cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 11 results. Next

A001067 Numerator of Bernoulli(2*n)/(2*n).

Original entry on oeis.org

1, -1, 1, -1, 1, -691, 1, -3617, 43867, -174611, 77683, -236364091, 657931, -3392780147, 1723168255201, -7709321041217, 151628697551, -26315271553053477373, 154210205991661, -261082718496449122051, 1520097643918070802691, -2530297234481911294093
Offset: 1

Views

Author

N. J. A. Sloane, Richard E. Borcherds (reb(AT)math.berkeley.edu)

Keywords

Comments

It was incorrectly claimed that a(n) is "also numerator of "modified Bernoulli number" b(2n) = Bernoulli(2*n)/(2*n*n!)"; actually, the numerators of these fractions and the numerators of "modified Bernoulli numbers" (see A057868 for details) differ from each other and from this sequence. - Andrey Zabolotskiy, Dec 03 2022
Ramanujan incorrectly conjectured that the sequence contains only primes (and 1). - Jud McCranie. See A112548, A119766.
a(n) = A046968(n) if n < 574; a(574) = 37 * A046968(574). - Michael Somos, Feb 01 2004
Absolute values give denominators of constant terms of Fourier series of meromorphic modular forms E_k/Delta, where E_k is the normalized k th Eisenstein series [cf. Gunning or Serre references] and Delta is the normalized unique weight-twelve cusp form for the full modular group (the generating function of Ramanujan's tau function.) - Barry Brent (barrybrent(AT)iphouse.com), Jun 01 2009
|a(n)| is a product of powers of irregular primes (A000928), with the exception of n = 1,2,3,4,5,7. - Peter Luschny, Jul 28 2009
Conjecture: If there is a prime p such that 2*n+1 < p and p divides a(n), then p^2 does not divide a(n). This conjecture is true for p < 12 million. - Seiichi Manyama, Jan 21 2017

Examples

			The sequence Bernoulli(2*n)/(2*n) (n >= 1) begins 1/12, -1/120, 1/252, -1/240, 1/132, -691/32760, 1/12, -3617/8160, ...
The sequence of modified Bernoulli numbers begins 1/48, -1/5760, 1/362880, -1/19353600, 1/958003200, -691/31384184832000, ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 259, (6.3.18) and (6.3.19); also p. 810.
  • L. V. Ahlfors, Complex Analysis, McGraw-Hill, 1979, p. 205
  • R. C. Gunning, Lectures on Modular Forms. Princeton Univ. Press, Princeton, NJ, 1962, p. 53.
  • R. Kanigel, The Man Who Knew Infinity, pp. 91-92.
  • J. W. Milnor and J. D. Stasheff, Characteristic Classes, Princeton, 1974, p. 285.
  • J.-P. Serre, A Course in Arithmetic, Springer-Verlag, 1973, p. 93.

Crossrefs

Similar to but different from A046968. See A090495, A090496.
Denominators given by A006953.

Programs

  • GAP
    List([1..25], n-> NumeratorRat(Bernoulli(2*n)/(2*n)));  # G. C. Greubel, Sep 19 2019
  • Magma
    [Numerator(Bernoulli(2*n)/(2*n)):n in [1..40]]; // Vincenzo Librandi, Sep 17 2015
    
  • Maple
    A001067_list := proc(n) 1/(1-1/exp(z)); series(%,z,2*n+4);
    seq(numer((2*i+1)!*coeff(%,z,2*i+1)),i=0..n) end:
    A001067_list(21); # Peter Luschny, Jul 12 2012
  • Mathematica
    Table[ Numerator[ BernoulliB[2n]/(2n)], {n, 1, 22}] (* Robert G. Wilson v, Feb 03 2004 *)
  • PARI
    {a(n) = if( n<1, 0, numerator( bernfrac(2*n) / (2*n)))}; /* Michael Somos, Feb 01 2004 */
    
  • Sage
    @CachedFunction
    def S(n, k) :
        if k == 0 :
            if n == 0 : return 1
            else: return 0
        return S(n, k-1) + S(n-1, n-k)
    def BernoulliDivN(n) :
        if n == 0 : return 1
        return (-1)^n*S(2*n-1,2*n-1)/(4^n-16^n)
    [BernoulliDivN(n).numerator() for n in (1..22)]
    # Peter Luschny, Jul 08 2012
    
  • Sage
    [numerator(bernoulli(2*n)/(2*n)) for n in (1..25)] # G. C. Greubel, Sep 19 2019
    

Formula

Zeta(1-2*n) = - Bernoulli(2*n)/(2*n).
G.f.: numerators of coefficients of z^(2*n) in z/(exp(z)-1). - Benoit Cloitre, Jun 02 2003
For 2 <= k <= 1000 and k != 7, the 2-order of the full constant term of E_k/Delta = 3 + ord_2(k - 7). - Barry Brent (barrybrent(AT)iphouse.com), Jun 01 2009
G.f. for Bernoulli(2*n)/(2*n) = a(n)/A006953(n): (-1)^n/((2*Pi)^(2*n)*(2*n))*integral(log(1-1/t)^(2*n) dt,t=0,1). - Gerry Martens, May 18 2011
E.g.f.: a(n) = numerator((2*n+1)!*[x^(2*n+1)](1/(1-1/exp(x)))). - Peter Luschny, Jul 12 2012
|a(n)| = numerator of Integral_{r=0..1} HurwitzZeta(1-n, r)^2 dr. More general: |Bernoulli(2*n)| = binomial(2*n,n)*n^2*I(n) for n >= 1 where I(n) denotes the integral. - Peter Luschny, May 24 2015

A090495 Numbers k such that numerator(Bernoulli(2*k)/(2*k)) is different from numerator(Bernoulli(2*k)/(2*k*(2*k-1))).

Original entry on oeis.org

574, 1185, 1240, 1269, 1376, 1906, 1910, 2572, 2689, 2980, 3238, 3384, 3801, 3904, 4121, 4570, 4691, 4789, 5236, 5862, 5902, 6227, 6332, 6402, 6438, 6568, 7234, 7900, 8113, 8434, 8543, 8557, 8566, 9232, 9611, 9670, 9824, 9891, 9898, 10564, 10587, 10754, 11230, 11247, 11535, 11691, 11896, 12562, 12965, 13019, 13228, 13246, 13355, 13484, 13894, 14560, 14714, 14957, 15176, 15226, 15346, 15892, 16558, 16668, 16944, 17035, 17224, 17387, 17890, 18379, 18406, 18534, 18556, 18761, 19222, 19598, 19888, 20090
Offset: 1

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Author

N. J. A. Sloane, Feb 03 2004

Keywords

Comments

Michael Somos (Feb 01 2004) discovered the remarkable fact that A001067 is different from A046968, even though they agree for the first 573 terms.
Numbers n such that A001067 is different from A046968, or alternatively, those n such that gcd(A001067(n),2n-1) is > 1.
If gcd(A000367(n), A000367(n+2)) <>1 then n = A090495(n) - (3*A090496(n) + 1)/2. - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Feb 08 2004
So far, all terms correspond to irregular primes. Notice that these numbers are generated by n=((2k+1)p+1)/2 where p is an irregular prime and k is some integer = 1,2,... . In the Excel spreadsheet provided at the link, you will notice that much larger firstborn irregular primes p tend to produce smaller values of k. E.g., p = 691, 683, 653, k = 5, 15, 23. So by some guessing we could test a given large irregular prime for the first few values of k. I found ip's 257, 293, 311 this way, but not the index. Also the spreadsheet shows the corresponding irregular primes where the Bacher forecast fails for firstborn irregular prime. - Cino Hilliard, Feb 15 2004

Crossrefs

Programs

  • Maple
    a := n->numer(bernoulli(2*n)/(2*n)): b := n->numer(bernoulli(2*n)/(2*n*(2*n-1))): for n from 1 to 2000 do if a(n)<>b(n) then print(n,a(n)/b(n)); fi; od:
  • Mathematica
    a[n_] := Numerator[BernoulliB[2n]/(2n)] (* A001067 *); b[n_] := Numerator[BernoulliB[2n]/(2n(2n-1))] (* A046968 *); For[n=1, n <= 580, n++, If[ a[n] != b[n], Print[n, " ", a[n]/b[n]] ] ]
    k = 1; lst = {}; While[k < 38001, b = BernoulliB[2 k]; If [Numerator[b/(2 k)] != Numerator[b/(2 k (2 k - 1))], AppendTo[lst, k]; Print[{k}]]; k++ ]; lst (* Robert G. Wilson v, Aug 19 2010 *)
  • PARI
    bern2(c,m1,m2) = { for(n=m1,m2, n2=n+n; a = numerator(bernfrac(n2)/(n2)); \ A001067 b = numerator(a/(n2-1)); if(a <> b,print("A("c") = "n","a/b);c++) ) } \\ Cino Hilliard

Extensions

a(1)-a(7) from Michael Somos and W. Edwin Clark, Feb 03 2004
a(8)-a(9) from Robert G. Wilson v, Feb 03 2004
a(10)-a(12) from Eric W. Weisstein, Feb 03 2004
a(13)-a(39) from Cino Hilliard, Feb 03 2004
a(40) from Eric W. Weisstein, Feb 04 2004
Many further terms from Cino Hilliard, Feb 15 2004

A090496 Ratio of numerator(Bernoulli(2*n)/(2*n)) to numerator(Bernoulli(2*n)/(2*n*(2*n-1))) for n's for which they are different.

Original entry on oeis.org

37, 103, 37, 59, 131, 37, 67, 37, 283, 59, 37, 101, 691, 37, 67, 37, 59, 157, 37, 617, 37, 593, 67, 59, 103, 37, 37, 37, 59, 101, 67, 157, 37, 37, 149, 233, 59, 131, 37, 37, 683, 67, 37, 271, 59, 103, 37, 37, 67, 263, 37, 59, 307, 101, 37, 37, 577, 59, 67, 37, 653, 37, 37, 59, 103, 157, 37, 67, 37, 59, 131, 101
Offset: 1

Views

Author

N. J. A. Sloane, Feb 03 2004

Keywords

Comments

A001067(n) / A046968(n) when they are different, or alternatively, gcd(A001067(n),2n-1) when that number is > 1.
These numbers are always products of irregular primes (A000928).
All values yielding 37 are of the form 574+666*k, k=0,1,2,3,4,... and form thus an arithmetic progression with step 666=18*37=((37-1)/2)*37. All values yielding 59 are of the form 1269+1711*k, k=0,1,2,3 and 1711=28*59=((59-1)/2)*59. The two values yielding 67 are at distance 2211=((67-1)/2)*67. Conjecture: all indices yielding a given prime p form an arithmetic progression of step ((p-1)/2)*p. See A092291. - Roland Bacher, Feb 04 2004
The positions where 37 occurs appear to coincide with A026352. - Mohammed Bouayoun, Feb 05 2004
Roland Bacher conjectures that values of n yielding the same quotient p form an arithmetic progression n0+d*k, where d = p(p-1)/2. Actual and conjectured values of n0 are in the sequence A092291.
Composite values do occur. An example is 2n = 272876, which yields a quotient of 37*59. This was found by tdn using the Kummer congruences and CRT: using the irregular pairs (37,32) and (59,44), we know that the following Diophantine equations must be solved for (k,l,m): 32+36*k = 44+58*l = 1+37*59*m. Some quotients are not possible, e.g., 37*67, 37*103. All quotients are the product of irregular primes A000928. Composite quotients imply there are missing terms in the arithmetic progression conjectured by Bacher. - T. D. Noe, Feb 12 2004

Crossrefs

Programs

  • Mathematica
    A090496 = {}; Do[ r = Numerator[ b = BernoulliB[2n]/(2n) ] / Numerator[ b/(2n-1) ]; If[ r > 1, Print[n, " ", r]; AppendTo[ A090496, r] ], {n, 1, 20000}]; A090496 (* Jean-François Alcover, Jan 24 2012 *)

Extensions

a(1)-a(7) from Michael Somos and W. Edwin Clark, Feb 03 2004
a(8), a(9) from Robert G. Wilson v, Feb 03 2004
a(10)-a(12) from Eric W. Weisstein, Feb 03 2004
a(13)-a(39) from Cino Hilliard, Feb 03 2004
a(40)-a(44) from Eric W. Weisstein, Feb 04 2004
Terms from a(45) onwards from David Wasserman, Dec 06 2005

A141590 a(n) = numerator of Bernoulli(2*n)/(2*n + 1)!. Bisection of A120082.

Original entry on oeis.org

1, 1, -1, 1, -1, 1, -691, 1, -3617, 43867, -174611, 77683, -236364091, 657931, -3392780147, 1723168255201, -7709321041217, 151628697551, -26315271553053477373, 154210205991661, -261082718496449122051, 1520097643918070802691, -2530297234481911294093
Offset: 0

Views

Author

Paul Curtz, Aug 20 2008

Keywords

Comments

Numerators of the Taylor expansion coefficients of the Debye function D(1,x) at the even powers of x.

Examples

			Note that a(34) = -125235502160125163977598011460214000388469 but A255505(34) = -4633713579924631067171126424027918014373353.
		

Crossrefs

Programs

Formula

a(n) = A120082(2*n).

Extensions

Edited and extended by R. J. Mathar, Sep 03 2009
Edited by Peter Luschny, Dec 03 2022

A046969 Denominators of coefficients in Stirling's expansion for log(Gamma(z)).

Original entry on oeis.org

12, 360, 1260, 1680, 1188, 360360, 156, 122400, 244188, 125400, 5796, 1506960, 300, 93960, 2492028, 505920, 396, 2418179400, 444, 21106800, 3109932, 118680, 25380, 104700960, 6468, 324360, 2283876, 382800, 40356, 201025024200, 732
Offset: 1

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Author

Douglas Stoll, dougstoll(AT)email.msn.com

Keywords

Comments

From Lorenzo Sauras Altuzarra, Oct 13 2020: (Start)
Conjecture I: if n > 2, then a(A005382(n))/12 is prime.
Conjecture II: if a(n)/12 is prime, then a(n-1)/12 - (n-1), a(n)/12 - n and a(n+2)/12 - (n+2) are multiples of 6. (End)

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math.Series 55, Tenth Printing, 1972, p. 257, Eq. 6.1.41.
  • L. V. Ahlfors, Complex Analysis, McGraw-Hill, 1979, p. 205

Crossrefs

Numerators are given in A046968. Cf. A005382.

Programs

  • Maple
    a := n -> denom(bernoulli(2*n)/(2*n*(2*n-1))): # Lorenzo Sauras Altuzarra, Oct 13 2020
  • Mathematica
    Table[ Denominator[ BernoulliB[2n]/(2n(2n - 1))], {n, 31}] (* Robert G. Wilson v, Sep 21 2006 *)
    s = LogGamma[z] + z - (z - 1/2) Log[z] - Log[2 Pi]/2 + O[z, Infinity]^62;
    DeleteCases[CoefficientList[s, 1/z], 0] // Denominator (* Jean-François Alcover, Jun 13 2017 *)
  • PARI
    a(n)=if(n<1,0,denominator(bernfrac(2*n)/(2*n)/(2*n-1)))

Formula

From denominator of Jk(z) = (-1)^(k-1)*Bk/(((2k)*(2k-1))*z^(2k-1)), so Gamma(z) = sqrt(2pi)*z^(z-0.5)*exp(-z)*exp(J(z)).

Extensions

More terms from Frank Ellermann, Jun 13 2001
Bayes reference from Henry Bottomley, Jun 03 2003

A255505 Numerator of Bernoulli(2n)/(2n!).

Original entry on oeis.org

1, 1, -1, 1, -1, 1, -691, 1, -3617, 43867, -174611, 77683, -236364091, 657931, -3392780147, 1723168255201, -7709321041217, 151628697551, -26315271553053477373, 154210205991661, -261082718496449122051, 1520097643918070802691
Offset: 0

Views

Author

Jean-François Alcover, Feb 24 2015

Keywords

Comments

This sequence is different from A001067 or A046968 or A141590, at least at a(52).

Examples

			The sequence Bernoulli(2n)/(2n!) (n >= 0) begins 1/2, 1/12, -1/120, 1/504, -1/1440, 1/3168, -691/3931200, 1/8640, -3617/41126400, ...
		

Crossrefs

Cf. A000367, A001067, A046968, A141590, A255506 (denominator).

Programs

  • Magma
    [Numerator(Bernoulli(2*n)/(2*Factorial(n))):n in [0..30]]; // Vincenzo Librandi, Feb 24 2015
    
  • Mathematica
    Table[Numerator[BernoulliB[2 n]/(2 n!)], {n, 0, 25}]
  • PARI
    a(n) = numerator(bernfrac(2*n)/(2*n!)); \\ Michel Marcus, Feb 24 2015
    
  • Sage
    [numerator(bernoulli(2*n)/(2*factorial(n))) for n in (0..25)] # Bruno Berselli, Feb 24 2015

A231273 Numerator of zeta(4n)/(zeta(2n) * Pi^(2n)).

Original entry on oeis.org

1, 1, 1, 691, 3617, 174611, 236364091, 3392780147, 7709321041217, 26315271553053477373, 261082718496449122051, 2530297234481911294093, 5609403368997817686249127547, 61628132164268458257532691681, 354198989901889536240773677094747
Offset: 0

Views

Author

Leo Depuydt, Nov 07 2013

Keywords

Comments

Integer component of the numerator of a close variant of Euler's infinite prime product zeta(2n) = Product_{k>=1} (prime(k)^(2n))/(prime(k)^(2n)-1), namely with all minus signs changed into plus signs, as follows: zeta(4n)/zeta(2n) = Product_{k>=1} (prime(k)^(2n))/(prime(k)^(2n)+1). The transcendental component is Pi^(2n).
For a detailed account of the results, including proof and relation to the zeta function, see Links for the PDF file submitted as supporting material.
The reference to Apostol is to a discussion of the equivalence of 1) zeta(2s)/zeta(s) and 2) a related infinite prime product, that is, Product_{sigma>1} prime(n)^s/(prime(n)^s + 1), with s being a complex variable such that s = sigma + i*t where sigma and t are real (following Riemann), using a type of proof different from the one posted below involving zeta(4n)/zeta(2n). On this, see also Hardy and Wright cited below. - Leo Depuydt, Nov 22 2013, Nov 27 2013
The background of the sequence is now described in the link below to L. Depuydt, The Prime Sequence ... . - Leo Depuydt, Aug 22 2014
From Robert Israel, Aug 22 2014: (Start)
Numerator of (-1)^n*B(4*n)*4^n*(2*n)!/(B(2*n)*(4*n)!), where B(n) are the Bernoulli numbers (see A027641 and A027642).
Not the same as abs(A001067(2*n)): they differ first at n=17.
(End)

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer, 1976, p. 231.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fourth Edition, Clarendon Press, 1960, p. 255.

Crossrefs

Cf. A231327 (corresponding denominator).
Cf. A114362 and A114363 (closely related results).
Cf. A001067, A046968, A046988, A098087, A141590, A156036 (same number sequence, though in various transformations (alternation of signs, intervening numbers, and so on)).

Programs

  • Maple
    seq(numer((-1)^n*bernoulli(4*n)*4^n*(2*n)!/(bernoulli(2*n)*(4*n)!)),n=0..100); # Robert Israel, Aug 22 2014
  • Mathematica
    Numerator[Table[Zeta[4n]/(Zeta[2n] * Pi^(2n)), {n, 0, 15}]] (* T. D. Noe, Nov 18 2013 *)

A231327 Denominator of rational component of zeta(4n)/zeta(2n).

Original entry on oeis.org

1, 15, 105, 675675, 34459425, 16368226875, 218517792968475, 30951416768146875, 694097901592400930625, 23383376494609715287281703125, 2289686345687357378035370971875, 219012470258383844016431785453125, 4791965046290912124048163518904807546875
Offset: 0

Views

Author

Leo Depuydt, Nov 07 2013

Keywords

Comments

Denominator of a close variant of Euler's infinite prime product zeta(2n) = Product_{k>=1} (prime(k)^(2n))/(prime(k)^(2n)-1), namely with all minus signs changed into plus signs, as follows: zeta(4n)/zeta(2n) = Product_{k>=1} prime(k)^(2n)/(prime(k)^(2n)+1).
For a detailed account of the results in question, including proof and relation to the zeta function, see the PDF file submitted as supporting material in A231273.
The reference to Apostol below is a discussion of the equivalence of 1) zeta(2s)/zeta(s) and 2) a related infinite prime product, that is, Product_{sigma>1} prime(n)^s/(prime(n)^s + 1), with s being a complex variable such that s = sigma + i*t where sigma and t are real (following Riemann), using a type of proof different from the one posted below involving zeta(4n)/zeta(2n). - Leo Depuydt, Nov 22 2013
Denominator of B(4*n)*4^n*(2*n)!/(B(2*n)*(4*n)!) where B(n) are the Bernoulli numbers (see A027641 and A027642). - Robert Israel, Aug 22 2014

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer, 1976, p. 231.

Crossrefs

Cf. A231273 (the corresponding numerator).
Cf. A114362 and A114363 (closely related results).
Cf. A001067, A046968, A046988, A098087, A141590, and A156036 (same number sequence as found in numerator, though in various transformations (alternation of sign, intervening numbers, and so on)).
Cf. A027641 and A027642.

Programs

  • Maple
    seq(denom(bernoulli(4*n)*4^n*(2*n)!/(bernoulli(2*n)*(4*n)!)),n=0..100); # Robert Israel, Aug 22 2014
  • Mathematica
    Denominator[Table[Zeta[4 n]/Zeta[2 n], {n, 0, 15}]] (* T. D. Noe, Nov 15 2013 *)

A057868 Denominator of "modified Bernoulli number" b(2n) = Bernoulli(2*n)/(4*n*(2*n)!).

Original entry on oeis.org

48, 5760, 362880, 19353600, 958003200, 31384184832000, 2092278988800, 341459930972160000, 183927391818153984000, 32114306507931648000000, 620448401733239439360000, 81303558563123696133734400000, 9678995067038535254016000000, 2122022878497528469090467840000000
Offset: 1

Views

Author

Keywords

Comments

Note that Weisstein gives the formula b(n) = B(n)/(2*n*n!), and a(n) is the denominator of b(2*n). Numerators seem to be A141590 (not A001067 or A046968 or A255505). - Andrey Zabolotskiy, Dec 03 2022

Examples

			The sequence of modified Bernoulli numbers begins 1/48, -1/5760, 1/362880, -1/19353600, 1/958003200, -691/31384184832000, ...
		

Crossrefs

Numerators seem to be A141590.
Cf. A001067.

Programs

  • Maple
    seq(denom(bernoulli(2*n)/((4*n)*(2*n)!)), n = 1..14); # Peter Luschny, Dec 03 2022
  • Mathematica
    a[n_] := Denominator[ BernoulliB[2n] / (8n^2*(2n-1)!)];
    Table[a[n], {n, 1, 12}] (* Jean-François Alcover, Jun 07 2012 *)

Extensions

Name edited by Andrey Zabolotskiy, Dec 03 2022

A230282 Largest k such that (k*n)! >= (k!)^(n+1).

Original entry on oeis.org

1, 1, 6, 64, 679, 8468, 126784, 2238565, 45605124, 1053117974, 27182818156, 775557529509, 24236473829015, 823299898542083, 30205566231626957, 1190319005015526817, 50143449209799256306, 2248672171655330927835
Offset: 0

Views

Author

Alex Ratushnyak, Oct 14 2013

Keywords

Examples

			Biggest k such that (3*k)! >= k!^4 is k = 64, so a(3) = 64.
a(10) = 27182818156 because k = 27182818156 satisfies the inequality (k*10)! >= (k!)^11, but k = 27182818157 does not. To verify this, note that taking the logarithm of each side of the inequality gives log((k*10)!) >= 11*log(k!), and use the series expression log(m!) = log(2*Pi*m)/2 + m*log(m) - m + (1/12)/m - (1/360)/m^3 + (1/1260)/m^5 - ... (where the numerators and denominators of the fractions 1/12, -1/360, 1/1260, etc., are from A046968 and A046969, respectively), to get, at k = 27182818156, log(271828181560!) = 6884982704601.26... for the left hand side of the inequality, and the slightly smaller result 11*log(27182818156!) = 6884982704600.83... for the right hand side; then repeat the calculations using k = 27182818157, and observe that this makes the right hand side slightly larger than the left hand side. - _Jon E. Schoenfield_, Oct 23 2013
		

Crossrefs

Programs

  • Mathematica
    Table[k = 0; While[(k n)! >= (k!)^(n + 1), k++]; k - 1, {n, 0, 4}] (* T. D. Noe, Oct 18 2013 *)
  • Python
    import math
    for n in range(8):
      for k in range(10000000):
        if math.factorial(n*k) < math.factorial(k)**(n+1):
          print(k-1, end=', ')
          break

Formula

For n > 1, a(n) = floor(e*(n^n) - ((n^2-1)*log(n) + n*(1+log(2*Pi)))/2) [conjectural, but verified for all n in 2..5000]. - Jon E. Schoenfield, Oct 22 2013

Extensions

a(7)-a(17) from Jon E. Schoenfield, Oct 22 2013
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