A109052 -id:A109052 - cp's OEIS frontend

A106612 a(n) = numerator(n/(n+11)).

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 2, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 3, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 4, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 5, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 6, 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 = 11 * 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 (0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,-1).

Cf. A109052, A137564 (differs, e.g., for n=100).

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 A106611 (k = 7 thru 10), A051724 (k = 12), A106614 thru A106621 (k = 13 thru 20).

List([0..80],n->NumeratorRat(n/(n+11))); # Muniru A Asiru, Feb 19 2019

[Numerator(n/(n+11)): n in [0..100]]; // Vincenzo Librandi, Apr 18 2011

seq(numer(n/(n+11)),n=0..80); # Muniru A Asiru, Feb 19 2019

f[n_]:=Numerator[n/(n+11)];Array[f,100,0] (* Vladimir Joseph Stephan Orlovsky, Feb 17 2011 *)
LinearRecurrence[{0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,-1},{0,1,2,3,4,5,6,7,8,9,10,1,12,13,14,15,16,17,18,19,20,21},80] (* Harvey P. Dale, Jul 05 2021 *)

vector(100, n, n--; numerator(n/(n+11))) \\ G. C. Greubel, Feb 19 2019

[lcm(n,11)/11 for n in range(0, 54)] # Zerinvary Lajos, Jun 09 2009

G.f.: x/(1-x)^2 - 10*x^11/(1-x^11)^2. - Paul D. Hanna, Jul 27 2005
a(n) = lcm(n,11)/11.
From R. J. Mathar, Apr 18 2011: (Start)
a(n) = A109052(n)/11.
Dirichlet g.f.: zeta(s-1)*(1-10/11^s). (End)
a(n) = 2*a(n-11) - a(n-22). - G. C. Greubel, Feb 19 2019
From Amiram Eldar, Nov 25 2022: (Start)
Multiplicative with a(11^e) = 11^(e-1), and a(p^e) = p^e if p != 11.
Sum_{k=1..n} a(k) ~ (111/242) * n^2. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = 21*log(2)/11. - Amiram Eldar, Sep 08 2023

A109042 Square array read by antidiagonals: A(n, k) = lcm(n, k) for n >= 0, k >= 0.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 3, 2, 3, 0, 0, 4, 6, 6, 4, 0, 0, 5, 4, 3, 4, 5, 0, 0, 6, 10, 12, 12, 10, 6, 0, 0, 7, 6, 15, 4, 15, 6, 7, 0, 0, 8, 14, 6, 20, 20, 6, 14, 8, 0, 0, 9, 8, 21, 12, 5, 12, 21, 8, 9, 0, 0, 10, 18, 24, 28, 30, 30, 28, 24, 18, 10, 0, 0, 11, 10, 9, 8, 35, 6, 35, 8, 9, 10, 11, 0

Offset: 0

Mitch Harris, Jun 18 2005

			Seen as an array:
  [0] 0, 0,  0,  0,  0,  0,  0,  0,  0,  0, ...
  [1] 0, 1,  2,  3,  4,  5,  6,  7,  8,  9, ...
  [2] 0, 2,  2,  6,  4, 10,  6, 14,  8, 18, ...
  [3] 0, 3,  6,  3, 12, 15,  6, 21, 24,  9, ...
  [4] 0, 4,  4, 12,  4, 20, 12, 28,  8, 36, ...
  [5] 0, 5, 10, 15, 20,  5, 30, 35, 40, 45, ...
  [6] 0, 6,  6,  6, 12, 30,  6, 42, 24, 18, ...
  [7] 0, 7, 14, 21, 28, 35, 42,  7, 56, 63, ...
  [8] 0, 8,  8, 24,  8, 40, 24, 56,  8, 72, ...
  [9] 0, 9, 18,  9, 36, 45, 18, 63, 72,  9, ...
.
Seen as a triangle T(n, k) = lcm(n - k, k).
  [0] 0;
  [1] 0, 0;
  [2] 0, 1,  0;
  [3] 0, 2,  2,  0;
  [4] 0, 3,  2,  3,  0;
  [5] 0, 4,  6,  6,  4,  0;
  [6] 0, 5,  4,  3,  4,  5, 0;
  [7] 0, 6, 10, 12, 12, 10, 6,  0;
  [8] 0, 7,  6, 15,  4, 15, 6,  7, 0;
  [9] 0, 8, 14,  6, 20, 20, 6, 14, 8, 0;

Alois P. Heinz, Antidiagonals n = 0..140, flattened

Rows A000027, A109043, A109044, A109045, A109046, A109047, A109048, A109049, A109050, A109051, A109052, A109053, A006580 (row sums of triangle), A001477 (main diagonal, central terms).

Variants: A003990 is (1, 1) based, A051173 (T(n,k) = lcm(n,k)).

T := (n, k) -> ilcm(n - k, k):
for n from 0 to 9 do seq(T(n, k), k = 0..n) od;  # Peter Luschny, Mar 24 2025

lcm(n, k) = n*k / gcd(n, k) for (n, k) != (0, 0).

cp's OEIS Frontend

A106612 a(n) = numerator(n/(n+11)).

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