A106615 -id:A106615 - cp's OEIS frontend

A106614 a(n) = numerator of n/(n+13).

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 2, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 3, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 4, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 5, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75

Offset: 0

N. J. A. Sloane, May 15 2005

In general, the numerators of n/(n+p) for prime p and n >= 0, form a sequence with the g.f.: x/(1-x)^2 - (p-1)*x^p/(1-x^p)^2. - Paul D. Hanna, Jul 27 2005

a(n) <> n iff n = 13 * 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 (0,0,0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,-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), A106615 thru A106621 (k = 14 thru 20).

List([0..80],n->NumeratorRat(n/(n+13))); # Muniru A Asiru, Feb 19 2019

[Numerator(n/(n+13)): n in [0..100]]; // Vincenzo Librandi, Apr 18 2011

seq(numer(n/(n+13)),n=0..80); # Muniru A Asiru, Feb 19 2019

f[n_]:=Numerator[n/(n+13)];Array[f,100,0] (* Vladimir Joseph Stephan Orlovsky, Feb 17 2011 *)

vector(100, n, n--; numerator(n/(n+13))) \\ G. C. Greubel, Feb 19 2019

a(n)=if(n%13,n,n/13) \\ Charles R Greathouse IV, Jan 24 2022

[lcm(n,13)/13for n in range(0, 100)] # Zerinvary Lajos, Jun 09 2009

G.f.: x/(1-x)^2 - 12*x^13/(1-x^13)^2. - Paul D. Hanna, Jul 27 2005
Dirichlet g.f.: zeta(s-1)*(1-12/13^s). - R. J. Mathar, Apr 18 2011
a(n) = 2*a(n-13) - a(n-26). - G. C. Greubel, Feb 19 2019
From Amiram Eldar, Nov 25 2022: (Start)
Multiplicative with a(13^e) = 13^(e-1), and a(p^e) = p^e if p != 13.
Sum_{k=1..n} a(k) ~ (157/338) * n^2. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = 25*log(2)/13. - Amiram Eldar, Sep 08 2023

A106617 Numerator of n/(n+16).

Original entry on oeis.org

0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, 3, 13, 7, 15, 1, 17, 9, 19, 5, 21, 11, 23, 3, 25, 13, 27, 7, 29, 15, 31, 2, 33, 17, 35, 9, 37, 19, 39, 5, 41, 21, 43, 11, 45, 23, 47, 3, 49, 25, 51, 13, 53, 27, 55, 7, 57, 29, 59, 15, 61, 31, 63, 4, 65, 33, 67, 17, 69, 35, 71, 9, 73, 37, 75, 19, 77, 39, 79

Offset: 0

N. J. A. Sloane, May 15 2005

A multiplicative sequence. Also, 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

			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) - 8*G(x^8) - 16*G(x^16), where G(x) = x*(1 + x)/(1 - x)^3.
Sum_{n >= 1} n^2*a(n)*x^n = H(x) - 2^2*H(x^2) - 4^2*H(x^4) - 8^2*H(x^8) - 16^2*H(x^16), where H(x) = x*(1 + 4*x + x^2)/(1 - x)^4. In general, the o.g.f. for Sum_{n >= 1} (n^k*a(n))*x^n for positive k involves the Eulerian polynomials.
In the other direction,
Sum_{n >= 1} (a(n)/n)*x^n = J(x) - (1/2)*J(x^2) - (1/4)*J(x^4) - (1/8)*J(x^8) - (1/16)*J(x^16), where J(x) = x/(1 - x).
Sum_{n >= 1} (a(n)/n^2)*x^n = L(x) - (1/2^2)*L(x^2) - (1/4^2)*L(x^4) - (1/8^2)*L(x^8) - (1/16^2)*L(x^16), where L(x) = log(1/(1 - x)). In general, the o.g.f. for Sum_{n >= 0} (a(n)/n^k)*x^n, for k >= 3, involves the polylogarithm Li_(k-1)(x).
Sum_{n >= 1} (1/a(n))*x^n = L(x) + (1/2)*L(x^2) + (1/2)*L(x^4) + (1/2)*L(x^8) + (1/2)*L(x^16). (End)

Nathaniel Johnston, Table of n, a(n) for n = 0..10000
Peter Bala, A note on the sequence of numerators of a rational function, 2019.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1).

Cf. A106614, A106615, A106616, 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..100],n->NumeratorRat(n/(n+16))); # Muniru A Asiru, Feb 19 2019

[Numerator(n/(n+16)): n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

seq(numer(n/(n+16)), n=0..100); # Nathaniel Johnston, Apr 18 2011

Numerator/@Table[n/(n+16),{n,0,10}] (* Harvey P. Dale, Apr 26 2011 *)

a(n)=n>>min(valuation(n,2),4) \\ Charles R Greathouse IV, Jun 17 2012

[lcm(n,16)/16 for n in range(0, 100)] # Zerinvary Lajos, Jun 12 2009

a(n) = 2*a(n-16) - a(n-32) for n > 31. - Paul Curtz, Apr 12 2011
Octosections: a(8*n) = A026741(n). a(2+8*n) = 1+4*n. a(4+8*n) = 1+2*n. a(6+8*n) = 3+4*n. Bisection: a(1+2*n) = 1+2*n. - Paul Curtz, Apr 12 2011
Dirichlet g.f.: zeta(s-1)*(1-1/2^s-1/4^s-1/8^s-1/16^s). - R. J. Mathar, Apr 18 2011
a(n) = numerator of n/(2^(2*n+1)). - Ralf Steiner, Feb 09 2017
The previous comment is incorrect, a(n) first differs from the numerator of n/(2^(2*n+1)) at n = 32. - Peter Bala, Feb 27 2019
From Peter Bala, Feb 21 2019: (Start)
a(n) = n/gcd(n,16), where gcd(n,16) = [1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 16, ...] is a periodic sequence of period 16: a(n) is thus quasi_polynomial in n.
O.g.f.: Sum_{n >= 0} a(n)*x^n = F(x) - F(x^2) - F(x^4) - F(x^8) - F(x^16), 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) + F(m,x^8) + F(m,x^16) ), 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 produces generating functions for the sequences ( (n^m)*a(n) )n>=1 for m in Z. Some examples are given below. (End)
From Amiram Eldar, Nov 25 2022: (Start)
Multiplicative with a(2^e) = 2^max(0,e-4), and a(p^e) = p^e if p>2.
Sum_{k=1..n} a(k) ~ (171/512) * n^2. (End)

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.

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)

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).

cp's OEIS Frontend

A106614 a(n) = numerator of n/(n+13).

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A106617 Numerator of n/(n+16).

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A106616 Numerator of n/(n+15).

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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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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.

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