A027611 -id:A027611 - cp's OEIS frontend

A096617 Numerator of n*HarmonicNumber(n).

1, 3, 11, 25, 137, 147, 363, 761, 7129, 7381, 83711, 86021, 1145993, 1171733, 1195757, 2436559, 42142223, 42822903, 275295799, 279175675, 56574159, 19093197, 444316699, 1347822955, 34052522467, 34395742267, 312536252003

Offset: 1

Eric W. Weisstein, Jul 01 2004

a(1) = 1, a(n) = Numerator( H(n) / H(n-1) ), where H(n) = HarmonicNumber(n) = A001008(n)/A002805(n). - Alexander Adamchuk, Oct 29 2004
Sampling a population of n distinct elements with replacement, n HarmonicNumber(n) is the expectation of the sample size for the acquisition of all n distinct elements. - Franz Vrabec, Oct 30 2004
p^2 divides a(p-1) for prime p>3. - Alexander Adamchuk, Jul 16 2006
It appears that a(n) = b(n) defined by b(n+1) = b(n)*(n+1)/g(n) + f(n), f(n) = n*f(n-1)/g(n) and g(n) = gcd(b(n)*(n+1), n*f(n-1)), b(1) = f(1) = g(1) = 1, i.e., the recurrent formula of A000254(n) where both terms are divided by their GCD at each step (and remain divided by this factor in the sequel). Is this easy to prove? - M. F. Hasler, Jul 04 2019

			1, 3, 11/2, 25/3, 137/12, 147/10, 363/20, 761/35, 7129/280, ...

W. Feller, An Introduction to Probability Theory and Its Applications, Vol. I, 2nd Ed. 1957, p. 211, formula (3.3)

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Complete Set

Cf. A027611, A001008, A002805.
Differs from A025529 at 7th term.
Cf. A193758.

[Numerator(n*HarmonicNumber(n)): n in [1..40]]; // Vincenzo Librandi, Feb 19 2014

ZL:=n->sum(sum(1/i, i=1..n), j=1..n): a:=n->floor(numer(ZL(n))): seq(a(n), n=1..27); # Zerinvary Lajos, Jun 14 2007

Numerator[Table[(Sum[(1/k), {k, 1, n}]/Sum[(1/k), {k, 1, n-1}]), {n, 1, 20}]] (* Alexander Adamchuk, Oct 29 2004 *)
Table[n*HarmonicNumber[n] // Numerator, {n, 1, 27}]  (* Jean-François Alcover, Feb 17 2014 *)

{h(n) = sum(k=1,n,1/k)};
for(n=1,50, print1(numerator(n*h(n)), ", ")) \\ G. C. Greubel, Sep 01 2018

A=List(f=1); for(k=1,999, t=[A[k]*(k+1),f*=k]; t/=gcd(t); listput(A,t[1]+f=t[2])) \\ Illustrate conjectured equality. - M. F. Hasler, Jul 04 2019

a(n) = abs(Stirling1(n+1, 2))/(n-1)!. - Vladeta Jovovic, Jul 06 2004

a(n) = numerator of Integral_{t=0..oo} 1-(1-exp(-t/n))^n dt. - Jean-François Alcover, Feb 17 2014

A081528 a(n) = n*lcm{1,2,...,n}.

Original entry on oeis.org

1, 4, 18, 48, 300, 360, 2940, 6720, 22680, 25200, 304920, 332640, 4684680, 5045040, 5405400, 11531520, 208288080, 220540320, 4423058640, 4655851200, 4888643760, 5121436320, 123147264240, 128501493120, 669278610000, 696049754400

Offset: 1

Amarnath Murthy, Mar 27 2003

Denominators in binomial transform of 1/(n + 1)^2. - Paul Barry, Aug 06 2004
Construct a sequence S_n from n sequences b_1, b_2, ..., b_n of periods 1, 2, ..., n, respectively, say, b_1 = [1, 1, ...], b_2 = [1, 2, 1, 2, ...], ..., b_n = [1, 2, 3, ..., n, 1, 2, 3, ..., n, ...], by taking S_n = [b_1(1), b_2(1), ..., b_n(1), b_1(2), b_2(2), ..., b_n(2), ..., b_1(n), b_2(n), ..., b_n(n), ...] (by listing the b_i sequences in rows and taking each column in turn as the next n terms of S_n). Then a(n) is the period of sequence S_n. - Rick L. Shepherd, Aug 21 2006
This is a sequence that goes in strictly ascending order. The related sequence A003418 also goes in ascending order but has consecutive repeated terms. Since n increases, then so too does a(n) even when A003418(n) doesn't. - Alonso del Arte, Nov 25 2012

			a(2) = 4 because the least common multiple of 1 and 2 is 2, and 2 * 2 = 4.
a(3) = 18 because lcm(1,2,3) = 6, and 3 * 6 = 18.
a(4) = 48 because lcm(1, 2, 3, 4) = 12, and 4 * 12 = 48.

Cf. A027612, A027611, A022819, A002944, A081530, A097344.

a(n) := (n + 1)*LCM(VECTOR(k + 1, k, 0, n)) " Paul Barry, Aug 06 2004 "

Table[n*LCM@@Range[n], {n, 30}] (* Harvey P. Dale, Oct 09 2012 *)

l=vector(35); l[1]=1; print1("1, "); for(n=2,35, l[n]=lcm(l[n-1],n); print1(n*l[n],", ")) \\ Rick L. Shepherd, Aug 21 2006

a(n) = A003418(n) * n. - Martin Fuller, Jan 03 2006

More terms from Paul Barry, Aug 06 2004

Entry revised by N. J. A. Sloane, Jan 15 2006

A124838 Denominators of third-order harmonic numbers (defined by Conway and Guy, 1996).

Original entry on oeis.org

1, 2, 6, 4, 20, 10, 70, 56, 504, 420, 4620, 3960, 3432, 6006, 90090, 80080, 1361360, 408408, 369512, 67184, 470288, 1293292, 29745716, 27457584, 228813200, 212469400, 5736673800, 5354228880, 155272637520, 291136195350, 273491577450

Offset: 1

Jonathan Vos Post, Nov 10 2006

Numerators are A124837. All fractions reduced. Thanks to Jonathan Sondow for verifying these calculations. He suggests that the equivalent definition in terms of first order harmonic numbers may be computationally simpler. We are happy with the description of A027612 Numerator of 1/n + 2/(n-1) + 3/(n-2) +...+ (n-1)/2 + n, but baffled by the description of A027611.

			a(1) = 1 = denominator of 1/1.
a(2) = 2 = denominator of 1/1 + 5/2 = 7/2.
a(3) = 6 = denominator of 7/2 + 13/3 = 47/6.
a(4) = 4 = denominator of 47/6 + 77/12 = 57/4.
a(5) = 20 = denominator of 57/4 + 87/10 = 549/20.
a(6) = 10 = denominator of 549/20 + 223/20 = 341/10
a(7) = 70 = denominator of 341/10 + 481/35 = 3349/70.
a(8) = 1260 = denominator of 3349/70 + 4609/280 = 88327/1260.
a(9) = 45 = denominator of 88327/1260 + 4861/252 = 3844/45.
a(10) = 504 = denominator of 3844/45 + 55991/2520 = 54251/504, or, untelescoping:
a(10) = 504 = denominator of 1/1 + 5/2 + 13/3 + 77/12 + 87/10 + 223/20 + 481/35 + 4609/252 + 4861/252 + 55991/2520 = 54251/504.

J. H. Conway and R. K. Guy, The Book of Numbers, New York: Springer-Verlag, pp. 143 and 258-259, 1996.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Harmonic Number. See equation for third-order harmonic numbers.

Cf. A027611, A027612, A124837.

a124838 n = a213999 (n + 2) (n - 1) -- Reinhard Zumkeller, Jul 03 2012

Table[Denominator[(n+2)!/2!/n!*Sum[1/k,{k,3,n+2}]],{n,1,40}] (* Alexander Adamchuk, Nov 11 2006 *)

A124837(n)/A124838(n) = Sum_{i=1..n} A027612(n)/A027611(n+1).
a(n) = denominator(Sum_{m=1..n} Sum_{L=1..m} Sum_{k=1..L} 1/k).
a(n) = denominator(((n+2)!/(2!*n!)) * Sum_{k=3..n+2} 1/k). - Alexander Adamchuk, Nov 11 2006
a(n) = A213999(n+2,n-1). - Reinhard Zumkeller, Jul 03 2012

Corrected and extended by Alexander Adamchuk, Nov 11 2006

A363000 a(n) = numerator(R(n, n, 1)), where R are the rational polynomials R(n, k, x) = Sum_{u=0..k} ( Sum_{j=0..u} x^j * binomial(u, j) * (j + 1)^n ) / (u + 1).

Original entry on oeis.org

1, 5, 19, 188, 1249, 125744, 283517, 303923456, 138604561, 599865008128, 118159023973, 7078040993755136, 155792758736921, 146303841678548271104, 294014633772018349, 64670474732430319157248, 752324747622089633569, 3224753626003393505960919040, 2507759850059601711479669

Offset: 0

Peter Luschny, May 12 2023

R(n, n, 0) are the (0-based) harmonic numbers, R(n, n, -1) are the Bernoulli numbers, and R(n, n, 1) is this sequence in its rational form.

			a(n) are the numerators of the terms on the main diagonal of the triangle:
[0] 1;
[1] 1,  5/2;
[2] 1,  7/2,  19/2;
[3] 1, 11/2, 121/6,  188/3;
[4] 1, 19/2,  95/2,  369/2,   1249/2;
[5] 1, 35/2, 721/6, 1748/3, 35164/15, 125744/15;
[6] 1, 67/2, 639/2, 3877/2,  18533/2,   76317/2, 283517/2;

Cf. A363001 (denominators), A362999 (odd-indexed denominators), A362998.

Cf. A027611, A027612, A062709.

# For better context we put A362998, A362999, A363000, and A363001 together here.
R := (n, k, x) -> add(add(x^j*binomial(u, j)*(j+1)^n, j=0..u)/(u + 1), u=0..k):
### x = 1 -> this sequence
 for n from 0 to 7 do [n], seq(R(n, k, 1), k = 0..n) od;
 seq(R(n, n, 1), n = 0..9);
 A363000 := n -> numer(R(n, n, 1)): seq(A363000(n), n = 0..10);
 A363001 := n -> denom(R(n, n, 1)): seq(A363001(n), n = 0..20);
 A362999 := n -> denom(R(2*n+1, 2*n+1, 1)): seq(A362999(n), n = 0..11);
 A362998 := n -> add(R(2*n, k, 1), k = 0..2*n): seq(A362998(n), n = 0..9);
### x = -1 -> Bernoulli(n, 1)
# for n from 0 to 9 do [n], seq(R(n, k,-1), k = 0..n) od;
# seq(R(n, n, -1), n = 0..12); seq(bernoulli(n, 1), n = 0..12);
### x = 0 -> Harmonic numbers
# for n from 0 to 9 do [n], seq(R(n, k, 0), k = 0..n) od;
# seq(R(n, n, 0), n = 0..9); seq(harmonic(n+1), n = 0..9);

Sum_{k=0..n} R(n, k, 0) = Sum_{j=0..n} (n-j+1)/(j+1) = (n+2)*Harmonic(n+1)-n-1.
Sum_{k=0..n} R(n, k,-1) = (n + 2 - 0^n) * Bernoulli(n, 1).
Sum_{k=0..2*n} R(2*n, k, 1) = A362998(n).
2*R(n, 1, 1) = A062709(n).

A128438 a(n) = floor((denominator of H(n))/n), where H(n) = Sum_{k=1..n} 1/k, the n-th harmonic number.

Original entry on oeis.org

1, 1, 2, 3, 12, 3, 20, 35, 280, 252, 2520, 2310, 27720, 25740, 24024, 45045, 720720, 226893, 4084080, 775975, 246341, 235144, 5173168, 14872858, 356948592, 343219800, 2974571600, 2868336900, 80313433200, 77636318760, 2329089562800

Offset: 1

Leroy Quet, Mar 03 2007

nonn

This is very similar to A027611, which is a different sequence. - N. J. A. Sloane, Nov 21 2008

Indices where a(n) differs from A027611 are terms of A074791. - Gary Detlefs, Sep 03 2011

			The sequence denominator(H(n))/n begins 1, 1, 2, 3, 12, 10/3, 20, 35, 280, 252, 2520, 2310, ..., so the present sequence begins 1, 1, 2, 3, 12, 3, 20, 35, 280, 252, 2520, 2310, ...

Amiram Eldar, Table of n, a(n) for n = 1..2310

Cf. A128437, A002805, A027611, A074791.

H:=n->sum(1/k,k=1..n): a:=n->floor(denom(H(n))/n): seq(a(n),n=1..34); # Emeric Deutsch, Mar 25 2007

seq = {}; s = 0; Do[s += 1/n; AppendTo[seq, Floor[Denominator[s]/n]], {n, 1, 30}]; seq (* Amiram Eldar, Sep 18 2021 *)
Table[Floor[Denominator[HarmonicNumber[n]]/n],{n,40}] (* Harvey P. Dale, Nov 24 2023 *)

from sympy import harmonic
def A128438(n): return harmonic(n).q//n # Chai Wah Wu, Sep 27 2021

More terms from Emeric Deutsch, Mar 25 2007

A354860 a(n) is the denominator of 1/prime(n) + 2/prime(n-1) + 3/prime(n-2) + ... + (n-1)/3 + n/2.

Original entry on oeis.org

2, 3, 30, 35, 2310, 15015, 34034, 4849845, 223092870, 154040315, 200560490130, 742073813481, 101416754509070, 6541380665835015, 55899071144408310, 5431526412865007455, 54936010004406075402, 4511091590746421960895, 2619440517026755685293030, 278970415063349480483707695

Offset: 1

Ilya Gutkovskiy, Jun 09 2022

Denominator of a second order prime harmonic number.

			1/2, 4/3, 71/30, 124/35, 11111/2310, 92402/15015, 257189/34034, ...

Cf. A002110, A027611, A354859 (numerators).

Table[Sum[(n - k + 1)/Prime[k], {k, 1, n}], {n, 1, 20}] // Denominator

from fractions import Fraction
from sympy import prime, primerange
def a(n): return sum(Fraction(n-i, p) for i, p in enumerate(primerange(1, prime(n)+1))).denominator
print([a(n) for n in range(1, 21)]) # Michael S. Branicky, Jun 09 2022

a(n) is the denominator of Sum_{j=1..n} Sum_{i=1..j} 1/prime(i).

A354895 a(n) is the denominator of the n-th hyperharmonic number of order n.

Original entry on oeis.org

1, 2, 6, 12, 20, 60, 210, 56, 504, 504, 660, 3960, 5148, 4004, 4290, 34320, 17680, 31824, 302328, 77520, 813960, 8953560, 2288132, 27457584, 49031400, 12498200, 168725700, 42948360, 10925460, 163881900, 2540169450, 645122400, 327523680, 5567902560, 1412149200

Offset: 1

Ilya Gutkovskiy, Jun 10 2022

			1, 5/2, 47/6, 319/12, 1879/20, 20417/60, 263111/210, 261395/56, 8842385/504, ...

J. H. Conway and R. K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996, p. 258.

Robert Israel, Table of n, a(n) for n = 1..3764
Eric Weisstein's World of Mathematics, Harmonic Number
Wikipedia, Hyperharmonic number

Cf. A001700, A002805, A027611, A058806, A076175, A124838, A354894 (numerators).

N:= 100: # for a(1)..a(N)
H:= ListTools:-PartialSums([seq(1/i,i=1..2*N-1)]):
f:= n -> denom(binomial(2*n-1,n-1)*(H[2*n-1]-H[n-1])):
f(1):= 1:
map(f, [$1..N]); # Robert Israel, Jul 10 2023

Table[SeriesCoefficient[-Log[1 - x]/(1 - x)^n, {x, 0, n}], {n, 1, 35}] // Denominator
Table[Binomial[2 n - 1, n - 1] (HarmonicNumber[2 n - 1] - HarmonicNumber[n - 1]), {n, 1, 35}] // Denominator

H(n) = sum(i=1, n, 1/i);
a(n) = denominator(binomial(2*n-1,n-1) * (H(2*n-1) - H(n-1))); \\ Michel Marcus, Jun 10 2022

from math import comb
from sympy import harmonic
def A354895(n): return (comb(2*n-1,n-1)*(harmonic(2*n-1)-harmonic(n-1))).q # Chai Wah Wu, Jun 18 2022

a(n) is the denominator of the coefficient of x^n in the expansion of -log(1 - x) / (1 - x)^n.
a(n) is the denominator of binomial(2*n-1,n-1) * (H(2*n-1) - H(n-1)), where H(n) is the n-th harmonic number.
A354894(n) / a(n) ~ log(2) * 2^(2*n-1) / sqrt(Pi * n).

A382812 Numerator of the n-th partial sum of the squares of the harmonic numbers.

Original entry on oeis.org

1, 13, 119, 1577, 3233, 8867, 141563, 2844129, 28119709, 335676251, 3968696491, 55023970333, 758025067309, 799020611041, 1676892996083, 59597395635137, 351844709221043, 2314823924364859, 9114392136427625, 628176680098075, 216039223801697, 5117413095318143, 363066107054194281, 27957386425926920257

Offset: 1

Gary Detlefs, Apr 05 2025

			The squares of the first three harmonic numbers are 1, 9/4, 121/36 which sum to 119/18 so a(3)=119.

Robert Israel, Table of n, a(n) for n = 1..1000

Cf. A001008, A002805, A382813 (denominators).

Cf. A027611, A027612

H2:= n-> add(harmonic(k)^2, k = 1..n): seq(numer(H2(n)), n=1..25);

Accumulate[HarmonicNumber[Range[30]]^2]//Numerator (* Harvey P. Dale, Aug 10 2025 *)

a(n) = numerator(sum(k=1, n, sum(i=1, k, 1/i)^2)); \\ Michel Marcus, Apr 07 2025

a(n) = numerator((n+1)*H(n)^2-(2*n+1)*H(n) + 2*n), where H(n) is the n-th harmonic number.

a(n) = numerator((S(n)*H(n)^2 + (2*n - 2*S(n) + 1)*H(n)-2*n)/(H(n)-1)), where S(n) is the n-th partial sum of H(n).

A382813 Denominator of the n-th partial sum of the squares of the harmonic numbers.

Original entry on oeis.org

1, 4, 18, 144, 200, 400, 4900, 78400, 635040, 6350400, 64033200, 768398400, 9275666400, 8657288640, 16232416200, 519437318400, 2779951574400, 16679709446400, 60213751101504, 3823095308032, 1216439416192, 26761667156224, 1769615240705312, 127412297330782464, 3062795608913040000

Offset: 1

Gary Detlefs, Apr 05 2025

All terms for n>1 are even.

			The squares of the first three harmonic numbers are 1, 9/4, 121/36 which sum to 119/18 so a(3) = 18.

Robert Israel, Table of n, a(n) for n = 1..1000

Cf. A001008, A002805, A382812 (numerators).

Cf. A027611, A027612.

H2:= n-> add(harmonic(k)^2, k = 1..n): seq(denom(H2(n)), n=1..25);

a(n) = denominator(sum(k=1, n, sum(i=1, k, 1/i)^2)); \\ Michel Marcus, Apr 07 2025

a(n) = denominator((n+1)*H(n)^2-(2*n+1)*H(n)+2*n), where H(n) is the n-th harmonic number.

a(n) = denominator((S(n)*H(n)^2+(2*n-2*S(n)+1)*H(n) - 2*n)/(H(n) - 1)), where S(n) = the n-th partial sum of H(n).

A079076 Numerator of Sum_{1 0} n/k.

Original entry on oeis.org

0, 0, 3, 4, 65, 27, 203, 236, 3489, 2845, 53471, 21341, 757913, 553973, 619181, 1040164, 29169263, 16276383, 193614199, 116220883, 32925391, 10628013, 320160667, 455451475, 22987116115, 19980510667, 193553388003, 154777722503

Offset: 1

Reinhard Zumkeller, Dec 21 2002

Cf. A027611 (denominator), A049820, A024816.

f[n_] := Block[{s = 0, d = 1}, While[d < n, If[ Mod[n, d] != 0, s = s + n/d]; d++ ]; s]; Numerator[ Table[ f[n], {n, 1, 28}]]

p divides a(p^k) for prime p and integer k>1. p divides a(p), a(2p) and a(2p^2) for prime p>2. - Alexander Adamchuk, Jul 27 2006

Edited and extended by Robert G. Wilson v, Dec 30 2002

cp's OEIS Frontend

A096617 Numerator of n*HarmonicNumber(n).

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A081528 a(n) = n*lcm{1,2,...,n}.

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A124838 Denominators of third-order harmonic numbers (defined by Conway and Guy, 1996).

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A363000 a(n) = numerator(R(n, n, 1)), where R are the rational polynomials R(n, k, x) = Sum_{u=0..k} ( Sum_{j=0..u} x^j * binomial(u, j) * (j + 1)^n ) / (u + 1).

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A128438 a(n) = floor((denominator of H(n))/n), where H(n) = Sum_{k=1..n} 1/k, the n-th harmonic number.

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A354860 a(n) is the denominator of 1/prime(n) + 2/prime(n-1) + 3/prime(n-2) + ... + (n-1)/3 + n/2.

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A354895 a(n) is the denominator of the n-th hyperharmonic number of order n.

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