cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 31 results. Next

A020882 Ordered hypotenuses (with multiplicity) of primitive Pythagorean triangles.

Original entry on oeis.org

5, 13, 17, 25, 29, 37, 41, 53, 61, 65, 65, 73, 85, 85, 89, 97, 101, 109, 113, 125, 137, 145, 145, 149, 157, 169, 173, 181, 185, 185, 193, 197, 205, 205, 221, 221, 229, 233, 241, 257, 265, 265, 269, 277, 281, 289, 293, 305, 305, 313, 317, 325, 325, 337, 349, 353, 365, 365
Offset: 1

Views

Author

Keywords

Comments

The largest member 'c' of the primitive Pythagorean triples (a,b,c) ordered by increasing c.
These are numbers of the form a^2 + b^2 where gcd(b-a, 2*a*b)=1. - M. F. Hasler, Apr 04 2010
Equivalently, numbers of the form a^2 + b^2 where gcd(a,b) = 1 and a and b are not both odd. To avoid double-counting, require a > b > 0. - Franklin T. Adams-Watters, Mar 15 2015
The density of such points in a circle with radius squared = a(n) is ~ Pi * a(n). Restricting to a > b > 0 reduces this by a factor of 1/8; requiring gcd(a,b)=1 provides a factor of 6/Pi^2; and a, b not both odd is a factor of 2/3. (2/3, not 3/4, because the case a, b both even has already been eliminated.) Multiplying, a(n) * Pi * 1/8 * 6/Pi^2 * 2/3 is a(n) / (2 * Pi). But n is approximately this number of points, so a(n) ~ 2 * Pi * n. Conjectured by David W. Wilson, proof by Franklin T. Adams-Watters, Mar 15 2015
Permutations are in A094194, A088511, A121727, A119321, A113482 and A081804. Entries of A024409 occur here more than once. - R. J. Mathar, Apr 12 2010
The distinct terms of this sequence seem to constitute a subset of the sequence defined as a(n) = (-1)^n + 6*n for n >= 1. - Alexander R. Povolotsky, Mar 15 2015
The terms in this sequence are given by f(m,n) = m^2 + n^2 where m and n are any two integers satisfying m > 1, n < m, the greatest common divisor of m and n is 1, and m and n are both not odd. E.g., f(m,n) = f(2,1) = 2^2 + 1^2 = 4 + 1 = 5. - Agola Kisira Odero, Apr 29 2016

References

  • M. de Frénicle, "Méthode pour trouver la solutions des problèmes par les exclusions", in: "Divers ouvrages de mathématiques et de physique, par Messieurs de l'Académie royale des sciences", Paris, 1693, pp 1-44.

Crossrefs

Cf. A004613, A008846, A020883-A020886, A046086, A046087, A222946 (as a number triangle).

Programs

  • Mathematica
    t={};Do[Do[a=Sqrt[c^2-b^2];If[a>b,Break[]];If[IntegerQ[a]&&GCD[a,b,c]==1,AppendTo[t,c]],{b,c-1,3,-1}],{c,400}];t (* Vladimir Joseph Stephan Orlovsky, Jan 21 2012 *)
    f[c_] := Block[{a = 1, b, lst = {}}, While[b = Sqrt[c^2 - a^2]; a < b, If[ IntegerQ@ b && GCD[a, b, c] == 1, AppendTo[lst, a]]; a++]; lst]
    Join @@ Table[ConstantArray[n, Length@f@n], {n, 1, 400, 4}] (* Robert G. Wilson v, Mar 16 2014; corrected by Andrey Zabolotskiy, Oct 31 2019 *)
  • PARI
    {my( c=0, new=[]); for( b=1,99, for( a=1, b-1, gcd(b-a,2*a*b) == 1 && new=concat(new,a^2+b^2)); new=vecsort(new); for( j=1,#new, new[j] > (b+1)^2 & (new=vecextract(new, Str(j,".."))) & next(2); write("b020882.txt",c++," "new[j])); new=[])} \\ M. F. Hasler, Apr 04 2010

Formula

a(n) = sqrt((A120681(n)^2 + A120682(n)^2)/2). - Lekraj Beedassy, Jun 24 2006
a(n) = sqrt(A046086(n)^2 + A046087(n)^2). - Zak Seidov, Apr 12 2011
a(n) ~ 2*Pi*n. - observation by David W. Wilson, proved by Franklin T. Adams-Watters (cf. comments), Mar 15 2015
a(n) = sqrt(A180620(n)^2 + A231100(n)^2). - Rui Lin, Oct 09 2019

Extensions

Edited by N. J. A. Sloane, May 15 2010

A020883 Ordered long legs of primitive Pythagorean triangles.

Original entry on oeis.org

4, 12, 15, 21, 24, 35, 40, 45, 55, 56, 60, 63, 72, 77, 80, 84, 91, 99, 105, 112, 117, 120, 132, 140, 143, 144, 153, 156, 165, 168, 171, 176, 180, 187, 195, 208, 209, 220, 221, 224, 231, 240, 247, 252, 253, 255, 260, 264, 272, 273, 275, 285, 288, 299, 304, 308, 312, 323
Offset: 1

Views

Author

Keywords

Comments

Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, (A, B) = 1, A < B); sequence gives values of B, sorted.
Any term in this sequence is given by f(m,n) = 2*m*n or g(m,n) = m^2 - n^2 where m and n are any two positive integers, m > 1, n < m, the greatest common divisor of m and n is 1, m and n are not both odd; e.g., f(m,n) = f(2,1) = 2*2*1 = 4. - Agola Kisira Odero, Apr 29 2016
All terms are composite. - Thomas Ordowski, Mar 12 2017
a(1) is the only power of 2. - Torlach Rush, Nov 08 2019
The first term appearing twice is 420 = a(75) = a(76) = A024410(1). - Giovanni Resta, Nov 11 2019
From Bernard Schott, May 05 2021: (Start)
Also, ordered sides a of primitive triples (a, b, c) for integer-sided triangles where side a is the harmonic mean of the 2 other sides b and c, i.e., 2/a = 1/b + 1/c with b < a < c.
Example: a(2) = 12, because the second triple is (12, 10, 15) with side a = 12, satisfying 2/12 = 1/10 + 1/15 and 15-12 < 10 < 15+12.
The first term appearing twice 420 corresponds to triples (420, 310, 651) and (420, 406, 435), the second one is 572 = a(101) = a(102) = A024410(2) and corresponds to triples (572, 407, 962) and (572, 455, 770). The terms that appear more than once in this sequence are in A024410.
For the corresponding primitive triples and miscellaneous properties and references, see A343891. (End)

References

  • V. Lespinard & R. Pernet, Trigonométrie, Classe de Mathématiques élémentaires, programme 1962, problème B-337 p. 179, André Desvigne.

Crossrefs

Triangles with 2/a = 1/b + 1/c: A343891 (triples), A020883 (side a), A343892 (side b), A343893 (side c), A343894 (perimeter).

Programs

  • Maple
    for a from 4 to 325 do
    for b from floor(a/2)+1 to a-1 do
    c := a*b/(2*b-a);
    if c=floor(c) and igcd(a,b,c)=1 and c-bBernard Schott, May 05 2021

Extensions

Extended and corrected by David W. Wilson

A020884 Ordered short legs of primitive Pythagorean triangles.

Original entry on oeis.org

3, 5, 7, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 20, 21, 23, 24, 25, 27, 28, 28, 29, 31, 32, 33, 33, 35, 36, 36, 37, 39, 39, 40, 41, 43, 44, 44, 45, 47, 48, 48, 49, 51, 51, 52, 52, 53, 55, 56, 57, 57, 59, 60, 60, 60, 61, 63, 64, 65, 65, 67, 68, 68, 69, 69, 71, 72, 73, 75, 75, 76, 76, 77
Offset: 1

Views

Author

Keywords

Comments

Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, (A, B) = 1, A <= B); sequence gives values of A, sorted.
Union of A081874 and A081925. - Lekraj Beedassy, Jul 28 2006
Each term in this sequence is given by f(m,n) = m^2 - n^2 or g(m,n) = 2mn where m and n are relatively prime positive integers with m > n, m and n not both odd. For example, a(1) = f(2,1) = 2^2 - 1^2 = 3 and a(4) = g(4,1) = 2*4*1 = 8. - Agola Kisira Odero, Apr 29 2016
All powers of 2 greater than 4 (2^2) are terms, and are generated by the function g(m,n) = 2mn. - Torlach Rush, Nov 08 2019

Crossrefs

Cf. A009004, A020882, A020883, A020885, A020886. Different from A024352.
Cf. A024359 (gives the number of times n occurs).
Cf. A037213.

Programs

  • Haskell
    a020884 n = a020884_list !! (n-1)
    a020884_list = f 1 1 where
       f u v | v > uu `div` 2        = f (u + 1) (u + 2)
             | gcd u v > 1 || w == 0 = f u (v + 2)
             | otherwise             = u : f u (v + 2)
             where uu = u ^ 2; w = a037213 (uu + v ^ 2)
    -- Reinhard Zumkeller, Nov 09 2012
  • Mathematica
    shortLegs = {}; amx = 99; Do[For[b = a + 1, b < (a^2/2), c = (a^2 + b^2)^(1/2); If[c == IntegerPart[c] && GCD[a, b, c] == 1, AppendTo[shortLegs, a]]; b = b + 2], {a, 3, amx}]; shortLegs (* Vladimir Joseph Stephan Orlovsky, Aug 07 2008 *)
    Take[Union[Sort/@({Times@@#,(Last[#]^2-First[#]^2)/2}&/@(Select[Subsets[Range[1,101,2],{2}],GCD@@#==1&]))][[;;,1]],80] (* Harvey P. Dale, Feb 06 2025 *)

Extensions

Extended and corrected by David W. Wilson

A005279 Numbers having divisors d, e with d < e < 2*d.

Original entry on oeis.org

6, 12, 15, 18, 20, 24, 28, 30, 35, 36, 40, 42, 45, 48, 54, 56, 60, 63, 66, 70, 72, 75, 77, 78, 80, 84, 88, 90, 91, 96, 99, 100, 102, 104, 105, 108, 110, 112, 114, 117, 120, 126, 130, 132, 135, 138, 140, 143, 144, 150, 153, 154, 156, 160, 162, 165, 168, 170, 174, 175, 176
Offset: 1

Views

Author

Keywords

Comments

The arithmetic and harmonic means of A046793(n) and a(n) are both integers.
n is in this sequence iff n is a multiple of some term in A020886.
a(n) is also a positive integer v for which there exists a smaller positive integer u such that the contraharmonic mean (uu+vv)/(u+v) is an integer c (in fact, there are two distinct values u giving with v the same c). - Pahikkala Jussi, Dec 14 2008
A174903(a(n)) > 0; complement of A174905. - Reinhard Zumkeller, Apr 01 2010
Also numbers n such that A239657(n) > 0. - Omar E. Pol, Mar 23 2014
Erdős (1948) shows that this sequence has a natural density, so a(n) ~ k*n for some constant k. It can be shown that k < 3.03, and by numerical experiments it seems that k is around 1.8. - Charles R Greathouse IV, Apr 22 2015
Numbers k such that at least one of the parts in the symmetric representation of sigma(k) has width > 1. - Omar E. Pol, Dec 08 2016
Erdős conjectured that the asymptotic density of this sequence is 1. The numbers of terms not exceeding 10^k for k = 1, 2, ... are 1, 32, 392, 4312, 45738, 476153, 4911730, 50359766, 513682915, 5224035310, ... - Amiram Eldar, Jul 21 2020
Numbers with at least one partition into two distinct parts (s,t), sWesley Ivan Hurt, Jan 16 2022
Appears to be the set of numbers x such that there exist numbers y and z satisfying the condition (x^2+y^2)/(x^2+z^2) = (x+y)/(x+z). For example, (15^2+10^2)/(15^2+3^2) = (15+10)/(15+3), so 15 is in the sequence. - Gary Detlefs, Apr 01 2023
From Bob Andriesse, Nov 26 2023: (Start)
Rewriting (x^2+y^2) / (x^2+z^2) = (x+y) / (x+z) as (x^2+y^2) / (x+y) = (x^2+z^2) / (x+z) has the advantage that the values on both sides of the = sign in the given example become integers. A possible sequence with the name: "k's for which r = (k^2+m^2) / (k+m) can be an integer while mA053629(n) and the r's being A009003(n). If (k^2+m^2) / (k+m) = r and m satisfies the divisibility condition, then r-m also does, because (k^2 + (r-m)^2) / (k + (r-m)) = r as well, confirming Pahikkala Jussi's comment about the existence of two distinct values for his u.
The fact that 15 is in the sequence is not so much because (15^2 + 10^2) / (15^2 + 3^2) = 1.3888... = (15+10) / (15+3), as indicated by Gary Detlefs, but rather because (15+10) | (15^2 + 10^2). And since r = (15^2 + 10^2) / (15+10) = 13, the second value that satisfies the divisibility condition is 13-10 = 3, so (15^2 + 3^2) / (15+3) = 13 as well.
Since (k+m)| (k^2 + m^2) is equivalent to (k+m) | 2*k^2 as well as to (k+m) | 2*m^2, both of these alternative divisibility conditions can be used to generate the same sequence too. (End)

References

  • R. K. Guy, Unsolved Problems in Number Theory, E3.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Subsequence of A024619 and hence of A002808.

Programs

  • Haskell
    a005279 n = a005279_list !! (n-1)
    a005279_list = filter ((> 0) . a174903) [1..]
    -- Reinhard Zumkeller, Sep 29 2014
    
  • Maple
    isA005279 := proc(n) local divs,d,e ; divs := numtheory[divisors](n) ; for d from 1 to nops(divs)-1 do for e from d+1 to nops(divs) do if divs[e] < 2*divs[d] then RETURN(true) ; fi ; od: od: RETURN(false) : end; for n from 3 to 300 do if isA005279(n) then printf("%d,",n) ; fi ; od : # R. J. Mathar, Jun 08 2006
  • Mathematica
    aQ[n_] := Select[Partition[Divisors[n], 2, 1], #[[2]] < 2 #[[1]] &] != {}; Select[Range[178], aQ] (* Jayanta Basu, Jun 28 2013 *)
  • PARI
    is(n)=my(d=divisors(n));for(i=3,#d,if(d[i]<2*d[i-1],return(1)));0 \\ Charles R Greathouse IV, Apr 22 2015
    
  • Python
    from sympy import divisors
    def is_A005279(n): D=divisors(n)[1:]; return any(e<2*d  for d,e in zip(D, D[1:]))
    # M. F. Hasler, Mar 20 2025

Formula

a(n) = A010814(n)/2. - Omar E. Pol, Dec 04 2016

A024364 Ordered perimeters of primitive Pythagorean triangles.

Original entry on oeis.org

12, 30, 40, 56, 70, 84, 90, 126, 132, 144, 154, 176, 182, 198, 208, 220, 234, 240, 260, 286, 306, 312, 330, 340, 374, 380, 390, 408, 418, 420, 442, 456, 462, 476, 494, 510, 532, 544, 546, 552, 570, 598, 608, 644, 646, 650, 672, 684, 690, 700, 714, 736, 756
Offset: 1

Views

Author

Keywords

Comments

Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, (A, B) = 1, A <= B); sequence gives perimeters A+B+C.
k is in this sequence iff A070109(k) > 0. This is a subsequence of A010814.
For the corresponding primitive Pythagorean triples see A103606. - Wolfdieter Lang, Oct 06 2014
Any term in this sequence can be generated by f(m,k) = 2*m*(m+k), where m and k are positive coprime integers and m > 1, k < m, and m and k are not both odd. For example: f(2,1) = 2*2*(2+1) = 12. - Agola Kisira Odero, Apr 29 2016

Crossrefs

Cf. A020886 (semiperimeters: a(n)/2), A024408 (terms with multiplicity > 1).

Programs

  • Maple
    isA024364 := proc(an) local r::integer,s::integer ; for r from floor((an/4)^(1/2)) to floor((an/2)^(1/2)) do for s from r-1 to 1 by -2 do if 2*r*(r+s) = an and gcd(r,s) < 2 then RETURN(true) ; fi ; if 2*r*(r+s) < an then break ; fi ; od ; od : RETURN(false) ; end : for n from 2 to 400 do if isA024364(n) then printf("%d,",n) ; fi ; od ; # R. J. Mathar, Jun 08 2006
  • Mathematica
    isA024364[an_] := Module[{r, s}, For[r = Floor[(an/4)^(1/2)], r <= Floor[(an/2)^(1/2)], r++, For[s = r - 1, s >= 1, s -= 2, If[2r(r + s) == an && GCD[r, s] < 2, Return[True]]; If[2r(r + s) < an, Break[]]]]; Return[False]];
    Select[Range[2, 1000], isA024364] (* Jean-François Alcover, May 24 2024, after R. J. Mathar *)
  • PARI
    select( {is_A024364(n)=my(k=valuation(n,2), o=n>>k); k && fordiv(o, r, r^2<<(k-1) >= o && return; r^2< o && gcd(r,o/r)==1 && return(1))}, [1..400]*2) \\ M. F. Hasler, Jul 08 2025

Formula

a(n) = 2*A020886(n).

A058529 Numbers whose prime factors are all congruent to +1 or -1 modulo 8.

Original entry on oeis.org

1, 7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97, 103, 113, 119, 127, 137, 151, 161, 167, 191, 193, 199, 217, 223, 233, 239, 241, 257, 263, 271, 281, 287, 289, 311, 313, 329, 337, 343, 353, 359, 367, 383, 391, 401, 409, 431, 433, 439, 449, 457, 463, 479, 487
Offset: 1

Views

Author

William Bagby (bagsbee(AT)aol.com), Dec 24 2000

Keywords

Comments

Numbers of the form x^2 - 2*y^2, where x is odd and x and y are relatively prime. - Franklin T. Adams-Watters, Jun 24 2011
Consider primitive Pythagorean triangles (a^2 + b^2 = c^2, gcd(a, b) = 1, a <= b); sequence gives values b-a, sorted with duplicates removed; terms > 1 in sequence give values of a + b, sorted. (See A046086 and A046087.)
Ordered set of (semiperimeter + radius of largest inscribed circle) of all primitive Pythagorean triangles. Semiperimeter of Pythagorean triangle + radius of largest circle inscribed in triangle = ((a+b+c)/2) + ((a+b-c)/2) = a + b.
The terms of this sequence are all of the form 6*N +- 1, since the prime divisors are, and numbers of this form are closed under multiplication. In fact, all terms are == 1, 7, 17, or 23 (mod 24). - J. T. Harrison (harrison_uk_2000(AT)yahoo.co.uk), Apr 28 2009, edited by Franklin T. Adams-Watters, Jun 24 2011
Is similar to A001132, but includes composites whose factors are in A001132. Can be generated in this manner.
Third side of primitive parallepipeds with square base; that is, integer solution of a^2 + b^2 + c^2 = d^2 with gcd(a,b,c) = 1 and b = c. - Carmine Suriano, May 03 2013
Other than -1, values of difference z-y that solve the Diophantine equation x^2 + y^2 = z^2 + 2. - Carmine Suriano, Jan 05 2015
For k > 1, k is in the sequence iff A330174(k) > 0. - Ray Chandler, Feb 26 2020

References

  • B Berggren, Pytagoreiska trianglar. Tidskrift för elementär matematik, fysik och kemi, 17:129-139, 1934.
  • Olaf Delgado-Friedrichs and Michael O’Keeffe, Edge-transitive lattice nets, Acta Cryst. (2009). A65, 360-363.

Crossrefs

Programs

  • Haskell
    a058529 n = a058529_list !! (n-1)
    a058529_list = filter (\x -> all (`elem` (takeWhile (<= x) a001132_list))
                                     $ a027748_row x) [1..]
    -- Reinhard Zumkeller, Jan 29 2013
    
  • Mathematica
    Select[Range[500], Union[Abs[Mod[Transpose[FactorInteger[#]][[1]], 8, -1]]] == {1} &] (* T. D. Noe, Feb 07 2012 *)
  • PARI
    is(n)=my(f=factor(n)[,1]%8); for(i=1,#f, if(f[i]!=1 && f[i]!=7, return(0))); 1 \\ Charles R Greathouse IV, Aug 01 2016

Formula

a(n) = |A-B|=|j^2-2*k^2|, j=(2*n-1), k,n in N, GCD(j,k)=1, the absolute difference between primitive Pythagorean triple legs (sides adjacent to the right angle). - Roger M Ellingson, Dec 09 2023

Extensions

More terms from Naohiro Nomoto, Jul 02 2001
Edited by Franklin T. Adams-Watters, Jun 24 2011
Duplicated comment removed and name rewritten by Wolfdieter Lang, Feb 17 2015

A020885 Ordered areas (divided by 6) of primitive Pythagorean triangles (with multiple entries).

Original entry on oeis.org

1, 5, 10, 14, 30, 35, 35, 55, 84, 91, 105, 140, 154, 165, 204, 220, 231, 260, 285, 286, 385, 390, 429, 455, 455, 506, 595, 650, 680, 715, 770, 819, 836, 935, 969, 1015, 1105, 1190, 1240, 1309, 1326, 1330, 1330, 1495, 1496, 1615, 1729, 1771, 1785, 1820, 1925
Offset: 1

Views

Author

Keywords

Comments

Since squares are 0 or 1 under both mod 3 and mod 4, for the Pythagorean equation A^2 + B^2 = C^2 to hold, each of 3 and 4 divides either of leg A or leg B, so that area A*B/2 is divisible by 3*4/2 = 6. - Lekraj Beedassy, Apr 30 2004
From Wolfdieter Lang, Jun 14 2015: (Start)
This sequence gives the area/6 (in some squared length unit) of primitive Pythagorean triangles with multiplicities modulo leg exchange. See the example.
This sequence also gives Fibonacci's congruous numbers divided by 24, with multiplicities and ordered nondecreasingly. See A258150.
(End)
It appears that this sequence gives the list of dimensions of irreducible unitary representations of the Lie group SO(5). - Antoine Bourget, Mar 30 2022

Examples

			a(6) = a(7) = 35 from the two Pythagorean triangles (A,B,C) = (21, 20, 29)  and (35, 12, 37) with area 210. Triangles (20, 21, 29) and (12, 35, 37) are not counted (leg exchange). - _Wolfdieter Lang_, Jun 14 2015
		

Crossrefs

Programs

  • Mathematica
    Take[Sort[(Times@@#)/12&/@({Times@@#,(Last[#]^2-First[#]^2)/2}&/@ Select[ Subsets[Range[1,41,2],{2}],GCD@@#==1&])],60] (* Harvey P. Dale, Feb 27 2012 *)

Formula

a(n) = A024406(n)/6.

Extensions

Extended and corrected by David W. Wilson

A155171 Numbers p such that if q = p+1 then (a = q^2-p^2, b = 2*p*q, c = q^2 + p^2) is a primitive Pythagorean triple with s-1 and s+1 primes, where s = a+b+c.

Original entry on oeis.org

1, 2, 7, 10, 20, 29, 44, 50, 65, 70, 76, 77, 101, 104, 107, 115, 154, 175, 197, 202, 226, 227, 247, 275, 371, 380, 412, 457, 490, 500, 574, 596, 647, 671, 682, 710, 764, 829, 926, 1052, 1085, 1102, 1127, 1186, 1204, 1205, 1225, 1256, 1280, 1324, 1325, 1331
Offset: 1

Views

Author

Keywords

Examples

			p=1,q=2,a=3,b=4,c=5,s=12-+1 primes.
		

Crossrefs

Programs

  • Mathematica
    lst={};Do[p=n;q=p+1;a=q^2-p^2;c=q^2+p^2;b=2*p*q;s=a+b+c;If[PrimeQ[s-1]&&PrimeQ[s+1],AppendTo[lst,n]],{n,8!}];lst

Extensions

Definition edited by N. J. A. Sloane, Jul 19 2022

A155173 Short leg A of primitive Pythagorean triangles such that perimeter s is average of twin prime pairs, q=p+1, A=q^2-p^2, C=q^2+p^2, B=2*p*q, s=A+B+C; s -/+ 1 are primes.

Original entry on oeis.org

3, 5, 15, 21, 41, 59, 89, 101, 131, 141, 153, 155, 203, 209, 215, 231, 309, 351, 395, 405, 453, 455, 495, 551, 743, 761, 825, 915, 981, 1001, 1149, 1193, 1295, 1343, 1365, 1421, 1529, 1659, 1853, 2105, 2171, 2205, 2255, 2373, 2409, 2411, 2451, 2513, 2561, 2649
Offset: 1

Views

Author

Keywords

Comments

With p=1, then q=2,a=3,b=4,c=5, and s=12-+1 (11, 13) both primes.

Crossrefs

Programs

  • Mathematica
    lst={};Do[p=n;q=p+1;a=q^2-p^2;c=q^2+p^2;b=2*p*q;s=a+b+c;If[PrimeQ[s-1]&&PrimeQ[s+1],AppendTo[lst,a]],{n,8!}];lst

Extensions

Name edited by Zak Seidov, Mar 21 2014

A057096 Saint-Exupéry numbers: ordered products of the three sides of Pythagorean triangles.

Original entry on oeis.org

60, 480, 780, 1620, 2040, 3840, 4200, 6240, 7500, 12180, 12960, 14760, 15540, 16320, 20580, 21060, 30720, 33600, 40260, 43740, 49920, 55080, 60000, 65520, 66780, 79860, 92820, 97440, 97500, 103680, 113400, 118080, 120120, 124320, 130560, 131820, 164640
Offset: 1

Views

Author

Henry Bottomley, Aug 01 2000

Keywords

Comments

It is an open question whether any two distinct Pythagorean Triples can have the same product of their sides.
From Amiram Eldar, Nov 22 2020: (Start)
Named after the French writer Antoine de Saint-Exupéry (1900-1944).
The problem of finding two distinct Pythagorean triples with the same product was proposed by Eckert (1984). It is equivalent of finding a nontrivial solution of the Diophantine equation x*y*(x^4-y^4) = z*w*(z^4-w^4) (Bremner and Guy, 1988). (End)

Examples

			a(1) = 3*4*5 = 60.
		

References

  • Richard K. Guy, "Triangles with Integer Sides, Medians and Area." D21 in Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, pp. 188-190, 1994.
  • Antoine de Saint-Exupéry, Problème du Pharaon, Liège : Editions Dynamo, 1957.

Crossrefs

Programs

  • Mathematica
    k=5000000; lst={}; Do[Do[If[IntegerQ[a=Sqrt[c^2-b^2]], If[a>=b, Break[]]; x=a*b*c; If[x<=k, AppendTo[lst,x]]], {b,c-1,4,-1}], {c,5,400,1}]; Union@lst (* Vladimir Joseph Stephan Orlovsky, Sep 05 2009 *)

Formula

a(n) = 60*A057097(n) = A057098(n)*A057099(n)*A057100(n).
Showing 1-10 of 31 results. Next