A188386 -id:A188386 - cp's OEIS frontend

A189383 a(n) = n + [ns/r] + [nt/r]; r=1, s=1/sqrt(2), t=1/sqrt(3).

1, 4, 6, 8, 10, 13, 15, 17, 20, 22, 24, 26, 29, 31, 33, 36, 38, 40, 42, 45, 47, 49, 52, 53, 56, 59, 61, 63, 65, 68, 69, 72, 75, 77, 79, 81, 84, 85, 88, 91, 92, 95, 97, 100, 101, 104, 107, 108, 111, 113, 116, 118, 120, 123, 124, 127, 129, 132, 134, 136, 139, 140, 143, 145, 147, 150, 152, 155, 156, 159, 161, 163, 166, 168, 171, 172, 175, 178, 179, 182, 184, 186, 188, 191

Offset: 1

Clark Kimberling, Apr 21 2011

nonn

This is one of three sequences that partition the positive integers. In general, suppose that r, s, t are positive real numbers for which the sets {i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1} are pairwise disjoint. Let a(n) be the rank of n/r when all the numbers in the three sets are jointly ranked. Define b(n) and c(n) as the ranks of n/s and n/t.  It is easy to prove that
a(n) = n + [n*s/r] + [n*t/r],
b(n) = n + [n*r/s] + [n*t/s],
c(n) = n + [n*r/t] + [n*s/t], where []=floor.
Taking r=1, s=1/sqrt(2), t=1/sqrt(3) gives
a=A189383, b=A189384, c=A189385.

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

Cf. A189384, A189385, A188386, A189361, A189386, A189395.

r=1; s=2^(-1/2); t=3^(-1/2);
a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
c[n_] := n + Floor[n*r/t] + Floor[n*s/t]
Table[a[n], {n, 1, 120}]  (*A189383*)
Table[b[n], {n, 1, 120}]  (*A189384*)
Table[c[n], {n, 1, 120}]  (*A189385*)

A213998 Numerators of the triangle of fractions read by rows: pf(n,0) = 1, pf(n,n) = 1/(n+1) and pf(n+1,k) = pf(n,k) + pf(n,k-1) with 0 < k < n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 11, 1, 1, 7, 13, 25, 1, 1, 9, 47, 77, 137, 1, 1, 11, 37, 57, 87, 49, 1, 1, 13, 107, 319, 459, 223, 363, 1, 1, 15, 73, 533, 743, 341, 481, 761, 1, 1, 17, 191, 275, 1879, 2509, 3349, 4609, 7129, 1, 1, 19, 121, 1207, 1627, 2131, 2761, 3601, 4861, 7381, 1

Offset: 0

Reinhard Zumkeller, Jul 03 2012

T(n,0) = 1;
T(n,1) = A005408(n-1) for n > 0;
T(n,2) = A188386(n-2) for n > 2;
T(n,n-3) = A124837(n-2) for n > 2;
T(n,n-2) = A027612(n-1) for n > 1;
T(n,n-1) = A001008(n) for n > 0;
T(n,n) = 1;
A214075(n,k) = floor(T(n,k) / A213999(n,k)).

			Start of triangle pf with corresponding triangles of numerators and denominators:
. 0:                            1
. 1:                         1    1/2
. 2:                     1     3/2    1/3
. 3:                  1    5/2    11/6    1/4
. 4:              1   7/2    13/3    25/12    1/5
. 5:           1    9/2   47/6    77/12   137/60   1/6
. 6:        1  11/2   37/3    57/4    87/10    49/20    1/7
. 7:     1  13/2  107/6  319/12  459/20   223/20  363/140   1/8
. 8:  1  15/2  73/3  533/12  743/15  341/10   481/35   761/280  1/9,
.
. 0:   numerators     1                          1    denominators
. 1:                1  1                        1  2       A213999
. 2:              1   3  1                     1 2  3
. 3:            1   5  11 1                   1 2 6  4
. 4:          1  7  13  25  1                1 2 3  12 5
. 5:        1  9  47  77 137  1             1 2 6 12  60 6
. 6:      1 11  37 57  87  49  1           1 2 3 4 10  20  7
. 7:    1 13 107 319 459 223 363 1        1 2 6 12 20 20 140 8
. 8:  1 15 73 533 743 341 481 761 1,     1 2 3 12 15 10 35 280 9.

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
Index entries for triangles and arrays related to Pascal's triangle

Cf. A005408, A188386 (columns).
Cf. A001008, A027612, A124837 (diagonals).
Cf. A213999 (denominators).

import Data.Ratio ((%), numerator, denominator, Ratio)
a213998 n k = a213998_tabl !! n !! k
a213998_row n = a213998_tabl !! n
a213998_tabl = map (map numerator) $ iterate pf [1] where
   pf row = zipWith (+) ([0] ++ row) (row ++ [-1 % (x * (x + 1))])
            where x = denominator $ last row

T[, 0] = 1; T[n, n_] := 1/(n + 1);
T[n_, k_] := T[n, k] = T[n - 1, k] + T[n - 1, k - 1];
Table[T[n, k] // Numerator, {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 10 2021 *)

A033931 a(n) = lcm(n,n+1,n+2).

Original entry on oeis.org

6, 12, 60, 60, 210, 168, 504, 360, 990, 660, 1716, 1092, 2730, 1680, 4080, 2448, 5814, 3420, 7980, 4620, 10626, 6072, 13800, 7800, 17550, 9828, 21924, 12180, 26970, 14880, 32736, 17952, 39270, 21420, 46620, 25308, 54834, 29640, 63960, 34440, 74046, 39732, 85140

Offset: 1

Jeff Burch

Also denominator of h(n+2) - h(n-1), where h(n) is the n-th harmonic number Sum_{k=1..n} 1/k, the numerator is A188386. - Reinhard Zumkeller, Jul 04 2012

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-6,0,4,0,-1).
Index entries for sequences related to lcm's.

Cf. A001008, A002805, A067046, A078637, A007947, A188386, A244009.

a033931 n = lcm n (lcm (n + 1) (n + 2))  -- Reinhard Zumkeller, Jul 04 2012

[Numerator((n^3-n)/(n^2+1)): n in [2..50]]; // Vincenzo Librandi, Aug 19 2014

a:= n-> ilcm($n..n+2):
seq(a(n), n=1..50);  # Alois P. Heinz, Jul 18 2025

LCM@@@Partition[Range[50],3,1] (* or *) LinearRecurrence[{0,4,0,-6,0,4,0,-1},{6,12,60,60,210,168,504,360},50] (* Harvey P. Dale, Jun 29 2019 *)

a(n) = lcm(n^2+n,n+2) \\ Charles R Greathouse IV, Sep 30 2016

a(n) = n*(n+1)*(n+2)*[3-(-1)^n]/4.
From Reinhard Zumkeller, Jul 04 2012: (Start)
a(n) = 6 * A067046(n).
A007947(a(n)) = A078637(n). (End)
From Amiram Eldar, Sep 29 2022: (Start)
Sum_{n>=1} 1/a(n) = 1 - log(2) (A244009).
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*log(2) - 2. (End)

A119947 Triangle of numerators in the square of the matrix A[i,j] = 1/i for j <= i, 0 otherwise.

Original entry on oeis.org

1, 3, 1, 11, 5, 1, 25, 13, 7, 1, 137, 77, 47, 9, 1, 49, 29, 19, 37, 11, 1, 363, 223, 153, 319, 107, 13, 1, 761, 481, 341, 743, 533, 73, 15, 1, 7129, 4609, 3349, 2509, 1879, 275, 191, 17, 1, 7381, 4861, 3601, 2761, 2131, 1627, 1207, 121, 19, 1, 83711, 55991, 42131, 32891, 25961

Offset: 1

Wolfdieter Lang, Jul 20 2006

The triangle of the corresponding denominators is A119948. The rationals appear in lowest terms (while in A027446 they are row-wise on the least common denominator).
The triangle with row number i multiplied with the least common multiple (LCM) of its denominators yields A027446.
First column is A001008. - Tilman Neumann, Oct 01 2008
Column 2 is A064169. - Clark Kimberling, Aug 13 2012
Third diagonal (11, 13, 47, ...) is A188386. - Clark Kimberling, Aug 13 2012

			The rationals are [1]; [3/4, 1/4]; [11/18, 5/18, 1/9]; [25/48, 13/48, 7/48, 1/16]; ... See the W. Lang link for more.
From _Clark Kimberling_, Aug 13 2012: (Start)
As a triangle given by f(n,m) = Sum_{h=m..n} 1/h, the first six rows are:
    1
    3    1
   11    5    1
   25   13    7    1
  137   77   47    9    1
   49   29   19   37   11    1
  363  223  153  319  107   13    1
(End)

Wolfdieter Lang, First ten rows and rationals.

Cf. A002024: i appears i times (denominators in row i of the matrix A).
Row sums give A119949. Row sums of the triangle of rationals always give 1.
For the cube of this matrix see the rational triangle A119935/A119932 and A027447; see A027448 for the fourth power.
Cf. A001008, A027446, A064169, A119948, A188386.

f[n_, m_] := Numerator[Sum[1/k, {k, m, n}]]
Flatten[Table[f[n, m], {n, 1, 10}, {m, 1, n}]]
TableForm[Table[f[n, m], {n, 1, 10}, {m, 1, n}]] (* Clark Kimberling, Aug 13 2012 *)

A119947_upto(n)={my(M=matrix(n,n,i,j,(j<=i)/i)^2);vector(n,r,apply(numerator,M[r,1..r]))} \\ M. F. Hasler, Nov 05 2019

a(i,j) = numerator(r(i,j)) with r(i,j):=(A^2)[i,j], where the matrix A has elements a[i,j] = 1/i if j<=i, 0 if j>i, (lower triangular).

Edited by M. F. Hasler, Nov 05 2019

A192359 Numerator of h(n+6) - h(n), where h(n) = Sum_{k=1..n} 1/k.

Original entry on oeis.org

49, 223, 341, 2509, 2131, 20417, 18107, 30233, 96163, 1959, 36177, 51939, 436511, 598433, 80507, 532541, 1388179, 1785181, 378013, 95003, 1181909, 4370849, 2671363, 3240049, 1560647, 9333997, 5547947, 2185691, 5138581, 1201967, 10493071, 12159157, 28060691, 32250013

Offset: 0

Gary Detlefs, Jun 28 2011

Numerator of (2*n+7)*(3*n^4 + 42*n^3 + 203*n^2 + 392*n + 252)/((n+1)*(n+2)*...*(n+6)).
(2*n+7)*(3*n^4 + 42*n^3 + 203*n^2 + 392*n + 252)/a(n) can be factored into 2^m(n)*3^p(n)*5^(q1(n) + q2(n)) where
m(n) is of period 4, repeating [2,2,3,3]
p(n) is of period 9, repeating [2,2,2,1,1,1,1,1,1]
q1(n) is of period 5, repeating [0,0,0,0,1]
q2(n) is of period 25, repeating [0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0].

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A188386, A189642, A189998, A021913.

List(List([0..35],n->Sum([1..n+6],k->(1/k))-Sum([1..n],k->(1/k))),NumeratorRat); # Muniru A Asiru, Oct 21 2018

[49] cat [Numerator(HarmonicNumber(n+6) - HarmonicNumber(n)): n in [1..40]]; // G. C. Greubel, Oct 20 2018

h:= n-> sum(1/k,k=1..n):seq(numer(h(n+6)-h(n)), n=0..33);
P:=(x,y,z,n)-> floor(((n+x)mod y)/z):
a:=n->(2*n+7)*(3*n^4+42*n^3+203*n^2+392*n+252)/(2^(P(0,4,2,n)+2)*3^(P(6,9,6,n)+1)*5^(P(0,5,4,n)+P(15,25,24,n))):
seq(a(n), n=0..25);

Numerator[Table[HarmonicNumber[n+6]-HarmonicNumber[n],{n,0,40}]] (* Harvey P. Dale, Mar 27 2015 *)

h(n) = sum(k=1, n, 1/k);
a(n) = numerator(h(n+6)-h(n)); \\ Michel Marcus, Apr 15 2017

a(n) = (2*n+7)*(3*n^4 + 42*n^3 + 203*n^2 + 392*n + 252)/(2^(P(0,4,2,n)+2) * 3^(P(6,9,6,n)+1)*5^(P(0,5,4,n)+P(15,25,24,n))), where P(x,y,z,n) = floor(((n+x)mod y)/z).

A192449 Numerator of h(n+7) - h(n), where h(n) = Sum_{k=1..n} 1/k.

Original entry on oeis.org

363, 481, 3349, 2761, 25961, 22727, 263111, 237371, 21635, 8837, 695089, 529331, 9407549, 679829, 641069, 6671911, 36404897, 4075097, 2159257, 1412139, 36516143, 35036093, 88771727, 3715069

Offset: 0

Gary Detlefs, Jul 01 2011

a(n) = numerator((7*n^6 + 168*n^5 + 1610*n^4 + 7840*n^3 + 20307*n^2 + 26264*n + 13068)/((n+1)*(n+2)*...*(n+7)));
(7*n^6 + 168*n^5 + 1610*n^4 + 7840*n^3 + 20307*n^2 + 26264*n + 13068)/a(n) can be factored into 2^m(n)*3^p(n)*5^(q1(n) + q2(n)) where
m(n) is of period 4, repeating [2,4,3,4]
p(n) is of period 9, repeating [2,2,2,1,1,2,1,1,2]
q1(n) is of period 5, repeating [0,0,0,1,1]
q2(n) is of period 25, repeating [0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A188386, A189642, A189998, A192359.

h:= n-> sum(1/k,k=1..n):seq(numer(h(n+7)-h(n)), n=0..23);
P:=(x,y,z,n)-> floor(((n+x) mod y)/z):
m:=n-> P(1,4,3,n)+2*P(0,2,1,n)+2:
p:=n-> P(0,3,2,n)+P(7,9,7,n)+1:
q:=n-> P(0,5,3,n)+P(15,15,23,n):
N7:=n->(7*n^6+168*n^5+1610*n^4+7840*n^3+20307*n^2+26264*n+13068): seq(N7(n)/(2^m(n)*3^p(n)*5^q(n)), n=0..23);
# Alternative implementation, R. J. Mathar, Jul 12 2011:
A192449 := proc(n) add(1/i,i=n+1..n+7) ; numer(%) ; end proc:

#[[8]]-#[[1]]&/@Partition[HarmonicNumber[Range[0,30]],8,1]//Numerator (* Harvey P. Dale, Jul 22 2024 *)

a(n) = (7*n^6 + 168*n^5 + 1610*n^4 + 7840*n^3 + 20307*n^2 + 26264*n + 13068)/ (2^m(n)*3^p(n)*5^q(n)) where
  m(n) = P(1,4,3,n) + 2*P(0,2,1,n) + 2,
  p(n) = P(0,3,2,n) + P(7,9,7,n) + 1,
  q(n) = P(0,5,3,n) + P(15,15,23,n),
  P(x,y,z,n) = floor(((n+x) mod y)/z).

Corrected and extended by Harvey P. Dale, Jul 22 2024

cp's OEIS Frontend

A189383 a(n) = n + [n*s/r] + [n*t/r]; r=1, s=1/sqrt(2), t=1/sqrt(3).

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A213998 Numerators of the triangle of fractions read by rows: pf(n,0) = 1, pf(n,n) = 1/(n+1) and pf(n+1,k) = pf(n,k) + pf(n,k-1) with 0 < k < n.

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A033931 a(n) = lcm(n,n+1,n+2).

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A119947 Triangle of numerators in the square of the matrix A[i,j] = 1/i for j <= i, 0 otherwise.

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A192359 Numerator of h(n+6) - h(n), where h(n) = Sum_{k=1..n} 1/k.

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A192449 Numerator of h(n+7) - h(n), where h(n) = Sum_{k=1..n} 1/k.

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A189383 a(n) = n + [ns/r] + [nt/r]; r=1, s=1/sqrt(2), t=1/sqrt(3).