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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A271834 a(n) = 2^n - Sum_{m=0..n} binomial(n/gcd(n,m), m/gcd(n,m)) = 2^n - A082906.

Original entry on oeis.org

0, 0, 0, 4, 0, 42, 0, 116, 162, 730, 0, 2458, 0, 11494, 16890, 32628, 0, 180960, 0, 554994, 931476, 2800534, 0, 11005898, 6643750, 43946838, 44738892, 136580910, 0, 720879712, 0, 2147450740, 3250382916, 10923409738, 11517062060, 45683761528, 0, 172783692982
Offset: 1

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Author

Stanislav Sykora, Apr 19 2016

Keywords

Comments

Compared to A082906, this sequence shows better the drop from 2^n upon replacing every binomial(n,m) in the Newton's expansion of (1+1)^n by the 'reduced' binomial(n/gcd(n,m), m/gcd(n,m)). For n > 1, a(n) is zero if and only if n is prime (no reduction, no drop). The ratio r(n) = a(n)/2^n is always smaller than 1 and presents considerable excursions. For composite n up to 5000, the minimum of 0.01471... occurs for n = 4489, and the maximum of 0.80849... occurs for n = 2310. This apparently large relative difference is actually surprisingly small: on log_2 scale it amounts to just about 5.78; a tiny fraction compared to the full scale, given by the values of n for the extrema. This insight suggests the following conjecture: there exists an average ratio r, defined as r = lim_{n->infinity} Sum_{m=1..n} r(m)/n. Its value appears to be approximately 0.3915+-0.0010, which can be interpreted as the average drop in a binomial value upon the 'reduction' of its arguments.

Examples

			Sum_{m=1..2500} r(m)/2500 = 0.391460...
Sum_{m=2501..5000} r(m)/2500 = 0.391975...
Sum_{m=1..5000} r(m)/5000 = 0.391718...
		

Crossrefs

Programs

  • Maple
    A271834:=n->2^n-add(binomial(n/gcd(n,m),m/gcd(n,m)),m=0..n): seq(A271834(n), n=1..50); # Wesley Ivan Hurt, Apr 19 2016
  • Mathematica
    Table[2^n - Sum[Binomial[n/GCD[n, m], m/GCD[n, m]], {m, 0, n}], {n, 40}] (* Wesley Ivan Hurt, Apr 19 2016 *)
  • PARI
    bcg(n,m)=binomial(n/gcd(n,m),m/gcd(n,m));
    a = vector(1000,n,2^n-vecsum(vector(n+1,m,bcg(n,m-1))))

Formula

For prime p, a(p) = 0.
For any n, a(n) < 2^n - n(n+1)/2.
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