

A039678


Smallest number m > 1 such that m^(p1)1 is divisible by p^2, where p = nth prime.


24



5, 8, 7, 18, 3, 19, 38, 28, 28, 14, 115, 18, 51, 19, 53, 338, 53, 264, 143, 11, 306, 31, 99, 184, 53, 181, 43, 164, 96, 68, 38, 58, 19, 328, 313, 78, 226, 65, 253, 259, 532, 78, 176, 276, 143, 174, 165, 69, 330, 44, 33, 332, 94, 263, 48, 79, 171, 747, 731, 20, 147, 91, 40
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OFFSET

1,1


COMMENTS

Using Fermat's little theorem twice, it is easy to see that m=p^21 solves this problem for all odd primes p. In fact, there appear to be exactly p1 values of m with 1 <= m <= p^2 for which m^(p1) == 1 (mod p^2). See A096082 for the related open problem.  T. D. Noe, Aug 24 2008
That there are exactly p1 values of 1 <= m <= p^2 for which m^(p1) == 1 (mod p^2) follows immediately from Hensel's lifting lemma and Fermat's little theorem  every solution mod p corresponds to a unique solution mod p^2.  Phil Carmody, Jan 10 2011
For n > 2, prime(n) does not divide a(n)^2  1, so a(n) is the smallest m > 1 such that (m^(prime(n)1)  1)/(m^2  1) == 0 (mod prime(n)^2).  Thomas Ordowski, Nov 24 2018


REFERENCES

P. Ribenboim, The New Book of Prime Number Records, Springer, 1996, 345349.


LINKS

T. D. Noe, Table of n, a(n) for n = 1..10000


FORMULA

a(n) = A185103(A000040(n)).


EXAMPLE

For n=3, p=5 is the third prime and 5^2 = 25 divides 7^4  1 = 2400.


MATHEMATICA

dpa[n_]:=Module[{p=Prime[n], a=2}, While[PowerMod[a, p1, p^2]!=1, a++]; a]; Array[dpa, 70] (* Harvey P. Dale, Sep 05 2012 *)


PROG

(PARI) a(n) = my(p=prime(n)); for(a=2, oo, if(Mod(a, p^2)^(p1)==1, return(a))) \\ Felix Fröhlich, Nov 24 2018


CROSSREFS

Cf. A185103.
Sequence in context: A053787 A314569 A314570 * A259234 A131040 A231786
Adjacent sequences: A039675 A039676 A039677 * A039679 A039680 A039681


KEYWORD

nonn,nice


AUTHOR

Felice Russo


EXTENSIONS

More terms from David W. Wilson
Definition adjusted by Felix Fröhlich, Jun 24 2014
Edited by Felix Fröhlich, Nov 24 2018


STATUS

approved



