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A063100 Compute the cototient function for the g(n) = p(n+1)-p(n)-1 composite numbers between two consecutive primes. Let the number of distinct cototient values be c(n). Then, a(n) = g(n)-c(n).

%I A063100 #6 Apr 18 2023 16:45:55
%S A063100 0,0,0,0,0,1,0,1,1,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,
%T A063100 0,0,0,1,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,2,0,0,0,0,1,0,
%U A063100 0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0
%N A063100 Compute the cototient function for the g(n) = p(n+1)-p(n)-1 composite numbers between two consecutive primes. Let the number of distinct cototient values be c(n). Then, a(n) = g(n)-c(n).
%C A063100 a(n) = 0 means that all cototients in the gap are different, while 1, 2, or more means that 1 or more times inside the gap equal cototients occur.
%C A063100 Unlike totient, where phi(n+1) = phi(n) may occur (see A001274), cototients of consecutive numbers are different for n<1000000. At cases x, of A001274, cototient(x+1) = 1+cototient(x).
%e A063100 Case 1: a(n) = 0; primes = {229, 233}; primes and gap = {229, 230, 231, 232, 233}; cototients = {1, 142, 111, 120, 1}, all cototients inside gap are different, thus a(n) = 0 for p(n) = p(40) = 229 prime.
%e A063100 Case 2: a(n) = 1; primes = {113, 127}; gap = {113, 114, 115, ..., 125, 126, 127}; cototients = {1, 78, 27, 60, 45, 60, 23, 88, 11, 62, 43, 64, 25, 90, 1}; seemingly 60 occurs twice, so a(n) = a(30) = g(n)-c(n) = 13-12 = 1.
%e A063100 Case 3: a(n) = 3, primes = {2861, 2879}, gap = {2861, 2862, ..., 2878, 2879}; cototients = {1, 1926, 415, 1440, 1345, 1434, 107, 1916, 169, 1910, 1191, 1440, 377, 1918, 675, 1440, 1245, 1440, 1}; observe that 1440 occurs four times, so a(n) = 3.
%Y A063100 Cf. A051953 (cototient function), A000010, A001274, A061106.
%K A063100 nonn
%O A063100 1,62
%A A063100 _Labos Elemer_, Aug 07 2001