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A099468
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Numbers n such that there are no primes < 2n in the sequence m(0)=n, m(k+1)=m(k)+4k.
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1
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1, 21, 45, 51, 81, 213, 249, 315, 477, 525, 681, 891, 1143, 1221, 1851, 1965, 2415, 5133
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OFFSET
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1,2
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COMMENTS
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No others < 10^8. Note that 3 divides all these n > 1. This sequence is conjectured to be complete. Related to a question posed in A036468 by Zhang Ming-Zhi. Let r=2s+1 be an odd number. If n = (s+1)^2+s^2, then the sequence m(0)=n, m(k+1)=m(k)+4k for k=0,1,...s calculates the s+1 distinct sums of two squares (r-i)^2+i^2.
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LINKS
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EXAMPLE
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45 is here because 45, 49, 57, 69 and 85 are all composite.
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MATHEMATICA
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lst={}; Do[n=m; found=False; k=0; While[n=n+4k; !found && n<2m, found=PrimeQ[n]; k++ ]; If[ !found, AppendTo[lst, m]], {m, 1, 10000, 2}]; lst
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CROSSREFS
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Cf. A036468 (number of ways to represent 2n+1 as a+b with a^2+b^2 prime).
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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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