A096617 -id:A096617 - cp's OEIS frontend

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

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]

A025529 a(n) = (1/1 + 1/2 + ... + 1/n)*lcm{1,2,...,n}.

Original entry on oeis.org

1, 3, 11, 25, 137, 147, 1089, 2283, 7129, 7381, 83711, 86021, 1145993, 1171733, 1195757, 2436559, 42142223, 42822903, 825887397, 837527025, 848612385, 859193865, 19994251455, 20217344325, 102157567401, 103187226801, 312536252003, 315404588903, 9227046511387

Offset: 1

Clark Kimberling

nonn

First column of A027446. - Eric Desbiaux, Mar 29 2013
From Amiram Eldar and Thomas Ordowski, Aug 07 2019: (Start)
By Wolstenholme's theorem, if p > 3 is a prime, then p^2 | a(p-1).
Conjecture: for n > 3, if n^2 | a(n-1), then n is a prime.
Note that if n = p^2 with prime p > 3, then n | a(n-1).
It seems that composite numbers n such that n | a(n-1) are only the squares n = p^2 of primes p > 3.
Primes p such that p^3 | a(p-1) are the Wolstenholme primes A088164.
The n-th triangular number n(n+1)/2 | a(n) for n = 1, 2, 6, 4422, ... (End)

Alois P. Heinz, Table of n, a(n) for n = 1..1000
Frank A. Haight, and Robert B. Jones., "A probabilistic treatment of qualitative data with special reference to word association tests." Journal of Mathematical Psychology 11.3 (1974): 237-244. [Denominators of fractions in Eq. 21.] [Annotated scanned copy]
Frank A. Haight and N. J. A. Sloane, Correspondence, 1975
Yilmaz Simsek, Derivation of Computational Formulas for certain class of finite sums: Approach to Generating functions arising from p-adic integrals and special functions, arXiv:2108.10756 [math.NT], 2021.

Differs from A096617 at 7th term.

Cf. A001008, A002805, A027446, A110566.

List([1..30],n->Sum([1..n],k->1/k)*Lcm([1..n])); # Muniru A Asiru, Apr 02 2018

[HarmonicNumber(n)*Lcm([1..n]):n in [1..30]]; // Marius A. Burtea, Aug 07 2019

a:= n-> add(1/k, k=1..n)*ilcm($1..n):
seq(a(n), n=1..30);  # Alois P. Heinz, Mar 14 2013

Table[HarmonicNumber[n]*LCM @@ Range[n], {n, 27}] (* Arkadiusz Wesolowski, Mar 29 2012 *)

a(n) = sum(k=1, n, 1/k)*lcm([1..n]); \\ Michel Marcus, Apr 02 2018

a(n) = A001008(n)*A110566(n). - Arkadiusz Wesolowski, Mar 29 2012

a(n) = Sum_{k=1..n} lcm(1,2,...,n)/k. - Thomas Ordowski, Aug 07 2019

A074791 Numbers k such that k does not divide the denominator of the k-th harmonic number.

Original entry on oeis.org

6, 18, 20, 21, 33, 42, 54, 63, 66, 77, 100, 110, 120, 156, 162, 189, 198, 272, 294, 336, 342, 363, 377, 435, 486, 500, 506, 559, 567, 594, 600, 610, 629, 685, 703, 812, 847, 880, 924, 930, 957, 1067, 1166, 1210, 1243, 1247, 1287, 1320, 1332, 1458, 1590, 1640

Offset: 1

Benoit Cloitre, Sep 07 2002

nonn

k such that A064169(k) is different from A027612(k).
Also k such that A096617(k) is different from A001008(k). - Alexander Adamchuk, Jun 26 2006
This sequence contains A036689(k) for all k > 1. - Wouter van Doorn, Nov 06 2024

Amiram Eldar, Table of n, a(n) for n = 1..2000

Cf. A027612, A064169.

Cf. A096617, A001008.

Select[ Range[1700], Mod[ Denominator[ HarmonicNumber[ # ]], # ] != 0 &] (* Robert G. Wilson v, Sep 28 2005 *)
seq = {}; s = 0; Do[s += 1/n; If[! Divisible[Denominator[s], n], AppendTo[seq, n]], {n, 1, 2000}]; seq (* Amiram Eldar, Dec 01 2020 *)

Is a(n) asymptotic to c*n^2 0.5

a(n) < 2*n^2*log(n)^2 for all n > 2. This follows from the fact that for all k > 1 there exists an n such that A036689(k) is equal to A074791(n). - Wouter van Doorn, Nov 06 2024

Better description and more terms from Robert G. Wilson v, Sep 28 2005

A193758 Denominator of H(n)/H(n-1), where H(n) is the n-th harmonic number = Sum_{k=1..n} 1/k.

Original entry on oeis.org

2, 9, 22, 125, 137, 343, 726, 6849, 7129, 81191, 83711, 1118273, 1145993, 1171733, 2391514, 41421503, 42142223, 271211719, 275295799, 55835135, 18858053, 439143531, 1332950097, 33695573875, 34052522467, 309561680403, 312536252003, 9146733078187, 9227046511387

Offset: 2

Gary Detlefs, Aug 04 2011

a(n) mod n^3 = 0 iff n is prime > 3. - Gary Detlefs, Jan 30 2013

Michael S. Branicky, Table of n, a(n) for n = 2..1001

Cf. A001008, A002805.

Numerators are in A096617.

H:= n-> add(1/k, k=1..n): seq(denom(H(n)/H(n-1)), n=2..25);

h[n_] := Sum[1/i, {i, n}]; Table[Denominator[h[n]/h[n - 1]], {n, 2, 50}] (* T. D. Noe, Aug 04 2011 *)
Denominator[#[[2]]/#[[1]]]&/@Partition[HarmonicNumber[Range[30]],2,1] (* Harvey P. Dale, Jul 05 2015 *)

from fractions import Fraction
def aupton(nn):
  Hnm1, alst = Fraction(1, 1), []
  for n in range(2, nn+1):
    Hn = Hnm1 + Fraction(1, n)
    alst.append((Hn/Hnm1).denominator)
    Hnm1 = Hn
  return alst
print(aupton(30)) # Michael S. Branicky, Feb 09 2021

a(n) = denominator(H(n)/H(n-1)), where H(n) = Sum_{k=1..n} 1/k.

a(n) = numerator(n*H(n))-denominator(n*H(n)). - Gary Detlefs, Sep 05 2011

A120299 Largest prime factor of Stirling numbers of first kind s(n,2) = A000254(n).

Original entry on oeis.org

3, 11, 5, 137, 7, 11, 761, 7129, 61, 863, 509, 919, 1117, 41233, 8431, 1138979, 39541, 7440427, 11167027, 18858053, 227, 583859, 467183, 312408463, 34395742267, 215087, 375035183, 4990290163, 17783, 2667653736673, 535919, 199539368321, 15088528003, 137121586897, 9059

Offset: 2

Alexander Adamchuk, Jul 11 2006

nonn

M. F. Hasler, Table of n, a(n) for n = 2..168

Cf. A000254, A002547, A001008, A002805, A006530.

Table[Max[FactorInteger[Sum[1/i,{i,1,n}]/Product[1/i,{i,1,n}]]],{n,2,40}]
FactorInteger[#][[-1,1]]&/@StirlingS1[Range[3,40],2] (* Harvey P. Dale, May 10 2018 *)

A120299(n)=A006530(A000254(n)) \\ Probably A000254 can be replaced by (much smaller) A096617. - M. F. Hasler, Jul 04 2019

a(n) = Max[FactorInteger[Sum[1/i,{i,1,n}]/Product[1/i,{i,1,n}]]].

a(n) = gpf(A096617(n)), where gpf = A006530 is the greatest prime factor, and A096617 is a "reduced" variant of A001008 and thus A000254. [Conjectured; true if this gpf is always > n.] - M. F. Hasler, Jul 04 2019

More terms from M. F. Hasler, Jul 04 2019

A120308 Numerator((p-1)*H(p-1))/p^2 for p = prime(n) > 3, where H(k) is k-th harmonic number A001008(k)/A002805(k).

Original entry on oeis.org

1, 3, 61, 509, 8431, 118623, 36093, 375035183, 9682292227, 40030624861, 1236275063173, 46600968591317, 2690511212793403, 5006621632408586951, 73077117446662772669, 4062642402613316532391, 139715526178793824689891

Offset: 3

Alexander Adamchuk, Jul 16 2006

G. C. Greubel, Table of n, a(n) for n = 3..344

Cf. A096617, A001008, A002805.

[Numerator((NthPrime(n)-1)*HarmonicNumber(NthPrime(n)-1)/NthPrime(n)^2): n in [3..25]]; // G. C. Greubel, Sep 02 2018

N:= 50: # to get the first N terms
Primes:= select(isprime,[seq(2*i+1,i=2..(ithprime(N+2)-1)/2)]):
H:= ListTools[PartialSums]([seq(1/i,i=1..Primes[-1]-1)]):
seq(numer((p-1)*H[p-1])/p^2, p=Primes); # Robert Israel, Sep 09 2014

Numerator[Table[(Prime[n]-1)*(Sum[(1/k), {k, 1, Prime[n]-1}]),{n,3,20}]]/Table[Prime[n]^2,{n,3,20}]
Table[((p-1)HarmonicNumber[p-1])/p^2,{p,Prime[Range[2,20]]}]//Numerator (* Harvey P. Dale, May 19 2021 *)

{a(n) = numerator((prime(n)-1)*sum(k=1,prime(n)-1, 1/k)/prime(n)^2)};
for(n=3,25, print1(a(n), ", ")) \\ G. C. Greubel, Sep 02 2018

a(n) = numerator((prime(n)-1)*(Sum_{k=1..prime(n)-1} 1/k))/prime(n)^2 for n > 2.

a(n) = A096617(p-1)/p^2 for p = prime(n) > 3.

Showing 1-6 of 6 results.

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A025529 a(n) = (1/1 + 1/2 + ... + 1/n)*lcm{1,2,...,n}.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Formula

A074791 Numbers k such that k does not divide the denominator of the k-th harmonic number.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A193758 Denominator of H(n)/H(n-1), where H(n) is the n-th harmonic number = Sum_{k=1..n} 1/k.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Python

Formula

A120299 Largest prime factor of Stirling numbers of first kind s(n,2) = A000254(n).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A120308 Numerator((p-1)*H(p-1))/p^2 for p = prime(n) > 3, where H(k) is k-th harmonic number A001008(k)/A002805(k).

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula