A025529 -id:A025529 - cp's OEIS frontend

A112809 Positions of records in A110566.

1, 6, 20, 21, 42, 120, 342, 506, 567, 594, 600, 610, 2184, 4896, 6108, 6162, 6498, 12760, 14067, 14157, 14201, 93942, 123462, 123519, 734413, 2451397, 4591010, 11571129, 13346540, 13619348, 13619790, 46180567

Offset: 1

Robert G. Wilson v, Sep 19 2005

Cf. A110566, A001008, A002805, A003418, A025529, A098464, A112810.

c = 0; a = h = 1; t = {}; Do[a = LCM[a, n]; h = h + 1/n; b = a/Denominator[h]; If[b > c, c = b; AppendTo[t, n]], {n, 10^6}]; t

lista(nn) = {rec = 0; for (n=1, nn, new = lcm(vector(n, k, k))/denominator(sum(k=1, n, 1/k)); if (new > rec, print1(n, ", "); rec = new););} \\ Michel Marcus, Mar 07 2018

a(27)-a(28) from Amiram Eldar, Dec 18 2018
a(29)-a(31) from Chai Wah Wu, Mar 08 2021
a(32) from Chai Wah Wu, Mar 14 2021

A120263 Ratio of the numerator of n*HarmonicNumber[n] to the numerator of HarmonicNumber[n]: A096617(n)/A001008(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 5, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 25, 1, 1, 1, 1

Offset: 1

Alexander Adamchuk, Jun 26 2006

a(n) is not equal to 1 when n belongs to A074791 - numbers n such that n does not divide the denominator of the n-th harmonic number.
a(n) is almost always equal to 1 except for n=6,18,20,21,33,42,54,.. when a(n) seems to be equal to a prime divisor of n.
a(n) could be equal to a squared prime divisor of n as for n=100,294,500,847,..

G. C. Greubel, Table of n, a(n) for n = 1..10000

Cf. A096617, A001008, A074791.

[Numerator(n*HarmonicNumber(n))/Numerator(HarmonicNumber(n)): n in [1..100]]; // G. C. Greubel, Sep 01 2018

Numerator[Table[n*Sum[1/i,{i,1,n}],{n,1,500}]]/Numerator[Table[Sum[1/i,{i,1,n}],{n,1,500}]]

{h(n) = sum(k=1,n,1/k)};
for(n=1,100, print1(numerator(n*h(n))/numerator(h(n)), ", ")) \\ G. C. Greubel, Sep 01 2018

a(n) = A096617(n)/A001008(n) = numerator[n*Sum[1/i,{i,1,n}]] / numerator[Sum[1/i,{i,1,n}]].
a(n) = n / gcd(denominator(H(n)),n), where H(n) = sum(1/k, k=1..n). [Gary Detlefs, Sep 05 2011]
a(n) = A096617(n)*A110566(n)/A025529(n). [Arkadiusz Wesolowski, Mar 29 2012]

A120435 Triangle read by rows: T(n,k) = lcm(1,2,3,4,...,n)/k (1 <= k <= n).

Original entry on oeis.org

1, 2, 1, 6, 3, 2, 12, 6, 4, 3, 60, 30, 20, 15, 12, 60, 30, 20, 15, 12, 10, 420, 210, 140, 105, 84, 70, 60, 840, 420, 280, 210, 168, 140, 120, 105, 2520, 1260, 840, 630, 504, 420, 360, 315, 280, 2520, 1260, 840, 630, 504, 420, 360, 315, 280, 252, 27720, 13860, 9240

Offset: 1

Leroy Quet, Jul 15 2006

T(n,1) = A003418(n). Row sums yield A025529. - Emeric Deutsch, Jul 24 2006

			Triangle starts:
   1;
   2,  1;
   6,  3,  2;
  12,  6,  4,  3;
  60, 30, 20, 15, 12;

John Tyler Rascoe, Rows n = 1..140, flattened

Cf. A003418, A025529, A094310.

T:=proc(n,k) if k<=n then lcm(seq(j,j=1..n))/k else 0 fi end: for n from 1 to 11 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form); # Emeric Deutsch, Jul 24 2006

from math import lcm
def A120435(maxrow):
    A,rlcm = [],1
    for n in range(1,maxrow+1):
        rlcm = lcm(n,rlcm)
        A.append(list(rlcm//k for k in range(1,n+1)))
    return A # John Tyler Rascoe, Nov 08 2024

More terms from Emeric Deutsch, Jul 24 2006

A175455 a(n) = H(n) * (lcm(1,2,...,n))^2, where H(n) = harmonic numbers (1/1 + 1/2 + ... + 1/n).

Original entry on oeis.org

1, 6, 66, 300, 8220, 8820, 457380, 1917720, 17965080, 18600120, 2320468920, 2384502120, 412970037480, 422245703880, 430902992520, 1756076802480, 516336630329520, 524676485052720, 192260441419366320, 194970060218934000, 197550649551855600, 200013939369644400

Offset: 1

Jaroslav Krizek, May 17 2010

nonn

			For n = 3, a(3) = (1/1 + 1/2 + 1/3) * (1*2*3)^2 = (11/6) * 36 = 66.

Andrew Howroyd, Table of n, a(n) for n = 1..200

Cf. A000142, A001008, A002805, A003418, A025527, A025529.

a(n)={sum(k=1, n, 1/k)*lcm([1..n])^2} \\ Andrew Howroyd, Jan 08 2020

a(n) = (A001008(n) / A002805(n)) * (A003418(n))^2.
a(n) = A000142(n) * A025529(n) / A025527(n) = A025529(n) * A003418(n).
a(n) = (1/1 + 1/2 + ... + 1/n) * (lcm(1,2,...,n))^2.

Terms a(13) and beyond from Andrew Howroyd, Jan 08 2020

A187064 Coefficients in numerator polynomial of: Sum (k=1 to n) of x^k/(1-x^k).

Original entry on oeis.org

1, 2, 1, 3, 4, 3, 1, 4, 5, 7, 5, 3, 1, 5, 11, 19, 24, 26, 22, 16, 9, 4, 1, 6, 7, 15, 18, 23, 21, 21, 15, 11, 6, 3, 1, 7, 15, 32, 52, 77, 99, 120, 128, 130, 119, 102, 79, 57, 36, 21, 10, 4, 1, 8, 17, 36, 58, 93, 125, 165, 193, 220, 229, 231, 213, 191, 157, 124

Offset: 1

Mats Granvik, Mar 07 2011

The number of elements per row begins: 1,2,4,6,10,12,18,... which appears to be A002088.

Row sums begin: 1,3,11,25,137,147,1089,... which appears to be A025529.

			Table begins:
1,
2,1,
3,4,3,1,
4,5,7,5,3,1,
5,11,19,24,26,22,16,9,4,1,
6,7,15,18,23,21,21,15,11,6,3,1,
7,15,32,52,77,99,120,128,130,119,102,79,57,36,21,10,4,1,
Polynomials begin:
-(1*x^1)/(x^1-1)
-(2*x^2+1*x)/(x^2-1)
-(3*x^4+4*x^3+3*x^2+1*x^1)/(x^4+x^3-x^1-1)
-(4*x^6+5*x^5+7*x^4+5*x^3+3*x^2+1*x^1)/(x^6+x^5+x^4-x^2-x^1-1)

row(n) = v = Vec(numerator(sum(k=1, n, x^k/(1-x^k)))); for (k=1, #v-1, print1(abs(v[k]), ", ")); /*print*/; \\ Michel Marcus, Jun 11 2014

More terms from Michel Marcus, Jun 11 2014

A309556 Composite numbers m such that m divides Sum_{k=1..m-1} (lcm(1,2,...,(m-1)) / k)^2.

Original entry on oeis.org

52781, 782957, 1395353, 2602439

Offset: 1

Amiram Eldar and Thomas Ordowski, Aug 07 2019

Composites m such that m | H_2(m-1) * lcm(1^2,2^2,...,(m-1)^2), where H_2(m) = 1/1^2 + 1/2^2 + ... + 1/m^2.
By Wolstenholme's theorem, if p > 3 is a prime, then p divides the numerator of H_2(p-1) and thus H_2(p-1) * lcm(1,2^2,...,(p-1)^2) == 0 (mod p). This sequence is formed by the pseudoprimes that are solutions of this congruence.
a(5) > 10^7 if it exists.

Eric Weisstein's World of Mathematics, Wolstenholme's Theorem.
Wikipedia, Wolstenholme's theorem.

Cf. A007406, A007407, A025529 (see our comment), A051418.

s = 0; c = 1; n = 1; seq = {}; Do[s += 1/n^2; c = LCM[c, n^2]; n++; If[CompositeQ[n] && Divisible[s*c, n], AppendTo[seq, n]], {2 * 10^6}]; seq

A331343 a(n) = lcm(1,2,...,n) * Sum_{k=1..n} (2^(k-1) - 1) / k.

Original entry on oeis.org

0, 1, 9, 39, 375, 685, 8575, 30485, 162855, 291627, 5785857, 10514427, 250200951, 461037291, 854622483, 3185234481, 101381371377, 190598779657, 6833215763803, 12935721409039, 24559552771039, 46750514134519, 2051664357879617, 3923102768811707, 37581323659852375

Offset: 1

Amiram Eldar and Thomas Ordowski, Jan 14 2020

nonn

By Wolstenholme's theorem, if p > 3 is a prime, then p^3 | a(p).
Conjecture: for n > 3, if n^3 | a(n), then n is prime. If so, there are no such pseudoprimes.
Problem: are there weak pseudoprimes m such that m^2 | a(m)? None up to 5*10^4.
Composite numbers m such that m | a(m) are 9, 25, 49, 99, 121, 125, 169, 221, 289, 343, 357, 361, 399, 529, 665, 841, 961, 1331, 1369, 1443, 1681, 1849, 2183, ...  Cf. A082180.
Prime numbers p such that p^4 | a(p) are probably only the Wolstenholme primes A088164.

Wikipedia, Wolstenholme's theorem.

Cf. A003418, A025529, A082180, A088164, A330718, A330719.

[Lcm([1..n])*&+[(2^(k-1)-1)/k:k in [1..n]]:n in [1..25]]; // Marius A. Burtea, Jan 14 2020

a[n_] := LCM @@ Range[n] * Sum[(2^(k-1) - 1) / k, {k, 1, n}]; Array[a, 25]

a(n) = lcm([1..n])*sum(k=1, n, (2^(k-1) - 1) / k); \\ Michel Marcus, Jan 14 2020

a(n) = A003418(n) * A330718(n) / A330719(n).

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

A112809 Positions of records in A110566.

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A120263 Ratio of the numerator of n*HarmonicNumber[n] to the numerator of HarmonicNumber[n]: A096617(n)/A001008(n).

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A120435 Triangle read by rows: T(n,k) = lcm(1,2,3,4,...,n)/k (1 <= k <= n).

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A175455 a(n) = H(n) * (lcm(1,2,...,n))^2, where H(n) = harmonic numbers (1/1 + 1/2 + ... + 1/n).

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A187064 Coefficients in numerator polynomial of: Sum (k=1 to n) of x^k/(1-x^k).

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A309556 Composite numbers m such that m divides Sum_{k=1..m-1} (lcm(1,2,...,(m-1)) / k)^2.

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Mathematica

A331343 a(n) = lcm(1,2,...,n) * Sum_{k=1..n} (2^(k-1) - 1) / k.

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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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