login
Successive record-setters for tau(n+1)*tau(n-1)/tau(n)^2, where tau(n) is the number of divisors of n.
1

%I #6 Aug 28 2019 19:06:13

%S 2,3,5,7,11,17,19,29,41,71,181,239,379,449,701,881,1429,1871,2729,

%T 3079,4159,10529,11969,23561,40699,51679,90271,104651,146719,226799,

%U 244529,252449,388961,403649,825551,906751,1276001,2408561,2648449,3807649,4058209,4406401

%N Successive record-setters for tau(n+1)*tau(n-1)/tau(n)^2, where tau(n) is the number of divisors of n.

%C Most terms are primes. These are numbers with few factors which are sandwiched between numbers with many factors. Terms <379 are same as those of A090481.

%H Amiram Eldar, <a href="/A094342/b094342.txt">Table of n, a(n) for n = 1..61</a>

%e tau(16)*tau(18)/tau(17)^2 = 5*6/2^2 = 15/2 and this is larger than for any n < 17, so 17 is in the sequence.

%p f := x -> tau(x-1)*tau(x+1)/tau(x)^2:?print m := 1: A := []: for k from 2 to 10^6 do if f(k) > m then m := f(k): A := [op(A), [k, f(k)]]: fi; od;

%t s = {}; d1 = 1; d2 = 2; rm = 0; Do[d3 = DivisorSigma[0, n]; r = d1*d3/d2^2; If[r > rm, rm = r; AppendTo[s, n - 1]]; d1 = d2; d2 = d3, {n, 3, 10000}]; s (* _Amiram Eldar_, Aug 28 2019 *)

%Y Cf. A090481.

%K easy,nonn

%O 1,1

%A Isabel C. Lugo (isabel(AT)mit.edu), Jun 04 2004

%E a(1) = 2 and more terms added by _Amiram Eldar_, Aug 28 2019