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 1-4 of 4 results.

A034602 Wolstenholme quotient W_p = (binomial(2p-1,p) - 1)/p^3 for prime p=A000040(n).

Original entry on oeis.org

1, 5, 265, 2367, 237493, 2576561, 338350897, 616410400171, 7811559753873, 17236200860123055, 3081677433937346539, 41741941495866750557, 7829195555633964779233, 21066131970056662377432067, 59296957594629000880904587621, 844326030443651782154010715715
Offset: 3

Views

Author

Keywords

Comments

Equivalently, (binomial(2p,p)-2)/(2*p^3) where p runs through the primes >=5.
The values of this sequence's terms are replicated by conjectured general formula, given in A223886 (and also added to the formula section here) for k=2, j=1 and n>=3. - Alexander R. Povolotsky, Apr 18 2013

Examples

			Binomial(10,5)-2 = 250; 5^3=125 hence a(5)=1.
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, Sect. B31.

Crossrefs

Cf. A177783 (alternative definition of Wolstenholme quotient), A072984, A092101, A092103, A092193, A128673, A217772, A223886, A263882.

Programs

  • Magma
    [(Binomial(2*p-1,p)-1) div p^3: p in PrimesInInterval(4,100)]; // Vincenzo Librandi, Nov 23 2015
  • Maple
    f:= proc(n) local p;
    p:= ithprime(n);
    (binomial(2*p-1,p)-1)/p^3
    end proc:
    map(f, [$3..30]); # Robert Israel, Dec 19 2018
  • Mathematica
    Table[(Binomial[2 Prime[n] - 1, Prime[n] - 1] - 1)/Prime[n]^3, {n, 3, 20}] (* Vincenzo Librandi, Nov 23 2015 *)

Formula

a(n) = (A088218(p)-1)/p^3 = (A001700(p-1)-1)/p^3 = (A000984(p)-2)/(2*p^3), where p=A000040(n).
a(n) = A087754(n)/2.
a(n) = (binomial(j*k*prime(n), j*prime(n)) - binomial(k*j, j)) / (k*prime(n)^3) for k=2, j=1, and n>=3. - Alexander R. Povolotsky, Apr 18 2013
a(n) = A263882(n)/prime(n) for n > 2. - Jonathan Sondow, Nov 23 2015
a(n) = numerator(tanh(Sum_{k=1..p-1} artanh(k/p)))/p^3, where p = prime(n) for n >= 3. - Thomas Ordowski, Apr 17 2025

Extensions

Edited by Max Alekseyev, May 14 2010
More terms from Vincenzo Librandi, Nov 23 2015

A092103 Number of values of k for which prime(n) divides A001008(k), the numerator of the k-th harmonic number.

Original entry on oeis.org

3, 3, 13, 638, 3, 3, 25, 3, 18, 26, 15, 3, 27, 24, 17, 23, 13, 3, 45, 3, 3, 43038, 7, 74, 44, 63, 3, 1273, 3, 3515, 7, 38, 3, 3, 7, 3, 74, 526, 288, 3, 19, 3, 3, 41, 11, 59, 3, 31, 65, 176, 3, 3, 3, 20, 3, 106, 55, 3, 3, 89, 3, 3, 3, 79, 3, 3, 3, 47, 3, 21, 253, 29, 7, 79, 41, 19, 701533, 13, 9, 703, 23, 3, 205, 105, 3, 3, 323, 3, 7, 3, 3, 3, 3, 3, 3, 13, 1763
Offset: 2

Views

Author

T. D. Noe, Feb 20 2004

Keywords

Comments

For p = prime(n), Boyd defines J_p to be the set of numbers k such that p divides A001008(k). This sequence gives the size of J_p. A072984 and A177734 give the smallest and largest elements of J_p, respectively.
A092101 gives primes prime(n) such that a(n) = 3 (i.e., a(A000720(A092101(m))) = 3 for all m). A092102 gives primes prime(n) such that a(n) > 3.
From Carlo Sanna, Apr 06 2016: (Start)
Eswarathasan and Levine conjectured that for any prime number p the set J_p is finite.
I proved that if J_p(x) is the number of integers in J_p that are less than x > 1, then J_p(x) < 129 p^(2/3) x^0.765 for any prime p. In particular, J_p has asymptotic density zero. (End)
Bing-Ling Wu and Yong-Gao Chen improved Sanna's (see previous comment) result showing that J_p(x) <= 3 x^(2/3 + 1/(25 log p)) for any prime p and any x > 1. - Carlo Sanna, Jan 12 2017

Examples

			a(2) = 3 because 3 divides A001008(k) for k = 2, 7, and 22.
a(4) = 13 because 7 divides A001008(k) for only the 13 values k = 6, 42, 48, 295, 299, 337, 341, 2096, 2390, 14675, 16731, 16735, and 102728. This is the 4th row in A229493.
		

Crossrefs

Cf. A092193 (number of generations for each prime).
Cf. A229493 (terms for each prime).

Extensions

a(8), a(15), and a(17) corrected by Max Alekseyev, Oct 23 2012
Terms a(23) onward from Carofiglio et al. (2025) added by Max Alekseyev, Apr 01 2025

A177734 Largest k such that prime(n) divides the numerator of the k-th harmonic number (=A001008(k)).

Original entry on oeis.org

22, 24, 102728, 1011849771855214912968404217247, 168, 288, 848874360, 528, 695552, 886725671, 50641, 1680, 2359785, 10776888210, 414839198, 42176361744, 226972, 4488, 9094138358932, 5328, 6240
Offset: 2

Views

Author

Max Alekseyev, May 12 2010

Keywords

Comments

For p = prime(n), Boyd defines J_p to be the set of numbers k such that p divides A001008(k). This sequence gives the largest element of J_p. The smallest element of J_p is given by A072984. The size of J_p is given by A092103.
Term a(23) is too large to include, see b-file. - Max Alekseyev, Apr 04 2025

Crossrefs

Formula

For p = prime(n) in A092101, a(n) = p^2 - 1.

Extensions

a(5) computed by Boyd.
a(8)-a(22) from Max Alekseyev, Oct 23 2012

A177783 Wolstenholme quotient of prime p=A000040(n), i.e., such integer m

Original entry on oeis.org

3, 6, 6, 7, 10, 14, 18, 20, 16, 24, 17, 38, 39, 19, 29, 28, 12, 53, 31, 19, 53, 58, 48, 42, 1, 33, 53, 37, 5, 81, 4, 17, 29, 13, 13, 72, 75, 70, 173, 159, 111, 150, 39, 178, 106, 163, 196, 163, 172, 30, 98, 24, 177, 261, 212, 223, 122, 147, 276, 17, 92, 111, 27, 209, 241
Offset: 3

Views

Author

Max Alekseyev, May 13 2010

Keywords

Comments

a(n) = 0 iff A000040(n) is a Wolstenholme prime (given by A088164).
For n>2 and p=A000040(n), H(p^2-p) == H(p^2-1) == a(n)*p (mod p^2).

Crossrefs

Programs

  • PARI
    { a(n) = my(p); p=prime(n); ((binomial(2*p-1,p)-1)/2/p^3)%p }

Formula

a(n) = H(p-1)/p^2 mod p = A001008(p-1)/A002805(p-1)/p^2 mod p = A034602(n)/2 mod p = (binomial(2*p-1,p)-1)/(2*p^3) mod p, where p = A000040(n).
a(n) = (-1/3)*B(p-3) mod p, with p=prime(n) and B(n) is the n-th Bernoulli number. - Michel Marcus, Feb 05 2016
a(n) = A087754(n)/4 mod A000040(n).

Extensions

Edited by Max Alekseyev, May 16 2010
Showing 1-4 of 4 results.