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A039678

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

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^2-1 solves this problem for all odd primes p. In fact, there appear to be exactly p-1 values of m with 1 <= m <= p^2 for which m^(p-1) == 1 (mod p^2). See A096082 for the related open problem. - T. D. Noe, Aug 24 2008

That there are exactly p-1 values of 1 <= m <= p^2 for which m^(p-1) == 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, 345-349.

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, p-1, 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)^(p-1)==1, return(a))) \\ Felix Fröhlich, Nov 24 2018