A188158 -id:A188158 - cp's OEIS frontend

A301384 Number of integer-sided triangles of area A188158(n).

1, 2, 2, 1, 2, 1, 2, 1, 4, 1, 1, 4, 2, 2, 2, 1, 4, 3, 1, 2, 1, 2, 4, 2, 2, 1, 1, 6, 3, 1, 5, 3, 3, 2, 2, 1, 4, 1, 2, 4, 8, 5, 1, 2, 1, 3, 1, 15, 2, 4, 2, 1, 5, 1, 6, 2, 1, 3, 4, 3, 1, 2, 2, 1, 2, 4, 5, 1, 5, 4, 1, 2, 3, 5, 1, 1, 1, 6, 2, 4, 2, 1, 2, 1, 17, 4, 1

Offset: 1

Michel Lagneau, Mar 20 2018

nonn

Nonzero terms of A051584.

			a(9) = 4 because A188158(9) = 60 corresponding to 4 triangles (a, b, c) = (6, 25, 29), (8, 15, 17), (10, 13, 13) and (13, 13, 24) of area 60.

Giovanni Resta, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Triangle

Cf. A188158, A024153 (distinct side lengths), A051516, A051584, A051585.

A301384 := proc(A::integer)
    local Asqr, s,a,b,c,sol ;
    sol := 0 ;
    Asqr := A^2 ;
    for s in numtheory[divisors](Asqr) do
        if s^2> A then
        for a from 1 to s-1 do
            if modp(Asqr,s-a) = 0 then
                for b from a to s-1 do
                    c := 2*s-a-b ;
                    if s*(s-a)*(s-b)*(s-c) = Asqr and c >= b then
                        sol := sol+1 ;
                    end if;
                end do:
            end if;
        end do:
        end if;
    end do:
    sol ;
end proc:
for n from 3 to 600 do
    a301384 := A301384(n) ;
    if a301384 > 0 then
        printf("%d,\n",a301384) ;
    end if;
end do: # R. J. Mathar, May 02 2018

nn=1000;lst={};lst2={};Do[s=(a+b+c)/2;If[IntegerQ[s],area2=s(s-a)(s-b)(s-c);If[0

A008585 a(n) = 3*n.

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141, 144, 147, 150, 153, 156, 159, 162, 165, 168, 171, 174, 177

Offset: 0

N. J. A. Sloane

If n != 1 and n^2+2 is prime then n is a member of this sequence. - Cino Hilliard, Mar 19 2007
Multiples of 3. Positive members of this sequence are the third transversal numbers (or 3-transversal numbers): Numbers of the 3rd column of positive numbers in the square array of nonnegative and polygonal numbers A139600. Also, numbers of the 3rd column in the square array A057145. - Omar E. Pol, May 02 2008
Numbers n for which polynomial 27*x^6-2^n is factorizable. - Artur Jasinski, Nov 01 2008
1/7 in base-2 notation = 0.001001001... = 1/2^3 + 1/2^6 + 1/2^9 + ... - Gary W. Adamson, Jan 24 2009
A165330(a(n)) = 153 for n > 0; subsequence of A031179. - Reinhard Zumkeller, Sep 17 2009
A011655(a(n)) = 0. - Reinhard Zumkeller, Nov 30 2009
A215879(a(n)) = 0. - Reinhard Zumkeller, Dec 28 2012
Moser conjectured, and Newman proved, that the terms of this sequence are more likely to have an even number of 1s in binary than an odd number. The excess is an undulating multiple of n^(log 3/log 4). See also Coquet, who refines this result. - Charles R Greathouse IV, Jul 17 2013
Integer areas of medial triangles of integer-sided triangles.
Also integer subset of A188158(n)/4.
A medial triangle MNO is formed by joining the midpoints of the sides of a triangle ABC. The area of a medial triangle is A/4 where A is the area of the initial triangle ABC. - Michel Lagneau, Oct 28 2013
From Derek Orr, Nov 22 2014: (Start)
Let b(0) = 0, and b(n) = the number of distinct terms in the set of pairwise sums {b(0), ... b(n-1)} + {b(0), ... b(n-1)}. Then b(n+1) = a(n), for n > 0.
Example: b(1) = the number of distinct sums of {0} + {0}. The only possible sum is {0} so b(1) = 1. b(2) = the number of distinct sums of {0,1} + {0,1}. The possible sums are {0,1,2} so b(2) = 3. b(3) = the number of distinct sums of {0,1,3} + {0,1,3}. The possible sums are {0, 1, 2, 3, 4, 6} so b(3) = 6. This continues and one can see that b(n+1) = a(n). (End)
Number of partitions of 6n into exactly 2 parts. - Colin Barker, Mar 23 2015
Partial sums are in A045943. - Guenther Schrack, May 18 2017
Number of edges in a maximal planar graph with n+2 vertices, n > 0 (see A008486 comments). - Jonathan Sondow, Mar 03 2018
Also numbers such that when the leftmost digit is moved to the unit's place the result is divisible by 3. - Stefano Spezia, Jul 08 2025

			G.f.: 3*x + 6*x^2 + 9*x^3 + 12*x^4 + 15*x^5 + 18*x^6 + 21*x^7 + ...

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
J. Coquet, A summation formula related to the binary digits, Inventiones Mathematicae 73 (1983), pp. 107-115.
Charles Cratty, Samuel Erickson, Frehiwet Negass, and Lara Pudwell, Pattern Avoidance in Double Lists, preprint, 2015.
A. S. Fraenkel, New games related to old and new sequences, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 4, Paper G6, 2004.
John Graham-Cumming, The hollow triangular numbers are divisible by three (2013)
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 315
Tanya Khovanova, Recursive Sequences
D. J. Newman, On the number of binary digits in a multiple of three, Proc. Amer. Math. Soc. 21 (1969) 719-721.
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014-2015.
Wikipedia, Maximal planar graphs
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Row / column 3 of A004247 and of A325820.
Cf. A016957, A057145, A139600, A139606, A001651 (complement), A032031 (partial products), A190944 (binary), A061819 (base 4).
Cf. A031179, A008486, A008591, A017197, A017233, A045943.

a008585 = (* 3)
a008585_list = iterate (+ 3) 0  -- Reinhard Zumkeller, Feb 19 2013

[3*n: n in [0..60]]; // Vincenzo Librandi, Jul 23 2011

Range[0, 500, 3] (* Vladimir Joseph Stephan Orlovsky, May 26 2011 *)

makelist(3*n,n,0,30); /* Martin Ettl, Nov 12 2012 */

a(n)=3*n \\ Charles R Greathouse IV, Jun 28 2013

G.f.: 3*x/(1-x)^2. - R. J. Mathar, Oct 23 2008
a(n) = A008486(n), n > 0. - R. J. Mathar, Oct 28 2008
G.f.: A(x) - 1, where A(x) is the g.f. of A008486. - Gennady Eremin, Feb 20 2021
a(n) = Sum_{k=0..inf} A030308(n,k)*A007283(k). - Philippe Deléham, Oct 17 2011
E.g.f.: 3*x*exp(x). - Ilya Gutkovskiy, May 18 2016
From Guenther Schrack, May 18 2017: (Start)
a(3*k) = a(a(k)) = A008591(n).
a(3*k+1) = a(a(k) + 1) = a(A016777(n)) = A017197(n).
a(3*k+2) = a(a(k) + 2) = a(A016789(n)) = A017233(n). (End)

Partially edited by Joerg Arndt, Mar 11 2010

A055595 Area of triangles with integer sides and positive integer area, ordered by longest side, then second longest side and finally shortest side.

Original entry on oeis.org

6, 12, 12, 24, 48, 30, 60, 54, 24, 84, 48, 36, 60, 120, 108, 66, 42, 96, 84, 126, 60, 108, 192, 90, 150, 84, 168, 120, 36, 204, 240, 210, 210, 60, 120, 216, 132, 300, 96, 336, 72, 192, 144, 240, 480, 294, 84, 252, 360, 432, 114, 156, 180, 210, 420, 120, 210, 420

Offset: 1

Henry Bottomley, May 26 2000

nonn

This is the ordering of triangles used for A316841.

Ray Chandler, Table of n, a(n) for n = 1..10000
A. Dendane, Heron's Formula For Area of a Triangle-Geometry Calculator

The sides are given by A055592, A055593, A055594.
Cf. A046131, A070786, A316841, A316853.
Range of values: A188158.

max = 42; triangles = Reap[Do[s = (a+b+c)/2; area = Sqrt[s*(s-a)*(s-b)*(s-c)]; If[IntegerQ[area] && area > 0, Sow[{a, b, c, area}]], {a, 1, max}, {b, a, max}, {c, b, max}]][[2, 1]]; A055595 = Sort[triangles, #1[[3]]*max^2 + #1[[2]]*max + #1[[1]] < #2[[3]]* max^2 + #2[[2]]*max + #2[[1]] &][[All, 4]](* Jean-François Alcover, Jun 12 2012 *)

a(n) = sqrt(s(n)*(s(n)-A055592(n))*(s(n)-A055593(n))*(s(n)-A055594(n))) where s(n) = (A055592(n)+A055593(n)+A055594(n))/2 i.e. half the perimeter of the triangle

A208984 Areas A of the triangles such that A, the sides, the circumradius and the inradius are integers.

Original entry on oeis.org

24, 96, 120, 168, 216, 240, 336, 384, 432, 480, 600, 624, 672, 720, 768, 840, 864, 960, 1080, 1176, 1320, 1344, 1512, 1536, 1560, 1680, 1728, 1848, 1920, 1944, 2016, 2040, 2160, 2184, 2304, 2376, 2400, 2496, 2520, 2688, 2856, 2880, 2904, 3000, 3024, 3072, 3240

Offset: 1

Michel Lagneau, Mar 04 2012

nonn

a(n) is divisible by 24, and the positive squares A000290(n) are included in the sequence a(n)/24 = {1, 4, 5, 7, 9, 10, 14, 16, 18, 20, 25, 26, 28, 30, 32, 35, 36, 40, 45, 49, 55, 56, 63, 64, 65, ...}.

The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2. The inradius r is given by r = A/s and the circumradius is given by R = abc/4A.

			a(1) = 24 because, for (a,b,c) = (6, 8, 10) => s = (6 + 8 + 10)/2 = 12, and
A = sqrt(12(12-6)(12-8)(12-10)) = sqrt(576) = 24;
R = abc/4A = 480/4*24 = 5;
r = A/p = 24/12 = 2.

Mohammad K. Azarian, Solution of problem 125: Circumradius and Inradius, Math Horizons, Vol. 16, No. 2 (Nov. 2008), p. 32.
Eric W. Weisstein, MathWorld: Circumradius
Eric W. Weisstein, MathWorld: Inradius

Cf. A120062, A188158.

with(numtheory):T:=array(1..1000):k:=0:nn:=250: for a from 1
to nn do: for b from a to nn  do: for c from b to nn  do: p:=(a+b+c)/2 : x:=p*(p-a)*(p-b)*(p-c): if x>0 then s:=sqrt(x) :if s=floor(s) and irem(a*b*c,4*s) = 0 and irem(s,p)=0 then k:=k+1:T[k]:= s: else fi:fi:od:od:od: L := [seq(T[i],i=1..k)]:L1:=convert(T,set):A:=sort(L1, `<`): print(A):

nn = 1000; lst = {}; Do[s = (a + b + c)/2; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); If[0 < area2 <= nn^2 && IntegerQ[Sqrt[area2]] && IntegerQ[a*b*c/(4* Sqrt[area2])] && IntegerQ[Sqrt[area2]/s], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst]

A016910 a(n) = (6*n)^2.

Original entry on oeis.org

0, 36, 144, 324, 576, 900, 1296, 1764, 2304, 2916, 3600, 4356, 5184, 6084, 7056, 8100, 9216, 10404, 11664, 12996, 14400, 15876, 17424, 19044, 20736, 22500, 24336, 26244, 28224, 30276, 32400, 34596, 36864, 39204, 41616, 44100, 46656, 49284, 51984, 54756, 57600, 60516, 63504, 66564, 69696, 72900

Offset: 0

N. J. A. Sloane

Areas A of two classes of triangles with integer sides (a,b,c) where a = 9k, b=10k and c = 17k, or a = 3k, b = 25k and c = 26k for k=0,1,2,... These areas are given by Heron's formula A = sqrt(s(s-a)(s-b)(s-c)) = (6k)^2, with the semiperimeter s = (a+b+c)/2. This sequence is a subsequence of A188158. - Michel Lagneau, Oct 11 2013

Sequence found by reading the line from 0, in the direction 0, 36, ..., in the square spiral whose vertices are the generalized 20-gonal numbers A218864. - Omar E. Pol, May 13 2018.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A089491, A188158, A243941.

Cf. similar sequences of the type k*n^2: A000290 (k=1), A001105 (k=2), A033428 (k=3), A016742 (k=4), A033429 (k=5), A033581 (k=6), A033582 (k=7), A139098 (k=8), A016766 (k=9), A033583 (k=10), A033584 (k=11), A135453 (k=12), A152742 (k=13), A144555 (k=14), A064761 (k=15), A016802 (k=16), A244630 (k=17), A195321 (k=18), A244631 (k=19), A195322 (k=20), A064762 (k=21), A195323 (k=22), A244632 (k=23), A195824 (k=24), A016850 (k=25), A244633 (k=26), A244634 (k=27), A064763 (k=28), A244635 (k=29), A244636 (k=30).

[(6*n)^2: n in [0..40]]; // Vincenzo Librandi, May 03 2011

(6*Range[0,40])^2 (* or *) LinearRecurrence[{3,-3,1},{0,36,144},40] (* Harvey P. Dale, Dec 25 2017 *)
Table[36 n^2, {n, 0, 45}] (* Omar E. Pol, Jun 07 2018 *)

a(n)=36*n^2 \\ Charles R Greathouse IV, Jun 10 2016

From Ilya Gutkovskiy, Jun 09 2016: (Start)
O.g.f.: 36*x*(1 + x)/(1 - x)^3.
E.g.f.: 36*x*(1 + x)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
Sum_{n>=1} 1/a(n) = Pi^2/216 = A086726. (End)
Product_{n>=1} a(n)/A136017(n) = Pi/3. - Fred Daniel Kline, Jun 09 2016
a(n) = t(9*n) - 9*t(n), where t(i) = i*(i+k)/2 for any k. Special case (k=1): a(n) = A000217(9*n) - 9*A000217(n). - Bruno Berselli, Aug 31 2017
a(n) = 36*A000290(n) = 18*A001105(n) = 12*A033428 = 9*A016742(n) = 6*A033581(n) = 4*A016766(n) = 3*A135453(n) = 2*A195321(n). - Omar E. Pol, Jun 07 2018
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/432. - Amiram Eldar, Jun 27 2020
From Amiram Eldar, Jan 25 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = sinh(Pi/6)/(Pi/6).
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/6)/(Pi/6) = 3/Pi (A089491). (End)

A024153 Number of integer-sided triangles with sides a,b,c, a

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 2, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 3, 0, 1, 0, 0, 0, 3, 0, 0, 0, 2, 0, 1, 0, 1, 0, 2, 0, 3, 0, 0, 0, 1, 0, 1, 0, 3, 0, 0, 0, 8, 0, 0, 0, 1, 0, 4, 0, 0, 0, 0, 0, 4, 0, 3, 0, 1

Offset: 1

Clark Kimberling

nonn

No such triangles with odd perimeter (see A051516).

Records occur at: 1, 12, 36, 54, 84, 108, 192, 216, 294, 324, 378, 420, 432, 540, 588, 756, 972, 1176, 1452, 1764, 1944, 2028, 2352, 2904, 2916, 3024, 3072, 3402, 3468, 3780, 3888, 4116, 5292, 6348, 6804, 8748, 10164, ... - Antti Karttunen, Sep 25 2018

Antti Karttunen, Table of n, a(n) for n = 1..12021 (terms 1..1000 from Seiichi Manyama)
Eric Weisstein's World of Mathematics, Heronian Triangle.

Cf. A051516, A070139, A070203, A188158, A301384.

A024153(n) = if(n%2,0,my(k=0, t, p=n/2); for(a=1,n,for(b=1+max(a,(p-a)),n-(a+1),my(c=n-(a+b)); if((c<=b),break); if(((t = (p*(p-a)*(p-b)*(p-c))) > 0)&&issquare(t),k++))); (k)); \\ Antti Karttunen, Sep 25 2018

a(100) corrected by Seiichi Manyama, Sep 13 2018

A120572 Smallest area of any triangle with integer sides a <= b <= c and inradius n.

Original entry on oeis.org

6, 24, 48, 84, 150, 192, 294, 336, 432, 540, 726, 756, 1014, 1134, 1170, 1344, 1734, 1710, 2166, 2100, 2310, 2640, 3174, 3000, 3750, 4056, 3888, 4116, 5046, 4680, 5766, 5376, 5808, 6936, 6510, 6804, 8214, 8664, 8112, 8400, 10086, 9240, 11094, 10164

Offset: 1

David W. Wilson, Jun 17 2006

nonn

a(n) == 0 (mod 6).
Empirically, 3*sqrt(3) < a(n)/n^2 <= 6. The lower bound is provably tight, the upper bound seems to be achieved infinitely often, e.g., for prime n >= 5.
From Michel Lagneau, Mar 02 2012: (Start)
Subset of A188158.
The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s(s-a)(s-b)(s-c)), where s = (a+b+c)/2. The radius of the incircle or inscribed circle (also known as the inradius, r) is given by r = A/s.
From n = 17, it is possible to find couples of triangles with the property: r1 > r2 and A1 < A2 where A1, A2 are the consecutive areas corresponding to the inradius r1, r2. For example, a(17) = 1734 with (a,b,c) = (51, 68, 85) and a(18) = 1710 with (a,b,c) = (57, 65, 68).
Another interesting property of this sequence is that a(n) is divisible by 6 and, except n = 3, a(n)/6 = n^2 if n prime, hence the proposition:
Among the set of triangles whose area and sides are integers and whose inradius r is a prime number other than 3, the smallest area A is given by A = 6r^2.
Example: if r = 5, the areas of the triangles are {150, 210, 270, 330, 390, 510, ...} and the smallest area of them is A = 6*5^2 = 150 because 5 is prime.
Proof: Let r be a number such that the sides of a triangle are a = 3r, b = 4r, c = 5r. Then s = (a+b+c)/2 = 6r and A = sqrt(s(s-a)(s-b)(s-c)) = sqrt(36r^4) = 6r^2 is a possible area. Is 6r^2 the smallest area? The response is no in the general case for the composite numbers.
Writing a = (m+n)/2, b = (n+l)/2, c = (l+m)/2, and using rs = A and Heron's formula for A, we find lmn = 4r^2(l+m+n). Since m, n and l have to be of the same parity for a, b and c to be integral, they must therefore be even. Setting l = 2u, m = 2v, and n = 2w, we have a = v+w, b = w+u, c = u+v, and uvw = r^2(u+v+w). Then r^2 = uvw/(u+v+w).
First case: If r = p is prime, we prove that A = 6p^2 is the smallest area of all the triangles whose inradius is p. Suppose A' < A with inradius(A') = p. The area A is the corresponding value of the triangle (u,v,w) = (1*p, 2*p, 3*p) because 6p^3/6p = p^2. However, inradius(A') = p => u'v'w'/(u'+v'+w') = p^2 => (u',v',w') = (u,v,w) and A is the smallest area.
Second case: If r = q is composite, the triangle (u,v,w) = (1*q, 2*q, 3*q) gives an area A with inradius(A) = q, but it is possible to find A' < A with inradius(A') = q; for example, if q = 10, the triangle (u,v,w) = (30, 20, 10) whose area is A = 600 gives sqrt{(30*20*10)/(30+20+10)} = sqrt(6000/60) = 10 and the triangle(u',v',w') = (24,15,15) whose area is A' = 540 gives sqrt{(24*15*15)/(24+15+15)} = sqrt(5400/54) = 10.
(End)

			a(4) = 84 because, for (a,b,c) = (13,14,15) => A = sqrt(21(21-13)(21-14)(21-15)) = 84 and r = 84/21 = 4.

David W. Wilson, Table of n, a(n) for n = 1..10000
Mohammad K. Azarian, Solution of problem 125: Circumradius and Inradius, Math Horizons, Vol. 16, No. 2 (Nov. 2008), p. 32.

See A120062 for sequences related to integer-sided triangles with integer inradius n.

T:=array(1..500):nn:=70: for n from 1 to 16 do:k:=0:ii:=0:for a from 1
to nn do: for b from a to nn  do: for c from b to nn  do: p:=(a+b+c)/2 : x:=p*(p-a)*(p-b)*(p-c): if x>0 then s:=sqrt(x) :if s=floor(s) and s/p = n then k:=k+1:T[k]:=s: else fi:fi:od:od:od: L := [seq(T[i], i=1..k)]:A:=sort(L, `<`): w:=A[1]: printf ( "%d %d \n", n, w):od: # Michel Lagneau, Mar 02 2012

A224301 Sorted areas of primitive integer Heronian triangles.

Original entry on oeis.org

6, 12, 12, 24, 30, 36, 36, 42, 60, 60, 60, 60, 66, 72, 84, 84, 84, 84, 90, 90, 114, 120, 120, 120, 126, 126, 126, 132, 156, 156, 168, 168, 168, 180, 180, 198, 204, 210, 210, 210, 210, 210, 210, 216, 234, 240, 252, 252, 252, 264, 264, 270, 288, 300, 300, 306

Offset: 1

Mihir Mathur, Apr 03 2013

nonn

The sequence gives the sorted areas of primitive triangles which have integer side lengths and integer areas.

			The smallest Heronian triangle is (3,4,5) as perimeter and area are integers. The first term of the sequence is thus the area of this triangle, which is 6.

Giovanni Resta, Table of n, a(n) for n = 1..10000
Michael Somos, Heronian Triangle Table

Cf. A120131, A120132, A120133, A072294, A188158.

AMax=400;
Do[
  c=p/b;
  a1=Sqrt[b^2+c^2+2Sqrt[b^2c^2-4A^2]];
  a2=Sqrt[b^2+c^2-2Sqrt[b^2c^2-4A^2]];
  If[IntegerQ[a2]&&GCD[a2,b,c]==1&&a1>a2>=b,A//Sow(*{A,a2,b,c}//Sow*)];
  If[IntegerQ[a1]&&GCD[a1,b,c]==1,A//Sow(*{A,a1,b,c}//Sow*)];
  ,{A,6,AMax,6}
  ,{p,4A^2//Divisors//Select[#,EvenQ[#]&&#>=2A&]&//#/2+2A^2/#&//Select[#,IntegerQ]&}
  ,{b,p//Divisors//Select[#,#^2>=p&]&}
]//Reap//Last//Last
{a1,a2,c}=.;
(* Albert Lau, May 20 2016 *)

Definition corrected by Giovanni Resta, Apr 03 2013

More terms from Giovanni Resta, Apr 03 2013

A185210 Areas A of the triangles such that A, the sides, the inradius and the radius of the three excircles are integers.

Original entry on oeis.org

6, 24, 30, 42, 48, 54, 60, 84, 96, 108, 120, 144, 150, 156, 168, 180, 192, 210, 216, 240, 270, 294, 330, 336, 378, 384, 390, 420, 432, 462, 480, 486, 504, 510, 528, 540, 546, 576, 594, 600, 624, 630, 672, 714, 720, 726, 750, 756, 768, 810, 840, 864, 924, 930

Offset: 1

Michel Lagneau, Mar 21 2012

nonn

Theorem 1: Consider a triangle whose area A, sides (a,b,c), inradius r and the radius of whose three excircles r1, r2, r3 are integers. Then the sum a^2 + b^2 + c^2 + r^2 + r1^2 + r2^2 + r3^2 is a perfect square equal to 16R^2, where R is the circumradius.
Proof: (r1 + r2 + r3 - r)^2 = r1^2 + r2^2 + r3^2 + r^2 + a^2 + b^2 + c^2 because: r1*r2 + r2*r3 + r3*r1 - r*r1 - r*r2 - r*r3 = (a^2 + b^2 + c^2)/2 (formula from Feuerbach - see the link). But r1 + r2 + r3 - r = 4*R (see the reference: Johnson 1929, pp. 190-191), hence the result. Remark: R is not necessarily an integer; for example, at a(1) = 6 with (a,b,c) = (3, 4, 5) we obtain r = 1, r1 = 2, r2 = 3, r3 = 6 and R = 5/2. Then 3^2 + 4^2 + 5^2 + 1^2 + 2^2 + 3^2 + 6^2 = 16*(5/2)^2 = 10^2. Nevertheless, if R is an integer, then r, r1, r2 and r3 are necessarily integers (see the following theorem). The subset of a(n) with R integer is A208984 = {24, 96, 120, 168, 216, 240, 336, 384, 432, 480, 600, ...}
Theorem 2: Consider a triangle whose area A, sides (a,b,c) and circumradius R are integers. Then the inradius r and the radius of the three excircles r1, r2, r3 are also integers.
Proof: Let s be the semiperimeter, let s*A = r1*r2*r3 be an integer, and let r*r1*r2*r3 = A^2 also be an integer => r is an integer. r1 = A/(s-a), r2 = A/(s-b), r3 = A/(s-c) => r1*r2 = s*(s-c), r1*r3=s*(s-b), r2*r3 = s*(s-a) are integers. Because r1*r2*r3 is an integer => r1, r2, r3 are integers.

			24 is in the sequence because for (a, b, c) = (6, 8, 10) => s =(6+8+10)/2 = 12; A = sqrt(12(12-6)(12-8)(12-10)) = sqrt(576) = 24; r = A/s = 2; r1 = 2*24(-6+8+10) = 4; r2 = 2*24(6-8+10) = 6; r3 = 2*24(6+8-10) = 12.

Mohammad K. Azarian, Circumradius and Inradius, Problem S125, Math Horizons, Vol. 15, Issue 4, April 2008, p. 32. Solution published in Vol. 16, Issue 2, November 2008, p. 32.
Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, 1929.

Michel Lagneau, Table of n, a(n) for n = 1..54
Eric W. Weisstein's World of Mathematics, Excircles
Eric W. Weisstein's World of Mathematics, Exradius
Eric W. Weisstein's World of Mathematics, Inradius

Cf. A120572, A188158, A208984, A210207.

nn = 1000; lst = {}; Do[s = (a + b + c)/2; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); If[0 < area2 <= nn^2 && IntegerQ[Sqrt[area2]] && IntegerQ[Sqrt[area2]/s] && IntegerQ[2*Sqrt[area2]/(-a+b+c)] &&  IntegerQ[2*Sqrt[area2]/(a-b+c)] && IntegerQ[2*Sqrt[area2]/(a+b-c)], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst]

A = sqrt(s*(p-a)*(s-b)*(s-c)) with s = (a+b+c)/2 (Heron's formula);
the inradius is r=A/s;
the exradii of the excircles are r1 = 2*A/(-a+b+c), x2 = 2*A*b/(a-b+c), and x3 = 2*A*c/(a+b-c).

A210643 Areas A of the triangles such that A, the sides and the three altitudes are integers.

Original entry on oeis.org

150, 300, 600, 1050, 1200, 1350, 2400, 2700, 3750, 4200, 4800, 5070, 5400, 7350, 7500, 9450, 9600, 10140, 10800, 12150, 14700, 15000, 16800, 17340, 18150, 19200, 20280, 21600, 24300, 25350, 26250, 29400, 30000, 33750, 34680, 36300, 37800

Offset: 1

Michel Lagneau, Mar 26 2012

nonn

Properties of this sequence :
There exists three class of numbers included into a(n) :
(i) A subset such that {150, 600, 1350, 2400, 3750, 5070,…} where the sides a h1 = b, h2 = a, h3 = a*b/c.
(ii) A subset such that a(n) = 300*n^2 = {300, 1200, 2700, 4800, …} where the triangles (a,b,c) are isosceles with a = b < c, and it is easy to check that a = b = 25*n,  c=30*n, h1 = h2 = 24*n and h3 = sqrt(b^2 - c^2/4).
(iii) A subset such that {1050, 4200, 9450,…} without the precedent properties.

			Primitive solutions follow:
Area,  ( a,   b,   c),  (h1,  h2,  h3), Case
  150,  (15,  20,  25),  (20,  15,  12), Right,
  300,  (25,  25,  30),  (24,  24,  20), Isosceles,
  300,  (25,  25,  40),  (24,  24,  15), Isosceles,
1050,  (35,  75, 100),  (60,  28,  21), Other,
5070,  (65, 156, 169), (156,  65,  60), Right,
10140, (130, 169, 169), (156, 120, 120), Isosceles,
10140, (169, 169, 312), (120, 120,  65), Isosceles,
17340, (136, 255, 289), (255, 136, 120), Right,
34680, (272, 289, 289), (255, 240, 240), Isosceles,
34680, (289, 289, 510), (240, 240, 136), Isosceles,
52500, (175, 600, 625), (600, 175, 168), Right,
82500, (275, 625, 750), (600, 264, 220), Other.

Ray Chandler, Table of n, a(n) for n = 1..64
Eric Weisstein, Isosceles Triangle
Eric Weisstein, Altitude

Cf. A188158.

with(numtheory):T:=array(1..1000):k:=0:nn:=500:for a from 1 to nn do: for b from a to nn do: for c from b to nn do:p:=(a+b+c)/2:s:=p*(p-a)*(p-b)*(p-c):if s>0 then s1:=sqrt(s): h1:=2*s1/a: h2:=2*s1/b:h3:=2*s1/c:if s1=floor(s1) and h1=floor(h1) and h2=floor(h2) and h3=floor(h3) then k:=k+1:T[k]:=s1:else fi:fi:od:od:od: L := [seq(T[i],i=1..k)]:L1:=convert(T,set):A:=sort(L1, `<`): print(A):

nn = 900; lst = {}; Do[s = (a + b + c)/2; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); If[0 < area2 <= nn^2 && IntegerQ[Sqrt[area2]]&&IntegerQ[(2*Sqrt[area2])/a] &&IntegerQ[(2*Sqrt[area2])/b] &&IntegerQ[(2*Sqrt[area2])/c], AppendTo[lst, Sqrt[area2]]]], {a, nn}, {b, a}, {c, b}]; Union[lst]

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