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%I #25 Mar 26 2024 04:16:28
%S 3,11,17,23,29,37,41,47,53,59,67,71,79,83,89,97,101,107,113,127,131,
%T 137,149,157,163,167,173,179,191,197,211,223,227,233,239,251,257,263,
%U 269,277,281,293,307
%N Primes q with result 1 under iterations of {r mod (max prime p < r)} starting at r = q.
%C Subsequence of A175071.
%C Union of a(n) and A175074 is A175071. - _Jaroslav Krizek_, Jan 30 2010
%C The terms in A025584 but not in here are 2, 2999, 3299, 5147, 5981, 8999, 9587, ... , apparently those listed in A175080. - _R. J. Mathar_, Feb 01 2010
%C a(n-1)=A156828(n) in the range n=3..348, but afterwards the sequences differ because numbers like 2999 and 3229 are in A156828 but not in here. - _R. J. Mathar_, Mar 01 2010
%C Conjecture: under this iteration procedure, all primes eventually will yield either a 2 or a 1. If a 2 results, all subsequent terms are zeros; if a 1 results, all subsequent terms are -1s. The conjecture is true for the first 2 million primes. - _Harvey P. Dale_, Jan 17 2014
%H R. J. Mathar, <a href="/A175073/b175073.txt">Table of n, a(n) for n = 1..10000</a>
%e Iteration procedure for a(2) = 11: 11 mod 7 = 4, 4 mod 3 = 1.
%p isA175073 := proc(p)
%p local r,rold;
%p if not isprime(p) then
%p return false;
%p end if;
%p r := p ;
%p while true do
%p rold :=r ;
%p if r = 2 then
%p return false ;
%p end if;
%p r := modp(r,prevprime(r)) ;
%p if r = 1 then
%p return true;
%p elif r= rold then
%p return false ;
%p end if;
%p end do:
%p end proc:
%p A175073 := proc(n)
%p option remember ;
%p if n= 1 then
%p 3;
%p else
%p for p from procname(n-1)+2 by 2 do
%p if isA175073(p) then
%p return p;
%p end if;
%p end do:
%p end if;
%p end proc:
%p seq(A175073(n),n=1..40) ; # _R. J. Mathar_, Mar 25 2024
%t r1Q[n_] := FixedPoint[Mod[#, NextPrime[#, -1]] &, n] == -1; Select[Prime[ Range[70]],r1Q] (* This program relies upon the conjecture described in the comments above *) (* _Harvey P. Dale_, Jan 17 2014 *)
%Y Note that all three of A025584, A156828, A175073 are different sequences. - _N. J. A. Sloane_, Apr 10 2011
%K nonn
%O 1,1
%A _Jaroslav Krizek_, Jan 23 2010
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