OFFSET
1,1
COMMENTS
Any run of 8 or more consecutive integers must include at least one number k of the form 8j+4; in the prime factorization of k, the prime factor 2 appears with multiplicity exactly 2, so the number of divisors of k is divisible by 3 (which is not a power of 2). Thus, there is no term a(8): the sequence is complete, ending with a(7). - Jon E. Schoenfield, Nov 12 2017
EXAMPLE
a(4)=613: q=613 (a prime, hence two divisors), q+1 = 614 = 2*307 (4 divisors), q+2 = 615 = 3*5*41 (8 divisors), and q+3 = 616 = 2^3 * 7 * 11 (16 divisors).
CROSSREFS
KEYWORD
nonn,fini,full
AUTHOR
Labos Elemer, Nov 19 2004
EXTENSIONS
a(6)-a(7) from Donovan Johnson, Mar 23 2011
Keywords fini and full added and Example section edited by Jon E. Schoenfield, Nov 12 2017
STATUS
approved