%I #7 Jul 06 2021 15:57:32

%S 3,4,3,6,4,6,4,4,8,4,4,8,6,6,8,12,7,4,12,6,5,4,4,6,8,8,12,4,16,4,4,8,

%T 15,12,8,12,8,8,10,18,8,14,8,12,4,4,9,12,8,6,20,8,4,12,8,16,4,6,8,18,

%U 18,4,16,12,15,4,12,12,8,6,6,8,8,4,4,4,8,10,12,24,8,20,6,9,4,4,8,16,8,4,8,4

%N Number of divisors of the average of consecutive odd primes.

%H Harvey P. Dale, <a href="/A078809/b078809.txt">Table of n, a(n) for n = 1..1000</a>

%e The first pair of consecutive odd primes is 3,5, with average = 4 and tau(4) = 3. Hence a(1) = 3. The second pair of consecutive odd primes is 5,7, with average = 6 and tau(6) = 4, so a(2) = 4.

%t Table[DivisorSigma[0, (Prime[i] + Prime[i + 1])/2], {i, 2, 101}]

%t DivisorSigma[0,#]&/@(Mean/@Partition[Prime[Range[2,100]],2,1]) (* _Harvey P. Dale_, Jul 06 2021 *)

%K easy,nonn

%O 1,1

%A _Joseph L. Pe_, Jan 11 2003