OFFSET
0,1
COMMENTS
From Jianing Song, Dec 14 2021: (Start)
Let a(n) = 2^n * k, then k must be odd, otherwise a(n)/2 is a totient number, which implies that a(n) is a totient.
Note that 271129 * 2^m is a nontotient for all m (see A058887), so k <= 271129. In fact, let p be smallest prime such that 2^e*p + 1 is composite for all 0 <= e <= n, then k <= p (since 2^n*p is a nontotient).
Actually, k is equal to p. To verify this, it suffices to show that k cannot be an odd composite number < 271129; that is to say, if 2^n * k is a nontotient for an odd composite number < 271129, then there exists k' < k such that 2^n * k' is a nontotient.
The case k < 383 can be easily checked. Let k be an odd composite number in the range (383, 271129), k * 2^n is a nontotient implies n < 2554 unless k = 98431 or 248959 (see the a-file below), then 383 * 2^n is a nontotient (the least n such that 383 * 2^n + 1 is prime is n = 6393). For k = 98431 or 248959, k * 2^n is a nontotient implies n < 7062, then 2897 * 2^n is a nontotient (the least n such that 2897 * 2^n + 1 is prime is n = 9715. (End)
REFERENCES
David Harden, Posting to Sequence Fans Mailing List, Sep 19 2010.
LINKS
FORMULA
a(n) = A058887(n)*2^n.
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Nov 18 2010
EXTENSIONS
Escape clause removed by Jianing Song, Dec 14 2021
STATUS
approved