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Numbers k such that binomial(2k-1, k-1) == 1 (mod k).
4

%I #19 Feb 09 2021 02:28:17

%S 2,3,5,7,9,11,13,17,19,23,25,29,31,37,41,43,47,49,53,59,61,67,71,73,

%T 79,83,89,97,101,103,107,109,113,121,125,127,131,137,139,149,151,157,

%U 163,167,169,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251

%N Numbers k such that binomial(2k-1, k-1) == 1 (mod k).

%C k such that A099905(k) = 1.

%C Contains primes, squares of odd primes and cubes of primes >= 5.

%C See A228562 for terms that are neither primes nor prime powers. [_Joerg Arndt_, Aug 27 2013]

%H McIntosh, R. J. (1995), <a href="http://matwbn.icm.edu.pl/ksiazki/aa/aa71/aa7144.pdf">On the converse of Wolstenholme's theorem</a>, Acta Arithm., LXXI.4 (1995), 381-389.

%e a(3) = 5 because binomial(9, 4) = 126 == 1 (mod 5).

%t Select[Range[300], Mod[Binomial[2# - 1, # - 1], #] == 1 &] (* _Alonso del Arte_, May 11 2014 *)

%o (PARI) isok(n) = (binomial(2*n-1, n-1) % n) == 1; \\ _Michel Marcus_, Aug 26 2013

%Y Cf. A099905.

%K nonn

%O 1,1

%A _Franz Vrabec_, Mar 26 2008