A106611 -id:A106611 - 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

A109051 a(n) = lcm(n,10).

Original entry on oeis.org

0, 10, 10, 30, 20, 10, 30, 70, 40, 90, 10, 110, 60, 130, 70, 30, 80, 170, 90, 190, 20, 210, 110, 230, 120, 50, 130, 270, 140, 290, 30, 310, 160, 330, 170, 70, 180, 370, 190, 390, 40, 410, 210, 430, 220, 90, 230, 470, 240, 490, 50, 510, 260, 530, 270, 110, 280

Offset: 0

Mitch Harris, Jun 18 2005

Amiram Eldar, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,-1).
Index entries for sequences related to lcm's.

Cf. A106611, A109013, A109042.

[Lcm(n,10): n in [0..100]]; // Vincenzo Librandi, Apr 18 2011

q:= [seq(10/igcd(i,10),i=1..10)]:
[0,seq(seq((10*i+j)*q[j],j=1..10),i=0..10)]; # Robert Israel, Feb 23 2016

a[n_] := LCM[n, 10]; Array[a, 60, 0] (* Amiram Eldar, Nov 26 2022 *)

a(n) = lcm(n, 10); \\ Michel Marcus, Feb 23 2016

[lcm(n,10)for n in range(0, 57)] # Zerinvary Lajos, Jun 07 2009

a(n) = n*10/gcd(n, 10).
a(n) = 10*n/A109013(n) = 10*A106611(n). - R. J. Mathar, Apr 18 2011
G.f.: 10*x*(1 +x +3*x^2 +2*x^3 +x^4 +3*x^5 +7*x^6 +4*x^7 +9*x^8 +x^9 +9*x^10 +4*x^11 +7*x^12 +3*x^13 +x^14 +2*x^15 +3*x^16 +x^17 +x^18)/(1 -x^10)^2. - Robert Israel, Feb 23 2016
Sum_{k=1..n} a(k) ~ (63/20) * n^2. - Amiram Eldar, Nov 26 2022

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).

A373917 Triangle read by rows: T(n,k) = k*10 mod n, with n >= 1, k >= 0.

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 0, 2, 0, 2, 0, 0, 0, 0, 0, 0, 4, 2, 0, 4, 2, 0, 3, 6, 2, 5, 1, 4, 0, 2, 4, 6, 0, 2, 4, 6, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 10, 8, 6, 4, 2, 0, 10, 8, 6, 4, 2, 0, 10, 7, 4, 1, 11, 8, 5, 2, 12, 9, 6, 3

Offset: 1

Paolo Xausa, Jun 26 2024

Each row n encodes a "division graph" used to determine m mod n (where m is an arbitrary nonnegative integer), using the method described in the Numberphile link (see also example).

			Triangle begins:
  n\k| 0  1  2  3  4  5  6  7  8  9
  ---------------------------------
   1 | 0;
   2 | 0, 0;
   3 | 0, 1, 2;
   4 | 0, 2, 0, 2;
   5 | 0, 0, 0, 0, 0;
   6 | 0, 4, 2, 0, 4, 2;
   7 | 0, 3, 6, 2, 5, 1, 4;
   8 | 0, 2, 4, 6, 0, 2, 4, 6;
   9 | 0, 1, 2, 3, 4, 5, 6, 7, 8;
  10 | 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
  ...
Suppose m = 3714289 and you want to determine m mod 7 (the example shown in the video).
Start with the first digit of m (3) and calculate T(7,3 mod 7) = T(7,3) = 2.
Add it to the next digit of m (7) and calculate T(7,(2+7) mod 7) = T(7,2) = 6.
Add it to the next digit of m (1) and calculate T(7,(6+1) mod 7) = T(7,0) = 0.
Add it to the next digit of m (4) and calculate T(7,(0+4) mod 7) = T(7,4) = 5.
Add it to the next digit of m (2) and calculate T(7,(5+2) mod 7) = T(7,0) = 0.
Add it to the next digit of m (8) and calculate T(7,(0+8) mod 7) = T(7,1) = 3.
Add it to the final digit of m (9) and calculate (3+9) mod 7 = 5, which corresponds to 3714289 mod 7.

John Tyler Rascoe, Rows n = 1..150, flattened
James Grime and Brady Haran, Solving Seven, Numberphile YouTube video, 2024.

Cf. A051127, A106611 (number of distinct terms in each row), A374195 (row sums).

Table[Mod[Range[0, 10*(n-1), 10], n], {n, 15}]

def A373917(n,k): return(k*10%n) # John Tyler Rascoe, Jul 02 2024

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

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A109051 a(n) = lcm(n,10).

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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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A373917 Triangle read by rows: T(n,k) = k*10 mod n, with n >= 1, k >= 0.

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