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.

A283208 Minimal exponent integer sequence associated with Vietoris sequence.

Original entry on oeis.org

0, 2, 3, 6, 7, 9, 10, 14, 15, 17, 18, 21, 22, 24, 25, 30, 31, 33, 34, 37, 38, 40, 41, 45, 46, 48, 49, 52, 53, 55, 56, 62, 63, 65, 66, 69, 70, 72, 73, 77, 78, 80, 81, 84, 85, 87, 88, 93, 94, 96, 97, 100, 101, 103, 104, 108, 109, 111, 112, 115, 116, 118, 119
Offset: 0

Views

Author

Isabel Cação and Maria Irene Falcão and Helmuth Malonek, Mar 08 2017

Keywords

Comments

a(n) is the least nonnegative integer such that 2^a(n) (2n+1)/(n+1) c(2n) is a nonnegative integer, where c(2n) = c(2n-1) = Pochhammer(1/2,n)/(n!) used in [Vietoris, Askey & Steinig, Ruscheweyh & Salinas] and named Vietoris sequence in [Cação et al.]. Also c(n) = A001405(n)/A000079(n).
Also 2^a(n) is the denominator of 2^(-2n)*A001700(n).
From Rogério Serôdio, Feb 19 2019: (Start)
Sum_{k = -1..2} (-1)^ceiling(k/2 + 1) * a(4*n + k) = 6, for n >= 1.
Sum_{k = 1..4} (-1)^ceiling((k-1 mod 3)/3) * a(2*n + k) = 0, for n >= 0. (End)

Examples

			For n=3, a(3)=6 and 2^a(n)(2n+1)/(n+1) c(2n) = (2^6)*7/4*c(6) = 64*35/64 = 35.
		

References

  • L. Vietoris. Über das Vorzeichen gewisser trigonometrischer Summen. Sitzungsber. Österr. Akad. Wiss., 167 (1958), 125-135.

Crossrefs

Programs

  • Mathematica
    a[n_]:=n+Sum[Floor[(n+1)/(2^j)],{j,1,Log2[n+1]}]
    (* or *)
    a[n_]:=Log2@Denominator[Binomial[2 n + 1, n+1] 2^(-2 n)]
  • PARI
    for(n=0, 62, print1(n + sum(j=1, logint(n + 1, 2), floor((n+1)/(2^j))), ", ")) \\ Indranil Ghosh, Mar 10 2017
    
  • Python
    def A283208(n): return n+sum((n+1)//(1<Chai Wah Wu, Jul 16 2022
  • R
    A283208 <- function(n){
    soma <- n
    for (i in c(1:ceiling(log2(n+1)))){soma <- soma + floor((n+1)/2^i)}
    print(soma)} # Rogério Serôdio, Feb 19 2019
    

Formula

a(n) = n + Sum_{j=1..k} floor((n+1)/(2^j)) where k = floor(log_2(n+1)) = A000523(1+n).
From Rogério Serôdio, Feb 19 2019: (Start)
a(n+4) = a(n) + 6 + b(n), where b(n) = min(k, floor((n+5)/4 mod 2^k) = 1), for n >= 0.
a(n+4) = a(n) + 6 + A001511(ceiling((n+2)/4)), for n >= 0.
G.f.: ((2*x + x^2 + 3*x^3 + x^4)/(1 - x) + Sum_{k >= 0} (Sum_{i = 0..3} x^(8*2^k-1+i))/(1 - x^(8*2^k)))/(1 - x^4). (End)
a(n) = n + log_2((n+1)!-((n+1)! AND (n+1)!-1)) (empirical). - Gary Detlefs, Apr 29 2019 [True: equivalent to a(n) = n + A011371(n+1), which is equivalent to the top formula here. - Andrey Zabolotskiy, Mar 26 2021]
a(n) = A005187(n+1) - 1. - Andrey Zabolotskiy, Mar 26 2021