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A222207 Morley quotients: (2^(2*p-2) - (-1)^((p-1)/2)*binomial(p-1,(p-1)/2)) / p^3, where p = prime(n) and n >= 3. 3
2, 12, 788, 7636, 874202, 10018884, 1445893544, 2954512034024, 38700329118256, 93229749133527532, 17540746936557672236, 243284404062970619608, 47694250379410432495952, 136236017676683906365850456, 404504597532158799519693872144, 5856120097210409121404621878992, 18102352585707069737371994385420772, 3894254646848417473467131712404310728 (list; graph; refs; listen; history; text; internal format)
OFFSET

3,1

COMMENTS

Morley (1894/95) proved 2^(2*p-2) == (-1)^((p-1)/2)*binomial(p-1,(p-1)/2) mod p^3 for all primes p > 3.

Morley quotients are even, since 2^(2*p-2) and binomial(p-1,(p-1)/2) are even and p^3 is odd.

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 3..200

C. Aebi, G. Cairns, Morley’s other miracle, Math. Mag., 85 (2012), 205-211.

F. Morley, Note on the Congruence 2^4n == (-1)^n*(2n)!/(n!)^2 where 2n+1 is a prime, Annals of Mathematics, Vol. 9 (1894 - 1895), pp. 168-170.

EXAMPLE

prime(3) = 5, so a(3) = (2^(2*5-2) - (-1)^((5-1)/2)*binomial(5-1,(5-1)/2))/5^3 = (2^8 - binomial(4,2))/5^3 = (256-6)/125 = 2.

MATHEMATICA

m[p_] := (2^(2*p-2) - (-1)^((p-1)/2)*Binomial[p-1, (p-1)/2])/p^3; Table[ m[ Prime[n]], {n, 3, 20}]

CROSSREFS

Cf. A007619, A007663, A034602, A197630, A197633.

Sequence in context: A060055 A061149 A191555 * A129933 A064320 A112373

Adjacent sequences:  A222204 A222205 A222206 * A222208 A222209 A222210

KEYWORD

nonn

AUTHOR

Jonathan Sondow, Feb 22 2013

STATUS

approved

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Last modified May 25 02:01 EDT 2020. Contains 334581 sequences. (Running on oeis4.)