A060789 -id:A060789 - cp's OEIS frontend

A106610 Numerator of n/(n+9).

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

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)

A262343 Numerator of 3*(1-2/n), for n >= 3.

Original entry on oeis.org

1, 3, 9, 2, 15, 9, 7, 12, 27, 5, 33, 18, 13, 21, 45, 8, 51, 27, 19, 30, 63, 11, 69, 36, 25, 39, 81, 14, 87, 45, 31, 48, 99, 17, 105, 54, 37, 57, 117, 20, 123, 63, 43, 66, 135, 23, 141, 72, 49, 75, 153, 26, 159, 81, 55, 84, 171, 29, 177, 90, 61, 93, 189, 32

Offset: 3

Kival Ngaokrajang, Sep 18 2015

Given a regular n-gon with side length s, draw a circular arc of radius s around each of the n-gon's vertices so as to connect that vertex's two nearest neighbors, drawing the arc on the shorter side of the circle; i.e., each arc will extend through an angle of Pi*(n-2)/n radians (see illustration). Connect the n arcs thus drawn into a single closed curve if n is odd, or into a pair of identical (but with one rotated by 2*Pi/n radians with respect to the other) overlapping closed curves if n is even. The arcs and the curve (or pair of curves) have the following properties:
(i) Since the length L(n) of each single arc is L(n) = s*Pi*(n-2)/n, the ratio of the length of a single arc for an n-gon to the length of a single arc for the n=3 case is L(n)/L(3) = (s*Pi*(n-2)/n)/(s*Pi*(3-2)/3) = 3(1-2/n). The numerator and denominator of 3(1-2/n) are a(n) and A060789(n) respectively.
(ii) Since the loop length (considering only one of the two loops when there are two overlapping loops) is L(n)*n when n is odd, or L(n)*n/2 when n is even, the ratio of the loop length for an n-gon to the loop length for the n=3 case is (L(n)*n)/(L(3)*3) = (s*Pi*(n-2))/(s*Pi) = n-2 when n is odd, or (L(n)*n/2)/(L(3)*3) = (s*Pi*(n-2)/2)/(s*Pi) = (n-2)/2 when n is even; thus, whether odd or even, that ratio is numerator(1-2/n) = A026741(n-2).
The moment generating function of p(x, m=1, n=2, mu=2) = 3*x*E(x, 1, 2), see A163931 and A274181, is given by M(a) = (3*a-6)/(a^2*(a-1)) + 6*log(1-a)/a^3. The series expansion of M(a) leads to the sequence given above. - Johannes W. Meijer, Jul 04 2016

Peter Kagey, Table of n, a(n) for n = 3..10000
Kival Ngaokrajang, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,2,0,0,0,0,0,-1).

Cf. A060789, A026741.

[Numerator(3*(1-2/n)): n in [3..80]]; // Vincenzo Librandi, Sep 19 2015

a:= proc(n): numer(3*(n-2)/n) end: seq(a(n), n=3..66); # Johannes W. Meijer, Jul 03 2016

Table[Numerator[3 (1 - 2/n)], {n, 3, 60}] (* Michael De Vlieger, Sep 18 2015 *)

{for(n=3, 100, a=numerator(3*(1-2/n)); print1 (a, ", "))}

Vec(x^3*(3*x^10+x^9+9*x^8+6*x^7+5*x^6+9*x^5+15*x^4+2*x^3+9*x^2+3*x+1)/((x-1)^2*(x+1)^2*(x^2-x+1)^2*(x^2+x+1)^2) + O(x^100)) \\ Colin Barker, Sep 20 2015

a(n) = numerator(3*(1-2/n)), for n >= 3.
From Peter Kagey, Sep 18 2015: (Start)
For integers k:
a(6k+0) =  3 * k - 1
a(6k+1) = 18 * k - 3
a(6k+2) =  9 * k + 1
a(6k+3) =  6 * k + 1
a(6k+4) =  9 * k + 3
a(6k+5) = 18 * k + 9
(End)
From Colin Barker, Sep 20 2015: (Start)
a(n) = 2*a(n-6) - a(n-12).
G.f.: x^3*(3*x^10+x^9+9*x^8+6*x^7+5*x^6+9*x^5+15*x^4+2*x^3+9*x^2+3*x+1) / ((x-1)^2*(x+1)^2*(x^2-x+1)^2*(x^2+x+1)^2).
(End)

More terms from Vincenzo Librandi, Sep 19 2015

A367824 Array read by ascending antidiagonals: A(n, k) is the numerator of (R(n) - k)/(n + k), where R(n) is the digit reversal of n, with A(0, 0) = 1.

Original entry on oeis.org

1, 1, -1, 1, 0, -1, 1, 1, -1, -1, 1, 1, 0, -1, -1, 1, 3, 1, -1, -3, -1, 1, 2, 1, 0, -1, -2, -1, 1, 5, 3, 1, -1, -3, -5, -1, 1, 3, 1, 1, 0, -1, -1, -3, -1, 1, 7, 5, 1, 1, -1, -1, -5, -7, -1, 1, 4, 3, 2, 1, 0, -1, -2, -3, -4, -1, 1, 0, 7, 5, 3, 1, -1, -3, -5, -7, -9, -1

Offset: 0

Stefano Spezia, Dec 02 2023

This array generalizes A367727.

			The array of the fractions begins:
  1,  -1,   -1,   -1,   -1,   -1,    -1,    -1, ...
  1,   0, -1/3, -1/2, -3/5, -2/3,  -5/7,  -3/4, ...
  1, 1/3,    0, -1/5, -1/3, -3/7,  -1/2,  -5/9, ...
  1, 1/2,  1/5,    0, -1/7, -1/4,  -1/3,  -2/5, ...
  1, 3/5,  1/3,  1/7,    0, -1/9,  -1/5, -3/11, ...
  1, 2/3,  3/7,  1/4,  1/9,    0, -1/11,  -1/6, ...
  1, 5/7,  1/2,  1/3,  1/5, 1/11,     0, -1/13, ...
  1, 3/4,  5/9,  2/5, 3/11,  1/6,  1/13,     0, ...
  ...
The array of the numerators begins:
  1, -1, -1, -1, -1, -1, -1, -1, ...
  1,  0, -1, -1, -3, -2, -5, -3, ...
  1,  1,  0, -1, -1, -3, -1, -5, ...
  1,  1,  1,  0, -1, -1, -1, -2, ...
  1,  3,  1,  1,  0, -1, -1, -3, ...
  1,  2,  3,  1,  1,  0, -1, -1, ...
  1,  5,  1,  1,  1,  1,  0, -1, ...
  1,  3,  5,  2,  3,  1,  1,  0, ...
  ...

Stefano Spezia, First 151 antidiagonals of the array

Cf. A367825 (denominator), A367826 (antidiagonal sums).

Cf. A004086, A026741, A051724, A060789, A060819, A106609, A106611, A106612, A106617, A106619, A153881 (n=0), A231190, A367727 (k=1).

A[0,0]=1; A[n_,k_]:=Numerator[(FromDigits[Reverse[IntegerDigits[n]]]-k)/(n+k)]; Table[A[n-k,k],{n,0,11},{k,0,n}]//Flatten

A(1, n) = -A026741(n-1) for n > 0.
A(2, n) = -A060819(n-2) for n > 2.
A(3, n) = -A060789(n-3) for n > 3.
A(4, n) = -A106609(n-4) for n > 3.
A(5, n) = -A106611(n-5) for n > 4.
A(6, n) = -A051724(n-6) for n > 5.
A(7, n) = -A106615(n-7) for n > 6.
A(8, n) = -A106617(n-8) = A231190(n) for n > 7.
A(9, n) = -A106619(n-9) for n > 8.
A(10, n) = -A106612(n-10) for n > 9.

A367825 Array read by ascending antidiagonals: A(n, k) is the denominator of (R(n) - k)/(n + k), where R(n) is the digit reversal of n, with A(0, 0) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 2, 1, 2, 1, 1, 5, 5, 5, 5, 1, 1, 3, 3, 1, 3, 3, 1, 1, 7, 7, 7, 7, 7, 7, 1, 1, 4, 2, 4, 1, 4, 2, 4, 1, 1, 9, 9, 3, 9, 9, 3, 9, 9, 1, 10, 5, 5, 5, 5, 1, 5, 5, 5, 5, 1, 1, 1, 11, 11, 11, 11, 11, 11, 11, 11, 11, 1, 4, 6, 12, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1

Offset: 0

Stefano Spezia, Dec 02 2023

This array generalizes A367728.

			The array of the fractions begins:
  1,  -1,   -1,   -1,   -1,   -1,    -1,    -1, ...
  1,   0, -1/3, -1/2, -3/5, -2/3,  -5/7,  -3/4, ...
  1, 1/3,    0, -1/5, -1/3, -3/7,  -1/2,  -5/9, ...
  1, 1/2,  1/5,    0, -1/7, -1/4,  -1/3,  -2/5, ...
  1, 3/5,  1/3,  1/7,    0, -1/9,  -1/5, -3/11, ...
  1, 2/3,  3/7,  1/4,  1/9,    0, -1/11,  -1/6, ...
  1, 5/7,  1/2,  1/3,  1/5, 1/11,     0, -1/13, ...
  1, 3/4,  5/9,  2/5, 3/11,  1/6,  1/13,     0, ...
  ...
The array of the denominators begins:
  1, 1, 1, 1,  1,  1,  1,  1, ...
  1, 1, 3, 2,  5,  3,  7,  4, ...
  1, 3, 1, 5,  3,  7,  2,  9, ...
  1, 2, 5, 1,  7,  4,  3,  5, ...
  1, 5, 3, 7,  1,  9,  5, 11, ...
  1, 3, 7, 4,  9,  1, 11,  6, ...
  1, 7, 2, 3,  5, 11,  1, 13, ...
  1, 4, 9, 5, 11,  6, 13,  1, ...
  ...

Stefano Spezia, First 151 antidiagonals of the array

Cf. A367824 (numerator), A367827 (antidiagonal sums).

Cf. A000012 (n=0), A004086, A026741, A051724, A060789, A060819, A106609, A106611, A106612, A106615, A106617, A231190, A367728 (k=1).

A[0,0]=1; A[n_,k_]:=Denominator[(FromDigits[Reverse[IntegerDigits[n]]]-k)/(n+k)]; Table[A[n-k,k],{n,0,12},{k,0,n}]//Flatten

A(1, n) = A026741(n+1).
A(2, n) = A060819(n+2).
A(3, n) = A060789(n+3).
A(4, n) = A106609(n+4).
A(5, n) = A106611(n+5).
A(6, n) = A051724(n+6).
A(7, n) = A106615(n+7).
A(8, n) = A106617(n+8) = A231190(n+16).
A(9, n) = A106619(n+9).
A(10, n) = A106612(n+10).

A168061 Denominator of (n+3) / ((n+2) * (n+1) * n).

Original entry on oeis.org

3, 24, 10, 120, 105, 112, 252, 720, 165, 1320, 858, 728, 1365, 3360, 680, 4896, 2907, 2280, 3990, 9240, 1771, 12144, 6900, 5200, 8775, 19656, 3654, 24360, 13485, 9920, 16368, 35904, 6545, 42840, 23310, 16872, 27417, 59280, 10660, 68880, 37023, 26488, 42570

Offset: 1

Vladimir Joseph Stephan Orlovsky, Nov 17 2009

Numerator of ((n+3)/(n+2)/(n+1)/n) = A060789(n).

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,4,0,0,0,0,0,-6,0,0,0,0,0,4,0,0,0,0,0,-1).

Cf. A060789.

List([1..10^3],n->DenominatorRat((n+3)/(n+2)/(n+1)/n)); # Muniru A Asiru, Feb 04 2018

seq(denom((n+3)/(n+2)/(n+1)/n), n=1..10^3); # Muniru A Asiru, Feb 04 2018

Table[Denominator[(n+3)/(n+2)/(n+1)/n],{n,60}]
LinearRecurrence[{0,0,0,0,0,4,0,0,0,0,0,-6,0,0,0,0,0,4,0,0,0,0,0,-1},{3,24,10,120,105,112,252,720,165,1320,858,728,1365,3360,680,4896,2907,2280,3990,9240,1771,12144,6900,5200},50] (* Harvey P. Dale, Apr 06 2017 *)

vector(50, n, denominator(((n+3)/(n+2)/(n+1)/n))) \\ Colin Barker, Feb 04 2018

Vec(x*(3 + 24*x + 10*x^2 + 120*x^3 + 105*x^4 + 112*x^5 + 240*x^6 + 624*x^7 + 125*x^8 + 840*x^9 + 438*x^10 + 280*x^11 + 375*x^12 + 624*x^13 + 80*x^14 + 336*x^15 + 105*x^16 + 40*x^17 + 30*x^18 + 24*x^19 + x^20) / ((1 - x)^4*(1 + x)^4*(1 - x + x^2)^4*(1 + x + x^2)^4) + O(x^60)) \\ Colin Barker, Feb 04 2018

a(n) = 4*a(n-6) -6*a(n-12) +4*a(n-18) -a(n-24) = A007531(n+2)/A089145(n). - R. J. Mathar, Nov 18 2009

G.f.: x*(3 + 24*x + 10*x^2 + 120*x^3 + 105*x^4 + 112*x^5 + 240*x^6 + 624*x^7 + 125*x^8 + 840*x^9 + 438*x^10 + 280*x^11 + 375*x^12 + 624*x^13 + 80*x^14 + 336*x^15 + 105*x^16 + 40*x^17 + 30*x^18 + 24*x^19 + x^20) / ((1 - x)^4*(1 + x)^4*(1 - x + x^2)^4*(1 + x + x^2)^4). - Colin Barker, Feb 04 2018

A168062 Denominators of ((n+3)/(n+2)/(n+1)/n) (sorted with no repeats).

Original entry on oeis.org

3, 10, 24, 105, 112, 120, 165, 252, 680, 720, 728, 858, 1320, 1365, 1771, 2280, 2907, 3360, 3654, 3990, 4896, 5200, 6545, 6900, 8775, 9240, 9920, 10660, 12144, 13485, 16215, 16368, 16872, 19656, 23310, 23426, 24360, 26488, 27417, 32509, 35904

Offset: 1

Vladimir Joseph Stephan Orlovsky, Nov 17 2009

nonn

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

Cf. A060789, A168061.

Take[Union@Table[Denominator[(n+3)/(n+2)/(n+1)/n],{n,200}],120]

A257106 Denominators of the inverse binomial transform of the Bernoulli numbers with B(1)=2/3.

Original entry on oeis.org

1, 3, 6, 2, 10, 6, 42, 6, 30, 2, 22, 6, 2730, 6, 6, 2, 170, 6, 798, 6, 330, 2, 46, 6, 2730, 6, 6, 2, 290, 6, 14322, 6, 510, 2, 2, 6, 1919190, 6, 6, 2, 4510, 6, 1806, 6, 690, 2, 94, 6, 46410, 6, 66, 2, 530, 6, 798, 6, 870, 2, 118, 6, 56786730, 6, 6, 2, 170, 6

Offset: 0

Paul Curtz, Apr 23 2015

nonn

Difference table of Bernoulli numbers with B(1)=2/3:
1,       2/3,    1/6,      0, -1/30,     0,  1/42, 0, ...
-1/3,   -1/2,   -1/6,  -1/30,  1/30,  1/42, -1/42, ...
-1/6,    1/3,   2/15,   1/15, -1/105, -1/21, ...
1/2,    -1/5,  -1/15, -8/105, -4/105, ...
-7/10,  2/15, -1/105,  4/105, ...
5/6,    -1/7,   1/21, ...
-41/42, 2/15, ...
7/6, ...
...
First column: 1, -1/3, -1/6, 1/2, -7/10, 5/6, -41/42, 7/6, -41/30, 3/2, -35/22, 11/6, ... . a(n) is the n-th term of the denominators.
Antidiagonal sums: 1, 1/3, -1/2, 2/3, -5/6, 1, -7/6, 4/3, -3/2, 5/3, -11/6, 2, ... . See A060789(n).
a(2n+2)/a(2n+1) = 2, 5, 7, 5, 11, 455, ... .
By definition, for B(1) = b, the inverse binomial transform is
Bi(b) =       1,  -1 + b, 7/6 - 2*b, -3/2 + 3*b, 59/30 + 4*b, ...
      = A176328(n)/A176591(n) - (-1)^n *n*b.
With Bic(b) = 0, -1/2 + b, 1 - 2*b,  -3/2 + 3*b,   2 + 4*b, ...
            = (-1)^n *(A001477(n)/2 - n*b),
Bi(b) = (-1)^n *(A164555(n)/A027642(n) + A001477(n)/2 - n*b) =
      = A027641(n)/A027642(n) + Bic(b) .

			a(0) = 1-0, a(1) = -1/2 +1/6 = -1/3, a(2) = 1/6 -1/3 = -1/6, a(3) = 0 +1/2.

Cf. A256595, A027641/A027642(n), A164555(n)/A027642(n), A060789, A176328/A176591, A001477, A109007.

max = 66; B[1] = 2/3; B[n_] := BernoulliB[n]; BB = Array[B, max, 0]; a[n_] := Differences[BB, n] // First // Denominator; Table[a[n], {n, 0, max-1}] (* Jean-François Alcover, May 11 2015 *)

def A257106_list(len, B1) :
    T = matrix(QQ, 2*len+1)
    for m in (0..2*len) :
        T[0, m] = bernoulli_polynomial(1, m) if m <> 1 else B1
        for k in range(m-1, -1, -1) :
            T[m-k, k] = T[m-k-1, k+1] - T[m-k-1, k]
    return [denominator(T[k, 0]) for k in (0..len-1)]
A257106_list(66, 2/3) # Peter Luschny, May 09 2015

Conjecture: a(2n+1) = 3 followed by period 3: repeat 2, 6, 6.
Conjecture: a(2n) = A002445(n)/(period 3: repeat 1, 1, 3).
a(n) = A027641(n)/A027642(n) - (-1)^n *n/6.

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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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A106616 Numerator of n/(n+15).

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A262343 Numerator of 3*(1-2/n), for n >= 3.

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A367824 Array read by ascending antidiagonals: A(n, k) is the numerator of (R(n) - k)/(n + k), where R(n) is the digit reversal of n, with A(0, 0) = 1.

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