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A131020
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For all cyclic quadrilaterals with four consecutive primes as sides that have an area that is prime after rounding, the sequence gives the first of these four consecutive primes.
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5
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3, 5, 13, 17, 61, 67, 97, 139, 157, 163, 173, 223, 271, 349, 353, 419, 479, 503, 541, 691, 701, 743, 877, 941, 1013, 1049, 1051, 1097, 1123, 1229, 1231, 1249, 1297, 1301, 1423, 1453, 1493, 1531, 1559, 1607, 1621, 1697, 1811, 1901, 1999, 2017, 2027, 2053, 2087
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OFFSET
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1,1
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COMMENTS
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The semiperimeters of cyclic quadrilaterals with four consecutive odd prime sides are given in A131019. This arises in the cyclic quadrilateral analog of A106171.
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REFERENCES
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Coxeter, H. S. M. and Greitzer, S. L. "Cyclic Quadrangles; Brahmagupta's Formula", Sect. 3.2 in Geometry Revisited. Washington, DC: Math. Assoc. Amer., pp. 56-60, 1967.
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LINKS
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FORMULA
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a(n) = prime(k) for some k such that, where S = semiperimeter = (prime(k) + prime(k+1) + prime(k+2) + prime(k+3))/2 is an element of A131019 and rounded area = round(sqrt((S-prime(k)*S-prime(k+1)*S-prime(k+2)*S-prime(k+3)) is prime.
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EXAMPLE
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a(5) = 61 because (61 + 67 + 71 + 73)/2 = 136 and sqrt((136 - 61)*(136 - 67)*(136 - 71)*(136 - 73)) = 4603.43622 and round(4603.43622) = 4603 is prime.
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MAPLE
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Digits := 80 : isA131020 := proc(p) local p2, p3, p4, s, area; if isprime(p) then p2 := nextprime(p) ; p3 := nextprime(p2) ; p4 := nextprime(p3) ; s := (p+p2+p3+p4)/2 ; area := round(sqrt((s-p)*(s-p2)*(s-p3)*(s-p4))) ; RETURN(isprime(area)) ; else false ; fi ; end: for n from 1 to 380 do if isA131020(ithprime(n)) then printf("%d, ", ithprime(n)) ; fi ; od;
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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