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.

A285388 a(n) = numerator of ((1/n) * Sum_{k=0..n^2-1} binomial(2k,k)/4^k).

Original entry on oeis.org

1, 35, 36465, 300540195, 79006629023595, 331884405207627584403, 22292910726608249789889125025, 11975573020964041433067793888190275875, 411646257111422564507234009694940786177843149765, 56592821660064550728377610673427602421565368547133335525825
Offset: 1

Views

Author

Ralf Steiner, Apr 18 2017

Keywords

Comments

Editorial comment: This sequence arose from Ralf Steiner's attempt to prove Legendre's conjecture that there is a prime between N^2 and (N+1)^2 for all N. - N. J. A. Sloane, May 01 2017

Crossrefs

Cf. A000079, A000265, A056220, A060757, A201555, A285389 (denominators), A285406, A280655 (similar), A190732 (2/sqrt(Pi)), A285738 (greatest prime factor), A285717, A285730, A285786, A286264, A000290 (n^2), A056220 (2*n^2 -1), A286127 (sum a(n-1)/a(n)).

Programs

  • Magma
    [Numerator( n*(n^2+1)*Catalan(n^2)/2^(2*n^2-1) ): n in [1..21]]; // G. C. Greubel, Dec 11 2021
    
  • Mathematica
    Table[Numerator[Sum[Binomial[2k,k]/4^k,{k,0,n^2-1}]/n],{n,1,10}]
    Numerator[Table[2^(1-2 n^2) n Binomial[2 n^2,n^2],{n,1,10}]] (* Ralf Steiner, Apr 22 2017 *)
  • PARI
    A285388(n) = numerator((2^(1 - 2*(n^2)))*n*binomial(2*(n^2), n^2)); \\ Antti Karttunen, Apr 27 2017
    
  • PARI
    a(n) = m=n*binomial(2*n^2, n^2);m>>valuation(m,2) \\ David A. Corneth, Apr 27 2017
    
  • Python
    from sympy import binomial, Integer
    def a(n): return (Integer(2)**(1 - 2*n**2)*n*binomial(2*n**2, n**2)).numerator # Indranil Ghosh, Apr 27 2017
    
  • Sage
    [numerator( n*(n^2+1)*catalan_number(n^2)/2^(2*n^2-1) ) for n in (1..20)] # G. C. Greubel, Dec 11 2021

Formula

a(n) is numerator of n*binomial(2 n^2, n^2)/2^(2*n^2 - 1). - Ralf Steiner, Apr 26 2017
a(n) = numerator(n*A201555(n) / (A060757(n)/2)) = n*A201555(n) / 2^(A285717(n)) = A000265(n*A201555(n)). [Using Ralf Steiner's formula and A285717(n) <= A056220(n), cf. A285406.] - Antti Karttunen, Apr 27 2017
Limit_{i->oo} a(i)*A285389(i+1)/(a(i+1)*A285389(i)) = 1. - Ralf Steiner, May 03 2017

Extensions

Edited (including the removal of the author's claim that this leads to a proof of the Legendre conjecture) by N. J. A. Sloane, May 01 2017
Formula section edited by M. F. Hasler, May 02 2017
Edited by N. J. A. Sloane, May 10 2017