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%I #16 Mar 22 2017 12:41:23
%S 5,43,43,5,43,5,7,43,43,43,43,41,131,43,43,13,43,43,43,41,13,17,43,41,
%T 43,131,43,137,43,43,131,43,43,43,43,43,23,43,137,43,131,43,41,67,151,
%U 29,43,131,43,41,131,137,131,43,29,137,41,137,41,151,43,131,43,137,73,43,37,43,43,131,43,47
%N Array read by antidiagonals: T(i,j) is the largest prime in the sequence defined by a(1) = prime(i), a(2) = prime(j), a(n) = A006530(a(n-1)+a(n-2)+1) for n>=3, or 0 if that sequence contains arbitrarily large primes.
%C Conjecture: the sequence always eventually repeats, so T(i,j) > 0.
%H Robert Israel, <a href="/A284170/b284170.txt">Table of n, a(n) for n = 1..14196</a> (first 168 antidiagonals, flattened)
%H MathOverflow, <a href="http://mathoverflow.net/questions/264848/if-p-n-is-the-largest-prime-factor-of-p-n-1p-n-2m-then-p-n-is-b/264922"> If p_n is the largest prime factor of p_{n-1}+p_{n-2}+m, then p_n is bounded </a>
%e T(1,2) = 43 because the sequence in this case starts 2,3,3,7,11,19,31,17,7, and then repeats 5,13,19,11,31,43,5,7,13,7,7 in a cycle.
%e Array starts
%e 5 43 5 43 43 13 137 43 151 29 ...
%e 43 43 43 43 41 43 43 67 43 73 ...
%e 5 43 131 43 131 43 41 131 137 137 ...
%e 7 41 43 43 43 43 137 43 131 67 ...
%e 43 43 41 43 131 131 131 43 131 151 ...
%e 13 43 131 43 41 43 43 43 73 73 ...
%e 17 43 137 43 151 47 43 41 41 131 ...
%e 43 43 131 41 43 41 43 41 67 137 ...
%e 23 43 137 131 43 151 137 137 197 137 ...
%e 29 41 43 137 73 43 131 41 131 389 ...
%p M:= 20: # to get the first M antidiagonals
%p with(queue):
%p backprop:= proc(r, p)
%p local t; global F;
%p for t in Parents[r] do
%p if F[t] < p then
%p F[t]:= p;
%p procname(t, p);
%p fi
%p od
%p end proc:
%p Verts:= {seq(seq([ithprime(i),ithprime(j)],i=1..M),j=1..M)}:
%p for v in Verts do F[v]:= max(v); Parents[v]:= {} od:
%p Agenda:= new(op(Verts)):
%p while not empty(Agenda) do
%p t:= dequeue(Agenda);
%p r:= [t[2],max(numtheory:-factorset(t[1]+t[2]+1))];
%p if member(r,Verts) then
%p Parents[r]:= Parents[r] union {t};
%p else
%p Verts:= Verts union {r};
%p Parents[r]:= {t};
%p enqueue(Agenda,r);
%p F[r]:= max(r);
%p fi;
%p backprop(r, F[r]);
%p od:
%p seq(seq(F[[ithprime(m-j),ithprime(j)]],j=1..m-1),m=2..M+1);
%Y Cf. A006530.
%K nonn,tabl
%O 1,1
%A _Robert Israel_, Mar 21 2017
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