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A253596
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Numbers k such that A002313(m) is the greatest prime divisor of k^2 + 1 and A002313(m+1) is the greatest prime divisor of (k+1)^2 + 1 for some m.
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0
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1, 7, 31, 293, 1936, 2244, 4158, 5744, 11573, 25242, 285202, 339354
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OFFSET
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1,2
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COMMENTS
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A002313 contains the primes congruent to 1 or 2 (mod 4).
The corresponding indices m in A002313 are 1, 2, 6, 13, 69, 65, 322, 199, 130, 46, 1471, 866, ...
The corresponding primes A002313(m) are 2, 5, 37, 101, 809, 761, 4877, 2777, 1709, 509, 26821, 14957, ...
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LINKS
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EXAMPLE
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31 is in the sequence because 31^2 + 1 = 2*13*37 and 32^2 + 1 = 5*5*41 with the property that 37 = A002313(6) and 41 = A002313(7).
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MAPLE
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with(numtheory): nn:=500000:print(1):
for n from 1 to nn do:
p:=n^2+1:x:=factorset(p):n0:=nops(x):p1:=x[n0]:
q:=(n+1)^2+1:y:=factorset(q):n1:=nops(y):p2:=y[n1]:ii:=0:
for j from 2 by 2 to 1000 while(ii=0) do:
pp:=p1+j:
if type(pp, prime)=true and irem(pp, 4)=1
then
p3:=pp:ii:=1:
else
fi:
od:
if p3=p2
then
print(n):
else
fi:
od:
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MATHEMATICA
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lst={}; Do[If[Mod[Prime[i], 4]==1||Mod[Prime[i], 4]==2, AppendTo[lst, Prime[i]]], {i, 1, 1000}]; Do[Do[If[FactorInteger[n^2+1][[-1]][[1]]==Part[lst, j]&&FactorInteger[(n+1)^2+1][[-1]][[1]]==Part[lst, j+1], Print[n]], {n, 1, 20000}], {j, 1, 999}]
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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STATUS
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approved
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