%I #21 Dec 11 2017 05:31:59

%S 29,47,89,101,107,113,139,151,181,199,211,229,233,263,281,307,331,347,

%T 349,353,401,421,461,509,521,541,557,563,619,661,677,691,709,743,761,

%U 769,797,809,811,829,859,881,911,919,941,953,967,977,991,1009,1021,1031,1049,1061,1069,1087,1097,1103,1109,1151,1217,1223,1229,1231,1249,1277

%N All primes p > 5 such that A001175(p) is smaller than the maximal value permitted by Wall's Theorems 6 and 7.

%C Included because A001175 is still a mystery (as are many sequences of the same type).

%C A222414 gives the corresponding values of A001175(a(n)).

%C The maximal value for a prime p > 5 is p-1 if p == 1 or 9 (mod 10) and 2*(p+1) if p == 3 or 7 (mod 10). See Wall's Theorems 6 and 7. These values are given in A253806. - _Wolfdieter Lang_, Jan 16 2015

%C Prime(n) is a member if and only if A296240(n) > 1. - _Jonathan Sondow_, Dec 10 2017

%H D. D. Wall, <a href="http://www.jstor.org/stable/2309169">Fibonacci series modulo m</a>, Amer. Math. Monthly, 67 (1960), 525-532.

%e From _Wolfdieter Lang_, Jan 16 2015: (Start)

%e a(1) = 29 because A001175(29) = 14 but the maximal value is 29 - 1 = 28.

%e a(2) = 47 because A001175(47) = 32 but the maximal value is 2*(47 + 1) = 96.

%e All other primes p > 5 have A001175(p) = maximal value for p.

%e E.g., p = 11 has A001175(11) = 11-1 = 10 and p = 7 has A001175(7) = 2*(7 + 1) = 16. (End)

%Y Cf. A001175, A060305, A222414, A296240.

%Y Cf. A001176, A001177. - _Wolfdieter Lang_, Jan 16 2015

%K nonn

%O 1,1

%A _N. J. A. Sloane_, Feb 28 2013

%E Name corrected by _Wolfdieter Lang_, Jan 16 2015