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 1-3 of 3 results.

A075135 Numerator of the generalized harmonic number H(n,3,1) described below.

Original entry on oeis.org

1, 5, 39, 209, 2857, 11883, 233057, 2632787, 13468239, 13739939, 433545709, 7488194853, 281072414761, 284780929571, 12393920563953, 288249495707519, 2038704876507433, 2058454144222533, 2077126179153173, 60750140156034617
Offset: 1

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Author

T. D. Noe, Sep 04 2002

Keywords

Comments

For integers a and b, H(n,a,b) is the sum of the fractions 1/(a i + b), i = 0,1,..,n-1. This database already contains six instances of generalized harmonic numbers. Partial sums of the harmonic series 1+1/2+1/3+1/4+... are given by the sequence of harmonic numbers H(n,1,1) = A001008(n) / A002805(n).
The Jeep problem gives rise to the series H(n,2,1) = A025550(n) / A025547(n). Recent additions to the database are 3 * H(n,3,1) = A074596(n) / A051536(n), 3 * H(n,3,2) = A074597(n) / A051540(n), 4 * H(n,4,1) = A074598(n) / A051539(n) and 4 * H(n,4,3) = A074637(n) / A074638(n) . The numerator of H(n,4,1) is A075136. The fractions H(n,5,1), H(n,5,2), H(n,5,3) and H(n,5,4) are in A075137-A075144.
The sequence H(n,a,b) is of interest only when a and b are relatively prime. The sequence can also be computed as H(n,a,b) = (PolyGamma[n+1+b/a] - PolyGamma[1+b/a])/a. The sequence H(n,a,b) diverges for all a and b.
According to Hardy and Wright, if p is an odd prime, then p divides the numerator of the harmonic number H(p-1,1,1). This result can be extended to generalized harmonic numbers: for odd integer n, let q = (n-2)a + 2b. If q is prime, then q divides the numerator of H(n-1,a,b). For this sequence (a=3, b=1) we conclude that 11 divides a(4), 17 divides a(6), 29 divides a(10) and 47 divides a(16).
Graham, Knuth and Patashnik define another type of generalized harmonic number as the sum of fractions 1/i^k, i=1,...,n. For k=2, the sequence of fractions is A007406(n) / A007407(n).

Examples

			a(3)=39 because 1 + 1/4 + 1/7 = 39/28.

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 263, 269, 272, 297, 302, 356.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 4th ed., Oxford Univ. Press, 1971, page 88.

Links

Crossrefs

Cf. A001008, A002805, A025550, A025547, A051536, A051539, A074637, A074638, A075136-A075144.
Cf. A051540, A074596, A074597, A074598.
Cf. A007406, A007407.

Programs

  • Mathematica
    a=3; b=1; maxN=20; s=0; Numerator[Table[s+=1/(a n + b), {n, 0, maxN-1}]]
Accumulate[1/Range[1,60,3]]//Numerator (* Harvey P. Dale, Dec 30 2019 *)

A074638 Denominator of 1/3 + 1/7 + 1/11 + ... + 1/(4n-1).

Original entry on oeis.org

3, 21, 231, 385, 7315, 168245, 4542615, 140821065, 28164213, 366134769, 15743795067, 739958368149, 12579292258533, 62896461292665, 3710891216267235, 3710891216267235, 248629711489904745, 17652709515783236895, 88263547578916184475, 6972820258734378573525
Offset: 1

Views

Author

Robert G. Wilson v, Aug 27 2002

Keywords

Comments

This s(n) := Sum_{j=0..n-1} 1/(4*j + 3), for n >= 1, equals (Psi(n + 3/4) - Psi(3/4))/4, with the digamma function Psi(z). See Abramowitz-Stegun, p. 258, eqs. 6.3.7 and 6.3.5, with z -> 3/4. A200134 = -Psi(3/4). - Wolfdieter Lang, Apr 06 2022

Links

Crossrefs

The numerators times 4 are A074637.
Cf. A075135, A200134.

Programs

  • Mathematica
    Table[ Denominator[ Sum[1/i, {i, 3/4, n}]], {n, 1, 20}]
  • PARI
    a(n) = denominator(sum(i=1, n, 1/(4*i-1))); \\ Michel Marcus, Mar 21 2021
  • Python
    from fractions import Fraction
def a(n): return sum(Fraction(1, 4*i-1) for i in range(1, n+1)).denominator
print([a(n) for n in range(1, 21)]) # Michael S. Branicky, Mar 21 2021

Formula

Denominator( (Psi(n + 3/4) - Psi(3/4))/4 ). See the comment above. - Wolfdieter Lang, Apr 05 2022

A352395 Denominator of Sum_{k=0..n} (-1)^k / (2*k+1).

Original entry on oeis.org

1, 3, 15, 105, 315, 3465, 45045, 45045, 765765, 14549535, 14549535, 334639305, 1673196525, 5019589575, 145568097675, 4512611027925, 4512611027925, 4512611027925, 166966608033225, 166966608033225, 6845630929362225, 294362129962575675, 294362129962575675, 13835020108241056725
Offset: 0

Views

Author

Wolfdieter Lang, Apr 06 2022

Keywords

Comments

This is not the sequence A025547(n+1)_{n>=0}, because a(32) = 1420993851085122917681925 but A025547(33) = 18472920064106597929865025. Hence it is also not the sequence A350670.
The alternating sum Sum_{k=0..n} (-1)^k/(2*k+1) = (Psi(n + 3/2) - Psi((2*n - (-1)^n)/4 + 1) - log(2) + Pi/2)/2, with the Digamma function Psi(z).
Proof by subtracting twice the negative fractions from Sum_{k=0..n} 1/(2*k+1) = A350669(n)/A350670(n) (Abramowitz-Stegun, p. 258, eq. 6.3.4.), using Sum_{j=0..k} 1/(4*j + 3) = A074637((k+1)/4)/A074638(k+1) (Abramowitz-Stegun, p. 258, eqs. 6.3.6. with 6.3.5.) and, finally, replacing in the results for the even and odd n cases the formula for Psi(3/4) = -A200134.

Links

Crossrefs

Cf. A007509 (numerators).
Cf. A025547, A074637, A074638, A200134, A350669, A350670.

Programs

  • Mathematica
    Denominator @ Accumulate @ Table[(-1)^k/(2*k + 1), {k, 0, 25}] (* Amiram Eldar, Apr 08 2022 *)
  • PARI
    a(n) = denominator(sum(k=0, n, (-1)^k / (2*k+1))); \\ Michel Marcus, Apr 07 2022
  • Python
    from fractions import Fraction
def A352395(n): return sum(Fraction(-1 if k % 2 else 1,2*k+1) for k in range(n+1)).denominator # Chai Wah Wu, May 18 2022

Formula

a(n) = denominator( (Psi(n + 3/2) - Psi((2*n - (-1)^n)/4 + 1) - log(2) + Pi/2)/2), for n >= 0, with the Digamma function. See the above comment.
a(n) = denominator(Pi/4 + (-1)^n * (Psi((n + 5/2)/2) - Psi((n + 3/2)/2))/4). - Vaclav Kotesovec, May 16 2022
Showing 1-3 of 3 results.

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