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A324669
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a(n) is the least k>0 such that A001359(n)+k^2 is in A001359.
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1
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6, 114, 162, 210, 24, 330, 6, 6, 18, 12, 30, 210, 6, 18, 120, 150, 330, 24, 6, 42, 30, 66, 96, 210, 180, 210, 42, 54, 60, 360, 6, 18, 630, 60, 210, 24, 30, 66, 24, 126, 30, 48, 1380, 24, 90, 102, 6, 30, 42, 18, 90, 90, 42, 54, 12, 36, 60, 186, 210, 12, 72, 24, 42, 24, 330, 60, 12
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OFFSET
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2,1
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COMMENTS
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Offset is 2 because 3+k^2 is never in A001359.
All terms are divisible by 6.
The generalized Bunyakovsky conjecture implies that a(n) always exists, for n >= 2.
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LINKS
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EXAMPLE
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a(3) = 114 because A001359(3)=11, 11+114^2=13007 is in A001359, and no smaller k works.
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MAPLE
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P:= select(isprime, {seq(i, i=3..10000, 2)}):
TP:= sort(convert(P intersect map(`-`, P, 2), list)):
f:= proc(p) local k;
for k from 6 by 6 do if isprime(p + k^2) and isprime(p + k^2 + 2) then return k fi od
end proc:
map(f, TP[2..-1]);
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MATHEMATICA
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With[{s = Select[Prime@ Range[3, 332], PrimeQ[# + 2] &]}, Array[Block[{k = 1}, While[! AllTrue[s[[#]] + k^2 + {0, 2}, PrimeQ], k++]; k] &, Length@ s]] (* Michael De Vlieger, Sep 03 2019 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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