A106171 -id:A106171 - cp's OEIS frontend

A131019 Semiperimeters of quadrilaterals whose sides are 4 consecutive odd primes.

13, 18, 24, 30, 36, 44, 51, 60, 69, 76, 84, 92, 101, 110, 120, 129, 136, 145, 153, 162, 174, 185, 195, 204, 210, 216, 228, 240, 254, 267, 278, 288, 298, 310, 319, 330, 341, 350, 362, 372, 381, 390, 400, 415, 430, 445, 456, 464, 471, 482, 494, 506, 520, 530

Offset: 1

Jonathan Vos Post, Jun 09 2007

nonn

(2+3+5+7)/2 = 8.5, not an integer. Hence we restrict to odd primes. The cyclic quadrilaterals whose areas, rounded, are prime are given in A131020. The prime semiperimeters begin: a(1) = 13, a(13) = 101. This arises in the cyclic quadrilateral analog of A106171.

			a(1) = (3 + 5 + 7 + 11)/2 = 13.

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.

J. L. Coolidge, A Historically Interesting Formula for the Area of a Quadrilateral, Amer. Math. Monthly 46, 345-347, 1939.
Eric Weisstein's World of Mathematics, Brahmagupta's Formula.

Cf. A106171, A131020.

Cf. A034963.

A131019 := proc(n) local i ; add( ithprime(n+i),i=1..4)/2 ; end: for n from 1 to 180 do printf("%d, ",A131019(n)) : od:

Plus@@@Partition[Prime[Range[2,6! ]],4,1]/2 (* Vladimir Joseph Stephan Orlovsky, Feb 18 2010 *)

a(n) = (prime(n) + prime(n+1) + prime(n+2) + prime(n+3))/2 for n>1.

a(n) = (prime(n+1) + prime(n+2) + prime(n+3) + prime(n+4))/2 = A034963(n)/2.

Edited by R. J. Mathar, Jun 12 2007

A131020 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.

Original entry on oeis.org

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

Offset: 1

Jonathan Vos Post, Jun 09 2007

The semiperimeters of cyclic quadrilaterals with four consecutive odd prime sides are given in A131019. This arises in the cyclic quadrilateral analog of A106171.

			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.

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.

J. L. Coolidge, A Historically Interesting Formula for the Area of a Quadrilateral, Amer. Math. Monthly 46, 345-347, 1939.
Eric Weisstein's World of Mathematics, Brahmagupta's Formula.

Cf. A000040, A106171, A131019.

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;

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.

Edited by R. J. Mathar, Jun 12 2007

A360790 Squared length of diagonal of right trapezoid with three consecutive prime length sides.

Original entry on oeis.org

8, 13, 41, 53, 137, 173, 305, 397, 533, 877, 977, 1373, 1697, 1885, 2245, 2813, 3517, 3737, 4493, 5077, 5345, 6277, 6953, 7937, 9413, 10217, 10613, 11465, 12077, 12785, 16165, 17165, 18869, 19325, 22237, 22837, 24665, 26605, 27925, 29933, 32141, 32765, 36497, 37253, 38953, 39745

Offset: 1

Aaron T Cowan, Feb 20 2023

nonn

The value d is the square of the length of the diagonal of a trapezoid with a height and bases that are consecutive primes, respectively. The diagonal length is calculated using the Pythagorean theorem, but this distance is squared so that the value is an integer.

			        p(2)=3
        _ _ _ _
a(1):  |        \  d^2=2^2+(5-3)^2=8
p(1)=2 |_ _ _ _ _\
        p(3)=5
        p(3)=5
        _ _ _ _ _ _
a(2):  |           \    d^2=3^2 + (7-5)^2 = 9+4 = 13
p(2)=3 |            \
       |_ _ _ _ _ _ _\
        p(4)=7
a(3)= 5^2+(11-7)^2 = 25+16 = 41
a(7)= 17^2+(23-19)^2=305 = 5*61

Aaron T Cowan, Table of n, a(n) for n = 1..500

Cf. A001248, A076821.

Cf. A131019, A106171.

%shorter 1 line version
arrayfun(@(p) p^2+(nextprime(nextprime(p+1)+1)-nextprime(p+1))^2,[primes(10^6)])

Map[(#[[1]]^2 + (#[[3]] - #[[2]])^2) &, Partition[Prime[Range[50]], 3, 1]] (* Amiram Eldar, Feb 24 2023 *)

a(n) = prime(n)^2 + (prime(n+2)-prime(n+1))^2; \\ Michel Marcus, Feb 23 2023

a(n) = prime(n)^2 + (prime(n+2)-prime(n+1))^2.

a(n) = A001248(n) + A076821(n+1). - Michel Marcus, Feb 23 2023

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A131019 Semiperimeters of quadrilaterals whose sides are 4 consecutive odd primes.

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A131020 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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A360790 Squared length of diagonal of right trapezoid with three consecutive prime length sides.

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