A131019 -id:A131019 - cp's OEIS frontend

A107373 a(n) = (n/2)*binomial(n-1, floor((n-1)/2)) - 2^(n-2).

0, 0, 1, 2, 7, 14, 38, 76, 187, 374, 874, 1748, 3958, 7916, 17548, 35096, 76627, 153254, 330818, 661636, 1415650, 2831300, 6015316, 12030632, 25413342, 50826684, 106853668, 213707336, 447472972, 894945944, 1867450648, 3734901296, 7770342787, 15540685574

Offset: 1

N. J. A. Sloane, Jul 20 2007

nonn

Total number of descents in all faro permutations of length n-1. Faro permutations are permutations avoiding the three consecutive patterns 231, 321 and 312. They are obtained by a perfect faro shuffle of two nondecreasing words of lengths differing by at most one. See also A340567, A340568 and A340569. - Sergey Kirgizov, Jan 11 2021

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Jean-Luc Baril, Alexander Burstein, and Sergey Kirgizov, Pattern statistics in faro words and permutations, arXiv:2010.06270 [math.CO], 2020. See Table 1.
F. Disanto and S. Rinaldi, Symmetric convex permutominoes and involutions, PU. M. A. 22:1 (2011), 39-60.
Igor Pak, The area of cyclic polygons: Recent progress on Robbins' Conjectures, Adv. Applied Math. 34 (2005), 690-696. Special issue in memory of David Robbins.

Cf. A131019, A131020, A131021.

[(n/2)*Binomial(n-1, Floor((n-1)/2)) - 2^(n-2): n in [1..40]]; // Vincenzo Librandi, Oct 01 2013

A056040 := n -> n!/iquo(n,2)!^2:
A133265 := n -> (n+2+(n-2)*(-1)^n)/2:
A107373 := n -> (A056040(n)*A133265(n)-2^n)/4:
seq(A107373(n),n=1..34); # Peter Luschny, Aug 30 2011

Table[(n/2) Binomial[n-1, Floor[(n-1)/2]] - 2^(n-2), {n, 1, 40}] (* Vincenzo Librandi, Oct 01 2013 *)

a(2*n) = 2*A000531(n-1); a(2*n+1) = A000531(n). - Max Alekseyev, Sep 30 2013

(1-n)*a(n) + 2*(n-1)*a(n-1) + 4*(n-2)*a(n-2) + 8*(-n+2)*a(n-3) = 0. - R. J. Mathar, May 26 2013

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

A131021 Prime semiperimeters of quadrilaterals with sides which are four consecutive primes.

Original entry on oeis.org

13, 101, 739, 751, 1301, 1459, 1471, 1483, 1583, 1619, 1877, 1889, 1949, 2213, 2239, 2351, 2381, 2441, 2473, 2549, 2741, 2917, 3271, 3413, 3863, 4133, 4567, 4987, 5081, 5279, 5347, 5783, 5813, 6719, 7027, 7369, 7459, 7607, 8233, 8291, 9151, 9187, 9397

Offset: 1

R. J. Mathar and Jonathan Vos Post, Jun 12 2007

These are the prime numbers in A131019(k), occurring at k=1, 13, 71, 72, 116, 127, 128, 129, 136, 138, 157, 158, 162, 183, 185, ....

Zak Seidov, Table of n, a(n) for n = 1..1000

for n from 1 to 1000 do a131019 := add(ithprime(n+i),i=1..4)/2 : if isprime(a131019) then printf("%d, ",a131019) ; fi ; od:

Select[(Total /@ Partition[Prime[Range[10^3]], 4, 1])/2, PrimeQ] (* Zak Seidov, Nov 01 2012 *)

p=3;q=5;r=7;forprime(s=11,1e4,if(isprime(t=(p+q+r+s)/2),print1(t", ")); p=q; q=r; r=s) \\ Charles R Greathouse IV, Nov 01 2012

A131019 INTERSECT A000040.

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

A330496 Squared area of quadrilateral with sides prime(n), prime(n+1), prime(n+2), prime(n+3) of odd primes configured as a cyclic quadrilateral. Sequence index starts at n=2 to omit the even prime.

Original entry on oeis.org

960, 5005, 17017, 46189, 96577, 212625, 394240, 765049, 1361920, 2027025, 3065857, 4385745, 6314112, 8973909, 12780049, 17116960, 21191625, 27428544, 33980800, 42600829, 56581525, 72382464, 89835424, 107972737, 121330189, 135745657, 167244385, 204917929

Offset: 2

Frank M Jackson, Dec 16 2019

nonn

If a, b, c, d are consecutive odd primes configured as a cyclic quadrilateral, then Brahmagupta's formula K = sqrt((a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d))/16 means that K^2 will always be an integer. The only cyclic quadrilateral with consecutive prime sides starting with side 2 has a rational squared area of 3003/16.

			a(2)=960 because cyclic quadrilateral with sides 3,5,7,11 has squared area = (3+5+7-11)(3+5-7+11)(3-5+7+11)(-3+5+7+11)/16 = 960.

Wikipedia, Cyclic quadrilateral.

Cf. A131019, A330096.

lst = {}; Do[{a, b, c, d} = {Prime[n], Prime[n+1], Prime[n+2], Prime[n+3]}; A2=(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d)/16; AppendTo[lst, A2], {n, 1, 100}]; lst

Area K of a cyclic quadrilateral with sides a, b, c, d is given by Brahmagupta's formula K = sqrt((s-a)(s-b)(s-c)(s-d)) where s = (a+b+c+d)/2.

cp's OEIS Frontend

A107373 a(n) = (n/2)*binomial(n-1, floor((n-1)/2)) - 2^(n-2).

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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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A131021 Prime semiperimeters of quadrilaterals with sides which are 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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A330496 Squared area of quadrilateral with sides prime(n), prime(n+1), prime(n+2), prime(n+3) of odd primes configured as a cyclic quadrilateral. Sequence index starts at n=2 to omit the even prime.

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