A197630 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.

0, 13, 1356, 123229034, 79417031713, 97237045496594199, 166710337513971577670, 993090310179794898808058068, 60995221345838813484944512721637147449, 332049278209768881045237587717723153006704, 120846039713576242385812868532189241842793944235993733

Offset: 2

Author

Jonathan Sondow, Oct 16 2011

Keywords

• nonn

Comments

Lerch proved that the Lerch quotient of any odd prime is an integer.

Is 13 the only Lerch quotient that is itself prime?

No other primes below 300,000 digits. - Charles R Greathouse IV, Nov 16 2011

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

Examples

			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.
		

Links

• Michel Marcus, Table of n, a(n) for n = 2..75
• J. B. Dobson A note on Lerch primes, arXiv:1311.2242 [math.NT], 2014.
• J. B. Dobson A Characterization of Wilson-Lerch Primes, Integers, 16 (2016), A51.
• M. Lerch, Zur Theorie des Fermatschen Quotienten (a^(p-1)-1)/p = q(a), Math. Ann. 60 (1905), 471-490.
• J. Sondow, Lerch Quotients, Lerch Primes, Fermat-Wilson Quotients, and the Wieferich-non-Wilson Primes 2, 3, 14771, in Proceedings of CANT 2011, arXiv:1110.3113 [math.NT], 2011-2012.
• J. Sondow, Lerch Quotients, Lerch Primes, Fermat-Wilson Quotients, and the Wieferich-non-Wilson Primes 2, 3, 14771, Combinatorial and Additive Number Theory, CANT 2011 and 2012, Springer Proc. in Math. & Stat., vol. 101 (2014), pp. 243-255.

Crossrefs

Cf. A007619, A197631, A197632.

Programs

Mathematica

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 *)

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

Formula

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.