

A058887


Smallest prime p such that (2^n)*p is a nontotient number.


4



3, 7, 17, 19, 19, 19, 31, 31, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47, 47
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OFFSET

0,1


COMMENTS

For n=8,9,...,582, a(n) = 47. Note that A040076(47)=583.
For n=583,584,...,6392, a(n) = 383. Note that A040076(383)=6393.
Subsequent primes are 2897, 3061, 5297, and 7013 (cf. A057192 and A071628).
Starting with some large N, a(n)=p for all n >= N. This prime p will likely be the first prime SierpiĆski number, which is conjectured to be 271129.
In particular, a(n) <= 271129 for all n.


REFERENCES

David Harden, Posting to Sequence Fans Mailing List, Sep 19 2010


LINKS

Table of n, a(n) for n=0..65.
D. Bressoud, CNT.m Computational Number Theory Mathematica package.


FORMULA

Min{pp is prime and card(invphi((2^n)*p))=0}.


EXAMPLE

For n=1, the initial segment of {2p} sequence is nops(invphi({2p}))={4, 4, 2, 0, 2, 0, 0, 0, 2, 2, ...}, where the position of the first 0 is 4, corresponding to p(4)=7, so a(1)=7.
For n=8 the same initial segment is: {11, 32, 23, 18, 24, 10, 11, 4, 9, 21, 2, 16, 9, 12, 0, 14, 5, 6, 12, ...}, where the first 0 is the 15th, corresponding to p(15)=47, thus a(8)=47.


MATHEMATICA

Needs["CNT`"]; Table[p=3; While[PhiInverse[p*2^n] != {}, p=NextPrime[p]]; p, {n, 0, 20}]


PROG

(PARI) a(n) = my(p=2); while(istotient(2^n*p), p=nextprime(p+1)); p; \\ Michel Marcus, May 14 2020


CROSSREFS

Cf. A005277, A007617, A002020, A000010, A051953.
Sequence in context: A191147 A227211 A271725 * A306355 A087749 A140863
Adjacent sequences: A058884 A058885 A058886 * A058888 A058889 A058890


KEYWORD

nonn


AUTHOR

Labos Elemer, Jan 08 2001


EXTENSIONS

Edited by T. D. Noe, Nov 15 2010
Edited by Max Alekseyev, Nov 19 2010


STATUS

approved



