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.

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A143415 Another sequence of Apery-like numbers for the constant 1/e: a(n) = 1/(n+1)!*Sum_{k = 0..n-1} C(n-1,k)*(2*n-k)!.

Original entry on oeis.org

0, 1, 5, 41, 481, 7421, 142601, 3288205, 88577021, 2731868921, 94969529101, 3675200329841, 156725471006105, 7302990263511541, 369216917569411601, 20130327811188977621, 1177435382675193700021, 73546210385434763486705
Offset: 0

Views

Author

Peter Bala, Aug 14 2008

Keywords

Comments

This sequence is a modified version of A143414.

Crossrefs

The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)

Programs

  • Maple
    a := n -> 1/(n+1)!*add (binomial(n-1,k)*(2*n-k)!,k = 0..n-1): seq(a(n),n = 0..19);
    # Alternative:
    A143415 := n -> `if`(n=0, 0, ((2*n)!/(n+1)!)*hypergeom([1-n], [-2*n], 1)):
    seq(simplify(A143415(n)), n = 0..17); # Peter Luschny, May 14 2020
  • Mathematica
    Table[(1/(n+1)!)*Sum[Binomial[n-1,k]*(2*n-k)!, {k,0,n-1}], {n,0,50}] (* G. C. Greubel, Oct 24 2017 *)
  • PARI
    for(n=0,25, print1((1/(n+1)!)*sum(k=0,n-1, binomial(n-1,k)*(2*n-k)!), ", ")) \\ G. C. Greubel, Oct 24 2017

Formula

a(n) = 1/(n+1)!*sum {k = 0..n-1} C(n-1,k)*(2*n-k)!.
a(n) = 1/(n*(n+1))*A143414(n) for n > 0.
Recurrence relation: a(0) = 0, a(1) = 1, (n-1)*(n+1)*a(n) - (n-2)*n*a(n-2) = (2*n-1)*(2*n^2-2*n+1)*a(n-1) for n >= 2. 1/e = 1/2 - 2 * Sum_{n = 1..inf} (-1)^(n+1)/(n*(n+2)*a(n)*a(n+1)) = 1/2 - 2*[1/(3*1*5) - 1/(8*5*41) + 1/(15*41*481) - 1/(24*481*7421) + ...] .
Conjectural congruences: for r >= 0 and prime p, calculation suggests the congruences a(p^r*(p+1)) == a(p^r) (mod p^(r+1)) may hold.
a(n) = ((2*n)!/(n+1)!)*hypergeom([1-n], [-2*n], 1) for n > 0. - Peter Luschny, May 14 2020

A219692 a(n) = Sum_{j=0..floor(n/3)} (-1)^j C(n,j) * C(2j,j) * C(2n-2j,n-j) * (C(2n-3j-1,n) + C(2n-3j,n)).

Original entry on oeis.org

2, 6, 54, 564, 6390, 76356, 948276, 12132504, 158984694, 2124923460, 28877309604, 398046897144, 5554209125556, 78328566695736, 1114923122685720, 15999482238880464, 231253045986317814, 3363838379489630916
Offset: 0

Views

Author

Jason Kimberley, Nov 25 2012

Keywords

Comments

This sequence is s_18 in Cooper's paper.
This is one of the Apery-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
Every prime eventually divides some term of this sequence. - Amita Malik, Aug 20 2017

Crossrefs

The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)

Programs

  • Magma
    s_18 := func where C is Binomial;
    
  • Mathematica
    Table[Sum[(-1)^j*Binomial[n,j]*Binomial[2j,j]*Binomial[2n-2j, n-j]* (Binomial[2n-3j-1,n] +Binomial[2n-3j,n]), {j,0,Floor[n/3]}], {n,0,20}] (* G. C. Greubel, Oct 24 2017 *)
  • PARI
    {a(n) = sum(j=0,floor(n/3), (-1)^j*binomial(n,j)*binomial(2*j,j)* binomial(2*n-2*j,n-j)*(binomial(2*n-3*j-1,n) +binomial(2*n-3*j,n)))}; \\ G. C. Greubel, Apr 02 2019
    
  • Sage
    [sum((-1)^j*binomial(n,j)*binomial(2*j,j)*binomial(2*n-2*j,n-j)* (binomial(2*n-3*j-1,n)+binomial(2*n-3*j,n)) for j in (0..floor(n/3))) for n in (0..20)] # G. C. Greubel, Apr 02 2019

Formula

1/Pi
= 2*3^(-5/2) Sum {k>=0} (n a(n)/18^n) [Cooper, equation (42)]
= 2*3^(-5/2) Sum {k>=0} (n a(n)/A001027(n)).
G.f.: 1+hypergeom([1/8, 3/8],[1],256*x^3/(1-12*x)^2)^2/sqrt(1-12*x). - Mark van Hoeij, May 07 2013
Conjecture D-finite with recurrence: n^3*a(n) -2*(2*n-1)*(7*n^2-7*n+3)*a(n-1) +12*(4*n-5)*(n-1)* (4*n-3)*a(n-2)=0. - R. J. Mathar, Jun 14 2016
a(n) ~ 3 * 2^(4*n + 1/2) / (Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Mar 08 2023

A260667 a(n) = (1/n^2) * Sum_{k=0..n-1} (2k+1)*S(k,n)^2, where S(k,x) denotes the polynomial Sum_{j=0..k} binomial(k,j)*binomial(x,j)*binomial(x+j,j).

Original entry on oeis.org

1, 37, 1737, 102501, 6979833, 523680739, 42129659113, 3572184623653, 315561396741609, 28807571694394593, 2701627814373536601, 259121323945378645947, 25330657454041707496017, 2516984276442279642274311, 253667099464270541534450025, 25884030861250181046253181349, 2670255662315910532447096232073
Offset: 1

Views

Author

Zhi-Wei Sun, Nov 14 2015

Keywords

Comments

Conjecture: For k = 0,1,2,... define S(k,x):= Sum_{j=0..k} binomial(k,j)*binomial(x,j)*binomial(x+j,j).
(i) For any integer n > 0, the polynomial (1/n^2) * Sum_{k=0..n-1}(2k+1)*S(k,x)^2 is integer-valued (and hence a(n) is always integral).
(ii) Let r be 0 or 1, and let x be any integer. Then, for any positive integers m and n, we have the congruence
Sum_{k=0..n-1} (-1)^(k*r)*(2k+1)*S(k,x)^(2m) == 0 (mod n).
(iii) For any odd prime p, we have Sum_{k=0..p-1} S(k,-1/2)^2 == (-1/p)(1-7*p^3*B_{p-3}) (mod p^4), where (a/p) is the Legendre symbol, and B_0,B_1,B_2,... are Bernoulli numbers. Also, for any prime p > 3 we have Sum_{k=0..p-1} S(k,-1/3)^2 == p - (14/3)*(p/3)*p^3*B_{p-2}(1/3) (mod p^4), where B_n(x) denotes the Bernoulli polynomial of degree n; Sum_{k=0..p-1} S(k,-1/4)^2 == (2/p)*p - 26*(-2/p)*p^3*E_{p-3} (mod p^4), where E_0,E_1,E_2,... are Euler numbers; Sum_{k=0..p-1} S(k,-1/6)^2 == (3/p)*p - (155/12)*(-1/p)*p^3*B_{p-2}(1/3) (mod p^4).
Our conjecture is motivated by a conjecture of Kimoto and Wakayama which states that Sum_{k=0..p-1} S(k,-1/2)^2 == (-1/p) (mod p^3) for any odd prime p. The Kimoto-Wakayama conjecture was confirmed by Long, Osburn and Swisher in 2014.
For more related conjectures, see Sun's paper arXiv.1512.00712. - Zhi-Wei Sun, Dec 03 2015

Examples

			a(2) = 37 since (1/2^2) * Sum_{k=0..1} (2k+1)*S(k,2)^2 = (S(0,2)^2 + 3*S(1,2)^2)/4 = (1^2 + 3*7^2)/4 = 148/4 = 37.
G.f. = x + 37*x^2 + 1737*x^3 + 102501*x^4 + 6979833*x^5 + 523680739*x^6 + ...
		

Crossrefs

The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692, A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)

Programs

  • Maple
    # Implementing Mark van Hoeij's formula.
    c := n -> binomial(2*n, n)/(n + 1):
    h := n -> simplify(hypergeom([-n,-n,-n], [1,-2*n], 1)):
    b := n -> c(n)^2*((n+11)*(2+4*n)^2*h(n+1)^2-2*(n+1)*(11*n+16)*(1+2*n)*h(n)*h(n+1)-h(n)^2*(n+1)^3)/(25*(n+2)):
    a := n -> b(n-1): seq(a(n), n = 1..17);  # Peter Luschny, Nov 11 2022
  • Mathematica
    S[k_,x_]:=S[k,x]=Sum[Binomial[k,j]Binomial[x,j]Binomial[x+j,j],{j,0,k}]
    a[n_]:=a[n]=Sum[(2k+1)*S[k,n]^2,{k,0,n-1}]/n^2
    Do[Print[n," ",a[n]],{n,1,17}]

Formula

a(n) ~ phi^(10*n + 3) / (10 * Pi^2 * n^3), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Nov 06 2021
Conjecture: a(p-1) == 1 (mod p^3) for all primes p >= 5. - Peter Bala, Aug 15 2022
a(n) = ((n+10)*A005258(n)^2 - (11*n+5)*A005258(n)*A005258(n-1) - n*A005258(n-1)^2)/(25*(n+1)). - Mark van Hoeij, Nov 11 2022
Previous Showing 31-33 of 33 results.