A061106 -id:A061106 - cp's OEIS frontend

A061108 Compute Euler totient function for the prime(n+1)-prime(n)-1 composite numbers between two consecutive primes; choose the largest.

2, 2, 6, 4, 8, 6, 12, 20, 8, 24, 24, 12, 24, 42, 40, 16, 48, 44, 24, 60, 54, 64, 72, 60, 32, 52, 36, 72, 110, 84, 108, 44, 120, 40, 120, 132, 82, 156, 120, 48, 160, 64, 96, 60, 180, 192, 120, 72, 120, 184, 64, 216, 220, 216, 208, 72, 200, 180, 92, 272, 264, 204, 96

Offset: 2

Labos Elemer, May 29 2001

nonn

Harry J. Smith, Table of n, a(n) for n = 2..1000

Cf. A000010, A061106.

{ n=1; q=3; forprime (p=5, prime(1003), a=0; for (i=q + 1, p - 1, a=max(eulerphi(i), a)); q=p; write("b061108.txt", n++, " ", a) ) } \\ Harry J. Smith, Jul 18 2009

a(n) = Max{phi(c); prime(n) < c < prime(n+1)}.

Offset corrected by Michel Marcus, Mar 21 2018

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).

Original entry on oeis.org

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, 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, 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

Offset: 1

Labos Elemer, Aug 07 2001

nonn

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.

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).

			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.
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.
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.

Cf. A051953 (cototient function), A000010, A001274, A061106.

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A061108 Compute Euler totient function for the prime(n+1)-prime(n)-1 composite numbers between two consecutive primes; choose the largest.

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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).

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