A025558 -id:A025558 - cp's OEIS frontend

A099946 a(n) = lcm{1, 2, ..., n}/(n*(n-1)), n >= 2.

1, 1, 1, 3, 2, 10, 15, 35, 28, 252, 210, 2310, 1980, 1716, 3003, 45045, 40040, 680680, 612612, 554268, 503880, 10581480, 9699690, 44618574, 41186376, 114406600, 106234700, 2868336900, 2677114440, 77636318760, 145568097675, 136745788725

Offset: 2

N. J. A. Sloane, Nov 12 2004

Alois P. Heinz, Table of n, a(n) for n = 2..1000 (first 99 terms from Luke March)

Cf. A002378, A002944, A003418, A025558.

a:= n-> ilcm(seq(k,k=1..n))/n/(n-1): seq(a(n), n=2..37); # Emeric Deutsch, Jun 13 2005

Table[LCM@@Range[n]/(n(n-1)), {n,2,40}] (* Harvey P. Dale, Jan 14 2011 *)

a(n) = lcm(vector(n, i, i))/(n*(n-1)); \\ Michel Marcus, Jul 25 2014

from math import gcd
def lcm(a, b):
    return (a * b) // gcd(a, b)
def f(lim):
    l = 1
    for n in range(2, lim + 1):
        l = lcm(n, l)
        print(n, l // (n * (n - 1)))
f(100) # Luke March, Jul 23 2014

a(n) = A003418(n)/(n*(n-1)) = A003418(n)/A002378(n-1), n >= 2.

More terms from Emeric Deutsch, Jun 13 2005

A091342 Given (1) f(h,j,a) = ( [ ((a/gcd(a,h)) / gcd(j+1,(a/gcd(a,h)))) * (h(j+1)) ] - [ ((a/gcd(a,h)) / gcd(j+1,(a/gcd(a,h)))) * (ja) ] ) / a then let (2) a(h) = d(h,j) = lcm( f(h,j,1) ... f(h,j,h) ).

Original entry on oeis.org

1, 3, 10, 105, 252, 2310, 25740, 45045, 680680, 11639628, 10581480, 223092870, 1029659400, 2868336900, 77636318760, 4512611027925, 4247163320400, 4011209802600, 140603459396400, 133573286426580, 5215718803323600

Offset: 1

Scott C. Macfarlan (scottmacfarlan(AT)covance.com), Mar 01 2004

Solves the following arithmetic problem. Let (3) q#(i/j)=(q+h) be defined thus: Given q, an unknown fraction, q=qd/d, of integer value (initially) and h equal to any desired integer value and j equal to any desired integer value and i equal to j(q+h)+h.
Let '#' denote the repeated addition of the denominator j to the denominator of q and the repeated addition of the numerator i to the numerator of q, each addition recursively replacing the prior q fractions numerator and denominator respectively. This results in a series of fractions of mostly non-integer values.
After the first three iterations for any case would result in the following fractions: 1) (qd+i) / (d+j) 2) ((qd+i)+i) / ((d+j)+j) 3) (((qd+i)+i)+i) / (((d+j)+j)+j)
The question is, what is the smallest initial denominator, d (of q=qd/d) that in the course of the repeated additions will result in fractions of integer value, where every integer, from the initial (q+1)...(q+h) will be formed?
For example, let q = 4, h = 5, so (q+h)=9 and j = 1, so that for i we have i = 14. So in terms of q#(i/j)=(q+h) we have 4#(14/1)=9.
In this sample, since h=5 and j=1 we get from (2) above that d(5,1) = 252 as the solution. Applying this solution, we see then that the initial numerator of q, or q*d, becomes 4*252 = 1008 and the initial denominator is 252. Alternatively, in terms of q#(i/j)=(q+h) we have (1008/252)#(14/1)=9. We see that the repeated additions yield:
1) 1008 +14 and 252 +1 ~ 1022/253
2) 1022 +14 and 253 +1 ~ 1036/254
...
27) 1372 +14 and 277 +1 ~ 1386/279
28) 1386 +14 and 278 +1 ~ 1400/280 =5
...
63) 1876 +14 and 314 +1 ~ 1890/315 =6
...
108) 2506 +14 and 359 +1 ~ 2520/360 =7
...
168) 3346 +14 and 419 +1 ~ 3360/420 =8
...
252) 4522 +14 and 503 +1 ~ 4536/504 =9
Note that h=5 and j=1 were chosen so that d(5,1) = a(5) of the sequence. Also note that by the definition of q#(i/j)=(q+h) all answers are only a function of h and j.
Note also that q=qd/d can also be expressed as 0=0d/d and that any q#(i/j)=(q+h) can be expressed as 0#(i/j)=h [after adjustments are made to i]. For instance 4#(14/1)=9 is the equivalent of 0#(10/1)=5 in terms of d(h,j).
Interestingly: For any q#(i/j)=p and r#(s/j)=t then (q+r)#((i+s)/j)=(p+t). Also for any q#(i/j)=p then (qr)#((ir)/j)=(pr).
The sequence A025558 can be calculated from this formula when h = j, in otherwords using the sequence of d(1, 1)...d(n, n). i.e. a(7) = 735 = d(7,7) = lcm(49,21,35,7,21,7,1) a(8) = 2240 = d(8,8) = lcm(64,28,16,10,32,416,1)
Denominator of the sum of all elements of n X n Hilbert Matrix M[i,j] with alternate signs. M[i,j] = 1/(i+j-1)(i,j = 1..n). - Alexander Adamchuk, Apr 11 2006

			a(5) = lcm(9,4,7,3) = 252
a(7) = lcm(13,6,11,5,9,4,1) = 25740
a(10)= lcm(19,9,17,4,3,7,13,3,11,1) = 11639628
a(14)= lcm(27,13,25,6,23,11,3,5,19,9,17,4,15,1) = 2868336900
n=2: HilbertMatrix[n,n]
1 1/2
1/2 1/3
so a(2) = Denominator[(1 - 1/2 - 1/2 + 1/3)] = Denominator[1/3] = 3.
The n X n Hilbert matrix begins:
1 1/2 1/3 1/4 1/5 1/6 1/7 1/8 ...
1/2 1/3 1/4 1/5 1/6 1/7 1/8 1/9 ...
1/3 1/4 1/5 1/6 1/7 1/8 1/9 1/10 ...
1/4 1/5 1/6 1/7 1/8 1/9 1/10 1/11 ...
1/5 1/6 1/7 1/8 1/9 1/10 1/11 1/12 ...
1/6 1/7 1/8 1/9 1/10 1/11 1/12 1/13 ...

Eric Weisstein's World of Mathematics, Hilbert Matrix.
Eric Weisstein's World of Mathematics, Harmonic Number.

Cf. A025558.

Cf. A098118, A086881, A005249, A001008, A002805.

Denominator[Table[Sum[Sum[(-1)^(i+j)*1/(i+j-1),{i,1,n}],{j,1,n}],{n,1,40}]] (* Alexander Adamchuk, Apr 11 2006 *)

a(n) = Denominator[Sum[Sum[(-1)^(i+j)*1/(i+j-1),{i,1,n}],{j,1,n}]]. - Alexander Adamchuk, Apr 11 2006

A097382 a(h) = d(h,j) = lcm( f(h,j,1) ... f(h,j,h) ), when j=2.

Original entry on oeis.org

1, 4, 35, 20, 3003, 560, 692835, 4620, 185910725, 48048, 136745788725, 1361360, 4281195077775, 22170720, 6541380665835015, 446185740, 1898924328582105825, 5949143200, 90048990529077755175

Offset: 1

Scott C Macfarlan (scottmacfarlan(AT)covance.com), Sep 18 2004

nonn

See A091342, A025558

			a(2) = 4 = lcm(4,1)
a(5) = 3003 = lcm(13,11,3,7,1)
a(6) = 560 = lcm(16,7,4,5,8,1)
a(13) = 4281195077775 = lcm(37,35,11,31,29,9,25,23,7,19,17,5,1)

Given (1) f(h, j, a) = ( [ ((a/gcd(a, h)) / gcd(j+1, (a/gcd(a, h)))) * (h(j+1)) ] - [ ((a/gcd(a, h)) / gcd(j+1, (a/gcd(a, h)))) * (ja) ]) / a then let (2) a(h) = d(h, j) = lcm( f(h, j, 1) ... f(h, j, h) )

A098135 a(h) = d(h,j) = lcm( f(h,j,1) ... f(h,j,h) ), when j=3.

Original entry on oeis.org

1, 5, 9, 455, 2618, 315, 271700, 56751695, 6930, 3708514810, 32152414840, 135135, 19749267715100, 314645828225300, 34918884, 113543477411038675, 473930651619825400, 669278610, 467655703593591713200

Offset: 0

Scott C. Macfarlan (scottmacfarlan(AT)covance.com), Sep 27 2004

nonn

			a(2) = 5 = lcm(5,1)
a(3) = 9 = lcm(9,3,1)
a(5) = 2618 = lcm(17,7,11,2,1)
a(11) = 32152414840 = (41,19,35,8,29,13,23,5,17,7,1)

Cf. A091342, A025558, A097382.

Given f(h, j, a) = ( [ ((a/gcd(a, h)) / gcd(j+1, (a/gcd(a, h)))) * (h(j+1)) ] - [ ((a/gcd(a, h)) / gcd(j+1, (a/gcd(a, h)))) * (ja) ]) / a then let a(h) = d(h, j) = lcm( f(h, j, 1) ... f(h, j, h) )

A379561 a(n) = A003418(n+1)*H(n), where H(n) = 1 + 1/2 + ... + 1/n is the n-th harmonic number.

Original entry on oeis.org

2, 9, 22, 125, 137, 1029, 2178, 6849, 7129, 81191, 83711, 1118273, 1145993, 1171733, 2391514, 41421503, 42142223, 813635157, 825887397, 837527025, 848612385, 19761458895, 19994251455, 101086721625, 102157567401, 309561680403, 312536252003, 9146733078187

Offset: 1

Miko Labalan, Dec 26 2024

nonn

abs(log(a(n)) - n - log(log(n))) < c*sqrt(n)*log(n)^(-1/2), where constant c = (2+A206431)*Pi/4. This also gives the upper bound of the squared error, (log(a(n)) - n - log(log(n)))^2 < (c^2)*n*log(n)^(-1).
A slightly better absolute error bound could be achieved by using the imaginary part of the nontrivial zeros of the Riemann zeta function, (zetazero(n)-1/2)/sqrt(-1) ~ (2*Pi)*n*LambertW(n/exp(1))^(-1). That bound would be, abs(log(a(n)) - n - log(log(n))) < sqrt(k)*sqrt(n)*LambertW(n/exp(1))^(-1/2), where constant k = 4*Pi/(1+2*A206431). This also gives the upper bound of the squared error, (log(a(n)) - n - log(log(n)))^2 < k*n*LambertW(n/exp(1))^(-1). The midline of the squared error would run along (4/(4+A206431))*n*LambertW(n/exp(1))^(-1).
Another slightly better absolute error bound but without relying on the properties of the zeta zeros would be, abs(log(a(n)) - n - log(log(n))) < n^(3/(9-10*A077761)).
log(a(n))-c*sqrt(n)*log(n)^(-1/2) is a lower bound of sigma_1(n) = A000203(n). Such that, n+log(log(n))-c*sqrt(n)*log(n)^(-1/2) < sigma_1(n) < H(n)+exp(H(n))*log(H(n)).
a(n) gives the total number of ordered pairs (k,m) where k in set {1,2,...,n}, m in set {1,2,...,A003418(n+1)}, and k divides m. Example: For n = 3, there are 22 ordered pairs (k,m) where k is {1,2,3} and m is a multiple of k up to 12. For k = 1, every m is a multiple of 1, m is {1,2,3,4,5,6,7,8,9,10,11,12} so there are 12 pairs. For k = 2, every m is a multiple of 2, m is {2,4,6,8,10,12} so there are 6 pairs. For k = 3, every m is a multiple of 3, m is {3,6,9,12} so there are 4 pairs. So the total ordered pairs is 12 + 6 + 4 = 22 = a(3). Each ordered pair (k,m) also represents an edge in a bipartite graph. Counting all such pairs gives the total number of edges in a graph.

			a(n)/A025558(n) = [ 2/1, 9/4, 22/9, 125/48, 137/50, 1029/360, 2178/735, ... ]
To evaluate the integral:
For n = 1: Integral_{x=0..1} Li_1(x^(1/2))/x^(1/2) dx = Integral_{x=0..1} -log(1-x^(1/2))/x^(1/2) dx = -2 * -(Sum_{x=1..oo} 1/(x*(x+1))) = -2 * -1 = 2.
For n = 2: Integral_{x=0..1} Li_1(x^(1/3))/x^(1/3) dx = Integral_{x=0..1} -log(1-x^(1/3))/x^(1/3) dx = -3 * -(Sum_{x=1..oo} 1/(x*(x+2))) = -3 * -((1/2)*(1+1/2)) = -3 * -3/4 = 9/4.
For n = 3: Integral_{x=0..1} Li_1(x^(1/4))/x^(1/4) dx = Integral_{x=0..1} -log(1-x^(1/4))/x^(1/4) dx = -4 * -(Sum_{x=1..oo} 1/(x*(x+3))) = -4 * -((1/3)*(1+1/2+1/3)) = -4 * -11/18 = 22/9.

Guilherme França and André LeClair, Statistical and other properties of Riemann zeros based on an explicit equation for the n-th zero on the critical line, arXiv:1307.8395 [math.NT], 2014.
J. C. Lagarias, An elementary problem equivalent to the Riemann hypothesis, Am. Math. Monthly 109 (6) (2002) 534-543. arXiv preprint, arXiv:math/0008177 [math.NT], 2000-2001.

Cf. A001008/A002805 (harmonic numbers).
Cf. A003418 (lcm).
Cf. A025558 (denominator).
Cf. A193758 (very similar sequence).
Cf. A000203, A025529, A027457, A077761, A152648, A206431.

a(n) = lcm(vector(n+1, i, i))*sum(i=1, n, 1/i); \\ Michel Marcus, Dec 28 2024

a(n) = A025558(n)*(Integral_{x=0..1} Li_1(x^(1/(n+1)))/x^(1/(n+1)) dx).
a(n) = A025558(n) + A027457(n+1).
Integral_{x=0..1} Li_1(x^(1/(n+1)))/x^(1/(n+1)) dx = ((n+1)/n)*H(n) = a(n)/A025558(n).
((n+1)/n)*H(n) ~ log(n) + gamma + (log(n)+gamma+1/2)/n + O(1/n^2).
log(a(n)) ~ n + log(log(n)) + O(c*sqrt(n)*log(n)^(-1/2)), (See comments for constant c).
G.f. for ((n+1)/n)*H(n): G(x) = Li_2(x)+(1/2)*log(1-x)^2-log(1-x)/(1-x), the lim_{x->oo} G(x) = -zeta(2).
Hyperbolic l.g.f. for ((n+1)/n)*H(n): LH(x) = Li_2(x)+(1/2)*log(1-x)^2+Li_3(x)-Li_3(1-x)+Li_2(1-x)*log(1-x)+(1/2)*log(x)*log(1-x)^2+zeta(3), the Integral_{x=0..1} LH(x) dx = 2*zeta(3) = A152648.
Dirichlet g.f. for ((n+1)/n)*H(n): zeta(s+1)*(zeta(s)+zeta(s+2)).

cp's OEIS Frontend

A099946 a(n) = lcm{1, 2, ..., n}/(n*(n-1)), n >= 2.

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A091342 Given (1) f(h,j,a) = ( [ ((a/gcd(a,h)) / gcd(j+1,(a/gcd(a,h)))) * (h(j+1)) ] - [ ((a/gcd(a,h)) / gcd(j+1,(a/gcd(a,h)))) * (ja) ] ) / a then let (2) a(h) = d(h,j) = lcm( f(h,j,1) ... f(h,j,h) ).

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A097382 a(h) = d(h,j) = lcm( f(h,j,1) ... f(h,j,h) ), when j=2.

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A098135 a(h) = d(h,j) = lcm( f(h,j,1) ... f(h,j,h) ), when j=3.

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A379561 a(n) = A003418(n+1)*H(n), where H(n) = 1 + 1/2 + ... + 1/n is the n-th harmonic number.

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