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.

User: Gene Ward Smith

Gene Ward Smith's wiki page.

Gene Ward Smith has authored 31 sequences. Here are the ten most recent ones:

A279702 a(n) = floor( exp(gamma) k log log k ) - sigma(k), where gamma is Euler's constant (A001620) and sigma(k) is sum of divisors of k (A000203), the n-th colossally abundant number (A004490).

Original entry on oeis.org

-5, -6, -9, -18, -26, -34, -123, -107, 3953, 90021, 203866, 678250, 3860926, 62168609, 1022130830, 22777519100, 46323907000, 1499885420000, 47625567000000, 318447820000000, 974228630000000, 36070436000000000
Offset: 2

Author

Gene Ward Smith, Dec 17 2016

Keywords

Comments

By Robin's theorem, if the Riemann hypothesis is true the only negative values this sequence attains are the first eight terms; if it is false, it becomes negative again somewhere farther on. Briggs conjectured, in effect, that this sequence is asymptotic to C k / sqrt(log(k)) for some constant C.

References

  • G. Robin, Grandes valeurs de la fonction somme des diviseurs et hypothese de Riemann, J. Math. Pures Appl. 63 (1984), 187-213.

Crossrefs

A279609 a(n) = floor(H(k) + exp(H(k))*log(H(k))) - sigma(k) where H(k) is the k-th harmonic number Sum_{j=1..k} 1/j and k is the n-th colossally abundant number A004490(n).

Original entry on oeis.org

0, 0, 0, 2, 6, 34, 207, 492, 9051, 143828, 306310, 963859, 5155084, 81053635, 1334916490, 29106956400, 58655156200, 1817551636000, 56466287472000, 376943525488000, 1144451930851200, 41803526752345600
Offset: 2

Author

Gene Ward Smith, Dec 15 2016

Keywords

Comments

By a theorem of J. C. Lagarias, the Riemann hypothesis is equivalent to the proposition that this sequence never takes a negative value. In fact, by inspection it appears to be monotone increasing; this conjecture implies the Riemann hypothesis but is not in any obvious way implied by it. Stronger conjectures are easy to formulate--for example, if F(n) is the function defined by this sequence, then F(n)/2^n also appears to be monotone increasing.

Crossrefs

A259632 a(n) = floor(exp(H_k)*log(H_k)) - sigma(k) where k is the n-th colossally abundant number (Sequence A079526 applied to the colossally abundant numbers (A004490).)

Original entry on oeis.org

-2, -2, -3, -2, 0, 28, 199, 483, 9040, 143814, 306295, 963844, 5155067, 81053615, 1334916470, 29106956400, 58655156000, 1817551640000, 56466287000000, 376943530000000, 1144451930000000, 41803527000000000
Offset: 1

Author

Gene Ward Smith, Dec 17 2016

Keywords

Comments

It follows easily from the work of Lagarias that the Riemann hypothesis is equivalent to this sequence's being nonnegative except for the first four terms.

Crossrefs

A217055 Prime numbers which are conductors of elliptic curves.

Original entry on oeis.org

11, 17, 19, 37, 43, 53, 61, 67, 73, 79, 83, 89, 101, 109, 113, 131, 139, 163, 179, 197, 229, 233, 269, 277, 307, 331, 347, 353, 359, 373, 389, 431, 433, 443, 467, 503, 557, 563, 571, 593, 643, 659, 677, 701, 709, 733, 739, 797, 811, 827, 829, 997, 1019, 1051
Offset: 1

Author

Gene Ward Smith, Sep 25 2012

Keywords

Comments

Taken from the data by Armand Brumer and Oisin McGuinness listing 310716 elliptic curves with prime conductor. Note that for some primes, there is more than one elliptic curve with that conductor.
All primes p of the form p = u^2 + 64 for some integer u are in this sequence, as Setzer (1975) proved that for such primes p that there are exactly two elliptic curves E/Q of conductor p. - Robin Visser, Sep 04 2024

Examples

			a(1) = 11, as there are no elliptic curves over Q of conductor less than 11, but there are exactly three elliptic curves over Q of conductor equal to 11, for example E : y^2 + y = x^3 - x^2. - _Robin Visser_, Sep 04 2024
		

Crossrefs

Programs

  • Sage
    # Uses Cremona's database of elliptic curves (works for all p < 500000)
    def is_A217055(p):
        if not Integer(p).is_prime(): return False
        return CremonaDatabase().number_of_curves(p) > 0
    print([p for p in range(1, 1000) if is_A217055(p)])  # Robin Visser, Sep 04 2024

A123479 Coefficients of series giving the best rational approximations to sqrt(6).

Original entry on oeis.org

20, 1980, 194040, 19013960, 1863174060, 182572043940, 17890197132080, 1753056746899920, 171781670999060100, 16832850701160989900, 1649447587042777950120, 161629030679491078121880, 15837995559003082877994140, 1551961935751622630965303860
Offset: 1

Author

Gene Ward Smith, Sep 28 2006

Keywords

Comments

The partial sums of the series 5/2 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(6), which constitute every second convergent of the continued fraction. The corresponding continued fractions are [2;2], [2;2,4,2], [2;2,4,2,4,2], [2;2,4,2,4,2,4,2] and so forth.
Sequence of numbers x=a(n) such 4*x+1 and 6*x+1 are both square, and their square roots are A138288(n) and A054320(n). - Paul Cleary, Jun 23 2014

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{99,-99,1},{0,20,1980},{2, 25}] (* Paul Cleary, Jun 23 2014 *)
  • PARI
    Vec(-20*x/((x-1)*(x^2-98*x+1)) + O(x^100)) \\ Colin Barker, Jun 23 2014

Formula

a(n+3) = 99*a(n+2) - 99*a(n+1) + a(n).
a(n) = -5/24 + (( + 2*6^(1/2))/48)*(49 + 20*6^(1/2))^n + ((5 - 2*6^(1/2))/48)*(49 - 20*6^(1/2))^n.
G.f.: -20*x / ((x-1)*(x^2-98*x+1)). - Colin Barker, Jun 23 2014

Extensions

More terms from Colin Barker, Jun 23 2014

A123482 Coefficients of the series giving the best rational approximations to sqrt(11).

Original entry on oeis.org

60, 23940, 9528120, 3792167880, 1509273288180, 600686976527820, 239071907384784240, 95150018452167599760, 37869468272055319920300, 15071953222259565160679700, 5998599512991034878630600360, 2387427534217209622129818263640, 950190160018936438572789038328420
Offset: 1

Author

Gene Ward Smith, Oct 02 2006

Keywords

Comments

The partial sums of the series 10/3 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(11), which constitute every second convergent of the continued fraction. The corresponding continued fractions are [3;3,6,3], [3;3,6,3,6,3], [3;3,6,3,6,3,6,3] and so forth.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[-60*x/((x - 1)*(x^2 - 398*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Oct 13 2017 *)
  • PARI
    Vec(-60*x/((x-1)*(x^2-398*x+1)) + O(x^100)) \\ Colin Barker, Jun 23 2014

Formula

a(n+3) = 399*a(n+2) - 399*a(n+1) + a(n).
a(n) = -5/33 + (5/66 + 1/44*11^(1/2))*(199 + 60*11^(1/2))^n + (5/66 - 1/44*11^(1/2))*(199 - 60*11^(1/2))^n.
G.f.: -60*x / ((x-1)*(x^2-398*x+1)). - Colin Barker, Jun 23 2014

Extensions

More terms from Colin Barker, Jun 23 2014

A123478 Coefficients of series giving the best rational approximations to sqrt(7).

Original entry on oeis.org

48, 12240, 3108960, 789663648, 200571457680, 50944360587120, 12939667017670848, 3286624478127808320, 834789677777445642480, 212033291530993065381648, 53855621259194461161296160, 13679115766543862141903843040, 3474441549080881789582414836048
Offset: 1

Author

Gene Ward Smith, Sep 28 2006

Keywords

Comments

The partial sums of the series 8/3 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(7), which constitute every fourth convergent of the continued fraction. The corresponding continued fractions are [2;1,1,1], [2;1,1,1,4,1,1,1], [2;1,1,1,4,1,1,1,4,1,1,1] and so forth.

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{255,-255,1},{48,12240,3108960},30] (* Harvey P. Dale, Nov 20 2016 *)
  • PARI
    Vec(-48*x/((x-1)*(x^2-254*x+1)) + O(x^100)) \\ Colin Barker, Jun 23 2014

Formula

a(n+3) = 255 a(n+2) - 255 a(n+1) + a(n).
a(n) = -4/21 + (2/21+1/28*7^(1/2))*(127+48*7^(1/2))^n + (2/21-1/28*7^(1/2))*(127-48*7^(1/2))^n.
G.f.: -48*x / ((x-1)*(x^2-254*x+1)). - Colin Barker, Jun 23 2014

Extensions

More terms from Colin Barker, Jun 23 2014

A123480 Coefficients of the series giving the best rational approximations to sqrt(3).

Original entry on oeis.org

4, 60, 840, 11704, 163020, 2270580, 31625104, 440480880, 6135107220, 85451020204, 1190179175640, 16577057438760, 230888624967004, 3215863692099300, 44791203064423200, 623860979209825504, 8689262505873133860, 121025814103014048540, 1685672134936323545704
Offset: 1

Author

Gene Ward Smith, Sep 28 2006

Keywords

Comments

The partial sums of the series 2 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(3), which constitute every second convergent of the continued fraction. The corresponding continued fractions are [1;1,2,1], [1;1,2,1,2,1], [1;1,2,1,2,1,2,1], [1;1,2,1,2,1,2,1,2,1] and so forth.

Programs

  • Mathematica
    CoefficientList[Series[-4*x/((x - 1)*(x^2 - 14*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Oct 13 2017 *)
  • PARI
    my(x='x+O('x^50)); Vec(-4*x/((x-1)*(x^2-14*x+1))) \\ G. C. Greubel, Oct 13 2017

Formula

a(n+3) = 15*a(n+2) - 15*a(n+1) + a(n).
a(n) = -1/3 + (1/6 + 1/12*3^(1/2))*(7 + 4*3^(1/2))^n + (1/6 - 1/12*3^(1/2))*(7 - 4*3^(1/2))^n.
a(n) = 4*A076139(n) = 2*A217855(n) = 1/2*A045899(n) = 4/3*A076140(n). - Peter Bala, Dec 31 2012
G.f.: -4*x/((x-1)*(x^2-14*x+1)). - Colin Barker, Jan 20 2013
a(n) = A001353(n)*A001353(n+1). - Antonio Alberto Olivares, Apr 06 2020

A118778 Total degree of the classical modular curve X_n(0). Also the degree of the classical modular polynomial.

Original entry on oeis.org

1, 4, 6, 9, 10, 18, 14, 20, 20, 30, 22, 38, 26, 42, 40, 42, 34, 62, 38, 60, 56, 66, 46, 82, 54, 78, 66, 84, 58, 122, 62, 88, 88, 102, 84, 126, 74, 114, 104, 126, 82, 168, 86, 132, 128, 138, 94, 172, 104, 166, 136, 156, 106, 198, 132, 170, 152, 174, 118, 254, 122, 186, 172
Offset: 1

Author

Gene Ward Smith, May 22 2006

Keywords

Comments

This is the total degree of the classical modular curve relating j(z) to j(nz), where j is the j-invariant, or elliptic modular function. If F_n(x, y) = 0 is the equation for the curve (the classical modular equation) then F_n(x, x) is the classical modular polynomial and the sequence is also the sequence of degrees for it. When n is a prime, the degree is 2n.

References

  • Serge Lang, ''Elliptic Functions'', Addison-Wesley, 1973.

Programs

  • Maple
    with(numtheory): degx := proc (n) # degree of the classical modular curve X0(n) local a, s; s := 0; for a in divisors(n) do if a^2 > n then s := s + 2*a*phi(igcd(a, n/a))/igcd(a, n/a) fi od; if issqr(n) then s := s+phi(sqrt(n)) fi; s end:
  • Mathematica
    degx[n_] := Module[{s = 0}, Do[ If[ a^2 > n, s = s + 2*a*EulerPhi[ GCD[a, n/a]] / GCD[a, n/a]], {a, Divisors[n]}]; If[ IntegerQ[ Sqrt[n]], s = s + EulerPhi[ Sqrt[n] ] ]; s]; Table[ degx[n], {n, 1, 63}] (* Jean-François Alcover, Jan 29 2013, translated from Maple *)
  • PARI
    a(n)=2*sumdiv(n,d,if(d^2>n, my(g=gcd(d,n/d)); d*eulerphi(g)/g)) + if(issquare(n,&n),eulerphi(n)) \\ Charles R Greathouse IV, Jan 29 2013

A122533 Coefficients of the series giving the best rational approximations to 1/e.

Original entry on oeis.org

57, 3667, 525153, 133794291, 53325113593, 30632012923107, 23965268215166337, 24499823488381227043, 31709265214216777648761, 50678828500275334077977523, 98023476146668402679417310817
Offset: 1

Author

Gene Ward Smith, Sep 17 2006

Keywords

Comments

The series giving the best rational approximations to 1/e is 1/e = 1/3 + 2/a(1) - 2/a(2) + 2/a(3) - ... The continued fraction for 1/e is [0;2,1, 2,1,1,4,1,1,6,1,1,8...] and the above best approximations give every third convergent, the convergents deriving from [0;2,1], [0;2,1,2, 1,1], [0;2,1,2,1,1,4,1,1] and so forth are the partial sums of the above infinite series.

Crossrefs

Programs

  • Magma
    I:=[57, 3667, 525153]; [n le 3 select I[n] else ((2*n-3)*(16*n^2 - 3)*Self(n-1) + (2*n+1)*(16*(n-1)^2 -3)*Self(n-2) - (2*n+1)*Self(n-3))/(2*n-3): n in [1..30]]; // G. C. Greubel, Oct 27 2024
    
  • Mathematica
    RecurrenceTable[{a[n]== ((2*n-3)*(16*n^2 -3)*a[n-1] +(2*n+1)*(16*(n-1)^2 - 3)*a[n-2] -(2*n+1)*a[n-3])/(2*n-3), a[1]==57, a[2]==3667, a[3]==525153}, a, {n,30}] (* G. C. Greubel, Oct 27 2024 *)
  • SageMath
    @CachedFunction
    def a(n): # a = A122533
        if n<4: return (0,57,3667,525153)[n]
        else: return ((2*n-3)*(16*n^2 -3)*a(n-1) +(2*n+1)*(16*(n-1)^2 -3)*a(n-2) -(2*n+1)*a(n-3))/(2*n-3)
    [a(n) for n in range(1,31)] # G. C. Greubel, Oct 27 2024

Formula

a(n+3) = (16*n^2 + 96*n + 141) * a(n+2) + (2*n+7)*(16*n^2 + 64*n + 61)/(2*n+3) * a(n+1) - (2*n+7)/(2*n+3) * a(n). This recurrence relationship is identical to A122523, for the best approximations to e.