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A240719 Numbers k such that 2^k == 1 (mod (k+1)^2). 9
1092, 3510 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
There are only two known terms.
If p is in A001220, then p-1 is in the sequence. If k is in the sequence and k+1 is composite, then any prime factor of k+1 is in A001220 (see fifth comment for a proof). In that case, k+1 could be called a 'Wieferich pseudoprime'.
Any further terms are greater than 1.2 * 10^17. - Charles R Greathouse IV, Apr 12 2014
Both known terms have a periodic binary representation (i.e., 1092 = 010001000100, 3510 = 110110110110), so they are terms of A242139. Also, the ratio between those numbers and their divisor sums is 112/39 in both cases (see Dobson's website in the links and also A239875). Are those facts just coincidences? - Felix Fröhlich, Apr 15 2014
Proof of second part of second comment above: Let q be any odd prime factor of (k+1). Since 2 and q^2 are coprime, it follows from Euler's totient theorem (also known as Euler's theorem or Fermat-Euler theorem) that 2^(phi(q^2)) == 1 (mod q^2). Writing phi(q^2) = q^2 - q = q(q-1), one gets 2^(q(q-1)) == 1 (mod q^2). Taking the q-th root of both sides of the congruence yields 2^(q-1) == 1 (mod q^2). Q.E.D. - Felix Fröhlich, Jun 08 2015
If a(3) exists, it corresponds to A001220(3) - 1, i.e., a(3) + 1 must be prime. This can be shown the following way: Assume that a(3) + 1 is composite. Then the theorem from previous comment implies that a(3) + 1 is of the form 1093^x * 3511^y for some x, y >= 0 and x, y not both 0. If x or y is an integer k > 1, then p = 1093 or p = 3511 satisfies 2^(p-1) == 1 (mod p^(2k)). A quick check with PARI shows that neither 1093 nor 3511 satisfies this congruence for any k > 1. This leaves the case where x = y = 1, which can be excluded as well, since 3837523 is not in A001567. Q.E.D. - Felix Fröhlich, Jun 08 2015
LINKS
J. Dobes and M. Kures, Search for Wieferich primes through the use of periodic binary strings, Serdica J. Computing, Volume 4, Number 3, 2010, 293-300.
W. Johnson, On the nonvanishing of Fermat quotients (mod p), J. reine angew. Math., Issue 292 (Jan 1977), 196-200.
L. C. Washington, On Fermat's last theorem, J. reine angew. Math., Issue 289 (Jan 1977), 115-117.
MATHEMATICA
fQ[n_] := PowerMod[2, n, (n + 1)^2] == 1; Select[ Range@ 3600, fQ] (* Robert G. Wilson v, Jun 17 2015 *)
PROG
(PARI) isok(n) = lift(Mod(2, (n+1)^2)^n) == 1; \\ Michel Marcus, Apr 12 2014
(PARI) test(lim)=my(t=1); for(i=0, log(lim)\log(1093), my(n=t); while(n<=lim, if(Mod(2, n^2)^(n-1)==1&&n>1, print(n-1)); n*=3511); t*=1093)
test(1.2e17) \\ Test up to the current search bound for Wieferich primes; Charles R Greathouse IV, Apr 12 2014
CROSSREFS
Sequence in context: A043856 A043864 A043873 * A239875 A091673 A288097
KEYWORD
hard,more,nonn,bref
AUTHOR
Felix Fröhlich, Apr 11 2014
STATUS
approved

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Last modified July 24 13:23 EDT 2024. Contains 374584 sequences. (Running on oeis4.)