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A217403
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a(n) is the smallest possible largest prime factor of the difference of two prime numbers q > p such that p + q = 2n.
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1
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2, 2, 2, 2, 3, 2, 3, 2, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 5, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 2, 5, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 5, 3, 3, 2, 3, 3, 5, 3, 3, 2, 5, 3, 2, 3, 3, 2, 3, 3, 5, 2, 5, 2, 3, 3, 2, 3, 3, 2, 3, 3, 5, 3, 3, 2, 5, 3, 5, 3, 3, 2, 3, 3, 7
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OFFSET
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4,1
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COMMENTS
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Test up to n=1000000 shows that,
when n is odd, a(n)<=prime(m+3) such that m is the number of prime factors of n that are smaller or equal to prime(m+3);
when n is even, a(n)<=prime(m+4) such that m is the number of prime factors of n that are smaller or equal to prime(m+4).
This is hypothesized true for all n >= 4.
The first appearance of prime(k) in this sequence is A217016.
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LINKS
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EXAMPLE
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For n<4, there is not a pair of different prime numbers such that p+q=2n.
When n=4, we have p=3 and q=5 such that 3+5=2n=8 (the only case). The largest prime factor of q-p=2 is 2, so a(4)=2;
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When n=90, we have prime number pairs (7,173) (13,167), (17,163), (23,157), (29,151), (31,149), (41,139), (43,137), (53,127), (67,113), (71,109), (73,107), (79,101), (83,97), totalled 14 pairs such that p1+p2=2n=180. The difference of each pairs are 173-7=166=2*83, 163-17=154=2*7*11, 163-17=146=2*73, 157-23=134=2*67, 151-29=122=2*61, 149-31=118=2*59, 139-41=98=2*7^2, 137-43=94=2*47, 127-53=74=2*37, 113-67=46=2*23, 109-71=38=2*19, 107-73=34=2*17, 101-79=22=2*11, and 97-83=14=2*7 respectively. Among these prime factorizations, the largest prime factors for each pairs are, 83, 11, 73, 67, 61, 59, 7, 47, 37, 23, 19, 17, 11, and 7 respectively. In the fourteen prime numbers, the smallest one is 7. So a(90)=7.
Additionally, 90=2*3^2*5, prime(3)=5, a(90)=7<prime(3+4)=17, consistent with the hypothesis.
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MATHEMATICA
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Table[a = i; Do[If[PrimeQ[i - j] && PrimeQ[i + j], f = Last[FactorInteger[j*2]][[1]]; If[a > f, a = f]], {j, 1, i - 1}]; a, {i, 4, 100}]
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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