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%I #52 Oct 16 2019 06:19:23
%S 0,13,1356,123229034,79417031713,97237045496594199,
%T 166710337513971577670,993090310179794898808058068,
%U 60995221345838813484944512721637147449,332049278209768881045237587717723153006704,120846039713576242385812868532189241842793944235993733
%N Lerch quotients of odd primes: ((Sum_{k=1..p-1} q_p(k)) - w_p)/p, where q_p(k) = (k^(p-1)-1)/p is a Fermat quotient, w_p = ((p-1)!+1)/p is a Wilson quotient, and p is the n-th prime, with n > 1.
%C Lerch proved that the Lerch quotient of any odd prime is an integer.
%C Is 13 the only Lerch quotient that is itself prime?
%C No other primes below 300,000 digits. - _Charles R Greathouse IV_, Nov 16 2011
%C Proof that a(n) is an integer for n >= 2: Note that ((p-1)!)^(p-1) = Product_{i=1..p-1} (1+i^(p-1)-1) == 1+Sum_{i=1..p-1} (i^(p-1)-1) (mod p^2). Write (p-1)! = kp-1, then ((p-1)!)^(p-1) == 1-(p-1)*kp == kp+1 == (p-1)!+2 (mod p^2). This gives Sum_{i=1..p-1} (i^(p-1)-1) == (p-1)!+1 (mod p^2), or Sum_{i=1..p-1} (i^(p-1)-1)/p == ((p-1)!+1)/p (mod p). - _Jianing Song_, Oct 15 2019
%H Michel Marcus, <a href="/A197630/b197630.txt">Table of n, a(n) for n = 2..75</a>
%H J. B. Dobson <a href="http://arxiv.org/abs/1311.2242">A note on Lerch primes</a>, arXiv:1311.2242 [math.NT], 2014.
%H J. B. Dobson <a href="http://www.integers-ejcnt.org/q51/q51.Abstract.html">A Characterization of Wilson-Lerch Primes</a>, Integers, 16 (2016), A51.
%H M. Lerch, <a href="http://gdz.sub.uni-goettingen.de/dms/resolveppn/?PPN=GDZPPN002260441">Zur Theorie des Fermatschen Quotienten (a^(p-1)-1)/p = q(a)</a>, Math. Ann. 60 (1905), 471-490.
%H J. Sondow, <a href="http://arxiv.org/abs/1110.3113">Lerch Quotients, Lerch Primes, Fermat-Wilson Quotients, and the Wieferich-non-Wilson Primes 2, 3, 14771</a>, in Proceedings of CANT 2011, arXiv:1110.3113 [math.NT], 2011-2012.
%H J. Sondow, <a href="http://dx.doi.org/10.1007/978-1-4939-1601-6_17">Lerch Quotients, Lerch Primes, Fermat-Wilson Quotients, and the Wieferich-non-Wilson Primes 2, 3, 14771</a>, Combinatorial and Additive Number Theory, CANT 2011 and 2012, Springer Proc. in Math. & Stat., vol. 101 (2014), pp. 243-255.
%F a(n) = ((Sum_{k=1..p-1} k^(p-1)) - p - (p-1)!)/p^2, where p is the n-th prime and n >= 2.
%e a(3) = 13 because the 3rd prime is 5 and ((Sum_{k=1..4} q_5(k)) - w_5)/5 = (0 + 3 + 16 + 51 - 5)/5 = 13.
%t f[n_] := Block[{p = Prime[n]}, (Sum[(k^(p - 1) - 1)/p, {k, p - 1}] - ((p - 1)! + 1)/p)/p]; Array[f, 12, 2] (* _Robert G. Wilson v_, Dec 01 2016 *)
%o (PARI) a(n)=my(p=prime(n),m=p-1); sum(k=1,m, k^m,-p-m!)/p^2 \\ _Charles R Greathouse IV_, Oct 18 2011
%Y Cf. A007619, A197631, A197632.
%K nonn
%O 2,2
%A _Jonathan Sondow_, Oct 16 2011