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

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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

A268589 a(n) = (2*C(3p,p) - 9*C(2p,p) + 12) / p^5, where p = prime(n).

Original entry on oeis.org

12, 2364, 43500, 20791626, 514377588, 373783661124, 9888937247184828, 312285010312512084, 11167980739981519994382, 13185583459205473525798038, 462369843775374621687338484, 588608385261717115044847555476, 28758863221144089886068560242560564, 1508365481231852329668720928730586740868
Offset: 4

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Author

Max Alekseyev, Feb 07 2016

Keywords

Comments

a(n) is an integer for all n>=4, see A268512.

Crossrefs

Programs

  • Mathematica
    Table[(2*Binomial[3p,p]-9*Binomial[2p,p]+12)/p^5,{p,Prime[Range[4,20]]}] (* Harvey P. Dale, Aug 27 2025 *)
  • PARI
    { A268589(n) = my(p=prime(n)); (12 - 9*binomial(2*p,p) + 2*binomial(3*p,p))/p^5; }

A268590 a(n) = (3*C(4p,p) - 20*C(3p,p) + 54*C(2p,p) - 60) / p^7, where p = prime(n).

Original entry on oeis.org

984, 27780, 32144568, 1269360060, 2470299005220, 316528131552725460, 17262503097511844124, 3329177348896984023277536, 12461979236231507288981559840, 783882118494853605112684502280, 3251723952081272231067929776337100, 959689034437453143807696476144553320100
Offset: 5

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Author

Max Alekseyev, Feb 07 2016

Keywords

Comments

a(n) is an integer for all n>=5, see A268512.

Crossrefs

Programs

  • PARI
    { A268590(n) = my(p=prime(n)); (-60 + 54*binomial(2*p,p) - 20*binomial(3*p,p) + 3*binomial(4*p,p))/p^7; }

A087754 a(n) = (C(2p,p)-2) / p^3, where p = prime(n).

Original entry on oeis.org

2, 10, 530, 4734, 474986, 5153122, 676701794, 1232820800342, 15623119507746, 34472401720246110, 6163354867874693078, 83483882991733501114, 15658391111267929558466, 42132263940113324754864134
Offset: 3

Views

Author

Henry Bottomley, Oct 02 2003

Keywords

Examples

			a(6)=4734 since 13 is the sixth prime and (C(26,13)-2)/13^3 = (10400600-2)/2197 = 4734.
		

Crossrefs

Programs

  • Mathematica
    Table[(Binomial[2p,p]-2)/p^3,{p,Prime[Range[3,20]]}] (* Harvey P. Dale, Oct 23 2017 *)

Formula

a(n) = A060842(n) / A000040(n).
a(n) = 2 * A034602(n).
Showing 1-4 of 4 results.