

A036236


Least inverse of A015910: smallest integer k > 0 such that 2^k mod k = n, or 0 if no such k exists.


68



1, 0, 3, 4700063497, 6, 19147, 10669, 25, 9, 2228071, 18, 262279, 3763, 95, 1010, 481, 20, 45, 35, 2873, 2951, 3175999, 42, 555, 50, 95921, 27, 174934013, 36, 777, 49, 140039, 56, 2463240427, 110, 477, 697, 91, 578, 623, 156, 2453, 540923, 55, 70, 345119, 287
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OFFSET

0,3


COMMENTS

a(1) = 0, that is, no n exists with 2^n mod n = 1. Proof: Assume that there exists such n > 1. Consider its smallest prime divisor p. Then 2^n == 1 (mod p) implying that the multiplicative order ord_p(2) divides n. However, since ord_p(2) < p and p is the smallest divisor of n, we have ord_p(2) = 1, that is, p divides 2^1  1 = 1 which is impossible.  Max Alekseyev
Labos Elemer asked on Sep 27 2001 if all numbers > 1 eventually appear in A015910, that is, if a(n) > 0 for n > 1.
Obviously, k > n.  Daniel Forgues, Jul 06 2015


REFERENCES

P. ErdÅ‘s and R. L. Graham, Old and new problems and results in combinatorial number theory, Monographies de L'Enseignement Mathematique, 28, 1980.
R. K. Guy, Unsolved Problems in Number Theory, Section F10.


LINKS

David W. Wilson, Table of n, a(n) for n = 0..1026 (from the Havermann file)
Joe K. Crump, 2^n mod n
Hans Havermann, Table of n, a(n) for n = 1..10000 with 1 for those entries where a(n) is unknown
Eric Weisstein's World of Mathematics, 2


FORMULA

It's obvious that for each k, a(k) > k and we can easily prove that 2^(3^n) = 3^n1 (mod 3^n). So 3^n is the least k with 2^k mod k = 3^n1. Hence for each n, a(3^n1) = 3^n.  Farideh Firoozbakht, Nov 14 2006


EXAMPLE

n = 0: 2^1 mod 1 = 0, a(0) = 1;
n = 1: 2^k mod k = 1, no such k exists, so a(1) = 0;
n = 2: 2^3 mod 3 = 2, a(2) = 3;
n = 3: 2^4700063497 mod 4700063497 = 3, a(3) = 4700063497.


MATHEMATICA

a = Table[0, {75} ]; Do[ b = PowerMod[2, n, n]; If[b < 76 && a[[b]] == 0, a[[b]] = n], {n, 1, 5*10^9} ]; a
(* Second program: *)
t = Table[0, {1000} ]; k = 1; While[ k < 6500000000, b = PowerMod[2, k, k]; If[b < 1001 && t[[b]] == 0, t[[b]] = k]; k++ ]; t


PROG

(PARI) a(n)=if(n==1, return(0)); my(k=n); while(lift(Mod(2, k)^k)!=n, k++); k \\ Charles R Greathouse IV, Oct 12 2011


CROSSREFS

Cf. A015910, A015948, A078457, A119678, A119679, A127816, A119715, A119714, A127817, A127818, A127819, A127820, A127821.
Bisections: A122182, A124977.
Sequence in context: A067481 A058433 A154998 * A235357 A260002 A058447
Adjacent sequences: A036233 A036234 A036235 * A036237 A036238 A036239


KEYWORD

nonn,nice


AUTHOR

David W. Wilson


EXTENSIONS

a(3) was first computed by the Lehmers.
More terms from Joe K. Crump (joecr(AT)carolina.rr.com), Sep 04 2000
a(69) = 887817490061261 = 29 * 37 * 12967 * 63809371.  Hagen von Eitzen, Jul 26 2009
Edited by Max Alekseyev, Jul 29 2011


STATUS

approved



