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A036236
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Least inverse of A015910: smallest integer k > 0 such that 2^k mod k = n, or 0 if no such k exists.
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68
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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
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0,3
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COMMENTS
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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.
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REFERENCES
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P. Erdős 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.
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LINKS
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Eric Weisstein's World of Mathematics, 2
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FORMULA
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It's obvious that for each k, a(k) > k and we can easily prove that 2^(3^n) = 3^n-1 (mod 3^n). So 3^n is the least k with 2^k mod k = 3^n-1. Hence for each n, a(3^n-1) = 3^n. - Farideh Firoozbakht, Nov 14 2006
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EXAMPLE
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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.
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MATHEMATICA
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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
nk[n_] := Module[ {k}, k = 1; While[PowerMod[2, k, k] != n, k++]; k]
Join[{1, 0}, Table[nk[i], {i, 2, 46}]] (* Robert Price, Oct 11 2018 *)
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PROG
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(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
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CROSSREFS
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Cf. A015910, A015948, A078457, A119678, A119679, A127816, A119715, A119714, A127817, A127818, A127819, A127820, A127821.
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KEYWORD
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nonn,nice
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AUTHOR
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EXTENSIONS
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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
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STATUS
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approved
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