

A271834


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


1



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
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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



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



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^nvecsum(vector(n+1, m, bcg(n, m1))))


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



