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 1-5 of 5 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; }

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

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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).

A268512 Triangle of coefficients c(n,i), 1<=i<=n, such that for each n>=2, c(n,i) are setwise coprime; and for all primes p>2n-1, the sum of (-1)^i*c(n,i)*binomial(i*p,p) is divisible by p^(2n-1).

Original entry on oeis.org

1, 2, 1, 12, 9, 2, 60, 54, 20, 3, 840, 840, 400, 105, 12, 2520, 2700, 1500, 525, 108, 10, 27720, 31185, 19250, 8085, 2268, 385, 30, 360360, 420420, 280280, 133770, 45864, 10780, 1560, 105, 720720, 864864, 611520, 321048, 127008, 36960, 7488, 945, 56, 12252240, 15036840, 11138400, 6297480, 2776032, 942480
Offset: 1

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Author

Max Alekseyev, Feb 06 2016

Keywords

Examples

			n=1: 1
n=2: 2, 1
n=3: 12, 9, 2
n=4: 60, 54, 20, 3
n=5: 840, 840, 400, 105, 12
...
For all primes p>3, p^3 divides 2 - binomial(2*p,p) (cf. A087754).
For all primes p>5, p^5 divides 12 - 9*binomial(2*p,p) + 2*binomial(3*p,p) (cf. A268589).
For all primes p>7, p^7 divides 60 - 54*binomial(2*p,p) + 20*binomial(3*p,p) - 3*binomial(4*p,p) (cf. A268590).
		

Crossrefs

Cf. A099996 (first column), A068550 (diagonal), A087754, A268589, A268590, A254593.

Programs

  • Mathematica
    a3418[n_] := LCM @@ Range[n];
    c[1, 1] = 1; c[n_, i_] := a3418[2(n-1)] Binomial[2n-1, n-i] ((2i-1)/i/ Binomial[2n-1, n]);
    Table[c[n, i], {n, 1, 10}, {i, 1, n}] // Flatten (* Jean-François Alcover, Dec 04 2018 *)
  • PARI
    { A268512(n,i) = lcm(vector(2*(n-1),i,i)) * binomial(2*n-1,n-i) * (2*i-1) / i / binomial(2*n-1,n) }

Formula

c(n,i) = A003418(2*(n-1))*binomial(2*n-1,n-i)*(2*i-1)/i/binomial(2*n-1,n).

A357955 a(n) = 3*binomial(4*n,n) - 20*binomial(3*n,n) + 54*binomial(2*n,n).

Original entry on oeis.org

37, 60, 108, 60, -660, 60, 82404, 1411848, 17540460, 191318820, 1952058108, 19175376324, 184118073828, 1743153802320, 16359157606200, 152693295412560, 1420516291306860, 13190159377278324, 122358232382484420, 1134645084249344400, 10522118980232969340
Offset: 0

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Author

Peter Bala, Oct 22 2022

Keywords

Comments

Conjectures:
1) a(p) == a(1) (mod p^7) for all primes p >= 3 except p = 7.
2) a(p^r) == a(p^(r-1)) ( mod p^(3*r+5) ) for r >= 2 and all primes p >= 3.
These are stronger supercongruences than those satisfied separately by the sequences {binomial(2*n,n)} = A000984, {binomial(3*n,n)} = A005809 and {binomial(4*n,n)} = A005810.
Conjecture 1) was proved by Aidagulov and Alekseyev; see the remarks following Corollary 2. - Peter Bala, Oct 29 2022

Examples

			Examples of supercongruences:
a(11) - a(1) = 19175376324 - 60 = (2^3)*3*(11^7)*41 == 0 (mod 11^7).
a(5^2) - a(5) = 726506045044361132812560 - 60 = (2^2)*3*(5^11)*41*30241552444123 == 0 (mod 5^11).
		

Crossrefs

Programs

  • Maple
    seq(3*binomial(4*n,n) - 20*binomial(3*n,n) + 54*binomial(2*n,n), n = 0..20);
  • Mathematica
    A357955[n_] := 3*Binomial[4*n, n] - 20*Binomial[3*n, n] + 54*Binomial[2*n, n];
    Array[A357955, 25, 0] (* Paolo Xausa, Jul 17 2024 *)
  • Python
    from math import comb
    def A357955(n): return 54*comb(m:=n<<1,n)+3*comb(m<<1,n)-20*comb(m+n,n) # Chai Wah Wu, Oct 24 2022

Formula

a(n) = 3*A005810(n) - 20*A005809(n) + 54*A000984(n).
a(k*p^r) == a(k*p^(r-1)) ( mod p^(3*r) ) for positive integers k and r and for all primes p >= 5 (see Meštrović, Section 6, equation 39).
a(p) == a(1) (mod p^6) for all primes p >= 7 (apply Helou and Terjanian, Section 3, Proposition 2).
From Stefano Spezia, Jul 17 2024: (Start)
G.f.: 54/sqrt(1-4*x) - 40*cos(arccos(1-27*x/2)/6)/sqrt(4-27*x) + 3*hypergeom([1/4, 1/2, 3/4], [1/3, 2/3], 4^4*x/3^3).
E.g.f.: 54*exp(2*x)*BesselI(0, 2*x) - 20*hypergeom([1/3, 2/3], [1/2, 1], 27*x/4) + 3*hypergeom([1/4, 1/2, 3/4], [1/3, 2/3, 1], 4^4*x/3^3).
a(n) ~ exp(2*n*arctanh(229/283))*sqrt(6/(n*Pi)). (End)
Showing 1-5 of 5 results.