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 10 results.

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

Views

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

A046968 Numerators of coefficients in Stirling's expansion for log(Gamma(z)).

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
Offset: 1

Views

Author

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

Keywords

Comments

A001067(n) = a(n) if n<574; A001067(574) = 37*a(574). - Michael Somos, Feb 01 2004

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

Denominators given by A046969.
Similar to but different from A001067. See A090495, A090496.

Programs

  • GAP
    List([1..25], n-> NumeratorRat(Bernoulli(2*n)/(2*n*(2*n-1))) ); # G. C. Greubel, Sep 19 2019
  • Magma
    [Numerator(Bernoulli(2*n)/(2*n*(2*n-1))): n in [1..25]]; // G. C. Greubel, Sep 19 2019
    
  • Maple
    seq(numer(bernoulli(2*n)/(2*n*(2*n-1))), n = 1..25); # G. C. Greubel, Sep 19 2019
  • Mathematica
    Table[ Numerator[ BernoulliB[2n]/(2n(2n - 1))], {n, 1, 22}] (* Robert G. Wilson v, Feb 03 2004 *)
    s = LogGamma[z] + z - (z - 1/2) Log[z] - Log[2 Pi]/2 + O[z, Infinity]^42; DeleteCases[CoefficientList[s, 1/z], 0] // Numerator (* Jean-François Alcover, Jun 13 2017 *)
  • PARI
    a(n)=if(n<1,0,numerator(bernfrac(2*n)/(2*n)/(2*n-1)))
    
  • Sage
    [numerator(bernoulli(2*n)/(2*n*(2*n-1))) for n in (1..25)] # G. C. Greubel, Sep 19 2019
    

Formula

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

Extensions

More terms from Frank Ellermann, Jun 13 2001

A092291 Let p = n-th irregular prime, A000928(n). Then a(n) = smallest value of m such that numerator(Bernoulli(2*m)/(2*m)) / numerator(Bernoulli(2*m)/(2*m*(2*m-1))) equals p.

Original entry on oeis.org

574, 1269, 1910, 3384, 1185, 1376, 9611, 4789, 9670, 20946, 13019, 11247, 2689, 22708, 13355, 45251, 48407, 32653, 18761, 38706, 76391, 25563, 50310, 79023, 44948, 29864, 21716, 71441, 104339, 22993, 73572, 61549, 14714, 26122, 6227, 179369, 159687, 5862, 132157, 24925, 76023, 15346, 73479, 136956, 212240, 10587, 3801, 137040, 108520, 194171, 98550, 282532, 87272, 133081, 220187, 305002, 41764, 27268, 380180, 70921, 184940, 241076, 73858, 80108, 250927
Offset: 1

Views

Author

N. J. A. Sloane, based on a suggestion of Roland Bacher, Feb 05 2004

Keywords

Comments

It was conjectured that a(n) = (1 + A000928(n) * (A035112(n) - 1))/2. However, Bernd Kellner's insightful paper shows that this formula first fails for the irregular prime 6449. - T. D. Noe, Feb 10 2004

Crossrefs

Term in A090495 corresponding to first occurrence of p in A090496.

Programs

  • Mathematica
    (* This program is not convenient for a large number of terms *) irregularPrimeQ[p_] := Module[{k = 1}, While[2*k <= p-3 && Mod[ Numerator[ BernoulliB[2*k]], p] != 0, k++]; 2*k <= p-3]; irregularPrime[1] = 37; irregularPrime[n_] := irregularPrime[n] = Module[{p}, For[p = NextPrime[ irregularPrime[n-1]], True, p = NextPrime[p], If[ irregularPrimeQ[p], Return[p]]]]; a[n_] := a[n] = For[m = 1, True, m++, If[ Numerator[BernoulliB[2*m]/(2*m)] / Numerator[ BernoulliB[2*m]/(2*m*(2*m-1))] == irregularPrime[n], Return[m]]]; Table[ Print[a[n]]; a[n], {n, 1, 15}] (* Jean-François Alcover, Sep 27 2013 *)

Extensions

Initial terms were computed by Roland Bacher, Feb 04 2004; further terms from Hans Havermann, Feb 05 2004 and T. D. Noe, Feb 06 2004
Offset modified by Jean-François Alcover, Sep 27 2013

A090177 Numbers n such that numerator(Bernoulli(2*n)/(2*n)) is different from numerator(Bernoulli(2*n)/(2*n*(2*n+1))).

Original entry on oeis.org

610, 1276, 1287, 1327, 1506, 1942, 1976, 2608, 2971, 3038, 3274, 3484, 3940, 4187, 4491, 4606, 4749, 4945, 5272, 5938, 6398, 6460, 6478, 6540, 6604, 6819, 7270, 7936, 8171, 8534, 8602, 8609, 8713, 9268, 9759, 9882, 9902, 9934, 10021
Offset: 1

Views

Author

mohammed bouayoun (bouyao(AT)wanadoo.fr), Feb 04 2004

Keywords

Comments

Coincides with A090495(n) + A090496(n) - 1, except for order of terms.

Crossrefs

Programs

  • Mathematica
    Do[ If[ Numerator[ BernoulliB[2n] / (2n)] != Numerator[ BernoulliB[2n] / (2n(2n + 1))], Print[n]], {n, 1, 10030}]

Extensions

a(6)-a(39) from Robert G. Wilson v, Feb 09 2004

A090790 Numbers r arising in A090791.

Original entry on oeis.org

30, 42, 56, 66, 22, 20, 128, 60, 82, 162, 98, 82, 18
Offset: 1

Views

Author

Cino Hilliard, Feb 16 2004

Keywords

Comments

These values of r correspond to the first 13 irregular primes produced by a/b.

Examples

			Given a, b as defined above and p=37, r=30, 52 = pk+r/2 = 37*1 + 30/2 is the smallest number that for a<>b a/b = 37.
		

Crossrefs

Programs

  • PARI
    bern3(m,r) = { for(i=m,m, p=irprime(i); /* use the Somos script below to get irregular prime */ for(k=1,p, if(r%2,n=p*k+(p+r)/2,n=p*k+r/2); n2=n+n; a = numerator(bernfrac(n2)/(n2));
    b = numerator(a/(n2-r)); v=a/b; if(a <> b && v==p,print(k","n","v);break) ) ) } /* A001067 */
    
  • PARI
    irprime(n) = { my(p); if(n<1, 0, p=irprime(n-1) + (n==1); while(p = nextprime(p+2), forstep(i=2, p-3, 2, if( numerator(bernfrac(i))%p == 0, break(2)))); p) }; /* compute irregular primes irprime from - Michael Somos, Feb 04 2004 */

Formula

Given a = numerator(Bernoulli(2*n)/(2*n)) and b = numerator(a/(2*n-r)) for integer r positive or negative, then n>0 n = p*k+(p+r)/2 if r is odd and n = p*k+r/2 if r is even where k = 1, 2.. For every irregular prime p there is an r such that n is minimum.

A090791 Minimal numbers n such that numerator(Bernoulli(2*n)/(2*n)) is different from numerator(Bernoulli(2*n)/(2*n*(2*n-r))) for some integer r.

Original entry on oeis.org

52, 80, 95, 134, 114, 141, 213, 187, 274, 338, 312, 312, 292
Offset: 1

Views

Author

Cino Hilliard, Feb 16 2004

Keywords

Comments

These values of n correspond to the first 13 irregular primes produced by a/b.

Examples

			Given a,b as defined above and p=37,r=30, n=pk+r/2 = 37*k + 30/2 = 37k+15 = 52 = the smallest number that for a<>b a/b = 37.
		

Crossrefs

Programs

  • PARI
    bern3(m,r) = { for(i=m,m, p=irprime(i); /* use the Somos script below to get irregular prime */ for(k=1,p, if(r%2,n=p*k+(p+r)/2,n=p*k+r/2); n2=n+n; a = numerator(bernfrac(n2)/(n2));
    b = numerator(a/(n2-r)); v=a/b; if(a <> b && v==p,print(k","n","v);break) ) ) } /* A001067 */
    
  • PARI
    irprime(n) = { my(p); if(n<1, 0, p = irprime(n-1) + (n==1); while(p = nextprime(p+2), forstep(i=2, p-3, 2, if( numerator(bernfrac(i))%p == 0, break(2)))); p) };  /* compute irregular primes irprime from - Michael Somos, Feb 04 2004 */

Formula

Given a = numerator(Bernoulli(2*n)/(2*n)) and b = numerator(a/(2*n-r)) for integer r positive or negative, then n>0 n = p*k+(p+r)/2 if r is odd and n = p*k+r/2 if r is even where k = 1, 2.. For every irregular prime p there is an r such that n is minimum.

A090798 Irregular primes in the ratio numerator(Bernoulli(2*n)/(2*n)) / numerator(Bernoulli(2*n)/(2*n*(2*n-r))) when these numerators are different and n is a minimum for some integer r. Duplication indicates irregularity index > 1.

Original entry on oeis.org

37, 59, 67, 101, 103, 131, 149, 157, 157, 233, 257, 263, 271, 283, 293, 307, 311, 347, 353, 353, 379, 379, 389, 401, 409, 421, 433, 461, 463, 467, 467, 491, 491, 491, 523, 541, 547, 547, 557, 577, 587, 587, 593, 607, 613, 617, 617, 617, 619, 631, 631, 647
Offset: 1

Views

Author

Cino Hilliard, Feb 16 2004

Keywords

Comments

Only even values of r need to be tested.
See Table A.3, "Calculated irregular pairs of order 10 of primes below 1000," in B. C. Kellner.

Crossrefs

Programs

  • Mathematica
    f[p_] := Block[{c = 0, k = 1}, While[ 2k <= p - 3, If[ Mod[ Numerator@ BernoulliB[ 2k], p] == 0, c++]; k++]; c]; p = 5; lst = {}; While[p < 1001, AppendTo[lst, Table[p, {f@ p}]]; p = NextPrime@ p]; Flatten@ lst
  • PARI
    \ prestore some ireg primes in iprime[] bernmin(m) = { for(x=1,m, p=iprime[x]; forstep(r=2,p,2, n=r/2+p; n2=n+n; a = numerator(bernfrac(n2)/(n2)); \ A001067 b = numerator(a/(n2-r)); \ if(a <> b,print(r","n","a/b)) if(a <> b,print1(a/b",")) ) ) }

Formula

Given a = numerator(Bernoulli(2*n)/(2*n)) and b = numerator(a/(2*n-r)) for integer r positive or negative, then n>0 n = p + r/2 For every irregular prime p there is an r such that n is minimum.

A090793 Minimal numbers n such that numerator(Bernoulli(2*n)/(2*n)) is different from numerator(Bernoulli(2*n)/(2*n*(2*n-r))) for some integer r and the first m irregular primes including irregularity index > 1.

Original entry on oeis.org

52, 80, 95, 134, 114, 141, 213, 187, 211, 274, 338, 312, 312, 292, 370, 350, 456, 486, 445, 502, 428, 465, 488, 591, 471, 540, 615, 558, 527, 513, 563, 636, 658, 659, 722, 583, 681, 789, 667, 602, 631, 632, 603, 902, 873, 626, 703, 785, 832, 670, 743, 764
Offset: 1

Views

Author

Cino Hilliard, Feb 16 2004

Keywords

Comments

Only even values of r are tested.

Crossrefs

Programs

  • PARI
    \ prestore some ireg primes in iprime[] bernmin(m) = { for(x=1,m, p=iprime[x]; forstep(r=2,p,2, n=r/2+p; n2=n+n; a = numerator(bernfrac(n2)/(n2)); \ A001067 b = numerator(a/(n2-r)); \ if(a <> b,print(r","n","a/b)) if(a <> b,print1(n",")) ) ) }

Formula

Given a = numerator(Bernoulli(2*n)/(2*n)) and b = numerator(a/(2*n-r)) for integer r positive or negative, then n>0 n = p + r/2 For every irregular prime p there is an r such that n is minimum.

A090800 r when numerator(Bernoulli(2*n)/(2*n)) and numerator(Bernoulli(2*n)/(2*n*(2*n-r))) are different and n is minimum for some integer r for the first i irregular primes. These include entries when the irregularity index > 1.

Original entry on oeis.org

30, 42, 56, 66, 22, 20, 128, 60, 108, 82, 162, 98, 82, 18, 154, 86, 290, 278, 184, 298, 98, 172, 198, 380, 124, 238, 364, 194, 128, 92, 192, 290, 334, 336, 398, 84, 268, 484, 220, 50, 88, 90, 20, 590, 520, 18, 172, 336, 426, 78, 224, 234, 240, 552, 46, 222, 406, 500
Offset: 2

Views

Author

Cino Hilliard, Feb 16 2004

Keywords

Comments

This is a generalization of the concept in A090495 and A090496. One can change the code below from p = iprime[x] to p = prime(x) and see that data for only irregular primes is generated.

Crossrefs

Programs

  • PARI
    \ prestore some ireg primes in iprime[] or use slower PARI BIF prime() bernmin(m) = { for(x=1,m, p=iprime[x]; forstep(r=2,p,2, n=r/2+p; n2=n+n; a = numerator(bernfrac(n2)/(n2)); \ A001067 b = numerator(a/(n2-r)); \ if(a <> b,print(r","n","a/b)) if(a <> b,print1(r",")) ) ) }

Formula

Given a = numerator(Bernoulli(2*n)/(2*n)) and b = numerator(a/(2*n-r)) for integer r positive or negative, then n>0 n = p + r/2. For every irregular prime p there is an r such that n is minimum.
Showing 1-10 of 10 results.