feb 19 2019

f[n_]:=Numerator[n/(n+9)]; Array[f,100,0] (* Vladimir Joseph Stephan Orlovsky, Feb 17 2011*)

vector(100, n, n--; numerator(n/(n+9))) \\ G. C. Greubel, Feb 19 2019

[lcm(n,9)/9for n in range(0, 100)] # Zerinvary Lajos, Jun 09 2009

From R. J. Mathar, Apr 18 2011: (Start)
a(n) = A109050(n)/9.
Dirichlet g.f. zeta(s-1)*(1-2/3^s-2/9^s).
Multiplicative with a(3^e) = 3^max(0,e-2), a(p^e) = p^e if p<>3. (End)
a(n) = 2*a(n-9) - a(n-18). - G. C. Greubel, Feb 19 2019
From Peter Bala, Feb 21 2019: (Start)
a(n) = n/gcd(n,9), where gcd(n,9) = [1, 1, 3, 1, 1, 3, 1, 1, 9, ...] is a periodic sequence of period 9: a(n) is thus quasi_polynomial in n.
O.g.f.: Sum_{n >= 0} a(n)*x^n = F(x) - 2*F(x^3) - 2*F(x^9), where F(x) = x/(1 - x)^2.
More generally, for m >= 1, Sum_{n >= 0} (a(n)^m)*x^n = F(m,x) + (1 - 3^m)*( F(m,x^3) + F(m,x^9) ), 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 to the o.g.f. for the sequence produces generating functions for the sequences n^m*a(n), m in Z. Some examples are given below. (End)
Sum_{k=1..n} a(k) ~ (61/162) * n^2. - Amiram Eldar, Nov 25 2022

A106618 a(n) = numerator of n/(n+17).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 1, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 2, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 3, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 4, 69, 70, 71, 72, 73, 74

Offset: 0

N. J. A. Sloane, May 15 2005

a(n) <> n iff n = 17 * k, in this case, a(n) = k. - Bernard Schott, Feb 19 2019

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

Cf. Sequences given by the formula numerator(n/(n + k)): A026741 (k = 2), A051176 (k = 3), A060819 (k = 4), A060791 (k = 5), A060789 (k = 6), A106608 thru A106612 (k = 7 thru 11), A051724 (k = 12), A106614 thru A106621 (k = 13 thru 20).

List([0..80],n->NumeratorRat(n/(n+17))); # Muniru A Asiru, Feb 19 2019

[Numerator(n/(n+17)): n in [0..100]]; // Vincenzo Librandi, Apr 18 2011

seq(numer(n/(n+17)),n=0..80); # Muniru A Asiru, Feb 19 2019

f[n_]:=Numerator[n/(n+17)];Array[f,100,0] (* Vladimir Joseph Stephan Orlovsky, Feb 17 2011 *)

vector(100, n, n--; numerator(n/(n+17))) \\ G. C. Greubel, Feb 19 2019

[lcm(n,17)/17 for n in range(0, 100)] # Zerinvary Lajos, Jun 12 2009

Dirichlet g.f.: zeta(s-1)*(1 - 16/17^s). - R. J. Mathar, Apr 18 2011
a(n) = 2*a(n-17) - a(n-34). - G. C. Greubel, Feb 19 2019
From Amiram Eldar, Nov 25 2022: (Start)
Multiplicative with a(17^e) = 17^(e-1), and a(p^e) = p^e if p != 17.
Sum_{k=1..n} a(k) ~ (273/578) * n^2. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = 33*log(2)/17. - Amiram Eldar, Sep 08 2023

A129202 Denominator of 3*(3+(-1)^n) / (n+1)^2.

Original entry on oeis.org

1, 2, 3, 8, 25, 6, 49, 32, 27, 50, 121, 24, 169, 98, 75, 128, 289, 54, 361, 200, 147, 242, 529, 96, 625, 338, 243, 392, 841, 150, 961, 512, 363, 578, 1225, 216, 1369, 722, 507, 800, 1681, 294, 1849, 968, 675, 1058, 2209, 384, 2401, 1250, 867, 1352, 2809, 486

Offset: 0

Paul Barry, Apr 03 2007

A divisibility sequence, that is, if n divides m then a(n) divides a(m). - Peter Bala, Feb 27 2019

G. C. Greubel, Table of n, a(n) for n = 0..1000
Peter Bala, A note on the sequence of numerators of a rational function , Feb 2019.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,3,0,0,0,0,0,-3,0,0,0,0,0,1).

Cf. A026741, A051176, A129196, A129197 (numerators), A060789.

[Denominator(3*(3+(-1)^n)/(n+1)^2): n in [0..50]]; // G. C. Greubel, Oct 26 2017

A129202:=n->numer((n+1)/2)*numer((n+1)/3): seq(A129202(n), n=0..100); # Wesley Ivan Hurt, Jul 18 2014

Table[Numerator[(n + 1)/2] Numerator[(n + 1)/3], {n, 0, 100}] (* Wesley Ivan Hurt, Jul 18 2014 *)
LinearRecurrence[{0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, -3, 0, 0, 0, 0, 0, 1}, {1, 2, 3, 8, 25, 6, 49, 32, 27, 50, 121, 24, 169, 98, 75, 128, 289, 54}, 60] (* Harvey P. Dale, Nov 20 2016 *)

for(n=0,50, print1(denominator(3*(3+(-1)^n)/(n+1)^2), ", ")) \\ G. C. Greubel, Oct 26 2017

a(n) = A129196(n)/(n+1).
(1/(2*Pi))*Integral_{t=0..2*Pi} exp(i*(n+1)*t)*((t-Pi)/i)^3 dt = (a(n)*Pi^2-A129203(n))/A129196(n), i=sqrt(-1).
a(n) = ( Numerator of (n+1)/2 ) * ( Numerator of (n+1)/3 ) = A026741(n+1) * A051176(n+1). - Wesley Ivan Hurt, Jul 18 2014
G.f.: -(x^16 +2*x^15 +3*x^14 +8*x^13 +25*x^12 +6*x^11 +46*x^10 +26*x^9 +18*x^8 +26*x^7 +46*x^6 +6*x^5 +25*x^4 +8*x^3 +3*x^2 +2*x +1) / ((x -1)^3*(x +1)^3*(x^2 -x +1)^3*(x^2 +x +1)^3). - Colin Barker, Jul 18 2014
a(n+18) = 3*a(n+12)-3*a(n+6)+a(n). - Robert Israel, Jul 18 2014
a(n) = 2*(n+1)^2 * (7-4*cos(2*Pi*(n+1)/3)) / (9*(3-(-1)^n)). - Vaclav Kotesovec, Jul 20 2014
From Peter Bala, Feb 27 2019: (Start)
The following remarks assume an offset of 1.
a(n) = n^2/gcd(n,6) = n*A060789(n).
a(n) = n^2/b(n), where b(n) is the purely periodic sequence [1,2,3,2,1,6,...] with period 6. Thus a(n) is a quasi-polynomial in n:
a(6*n+1) = (6*n + 1)^2;
a(6*n+2) = 2*(3*n + 1)^2;
a(6*n+3) = 3*(2*n + 1)^2;
a(6*n+4) = 2*(3*n + 2)^2;
a(6*n+5) = (6*n + 5)^2;
a(6*n)   = 6*n^2.
O.g.f.: F(x) - 2*F(x^2) - 6*F(x^3) + 12*F(x^6), where F(x) = x*(1 + x)/(1 - x)^3 is the generating function for the squares. (End)
Sum_{n>=0} 1/a(n) = 55*Pi^2/216. - Amiram Eldar, Sep 27 2022

More terms from Wesley Ivan Hurt, Jul 18 2014

A194767 Denominator of the fourth increasing diagonal of the autosequence of second kind from (-1)^n / (n+1).

Original entry on oeis.org

2, 2, 12, 20, 10, 42, 56, 24, 90, 110, 44, 156, 182, 70, 240, 272, 102, 342, 380, 140, 462, 506, 184, 600, 650, 234, 756, 812, 290, 930, 992, 352, 1122, 1190, 420, 1332, 1406, 494, 1560, 1640, 574, 1806, 1892, 660, 2070, 2162, 752, 2352, 2450, 850, 2652, 2756, 954, 2970, 3080, 1064, 3306, 3422, 1180, 3660

Offset: 0

Paul Curtz, Sep 02 2011

The autosequence of first kind from (-1)^n/(n+1) is A189733.
For the second kind (the second increasing diagonal is (-1)^n/(n+1), half of the main one):
      2,     1,     0,  -1/2,  -1/3,   1/6,   1/2,  5/12,
     -1,    -1,  -1/2,   1/6,   1/2,   1/3, -1/12, -7/20,
      0,   1/2,   2/3,   1/3,  -1/6, -5/12, -4/15,  1/12,
    1/2,   1/6,  -1/3,  -1/2,  -1/4,  3/20,  7/20, 13/60,
   -1/3,  -1/2,  -1/6,   1/4,   2/5,   1/5, -2/15, -3/10,
   -1/6,   1/3,  5/12,  3/20,  -1/5,  -1/3,  -1/6,  5/42,
    1/2,  1/12, -4/15, -7/20, -2/15,   1/6,   2/7,   1/7,
  -5/12, -7/20, -1/12, 13/60,  3/10,  5/42,  -1/7,  -1/4.
Main diagonal: (period 2:repeat 2, -1)/A026741(n+1).
Second (increasing) diagonal: (-1)^n / (n+1).
Third (increasing) diagonal: (-1)^(n+1)*A026741(n) / A045896(n).
Fourth (increasing) diagonal: (-1)^(n+1)*A146535(n)/ a(n).

OEIS Wiki, Autosequence.
Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-3,0,0,1).

Cf. A001504 = 2*A060544, A049450 = 2*A000326, A045945 = 6*A005449.

Cf. A026741, A045896, A051176, A146535, A189733, A306368.

c = Table[1/9 (7 n + 7 n^2 + 2 n Cos[2 n *Pi/3] + 2 n^2 Cos[2 n *Pi/3] + 2 Sqrt[3] n Sin[2 n *Pi/3] + 2 Sqrt[3] n^2 Sin[2 n *Pi/3]), {n, 1, 50}] (* Roger Bagula, Mar 25 2012 *)
a[n_] := (n+1) * Numerator[(n+2)/3]; Array[a, 60, 0] (* Amiram Eldar, Sep 17 2023 *)
LinearRecurrence[{0,0,3,0,0,-3,0,0,1},{2,2,12,20,10,42,56,24,90},60] (* Harvey P. Dale, May 15 2025 *)

a(3*n) = (3*n+1)*(3*n+2), a(3*n+1) = (n+1)*(3*n+2), a(3*n+2) = 3*(n+1)*(3*n+4).
G.f.: 2*(1+x+6*x^2+7*x^3+2*x^4+3*x^5+x^6)/(1-x^3)^3. - Jean-François Alcover, Nov 11 2016
a(n+2) = 2 * A306368(n) for n >= 0. - Joerg Arndt, Aug 25 2023
a(n) = (n+1) * A051176(n+2) for n >= 0. - Paul Curtz, Sep 13 2023
Sum_{n>=0} 1/a(n) = 1 + log(3) - Pi/(3*sqrt(3)). - Amiram Eldar, Sep 17 2023

A214392 If n mod 4 = 0 then a(n) = n/4, otherwise a(n) = n.

Original entry on oeis.org

0, 1, 2, 3, 1, 5, 6, 7, 2, 9, 10, 11, 3, 13, 14, 15, 4, 17, 18, 19, 5, 21, 22, 23, 6, 25, 26, 27, 7, 29, 30, 31, 8, 33, 34, 35, 9, 37, 38, 39, 10, 41, 42, 43, 11, 45, 46, 47, 12, 49, 50, 51, 13, 53, 54, 55, 14, 57, 58

Offset: 0

Jeremy Gardiner, Jul 15 2012

Equivalent to A065883 for n mod 16 != 0. - Peter Kagey, Sep 02 2015

			a(16) = 16/4 = 4;
a(17) = 17.

Jeremy Gardiner, 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).

Cf. A026741, A051176, A186646, A065883, A038754.

Table[If[Mod[n, 4] == 0, n/4, n], {n, 0, 50}] (* G. C. Greubel, Oct 26 2017 *)
LinearRecurrence[{0,0,0,2,0,0,0,-1},{0,1,2,3,1,5,6,7},60] (* Harvey P. Dale, Mar 30 2018 *)

a(n)=if(n%4,n,n/4) \\ Charles R Greathouse IV, Oct 16 2015

From Bruno Berselli, Oct 16 2012: (Start)
G.f.: x*(1+2*x+3*x^2+x^3+3*x^4+2*x^5+x^6)/(1-x^4)^2.
a(n) = ( 1 - (3/16)*(1+(-1)^n)*(1+i^(n(n+1))) )*n, where i=sqrt(-1).
a(n) = a(-n) = 2*a(n-4) - a(n-8). (End)
From Werner Schulte, Jul 08 2018: (Start)
a(n) for n > 0 is multiplicative with a(2^e) = 2^e if e < 2 and a(2^e) = 2^(e-2) if e > 1 otherwise a(p^e) = p^e for prime p > 2 and e >= 0.
Dirichlet g.f.: Sum_{n>0} a(n)/n^s = (1-3/4^s)*zeta(s-1).
Dirichlet inverse b(n) is multiplicative with b(2^e) = (-1)^e * A038754(e), e >= 0, and for prime p > 2: b(p) = -p and b(p^e) = 0 if e > 1. (End)
Sum_{k=1..n} a(k) ~ (13/32) * n^2. - Amiram Eldar, Nov 28 2022

A106616 Numerator of n/(n+15).

Original entry on oeis.org

0, 1, 2, 1, 4, 1, 2, 7, 8, 3, 2, 11, 4, 13, 14, 1, 16, 17, 6, 19, 4, 7, 22, 23, 8, 5, 26, 9, 28, 29, 2, 31, 32, 11, 34, 7, 12, 37, 38, 13, 8, 41, 14, 43, 44, 3, 46, 47, 16, 49, 10, 17, 52, 53, 18, 11, 56, 19, 58, 59, 4, 61, 62, 21, 64, 13, 22, 67, 68, 23, 14, 71, 24, 73, 74, 5, 76, 77, 26

Offset: 0

N. J. A. Sloane, May 15 2005

Multiplicative and also a strong divisibility sequence: gcd(a(n),a(m)) = a(gcd(n,m)) for n, m >= 1. - Peter Bala, Feb 24 2019

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

Cf. A000010, A106614, A106615, A106617, A106618, A106619, A106620, A106621.

Cf. Other sequences given by the formula numerator(n/(n + k)): A026741 (k = 2), A051176 (k = 3), A060819 (k = 4), A060791 (k = 5), A060789 (k = 6), A106608 thru A106612 (k = 7 thru 11), A051724 (k = 12), A106614 thru A106621 (k = 13 thru 20).

List([0..80],n->NumeratorRat(n/(n+15))); # Muniru A Asiru, Feb 19 2019

[Numerator(n/(n+15)): n in [0..100]]; // Vincenzo Librandi, Apr 18 2011

seq(numer(n/(n+15)),n=0..100); # Nathaniel Johnston, Apr 18 2011

f[n_]:=Numerator[n/(n+15)]; Array[f,100,0] (* Vladimir Joseph Stephan Orlovsky, Feb 17 2011*)

a(n) = numerator(n/(n+15)); \\ Michel Marcus, Feb 19 2017

[lcm(n,15)/15 for n in range(0, 79)] # Zerinvary Lajos, Jun 09 2009

Dirichlet g.f.: zeta(s-1)*(1-4/5^s-2/3^s+8/15^s). - R. J. Mathar, Apr 18 2011
a(n) = gcd((n-2)*(n-1)*n*(n+1)*(n+2)/15, n) for n>=1. - Lechoslaw Ratajczak, Feb 19 2017
From Peter Bala, Feb 24 2019: (Start)
a(n) = n/gcd(n,15), a quasi-polynomial in n since gcd(n,15) is a purely periodic sequence of period 15.
O.g.f.: F(x) - 2*F(x^3) - 4*F(x^5) + 8*F(x^15), where F(x) = x/(1 - x)^2.
O.g.f. for reciprocals: Sum_{n >= 1} x^n/a(n) = Sum_{d divides 15} (phi(d)/d) * log(1/(1 - x^d)) = log(1/(1 - x)) + (2/3)*log(1/(1 - x^3)) + (4/5)*log(1/(1 - x^5)) + (8/15)*log(1/(1 - x^15)), where phi(n) denotes the Euler totient function A000010. (End)
From Amiram Eldar, Nov 25 2022: (Start)
Multiplicative with a(3^e) = 3^max(0,e-1), a(5^e) = 5^max(0,e-1), and a(p^e) = p^e otherwise.
Sum_{k=1..n} a(k) ~ (49/150) * n^2. (End)

A194880 The numerators of the inverse Akiyama-Tanigawa algorithm from A001045(n).

Original entry on oeis.org

0, -1, -1, -4, -5, -2, -7, -8, -3, -10, -11, -4, -13, -14, -5, -16, -17, -6, -19, -20, -7, -22, -23, -8, -25, -26, -9, -28, -29, -10, -31, -32, -11, -34, -35, -12, -37, -38, -13, -40, -41, -14, -43, -44, -15, -46, -47, -16, -49, -50, -17, -52, -53, -18, -55, -56, -19, -58, -59, -20

Offset: 0

Paul Curtz, Sep 07 2011

0,  -1, -1, -4/3, -5/3, -2, -7/3, -8/3, -3, -10/3, -11/3, -4, -13/4, -14/3, -5,    = a(n)/b(n),
1,    0,  1,  4/3,  5/3,  2,  7/3,  8/3,  3,
1,   -2, -1, -4/3, -5/3, -2, -7/3, -8/3, -3,
3,   -2,  1,  4/3,  5/3,  2,  7/3,  8/3,  3,
5,   -6, -1, -4/3, -5/3, -2, -7/3, -8/3, -3,
11, -10,  1,  4/3,  5/3,  2,  7/3,  8/3,  3,
21, -22, -1, -4/3, -5/3, -2, -7/3, -8/3, -3,
Vertical: A001045(n), -A078008(n), (-1)^(n+1)*A000012(n), (-1)^(n+1)*A010709(n)/A010701(n), (-1)^(n+1)*A010716(n+1)/A010701(n), A007395(n), .. .
a(n)=0, 1 before (-A145064(n+1)=-A051176(n+3).
b(n)=1, 1 before A169609(n). b(n)=1, 1, 1 before A144437(n+1).
a(n+5)-a(n+2)=b(n+5)  (=-1,-3,-3,=-A169609(n)).

G. C. Greubel, Table of n, a(n) for n = 0..5000
D. Merlini, R. Sprugnoli, and M. C. Verri, The Akiyama-Tanigawa Transformation, Integers, 5 (1) (2005) #A05.
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

a[0]=0; a[1]=-1; a[n_] := (-n-1)/Max[1, 2*Mod[n, 3]-1]; Table[a[n], {n, 0, 59}] (* Jean-François Alcover, Sep 18 2012 *)

a(3*n)=-3*n-1 except a(0)=0; a(3*n+1)=-3*n-2 except a(1)=-1; a(3*n+2)=-n-1.
From Chai Wah Wu, May 07 2024: (Start)
a(n) = 2*a(n-3) - a(n-6) for n > 7.
G.f.: x*(x^6 + x^5 - 3*x^3 - 4*x^2 - x - 1)/(x^6 - 2*x^3 + 1). (End)

A283971 a(n) = n except a(4n + 2) = 2n + 1.

Original entry on oeis.org

0, 1, 1, 3, 4, 5, 3, 7, 8, 9, 5, 11, 12, 13, 7, 15, 16, 17, 9, 19, 20, 21, 11, 23, 24, 25, 13, 27, 28, 29, 15, 31, 32, 33, 17, 35, 36, 37, 19, 39, 40, 41, 21, 43, 44, 45, 23, 47, 48, 49, 25, 51, 52, 53, 27, 55, 56, 57, 29, 59, 60, 61, 31, 63, 64, 65, 33, 67

Offset: 0

Paul Curtz, Mar 18 2017

From Federico Provvedi, Nov 13 2018: (Start)
For n > 1, a(n) is also the cycle length generated by the cycle lengths of the digital roots, in base n, of the powers of k, with k > 0.
Example for n=10 (decimal base): for every h >= 0, the digital roots of 2^h generate a periodic cycle {1,2,4,8,7,5} with period 6; 3^h generates {1,3,9,9,9,9,...} so the periodic cycle {9} has period 1; 4^h generates the periodic cycle {1,4,7} with period 3; etc. So, for n=10 (decimal base representation) the sequence generated by the periods of the digital roots of powers of k (with k > 0) is also periodic {1,6,1,3,6,1,3,2,1} with period 9, hence a(10) = 9. (End)

Colin Barker, 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).

Cf. A005408, A010121, A022998, A051176, A060819, A186646.

a:=[0,1,1,3,4,5,3,7];; for n in [9..85] do a[n]:=2*a[n-4]-a[n-8]; od; a; # Muniru A Asiru, Jul 20 2018

seq(coeff(series(x*(1+x+3*x^2+4*x^3+3*x^4+x^5+x^6)/((1-x)^2*(1+x)^2*(1+x^2)^2), x,n+1),x,n),n=0..80); # Muniru A Asiru, Jul 20 2018

Table[If[Mod[n, 4] == 2, (n - 2)/2 + 1, n], {n, 67}] (* or *)
CoefficientList[Series[x (1 + x + 3 x^2 + 4 x^3 + 3 x^4 + x^5 + x^6)/((1 - x)^2*(1 + x)^2*(1 + x^2)^2), {x, 0, 67}], x] (* Michael De Vlieger, Mar 19 2017 *)
LinearRecurrence[{0, 0, 0, 2, 0, 0, 0, -1}, {0, 1, 1, 3, 4, 5, 3, 7}, 70] (* Robert G. Wilson v, Jul 23 2018 *)
Table[Length[FindTransientRepeat[(Length[FindTransientRepeat[Mod[#1^Range[b]-1,b-1]+1,2][[2]]]&)/@Range[2, 2*b], 2][[2]]], {b, 2, 100}] (* Federico Provvedi, Nov 13 2018 *)

a(n)=if(n%4==2, n\4*2 + 1, n) \\ Charles R Greathouse IV, Mar 18 2017

concat(0, Vec(x*(1 + x + 3*x^2 + 4*x^3 + 3*x^4 + x^5 + x^6) / ((1 - x)^2*(1 + x)^2*(1 + x^2)^2) + O(x^40))) \\ Colin Barker, Mar 19 2017

def A283971(n): return n if (n-2)&3 else n>>1 # Chai Wah Wu, Jan 10 2023

a(2*n) = A022998(n), a(1+2*n) = 1 + 2*n.
a(n) = 2*a(n-4) - a(n-8).
From Colin Barker, Mar 19 2017: (Start)
G.f.: x*(1 + x + 3*x^2 + 4*x^3 + 3*x^4 + x^5 + x^6) / ((1 - x)^2*(1 + x)^2*(1 + x^2)^2).
a(n) = -((-1)^n - (-i)^n - i^n - 7)*n/8, where i = sqrt(-1).
(End)
a(n) = A060819(n) * periodic sequence of length 4: repeat [4, 1, 1, 1].
a(n) = a(n-4) + periodic sequence of length 4: repeat [4, 4, 2, 4].
From Werner Schulte, Jul 08 2018: (Start)
For n > 0, a(n) is multiplicative with a(p^e) = p^e for prime p >= 2 and e >= 0 except a(2^1) = 1.
Dirichlet g.f.: (1 - 1/2^s - 1/2^(2*s-1)) * zeta(s-1).
(End)
a(n) = n*(7 + cos(n*Pi/2) - cos(n*Pi) + cos(3*n*Pi/2))/8. - Wesley Ivan Hurt, Oct 04 2018
E.g.f.: (1/4)*x*(4*cosh(x) - sin(x) + 3*sinh(x)). - Franck Maminirina Ramaharo, Nov 13 2018
Sum_{k=1..n} a(k) ~ (7/16) * n^2. - Amiram Eldar, Nov 28 2022

A130770 One third of the least common multiple of 3 and n^2+n+1.

Original entry on oeis.org

1, 1, 7, 13, 7, 31, 43, 19, 73, 91, 37, 133, 157, 61, 211, 241, 91, 307, 343, 127, 421, 463, 169, 553, 601, 217, 703, 757, 271, 871, 931, 331, 1057, 1123, 397, 1261, 1333, 469, 1483, 1561, 547, 1723, 1807, 631, 1981, 2071, 721, 2257, 2353, 817, 2551, 2653, 919

Offset: 0

W. Neville Holmes, Jul 14 2007

This is a subset of A051176 and is also one third of A130723.

			a(4)=7 because 4^2+4+1 =21, the LCM of 3 and 21 is 21 and 21/3=7.

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

Cf. A051176, A130723.

[Lcm(3,n^2+n+1)/3: n in [0..50]]; // G. C. Greubel, Oct 26 2017

seq(denom((n-1)^2/(n^2+n+1)), n=0..52) ; # Zerinvary Lajos, Jun 04 2008

Table[LCM[3,n^2+n+1]/3,{n,0,60}] (* or *) LinearRecurrence[ {0,0,3,0,0,-3,0,0,1},{1,1,7,13,7,31,43,19,73},60] (* Harvey P. Dale, Apr 10 2014 *)

for(n=0,50, print1(lcm(3, n^2 + n +1)/3, ", ")) \\ G. C. Greubel, Oct 26 2017

Conjecture: a(n) = A046163(n), n>0. - R. J. Mathar, Jun 13 2008

a(n) = 3*a(n-3)-3*a(n-6)+a(n-9), with a(0)=1, a(1)=1, a(2)=7, a(3)=13, a(4)=7, a(5)=31, a(6)=43, a(7)=19, a(8)=73. - Harvey P. Dale, Apr 10 2014

cp's OEIS Frontend

A106448 Table of (x+y)/gcd(x,y) where (x,y) runs through the pairs (1,1), (1,2), (2,1), (1,3), (2,2), (3,1), ...

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A106610 Numerator of n/(n+9).

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A106618 a(n) = numerator of n/(n+17).

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A129202 Denominator of 3*(3+(-1)^n) / (n+1)^2.

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A194767 Denominator of the fourth increasing diagonal of the autosequence of second kind from (-1)^n / (n+1).

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A214392 If n mod 4 = 0 then a(n) = n/4, otherwise a(n) = n.

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A106616 Numerator of n/(n+15).

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A283971 a(n) = n except a(4n + 2) = 2n + 1.