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A284170
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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.
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1
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5, 43, 43, 5, 43, 5, 7, 43, 43, 43, 43, 41, 131, 43, 43, 13, 43, 43, 43, 41, 13, 17, 43, 41, 43, 131, 43, 137, 43, 43, 131, 43, 43, 43, 43, 43, 23, 43, 137, 43, 131, 43, 41, 67, 151, 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
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OFFSET
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1,1
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COMMENTS
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Conjecture: the sequence always eventually repeats, so T(i,j) > 0.
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LINKS
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EXAMPLE
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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.
Array starts
5 43 5 43 43 13 137 43 151 29 ...
43 43 43 43 41 43 43 67 43 73 ...
5 43 131 43 131 43 41 131 137 137 ...
7 41 43 43 43 43 137 43 131 67 ...
43 43 41 43 131 131 131 43 131 151 ...
13 43 131 43 41 43 43 43 73 73 ...
17 43 137 43 151 47 43 41 41 131 ...
43 43 131 41 43 41 43 41 67 137 ...
23 43 137 131 43 151 137 137 197 137 ...
29 41 43 137 73 43 131 41 131 389 ...
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MAPLE
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M:= 20: # to get the first M antidiagonals
with(queue):
backprop:= proc(r, p)
local t; global F;
for t in Parents[r] do
if F[t] < p then
F[t]:= p;
procname(t, p);
fi
od
end proc:
Verts:= {seq(seq([ithprime(i), ithprime(j)], i=1..M), j=1..M)}:
for v in Verts do F[v]:= max(v); Parents[v]:= {} od:
Agenda:= new(op(Verts)):
while not empty(Agenda) do
t:= dequeue(Agenda);
r:= [t[2], max(numtheory:-factorset(t[1]+t[2]+1))];
if member(r, Verts) then
Parents[r]:= Parents[r] union {t};
else
Verts:= Verts union {r};
Parents[r]:= {t};
enqueue(Agenda, r);
F[r]:= max(r);
fi;
backprop(r, F[r]);
od:
seq(seq(F[[ithprime(m-j), ithprime(j)]], j=1..m-1), m=2..M+1);
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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