

A039678


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


34



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



FORMULA



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
(Python)
from sympy import prime
from sympy.ntheory.residue_ntheory import nthroot_mod
def A039678(n): return 2**2+1 if n == 1 else int(nthroot_mod(1, (p:= prime(n))1, p**2, True)[1]) # Chai Wah Wu, May 18 2022


CROSSREFS



KEYWORD

nonn,nice


AUTHOR



EXTENSIONS



STATUS

approved



