OFFSET
1,4
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.
LINKS
Stanislav Sykora, Table of n, a(n) for n = 1..1000
Stanislav Sykora, Ratios A271834(n)/2^n for n=1..5000
FORMULA
For prime p, a(p) = 0.
For any n, a(n) < 2^n - n(n+1)/2.
EXAMPLE
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...
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 *)
PROG
(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))))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Stanislav Sykora, Apr 19 2016
STATUS
approved