A160014 -id:A160014 - cp's OEIS frontend

A346465 Numbers k such that (4^k - 2)(4^k - 1)/Clausen(2k, 1) is not squarefree, where Clausen(n, m) = A160014(n, m).

9, 11, 18, 27, 32, 36, 45, 50, 53, 54, 63, 68, 72, 74, 78, 81, 90, 95, 99, 100, 108, 116, 117, 126, 127, 135, 137, 144, 147, 150, 153, 155, 158, 162, 171, 179, 180, 182, 189, 198, 200, 204, 207, 216, 221, 225, 233, 234, 242, 243, 250, 252, 261, 263, 270, 279

Offset: 1

Peter Luschny, Jul 20 2021

nonn

Also numbers k such that 6*GaussBinomial(2*k, 2, 2)/denominator(Bernoulli(2*k, 1)) is not squarefree.

Cf. A006095, A002445, A007947, A160014, A346463, A346464.

with(NumberTheory): isa := n -> not IsSquareFree(((4^n - 2)*(4^n - 1))/
mul(i, i = select(isprime, map(i -> i+1, Divisors(2*n))))):
select(isa, [$(1..100)]);

q[n_] := Product[k, {k, Select[Table[d + 1, {d, Divisors[2 n]}], PrimeQ]}];
isA[n_] := ! SquareFreeQ[((4^n - 2) (4^n -1)) / q[n]];
Select[Range[50],  isA]

The positive multiples of 9 form a subsequence.

k is a term if and only if A346463(k) > A007947(A346463(k)).

More terms from Jinyuan Wang, Jul 23 2021

A363523 k is a term of this sequence if and only if Clausen(k, 0) divides Clausen(k, 2). (Clausen = A160014.)

Original entry on oeis.org

1, 3, 9, 15, 27, 45, 75, 81, 99, 105, 135, 225, 243, 255, 297, 315, 375, 405, 495, 525, 675, 729, 735, 765, 783, 891, 945, 1089, 1125, 1215, 1275, 1287, 1485, 1575, 1785, 1875, 2025, 2115, 2187, 2205, 2295, 2349, 2415, 2475, 2625, 2673, 2835, 3267, 3375, 3465

Offset: 1

Peter Luschny, Jun 08 2023

nonn

k is a term of A124240 if and only if Clausen(k, 0) divides Clausen(k, 1).

Cf. A160014, A124240.

# Using function 'Clausen' from A160014.
aList := m -> select(k -> irem(Clausen(k, 2), Clausen(k, 0)) = 0, [seq(1..m)]):
aList(3500);

A345365 a(n) = (2n)!Pi^(-2n)PolyLog(2n, 1)Clausen(2*n - 1)/2, where Clausen(n) = A160014(n, 1).

Original entry on oeis.org

1, 4, 16, 64, 1280, 707584, 28672, 59260928, 2874867712, 45773225984, 896021823488, 991382852337664, 143497256501248, 1593799350268461056, 2312797281748024033280, 8277820436597920759808, 11071085050982544965632, 452092922822895257024443973632

Offset: 1

Peter Luschny, Aug 21 2021

nonn

Cf. A160014.

Clausen[n_] := Times @@ (Select[Divisors[n + 1], PrimeQ[# + 1] &] + 1);
a[n_] := (2 n)! PolyLog[2 n, 1] Pi^(-2 n) Clausen[2 n - 1] / 2;
Table[a[n], {n, 1, 18}]

A002445 Denominators of Bernoulli numbers B_{2n}.

Original entry on oeis.org

1, 6, 30, 42, 30, 66, 2730, 6, 510, 798, 330, 138, 2730, 6, 870, 14322, 510, 6, 1919190, 6, 13530, 1806, 690, 282, 46410, 66, 1590, 798, 870, 354, 56786730, 6, 510, 64722, 30, 4686, 140100870, 6, 30, 3318, 230010, 498, 3404310, 6, 61410, 272118, 1410, 6, 4501770, 6, 33330, 4326, 1590, 642, 209191710, 1518, 1671270, 42

Offset: 0

N. J. A. Sloane

From the von Staudt-Clausen theorem, denominator(B_2n) = product of primes p such that (p-1)|2n.
Row products of A138239. - Mats Granvik, Mar 08 2008
Equals row products of even rows in triangle A143343. In triangle A080092, row products = denominators of B1, B2, B4, B6, ... . - Gary W. Adamson, Aug 09 2008
Julius Worpitzky's 1883 algorithm for generating Bernoulli numbers is shown in A028246. - Gary W. Adamson, Aug 09 2008
There is a relation between the Euler numbers E_n and the Bernoulli numbers B_{2*n}, for n>0, namely, B_{2*n} = A000367(n)/a(n) = ((-1)^n/(2*(1-2^{2*n}))) * Sum_{k = 0..n-1} (-1)^k*2^{2*k}*C(2*n,2*k)*A000364(n-k)*A000367(k)/a(k). (See Bucur, et al.) - L. Edson Jeffery, Sep 17 2012
a(n) is the product of all primes of the form (k + n)/(k - n). - Thomas Ordowski, Jul 24 2025

			B_{2n} = [ 1, 1/6, -1/30, 1/42, -1/30, 5/66, -691/2730, 7/6, -3617/510, ... ].

Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 136.
G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
See A000367 for further references and links (there are a lot).

T. D. Noe, Table of n, a(n) for n = 0..10000
Amelia Bucur, José Luis López-Bonilla, and Jaime Robles-García, A note on the Namias identity for Bernoulli numbers, Journal of Scientific Research (Banaras Hindu University, Varanasi), Vol. 56 (2012), 117-120.
Suyuong Choi and Younghan Yoon, A decomposition of graph a-numbers, arXiv:2508.06855 [math.CO], 2025. See p. 13.
G. Everest, A. J. van der Poorten, Y. Puri and T. Ward, Integer Sequences and Periodic Points, Journal of Integer Sequences, Vol. 5 (2002), Article 02.2.3
Shizuo Kaji, Toshiaki Maeno, Koji Nuida, and Yasuhide Numata, Polynomial Expressions of Carries in p-ary Arithmetics, arXiv preprint arXiv:1506.02742 [math.CO], 2015.
Takao Komatsu, Florian Luca, and Claudio de J. Pita Ruiz V. , A note on the denominators of Bernoulli numbers, Proc. Japan Acad., 90, Ser. A (2014), p. 71-72.
Guo-Dong Liu, H. M. Srivastava, and Hai-Quing Wang, Some Formulas for a Family of Numbers Analogous to the Higher-Order Bernoulli Numbers, J. Int. Seq. 17 (2014) # 14.4.6
Hong-Mei Liu, Shu-Hua Qi, and Shu-Yan Ding, Some Recurrence Relations for Cauchy Numbers of the First Kind, JIS 13 (2010) # 10.3.8.
Romeo Meštrović, On a Congruence Modulo n^3 Involving Two Consecutive Sums of Powers, Journal of Integer Sequences, Vol. 17 (2014), 14.8.4.
Niels Nielsen, Traité élémentaire des nombres de Bernoulli, Gauthier-Villars, 1923, pp. 398.
Niels Erik Nörlund, Vorlesungen über Differenzenrechnung, Springer-Verlag, Berlin, 1924 [Annotated scanned copy of pages 144-151 and 456-463]
Ronald Orozco López, Solution of the Differential Equation y^(k)= e^(a*y), Special Values of Bell Polynomials and (k,a)-Autonomous Coefficients, Universidad de los Andes (Colombia 2021).
Simon Plouffe, The First 498 Bernoulli numbers [Project Gutenberg Etext]
Jan W. H. Swanepoel, A Short Simple Probabilistic Proof of a Well Known Identity and the Derivation of Related New Identities Involving the Bernoulli Numbers and the Euler Numbers, Integers (2025) Vol. 25, Art. No. A50. See p. 2.
Index entries for sequences related to Bernoulli numbers.

Cf. A090801 (distinct numbers appearing as denominators of Bernoulli numbers)
B_n gives A027641/A027642. See A027641 for full list of references, links, formulas, etc.
See A000367 for numerators. Cf. A027762, A027641, A027642, A002882, A003245, A127187, A127188, A138239, A028246, A143343, A080092, A001897, A277087.
Cf. A160014 for a generalization.

[Denominator(Bernoulli(2*n)): n in [0..60]]; // Vincenzo Librandi, Nov 16 2014

A002445 := n -> mul(i,i=select(isprime,map(i->i+1,numtheory[divisors] (2*n)))): seq(A002445(n),n=0..40); # Peter Luschny, Aug 09 2011
# Alternative
N:= 1000: # to get a(0) to a(N)
A:= Vector(N,2):
for p in select(isprime,[seq(2*i+1,i=1..N)]) do
  r:= (p-1)/2;
  for n from r to N by r do
    A[n]:= A[n]*p
  od
od:
1, seq(A[n],n=1..N); # Robert Israel, Nov 16 2014

Take[Denominator[BernoulliB[Range[0,100]]],{1,-1,2}] (* Harvey P. Dale, Oct 17 2011 *)

a(n)=prod(p=2,2*n+1,if(isprime(p),if((2*n)%(p-1),1,p),1)) \\ Benoit Cloitre

A002445(n,P=1)=forprime(p=2,1+n*=2,n%(p-1)||P*=p);P \\ M. F. Hasler, Jan 05 2016

a(n) = denominator(bernfrac(2*n)); \\ Michel Marcus, Jul 16 2021

def A002445(n):
    if n == 0:
        return 1
    M = (i + 1 for i in divisors(2 * n))
    return prod(s for s in M if is_prime(s))
[A002445(n) for n in (0..57)] # Peter Luschny, Feb 20 2016

E.g.f: x/(exp(x) - 1); take denominators of even powers.
B_{2n}/(2n)! = 2*(-1)^(n-1)*(2*Pi)^(-2n) Sum_{k=1..inf} 1/k^(2n) (gives asymptotics) - Rademacher, p. 16, Eq. (9.1). In particular, B_{2*n} ~ (-1)^(n-1)*2*(2*n)!/ (2*Pi)^(2*n).
If n>=3 is prime,then a((n+1)/2)==(-1)^((n-1)/2)*12*|A000367((n+1)/2)|(mod n). - Vladimir Shevelev, Sep 04 2010
a(n) = denominator(-I*(2*n)!/(Pi*(1-2*n))*integral(log(1-1/t)^(1-2*n) dt, t=0..1)). - Gerry Martens, May 17 2011
a(n) = 2*denominator((2*n)!*Li_{2*n}(1)) for n > 0. - Peter Luschny, Jun 28 2012
a(n) = gcd(2!S(2n+1,2),...,(2n+1)!S(2n+1,2n+1)). Here S(n,k) is the Stirling number of the second kind. See the paper of Komatsu et al. - Istvan Mezo, May 12 2016
a(n) = 2*A001897(n) = A027642(2*n) = 3*A277087(n) for n>0. - Jonathan Sondow, Dec 14 2016

A091137 The Hirzebruch numbers. a(n) = Product_{2 <= p <= n+1, p prime} p^floor(n / (p - 1)).

Original entry on oeis.org

1, 2, 12, 24, 720, 1440, 60480, 120960, 3628800, 7257600, 479001600, 958003200, 2615348736000, 5230697472000, 31384184832000, 62768369664000, 32011868528640000, 64023737057280000, 51090942171709440000, 102181884343418880000, 33720021833328230400000, 67440043666656460800000

Offset: 0

Henry Bottomley, Dec 19 2003

nonn

Largest number m such that number of times m divides k! is almost k/n for large k, i.e., largest m with A090624(m) = n.
This is always a relatively small multiple of n!, since the multiplicity with which a prime p divides n! is always <= n/(p-1); it is equal to floor(n/(p-1)) at least when n is a power of p. - Franklin T. Adams-Watters, May 31 2010
At least for most small n, a(n) = A002790(n) * n!; the first difference is n=15. It appears that A002790(n) * n! always divides a(n).
Conjecture: The denominators of the series reversion of the sequence with e.g.f. Polylog(2,x). - Benedict W. J. Irwin, Jan 05 2017
Not only is a(n) divisible by n!; a(n) is divisible by (n + 1)! as has been observed by Bedhouche and Bakir (see links and A363596). - Hal M. Switkay, Aug 15 2025

			Let n = 4. The partitions of 4 are [[4], [3, 1], [2, 2], [2, 1, 1], [1, 1, 1, 1]]. Thus a(4) = lcm([5, 4*2, 3*3, 3*2*2, 2*2*2*2]) = 720.

P. Curtz, Integration numérique ..., Note 12, C.C.S.A., Arcueil, 1969; see pp. 36, 56.
F. Hirzebruch, Topological Methods in Algebraic Geometry, Springer, 3rd. ed., 1966; Lemma 1.7.3, p. 14. [From N. J. A. Sloane, Sep 06 2010]

Michael De Vlieger, Table of n, a(n) for n = 0..443
Abdelmalek Bedhouche and Bakir Farhi, On some products taken over the prime numbers, arXiv:2207.07957 [math.NT], 2022. See sigma_n p. 3.
Victor M. Buchstaber and Alexander P. Veselov, Todd polynomials and Hirzebruch numbers, arXiv:2310.07383 [math.AT], Oct. 2023.
Donghyun Kim and Jaeseong Oh, Extending the science fiction and the Loehr--Warrington formula, arXiv:2409.01041 [math.CO], 2024. See p. 32.

Starts similarly to A002207 especially for even n and all values of A002207 seen so far seem to divide a(n).

Cf. A002790, A000142, A090622, A090624, A091136, A171080, A238963, A275314, A368093, A368116, A363596.

A091137 := proc(n) local a,i,p ; a := 1 ; for i from 1 do p := ithprime(i) ; if p > n+1 then break; fi; a := a*p^floor(n/(p-1)) ; od: a ; end:
seq(A091137(n), n = 0..47); # R. J. Mathar, Feb 23 2009

A027760[n_] := Product[d, {d, Select[ Divisors[n] + 1, PrimeQ]}]; a[n_] := a[n] = A027760[n]*a[n-1]; a[0] = 1; Table[ a[n], {n, 0, 18}] (* Jean-François Alcover, Oct 04 2011 *)

a(n) = local(r); r=1; forprime(p=2, n+1, r*=p^(n\(p-1))); r
\\ Franklin T. Adams-Watters, May 31 2010

from math import prod
from sympy import primerange
def A091137(n): return prod(p**(n//(p-1)) for p in primerange(n+2))
# Chai Wah Wu, Apr 28 2023

def a(n): return lcm(product(r + 1 for r in p) for p in Partitions(n))
# Or, more efficient:
from functools import cache
@cache
def a_rec(n):
    if n == 0: return 1
    p = mul(s for s in map(lambda i: i + 1, divisors(n)) if is_prime(s))
    return p * a_rec(n - 1)
print([a_rec(n) for n in range(22)]) # Peter Luschny, Dec 12 2023

a(n) = Product_p {p prime} p^floor(n/(p-1)).
a(2n+1) = 2*a(2n).
a(n+1) = A027760(n+1)*a(n). - Paul Curtz, Aug 01 2008
From Peter Luschny, Dec 11 2023: (Start)
a(n) = lcm_{p in P(n)} Product_{r in p}(r + 1), where P(n) are the partitions of n.
a(n) = lcm(A238963row(n)).
a(n) = A368116(1, n), seen as the lcm of the product of the 1-shifted partitions.
a(n) = A368093(1, n), seen as the cumulative product of the Clausen numbers A160014(1, n).  (End)
a(n) = lcm({k: A275314(k) = n+1}). - Hal M. Switkay, Aug 13 2025
a(n) = (n + 1)! * A363596(n). - Hal M. Switkay, Aug 15 2025

New name using a formula of the author by Peter Luschny, Dec 11 2023

A124240 Numbers n such that lambda(n) divides n, where lambda is Carmichael's function (A002322).

Original entry on oeis.org

1, 2, 4, 6, 8, 12, 16, 18, 20, 24, 32, 36, 40, 42, 48, 54, 60, 64, 72, 80, 84, 96, 100, 108, 120, 126, 128, 144, 156, 160, 162, 168, 180, 192, 200, 216, 220, 240, 252, 256, 272, 288, 294, 300, 312, 320, 324, 336, 342, 360, 378, 384, 400, 420, 432, 440, 468, 480

Offset: 1

Alexander Adamchuk, Oct 22 2006

nonn

Numbers n such that A124239(n) is divisible by n.
If k is in the sequence then 2k is also in the sequence, but if 2m is in the sequence m is not necessarily a term of the sequence.
This sequence is a subsequence of A068563. The first term that is different is A068563(27) = 136. The terms of A068563 that are not the terms of a(n) are listed in A124241.
Also, the sequence of numbers n such that p-1 divides n for all primes p that divide n. - Leroy Quet, Jun 27 2008
Numbers n such that b^n == 1 (mod n) for every b coprime to n. - Thomas Ordowski, Jun 23 2017
Numbers m such that every divisor < m is the difference between two divisors of m. - Michel Lagneau, Aug 11 2017
All terms > 1 in this sequence are even. Furthermore, either 4 or 6 divides a(n) for n > 3. 1806 is the largest squarefree term. - Paul Vanderveen, Apr 24 2022

			a(1) = 1 because 1 divides A124239(1) = 1.
a(2) = 2 because 2 divides A124239(2) = 14.
a(3) = 4 because 4 divides A124239(4) = 3704, but 3 does not divide A124239(3) = 197.

Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
Alexander Kalmynin, On Novák numbers, arXiv:1611.00417 [math.NT], 2016. See Theorem 6 p. 11 where these numbers are called Novák-Carmichael numbers.
Eric Weisstein's World of Mathematics, Carmichael Function

Cf. A002322, A124239, A124241, A068563, A027748, A140470, A141766, A160014, A363523.

a124240 n = a124240_list !! (n-1)
a124240_list = filter
   (\x -> all (== 0) $ map ((mod x) . pred) $ a027748_row x) [1..]
-- Reinhard Zumkeller, Aug 27 2013

a:= proc(n) option remember; local k;
       for k from `if`(n=1, 0, a(n-1))+1 while
       irem(k, numtheory[lambda](k))>0 do od: k
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Jul 04 2021
# Using function 'Clausen' from A160014:
aList := m -> select(k -> irem(Clausen(k, 1), Clausen(k, 0)) = 0, [seq(1..m)]):
aList(480); # Peter Luschny, Jun 08 2023

Do[f=n + Sum[ (2k-1)((2k-1)^n-1) / (2(k-1)), {k,2,n} ]; If[IntegerQ[f/n],Print[n]],{n,1,900}]
Flatten[Position[Table[n/CarmichaelLambda[n], {n, 440}], Integer]] (* _T. D. Noe, Sep 11 2008 *)

is(n)=n%lcm(znstar(n)[2])==0 \\ Charles R Greathouse IV, Feb 11 2015

from itertools import islice, count
from sympy.ntheory.factor_ import reduced_totient
def A124240gen(): return filter(lambda n:n % reduced_totient(n) == 0,count(1))
A124240_list = list(islice(A124240gen(),20)) # Chai Wah Wu, Dec 11 2021

k is in a <=> Clausen(k, 0) divides Clausen(k, 1), (Clausen = A160014). - Peter Luschny, Jun 08 2023

New definition from T. D. Noe, Aug 31 2008

Edited by Max Alekseyev, Aug 25 2013

A001897 Denominators of cosecant numbers: -2(2^(2n-1)-1)Bernoulli(2n).

Original entry on oeis.org

1, 3, 15, 21, 15, 33, 1365, 3, 255, 399, 165, 69, 1365, 3, 435, 7161, 255, 3, 959595, 3, 6765, 903, 345, 141, 23205, 33, 795, 399, 435, 177, 28393365, 3, 255, 32361, 15, 2343, 70050435, 3, 15, 1659, 115005, 249, 1702155, 3, 30705, 136059, 705, 3, 2250885, 3, 16665, 2163

Offset: 0

N. J. A. Sloane

Same as half the denominators of the even-indexed Bernoulli numbers B_{2*n} for n>0, by the von Staudt-Clausen theorem and Fermat's little theorem. - Bernd C. Kellner and Jonathan Sondow, Jan 02 2017 [This is implemented in the second Maple program. - Peter Luschny, Aug 21 2021]

			Cosecant numbers {-2*(2^(2*n-1)-1)*Bernoulli(2*n)} are 1, -1/3, 7/15, -31/21, 127/15, -2555/33, 1414477/1365, -57337/3, 118518239/255, -5749691557/399, 91546277357/165, -1792042792463/69, 1982765468311237/1365, -286994504449393/3, 3187598676787461083/435, ... = A001896/A001897.

H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 187.
S. A. Joffe, Sums of like powers of natural numbers, Quart. J. Pure Appl. Math. 46 (1914), 33-51.
N. E. Nörlund, Vorlesungen über Differenzenrechnung. Springer-Verlag, Berlin, 1924, p. 458.
J. Riordan, Combinatorial Identities, Wiley, 1968, p. 199. See Table 3.3.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Hector Blandin and Rafael Diaz, Compositional Bernoulli numbers, arXiv:0708.0809 [math.CO], 2007-2008, Page 7, 3rd table, (B^sin)_1,n is identical to |A001896| / A001897.
S. A. Joffe, Sums of like powers of natural numbers, Quart. J. Pure Appl. Math. 46 (1914), 33-51. [Annotated scanned copy of pages 38-51 only, plus notes]
Masanobu Kaneko, Maneka Pallewatta, and Hirofumi Tsumura, On Polycosecant Numbers, J. Integer Seq. 23 (2020), no. 6, 17 pp. See line k=1 of Table 1 p. 3.
D. H. Lehmer, Lacunary recurrence formulas for the numbers of Bernoulli and Euler, Annals Math., 36 (1935), 637-649.
N. E. Nörlund, Vorlesungen über Differenzenrechnung, Springer 1924, p. 27.
N. E. Nörlund, Vorlesungen über Differenzenrechnung, Springer-Verlag, Berlin, 1924 [Annotated scanned copy of pages 144-151 and 456-463]

Cf. A001896 (numerators), A027642, A033469, A160014, A241171, A277087, A340556.

Cf. A132092-A132099,

[Denominator(2*(1-2^(2*n-1))*Bernoulli(2*n)): n in [0..55]]; // G. C. Greubel, Apr 06 2019

b := n -> bernoulli(n)*2^add(i,i=convert(n,base,2));
a := n -> denom(b(2*n)); # Peter Luschny, May 02 2009
# Alternative :
Clausen := proc(n) local i,S; map(i->i+1, numtheory[divisors](n));
S := select(isprime, %); if S <> {} then mul(i,i=S) else NULL fi end:
A001897_list := n -> [1,seq(Clausen(2*i)/2,i=1..n-1)];
A001897_list(52); # Peter Luschny, Oct 03 2011

a[n_] := Denominator[-2*(2^(2*n-1)-1)*BernoulliB[2*n]]; Table[a[n], {n, 0, 55}] (* Jean-François Alcover, Sep 11 2013 *)

a(n) = denominator(-2*(2^(2*n-1)-1)*bernfrac(2*n)); \\ Michel Marcus, Apr 06 2019

def A001897(n):
    if n == 0:
        return 1
    M = (d + 1 for d in divisors(2 * n))
    return prod(s for s in M if is_prime(s)) / 2
[A001897(n) for n in range(55)]  # Peter Luschny, Feb 20 2016

a(0)=1, a(n)=(1/2)*A002445(n) for n>=1. - Joerg Arndt, May 07 2012
a(n) = denominator((2*n)!*Li_{2*n}(1)) for n > 0. - Peter Luschny, Jun 29 2012
a(0)=1, a(n) = (1/2)*A027642(2*n) = (3/2)*A277087(n) for n>=1. - Jonathan Sondow, Dec 14 2016
From Peter Luschny, Sep 06 2017: (Start)
a(n) = denominator(r(n)) where r(n) = Sum_{0..n} (-1)^(n-k)*A241171(n, k)/(2*k+1).
a(n) = denominator(bernoulli(2*n, 1/2))/4^n = A033469(n)/4^n. (End)
Apparently a(n) = denominator(Sum_{k=0..2*n-2} (-1)^k*E2(2*n-1, k+1)/binomial(4*n-1, k+1)), where E2(n, k) denotes the second-order Eulerian numbers A340556. - Peter Luschny, Feb 17 2021

cp's OEIS Frontend

A268432 a(n) = Pochhammer(n+1, n)/Clausen(n, 1) = A001813(n) / A160014(n, 1).

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A368092 a(n) = A160014(m, n) * a(n - 1) for m = 2 and n > 0, a(0) = 1.

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A325871 a(n) = n!*ClausenNumber(n, 1)/(n + 1), Clausen numbers defined in A160014.

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A346465 Numbers k such that (4^k - 2)*(4^k - 1)/Clausen(2*k, 1) is not squarefree, where Clausen(n, m) = A160014(n, m).

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Mathematica

Formula

Extensions

A363523 k is a term of this sequence if and only if Clausen(k, 0) divides Clausen(k, 2). (Clausen = A160014.)

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Maple

A345365 a(n) = (2*n)!*Pi^(-2*n)*PolyLog(2*n, 1)*Clausen(2*n - 1)/2, where Clausen(n) = A160014(n, 1).

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Mathematica

A002445 Denominators of Bernoulli numbers B_{2n}.

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PARI

PARI

PARI

Sage

Formula

A091137 The Hirzebruch numbers. a(n) = Product_{2 <= p <= n+1, p prime} p^floor(n / (p - 1)).

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A346465 Numbers k such that (4^k - 2)(4^k - 1)/Clausen(2k, 1) is not squarefree, where Clausen(n, m) = A160014(n, m).

A345365 a(n) = (2n)!Pi^(-2n)PolyLog(2n, 1)Clausen(2*n - 1)/2, where Clausen(n) = A160014(n, 1).

A001897 Denominators of cosecant numbers: -2(2^(2n-1)-1)Bernoulli(2n).