A103606 -id:A103606 - cp's OEIS frontend

A198453 Consider triples a<=b

2, 2, 3, 3, 5, 6, 4, 9, 10, 5, 6, 8, 5, 14, 15, 6, 9, 11, 6, 20, 21, 7, 27, 28, 8, 10, 13, 8, 35, 36, 9, 13, 16, 9, 21, 23, 9, 44, 45, 10, 26, 28, 10, 54, 55, 11, 14, 18, 11, 20, 23, 11, 65, 66, 12, 17, 21, 12, 24, 27

Offset: 1

Charlie Marion, Oct 25 2011

The definition can be generalized to define Pythagorean k-triples a<=b

If a, b and c form a Pythagorean k-triple, then n*a, n*b and n*c form a Pythagorean n*k-triple.

A triangle is defined to be a Pythagorean k-triangle if its sides form a Pythagorean k-triple.

If a, b and c are the sides of a Pythagorean k-triangle ABC with a<=b0 and acute when k<0. When k=0, the triangles are Pythagorean, as in the Beiler reference and Ron Knott's link.

For all k, the area of a Pythagorean k-triangle ABC with a<=b

Define a Pythagorean k-triple to be primitive if and only if there are no integers r>1, s>0 such that = for some Pythagorean s-triple . Thus, every Pythagorean 1-triple is primitive. For every k>1, the set of Pythagorean k-triples contains some non-primitive triples.

In particular, for d a proper divisor of k, it includes (k/d)*(a,b,c), where (a,b,c) is a Pythagorean d-triple. - Franklin T. Adams-Watters, Dec 01 2011

			2*3 + 2*3 = 3*4
3*4 + 5*6 = 6*7
4*5 + 9*10 = 10*11
5*6 + 6*7 = 8*9
5*6 + 14*15 = 15*16
6*7 + 9*10 = 11*12

A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.

Ron Knott, Pythagorean Triples and Online Calculators

Cf. A103606, A198454-A198469.

A103605 Pythagorean triples in increasing order of perimeter (a+b+c). If two successive perimeters are equals, then in order of decreasing areas; each triple [a(i), a(j), a(k)] (with k multiple of 3, j=k-1, i=k-2) in increasing order.

Original entry on oeis.org

3, 4, 5, 6, 8, 10, 5, 12, 13, 9, 12, 15, 8, 15, 17, 12, 16, 20, 7, 24, 25, 15, 20, 25, 10, 24, 26, 20, 21, 29, 18, 24, 30, 16, 30, 34, 21, 28, 35, 12, 35, 37, 15, 36, 39, 9, 40, 41, 24, 32, 40, 27, 36, 45, 14, 48, 50, 20, 48, 52, 24, 45, 51, 30, 40, 50, 28, 45, 53

Offset: 1

Alexandre Wajnberg, Mar 24 2005

The corresponding perimeters A009096. - Wolfdieter Lang, Oct 06 2014

Andreas Boe, Table of n, a(n) for n = 1..30000
Ron Knott, The Pythagorean triples.

For primitive triples see A103606.

Cf. A009096, A376608.

In the name second 'increasing' -> 'decreasing' (observed by A. Boe). - Wolfdieter Lang, Oct 06 2014

A335893 Primitive triples for integer-sided triangles whose angles A < B < C are in arithmetic progression.

Original entry on oeis.org

3, 7, 8, 5, 7, 8, 7, 13, 15, 8, 13, 15, 5, 19, 21, 16, 19, 21, 11, 31, 35, 24, 31, 35, 7, 37, 40, 33, 37, 40, 13, 43, 48, 35, 43, 48, 16, 49, 55, 39, 49, 55, 9, 61, 65, 56, 61, 65, 32, 67, 77, 45, 67, 77, 17, 73, 80, 63, 73, 80, 40, 79, 91, 51, 79, 91, 11, 91, 96

Offset: 1

Bernard Schott, Jun 29 2020

The triples are displayed in nondecreasing order of middle side, and if middle sides coincide then by increasing order of the largest side, hence, each triple (a, b, c) is in increasing order.
These three properties below are equivalent:
-> integer-sided triangles whose angles A < B < C are in arithmetic progression,
-> integer-sided triangles such that B = (A+C)/2 with A < C,
-> integer-sided triangles such that A < B < C with B = Pi/3.
When A < B < C are in arithmetic progression with B = A + phi and C = B + phi, then 0 < phi < Pi/3.
The corresponding metric relation between sides is b^2 = a^2 - a*c + c^2.
There exists such primitive triangle iff b^2 is an odd square term of A024612. Hence, the first few middle sides b are 7, 13, 19, 31, 37, 43, 49, 61, 67, ... and b is a term of A004611 \ {1}. Indeed, b cannot be even if the triple is primitive.
As B = Pi/3 and C runs from Pi/3 to 2*Pi/3, sin(C) gets a maximum when C = Pi/2 with sin(C) = 1, hence, from law of sines (see link): b/sin(B) = c/sin(C), and c < b/sin(Pi/3) = b * 2/sqrt(3) < 6*b/5. This bound is used in the PARI and Maple programs below.
When triple (a, b, c) is solution, then triple (c-a, b, c) is another solution. Hence, for each b odd solution, there exist 2 triples with same middle side b and same largest side c.
The common tangent to the nine-point circle and the incircle of a triangle ABC is parallel to the Euler line iff angles A < B < C are in arithmetic progression (see Crux Mathematicorum for Indian team selection). - Bernard Schott, Apr 14 2022
These triples are called (primitive) Eisenstein triples (Wikipedia). - Bernard Schott, Sep 21 2022

			(3, 7, 8) is a triple for this sequence because from law of cosines (see link), cos(A) = (7^2 + 8^2 - 3^2)/(2*7*8) = 13/14, cos(B) = (8^2 + 3^2 - 7^2)/(2*8*3) = 1/2 and cos(C) = (3^2 + 7^2 - 8^2)/(2*3*7) = -1/7; then, (A+C)/2 = ( arccos(13/14) + arccos(-1/7) )/2 = Pi/3 = B.
Also, arccos(13/14) ~ 21.787 degrees, arccos(1/2) = 60 degrees, arccos(-1/7) ~ 98.213 degrees, so B-A = C-B ~ 38.213 degrees, hence (A, B, C) are in arithmetic progression.
5^2 - 5*8 + 8^2 = 7^2, hence (5, 7, 8) is another triple for triangle whose angles A < B < C are in arithmetic progression.

V. Lespinard & R. Pernet, Trigonométrie, Classe de Mathématiques élémentaires, programme 1962, problème B-298 p. 124, André Desvigne.

Bill Sands, Indian team selection test 2007, question 5, Crux Mathematicorum, Vol. 36, No. 5 (2010), p. 278.
Eric Weisstein's World of Mathematics, Law of Cosines.
Eric Weisstein's World of Mathematics, Law of Sines.
Wikipedia, Eisenstein triple.
Index to sequences related to Olympiads.

Cf. A004611, A024612.
Cf. A335894 (smallest side), A335895 (middle side), A335896 (largest side), A335897 (perimeter).
Cf. A103606 (primitive Pythagorean triples), A335034 (primitive triples for triangles with two perpendicular medians).

for b from 3 to 250 by 2 do
for c from b+1 to 6*b/5 do
a := (c - sqrt(4*b^2-3*c^2))/2;
if gcd(a,b)=1 and issqr(4*b^2-3*c^2) then print(a,b,c,c-a,b,c); end if;
end do;
end do;

lista(nn) = {forstep(b=1, nn, 2, for(c=b+1, 6*b\5, if (issquare(d=4*b^2 - 3*c^2), my(a = (c - sqrtint(d))/2); if ((denominator(a)==1) && (gcd(a, b) == 1), print(a, ", ", b, ", ", c, ", "); print(c-a, ", ", b, ", ", c, ", ");););););} \\ Michel Marcus, Jul 15 2020

A335034 Primitive triples, in nondecreasing order of perimeter, for integer-sided triangles with two perpendicular medians; each triple is in increasing order, and if perimeters coincide then increasing order of the smallest side.

Original entry on oeis.org

13, 19, 22, 17, 22, 31, 25, 38, 41, 37, 58, 59, 41, 58, 71, 53, 62, 101, 61, 82, 109, 65, 79, 122, 85, 118, 149, 89, 121, 158, 101, 139, 178, 109, 122, 211, 113, 142, 209, 145, 178, 271, 145, 191, 262, 149, 229, 242, 157, 179, 302, 173, 269, 278, 181, 218, 341

Offset: 1

Bernard Schott, May 20 2020

The study of these integer triangles that have two perpendicular medians was proposed in the problem of Concours Général in 2007 in France (see links).
If medians drawn from A and B are perpendicular in centroid G, then a^2 + b^2 = 5 * c^2, hence c is always the smallest side.
Some theoretical results and geometrical properties:
-> 1/2 < a/b < 2 and 1 < a/c < 2 (also 1 < b/c < 2).
-> a, b, c are pairwise coprimes.
-> a et b have different parities, so c is always odd.
-> a and b are not divisible nor by 3 nor by 4 neither by 5.
-> The odd prime factors of the even term a' (a' = a or b) are all of the form 10*k +- 1 (see formula for a').
-> The prime factors of the largest odd side b' (b' = a or b) are all of the form 10*k +- 1 (see formula)
-> Consequence: in each increasing triple (c,a,b), c is always the smallest odd side, but a can be either the even side or the largest odd side (see formulas and examples for explanations).
-> cos(C) >= 4/5 (or tan(C) <= 3/4), hence C <= 36.86989...° = A235509 (see Maths Challenge link with picture).
-> CG = c (see Mathematics Stack Exchange link).
-> Area(ABC) = (2/3) * m_a * m_b with m_a (resp. m_b) is the length of median AA' (resp. BB') (see Mathematics Stack Exchange link).
-> cot(A) + cot(B) >= 2/3 (see IMO Compendium link and Doob reference). - Bernard Schott, Dec 02 2021

			For 1st class, u/v > 3: (u,v) = (4,1), then (c,a,b) = (c,a',b') = (17,22,31) is the second triple and 22^2 + 31^2 = 5 * 17^2 = 1445.
For 2nd class, 1 < u/v < 2: (u,v) = (3,2), then (c,a,b) = (c,b',a') = (13,19,22) is the first triple and 19^2 + 22^2 = 5 * 13^2 = 845.

Michael Doob, The Canadian Mathematical Olympiad & L'Olympiade Mathématique du Canada 1969-1993, Canadian Mathematical Society & Société Mathématique du Canada, Problem 3, 1993, page 253, 1993.

Annales Concours Général, Corrigé Concours Général 2007.
Annales Concours Général, Sujet Concours Général 2007.
The IMO Compendium, Problem 3, 25-th Canadian Mathematical Olympiad 1993.
Maths Challenge, Perpendicular medians (Problem with picture).
Mathematics Stack Exchange, Perpendicular medians (Proves that AB = GC).
Mathematics Stack Exchange, Solving area of a triangle where medians are perpendicular.

Cf. A103606 (primitive Pythagorean triples), A235509 = arccos(4/5).

Cf. A335035 (corresponding perimeters), A335036 (smallest side and 1st trisection), A335347 (middle side and 2nd trisection), A335348 (largest side and 3rd trisection), A335273 (even side).

mycmp(x, y) = {my(xp = vecsum(x), yp = vecsum(y)); if (xp!=yp, return (xp-yp)); return (x[1] - y[1]);}
lista(nn) = {my(vm = List(), vt); for (u=1, nn, for (v=1, nn, if (gcd(u, v) == 1, vt = 0; if ((u/v > 3) && ((u-3*v) % 5), vt = [2*(u^2-u*v-v^2), u^2+4*u*v-v^2, u^2+v^2]); if ((u/v > 1) && (u/v < 2) && ((u-2*v) % 5), vt = [2*(u^2+u*v-v^2), -u^2+4*u*v+v^2, u^2+v^2]); if (gcd(vt) == 1, listput(vm, vt));););); vecsort(apply(vecsort, Vec(vm));, mycmp);} \\ Michel Marcus, May 21 2020

There exist two disjoint classes of such triangles, obtained with two distinct families of formulas: let u > v > 0 , u and v with different parities, gcd(u,v) = 1;  a' is the even side and b' the largest odd side.
1st class of triangles: (a',b',c) = (2*(u^2-uv-v^2), u^2+4*u*v-v^2, u^2+v^2) with u/v > 3 and 5 doesn't divide u-3v.
2nd class of triangles: (a',b',c) = (2*(u^2+uv-v^2), -u^2+4*u*v+v^2, u^2+v^2) with 1 < u/v < 2 and 5 doesn't divide u-2v.
For n>=1, a(3n-2) = A335036(n), a(3n-1) = A335347(n), a(3n) = A335348(n).

A336755 Primitive triples for integer-sided triangles whose sides a < b < c are in arithmetic progression.

Original entry on oeis.org

2, 3, 4, 3, 4, 5, 3, 5, 7, 4, 5, 6, 5, 6, 7, 4, 7, 10, 5, 7, 9, 6, 7, 8, 5, 8, 11, 7, 8, 9, 5, 9, 13, 7, 9, 11, 8, 9, 10, 7, 10, 13, 9, 10, 11, 6, 11, 16, 7, 11, 15, 8, 11, 14, 9, 11, 13, 10, 11, 12, 7, 12, 17, 11, 12, 13, 7, 13, 19, 8, 13, 18, 9, 13, 17, 10, 13, 16, 11, 13, 15, 12, 13, 14

Offset: 1

Bernard Schott, Sep 07 2020

The triples are displayed in increasing order of perimeter (equivalently in increasing order of middle side) and if perimeters coincide then by increasing order of the smallest side; also, each triple (a, b, c) is in increasing order.
When b is prime, all the corresponding triples in A336750 are primitive triples.
The only right integer triangle in the data corresponds to the triple (3, 4, 5).
The number of primitive such triangles whose middle side = b is equal to A023022(b) for b >= 3.
For all the triples (primitive or not), miscellaneous properties and references, see A336750.

			The table begins:
  2, 3, 4;
  3, 4, 5;
  3, 5, 7;
  4, 5, 6;
  5, 6, 7;
  4, 7, 10;
  5, 7, 9;
  6, 7, 8;
The smallest such primitive triple is (2, 3, 4).
The only triangle with perimeter = 12 corresponds to the Pythagorean triple: (3, 4, 5).
There exist two triangles with perimeter = 15 corresponding to triples (3, 5, 7) and (4, 5, 6).
There exists only one primitive triangle with perimeter = 18 whose triple is (5, 6, 7), because (4, 6, 8) is not a primitive triple.

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000

Cf. A336750 (triples, primitive or not), this sequence (primitive triples), A336756 (perimeter of primitive triangles), A336757 (number of such primitive triangles whose perimeter = n).
Cf. A103606 (similar for primitive Pythagorean triples).
Cf. A023022.

for b from 3 to 20 do
for a from b-floor((b-1)/2) to b-1 do
c := 2*b - a;
if gcd(a,b)=1 and gcd(b,c)=1 then print(a,b,c); end if;
end do;
end do;

Select[Flatten[Table[{a, b, 2*b-a}, {b, 3, 20}, {a, b-Floor[(b-1)/2], b-1}], 1], GCD @@ # == 1 &] (* Paolo Xausa, Feb 28 2024 *)

tabf(nn) = {for (b = 3, nn, for (a = b-floor((b-1)/2), b-1, my(c = 2*b - a); if (gcd([a, b, c]) == 1, print(a, " ", b, " ", c););););} \\ Michel Marcus, Sep 08 2020

A024408 Perimeters of more than one primitive Pythagorean triangle.

Original entry on oeis.org

1716, 2652, 3876, 3960, 4290, 5244, 5700, 5720, 6900, 6930, 8004, 8700, 9300, 9690, 10010, 10788, 11088, 12180, 12876, 12920, 13020, 13764, 14280, 15252, 15470, 15540, 15960, 16380, 17220, 17480, 18018, 18060, 18088, 18204, 19092, 19320, 20592, 20868

Offset: 1

David W. Wilson

nonn

a(23) = 14280 is the first perimeter of 3 primitive Pythagorean triangles: {119, 7080, 7081}, {168, 7055, 7057} and {3255, 5032, 5993}. - Jean-François Alcover, Mar 14 2012

			a(1) = 1716 with precisely two primitive Pythagorean triangles (with increasing entries): {195, 748, 773} and {364, 627, 725}. From Ron Knott's link. This is the first example of the family of perimeters 12*b(k)*(b(k) + 2) with b(k) = A007528(k), for k >= 2. See the Bernstein link, p. 234, Theorem 5. a). - _Wolfdieter Lang_, Sep 24 2019

David A. Corneth, Table of n, a(n) for n = 1..10000
Leon Bernstein, Primitive Pythagorean Triples, The Fibonacci Quarterly 20.3 (1982) 227-241.
Ron Knott, Pythagorean Triples and Online Calculators
Lindsay Witcosky, Perimeters of primitive Pythagorean triangles

Cf. A007528, A024364, A103606.

A198457 Consider triples (a, b, c) where a <= b < c and (a^2+b^2-c^2)/(c-a-b) = 2, ordered by a and then b; sequence gives a, b and c values in that order.

Original entry on oeis.org

3, 6, 7, 4, 4, 6, 5, 16, 17, 6, 10, 12, 7, 8, 11, 7, 30, 31, 8, 18, 20, 9, 14, 17, 9, 48, 49, 10, 12, 16, 10, 28, 30, 11, 70, 71, 12, 18, 22, 12, 40, 42, 13, 16, 21, 13, 30, 33, 13, 96, 97, 14, 25, 29, 14, 54, 56, 15, 22, 27, 15, 40, 43, 15, 126, 127, 16, 20, 26

Offset: 1

Charlie Marion, Nov 09 2011

See A198453.

Because either all sides or only one side of a Pythagorean (-+2)-triangle ABC is even their sum is always even. Thus csc(C) = -(a+b+c+k)/k is an integer. So ((a+2)^2 + (b+2)^2 - (c+2)^2)|(2*(a+2)*(b+2)) resp. (a^2 + b^2 - c^2)|(2*a*b). - Ralf Steiner, Sep 18 2019

			3*5 +  6*8  =  7*9;
4*6 +  4*6  =  6*8;
5*7 + 16*17 = 17*18;
6*8 + 10*12 = 12*14;
7*9 +  8*10 = 11*13;
7*9 + 30*32 = 31*33.

A. H. Beiler, Recreations in the Theory of Numbers, Dover, New York, 1964, pp. 104-134.

David A. Corneth, Table of n, a(n) for n = 1..9999
Ron Knott, Pythagorean Triples and Online Calculators.

Cf. A103606, A198453-A198456, A198458-A198468.

More terms from David A. Corneth, Sep 22 2019

A263728 Primitive Pythagorean triples (a, b, c) in lexicographic order, with a < b < c.

Original entry on oeis.org

3, 4, 5, 5, 12, 13, 7, 24, 25, 8, 15, 17, 9, 40, 41, 11, 60, 61, 12, 35, 37, 13, 84, 85, 15, 112, 113, 16, 63, 65, 17, 144, 145, 19, 180, 181, 20, 21, 29, 20, 99, 101, 21, 220, 221, 23, 264, 265, 24, 143, 145, 25, 312, 313, 27, 364, 365, 28, 45, 53

Offset: 1

Colin Barker, Nov 20 2015

a(3*k+1)*a(3*k+2) / (a(3*k+1)+a(3*k+2)+a(3*k+3)) is always an integer for k >= 0. Also note that a(3*k+1)*a(3*k+2)/2 is never a perfect square. - Altug Alkan, Apr 08 2016

			The first few triples are [3, 4, 5], [5, 12, 13], [7, 24, 25], [8, 15, 17], [9, 40, 41], [11, 60, 61], [12, 35, 37], [13, 84, 85], [15, 112, 113], [16, 63, 65], [17, 144, 145], [19, 180, 181], [20, 21, 29], [20, 99, 101], ... - _N. J. A. Sloane_, Dec 15 2015

H. M. Stark, An Introduction to Number Theory. Markham, Chicago, 1970, Chapter 5, Section 5.3.

Colin Barker, Table of n, a(n) for n = 1..10000
Yoshinosuke Hirakawa and Hideki Matsumura, A unique pair of triangles, Journal of Number Theory, Volume 194, January 2019, Pages 297-302. About the remarkable (135,352,377) Pythagorean triple.
Eric Weisstein's World of Mathematics, Pythagorean Triple
Wikipedia, Pythagorean triple

Cf. A008846, A020882, A046086, A046087, A103606, A139794, A156678, A156679.

A378148 a(n) is the number of distinct trapezoids having integer sides and height with exactly one pair of parallel sides and area n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 2, 1, 0, 3, 0, 1, 2, 1, 0, 3, 0, 1, 2, 1, 0, 4, 0, 2, 2, 1, 1, 5, 0, 1, 2, 3, 0, 5, 0, 2, 3, 1, 0, 6, 0, 2, 2, 2, 0, 7, 1, 3, 2, 1, 0, 9, 0, 1, 3, 3, 2, 8, 0, 3, 2, 3, 0, 10, 0, 1, 5, 3, 0, 9, 0, 6, 3, 1, 0, 10, 2, 1, 2

Offset: 1

Felix Huber, Dec 02 2024

nonn

The number of trapezoids having integer sides and height, which are neither right-angled nor isosceles, is a(n) - A378149(n) - A378150(n). The first trapezoid, which is neither right-angled nor isosceles, appears at a(36).
a(p) = 0 for prime p. Proof: Suppose there is a trapezoid with integer sides and prime area p. Then in p = m*h (m is the average of the parallel sides and h is the height of the trapezoid) m = p and h = 1 or m = p/2 and h = 2. At least one nonparallel side of the trapezoid is the hypotenuse of a right triangle with leg h. Legs in integer right triangles are >= 3. This is a contradiction and therefore a(p) = 0.
A214602 is the index of the positive terms in this sequence.
There are also integer-sided trapezoids with integer area that do not have an integer height. For example, the trapezoid with sides p = 630, d = 615, q = 5, f = 40 (p and q are parallel) has an area of 12192 and a height of h = 38.4.

			a(54) = 7 because there are 7 distinct trapezoids [p, d, q, f, h] (p and q are parallel, height h) having integer sides and height with area 54:[17, 10, 1, 10, 6], [13, 6, 5, 10, 6], [22, 5, 14, 5, 3], [20, 3, 16, 5, 3], [8, 15, 1, 20, 12], [7, 12, 2, 13, 12], [15, 4, 12, 5, 4].
For a(54) = 7 and (92) = 4 see the linked illustrations.
See also the linked Maple program "Trapezoids having integer sides and height with area n".

Felix Huber, Table of n, a(n) for n = 1..10000
Felix Huber, Illustration of a(54) = 7
Felix Huber, Illustration of a(92) = 4
Felix Huber, Trapezoids having integer sides and height with area n
Eric Weisstein's World of Mathematics, Trapezoid

Cf. A024406, A027750, A103606, A214602, A340858, A340859, A340860, A365049, A374594, A378149, A378150.

A378148:=proc(n)
   local a,m,p,q,h,x,y,M;
   a:=0;
   M:=map(x->x/2,NumberTheory:-Divisors(2*n) minus {1,2});
   for m in M do
      for q from 1 to m-1/2 do
         p:=2*m-q;
         h:=n/m;
         for x from max(3,floor((p-q+1)/2)) to (h^2-1)/2 do
            y:=p-q-x;
            if issqr(x^2+h^2) and issqr(y^2+h^2) then
               a:=a+1
            fi
         od
      od
   od;
   return a
end proc;
seq(A378148(n),n=1..87);

a(p) = 0 for prime p.

Showing 1-10 of 29 results. Next

cp's OEIS Frontend

A024364 Ordered perimeters of primitive Pythagorean triangles.

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A198453 Consider triples a<=b

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A103605 Pythagorean triples in increasing order of perimeter (a+b+c). If two successive perimeters are equals, then in order of decreasing areas; each triple [a(i), a(j), a(k)] (with k multiple of 3, j=k-1, i=k-2) in increasing order.

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A335893 Primitive triples for integer-sided triangles whose angles A < B < C are in arithmetic progression.

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A335034 Primitive triples, in nondecreasing order of perimeter, for integer-sided triangles with two perpendicular medians; each triple is in increasing order, and if perimeters coincide then increasing order of the smallest side.

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A336755 Primitive triples for integer-sided triangles whose sides a < b < c are in arithmetic progression.

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Maple

Mathematica

PARI

A024408 Perimeters of more than one primitive Pythagorean triangle.

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A198457 Consider triples (a, b, c) where a <= b < c and (a^2+b^2-c^2)/(c-a-b) = 2, ordered by a and then b; sequence gives a, b and c values in that order.

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