A009111 -id:A009111 - cp's OEIS frontend

A033581 a(n) = 6*n^2.

0, 6, 24, 54, 96, 150, 216, 294, 384, 486, 600, 726, 864, 1014, 1176, 1350, 1536, 1734, 1944, 2166, 2400, 2646, 2904, 3174, 3456, 3750, 4056, 4374, 4704, 5046, 5400, 5766, 6144, 6534, 6936, 7350, 7776, 8214, 8664, 9126, 9600, 10086, 10584, 11094, 11616

Offset: 0

N. J. A. Sloane

Number of edges of a complete 4-partite graph of order 4n, K_n,n,n,n. - Roberto E. Martinez II, Oct 18 2001
Number of edges of the complete bipartite graph of order 7n, K_n, 6n. - Roberto E. Martinez II, Jan 07 2002
Number of edges in the line graph of the product of two cycle graphs, each of order n, L(C_n x C_n). - Roberto E. Martinez II, Jan 07 2002
Total surface area of a cube of edge length n. See A000578 for cube volume. See A070169 and A071399 for surface area and volume of a regular tetrahedron and links for the other Platonic solids. - Rick L. Shepherd, Apr 24 2002
a(n) can represented as n concentric hexagons (see example). - Omar E. Pol, Aug 21 2011
Sequence found by reading the line from 0, in the direction 0, 6, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. Opposite numbers to the members of A003154 in the same spiral. - Omar E. Pol, Sep 08 2011
Together with 1, numbers m such that floor(2*m/3) and floor(3*m/2) are both squares. Example: floor(2*150/3) = 100 and floor(3*150/2) = 225 are both squares, so 150 is in the sequence. - Bruno Berselli, Sep 15 2014
a(n+1) gives the number of vertices in a hexagon-like honeycomb built from A003215(n) congruent regular hexagons (see link). Example: a hexagon-like honeycomb consisting of 7 congruent regular hexagons has 1 core hexagon inside a perimeter of six hexagons. The perimeter has 18 vertices. The core hexagon has 6 vertices. a(2) = 18 + 6 = 24 is the total number of vertices. - Ivan N. Ianakiev, Mar 11 2015
a(n) is the area of the Pythagorean triangle whose sides are (3n, 4n, 5n). - Sergey Pavlov, Mar 31 2017
More generally, if k >= 5 we have that the sequence whose formula is a(n) = (2*k - 4)*n^2 is also the sequence found by reading the line from 0, in the direction 0, (2*k - 4), ..., in the square spiral whose vertices are the generalized k-gonal numbers. In this case k = 5. - Omar E. Pol, May 13 2018
The sequence also gives the number of size=1 triangles within a match-made hexagon of size n. - John King, Mar 31 2019
For hexagons, the number of matches required is A045945; thus number of size=1 triangles is A033581; number of larger triangles is A307253 and total number of triangles is A045949. See A045943 for analogs for Triangles; see A045946 for analogs for Stars. - John King, Apr 04 2019

			From _Omar E. Pol_, Aug 21 2011: (Start)
Illustration of initial terms as concentric hexagons:
.
.                                 o o o o o o
.                                o           o
.              o o o o          o   o o o o   o
.             o       o        o   o       o   o
.   o o      o   o o   o      o   o   o o   o   o
.  o   o    o   o   o   o    o   o   o   o   o   o
.   o o      o   o o   o      o   o   o o   o   o
.             o       o        o   o       o   o
.              o o o o          o   o o o o   o
.                                o           o
.                                 o o o o o o
.
.    6            24                   54
.
(End)

Nathaniel Johnston, Table of n, a(n) for n = 0..10000
Bruno Berselli, An interpretation of initial terms.
Ivan N. Ianakiev, Hexagon-like honeycomb built form regular congruent hexagons.
Leo Tavares, Illustration: Diamond Star Rays
Eric Weisstein's World of Mathematics, Platonic Solid.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A032528. Central column of triangle A001283.
Cf. A000217, A000290, A001105, A009111, A033428, A033583, A085250, A086729, A152734, A152751.
Cf. A000578, A001318, A003215, A005897, A045943, A045945, A045949, A070169, A071399.
Cf. A017593 (first differences).

a033581 = (* 6) . (^ 2)  -- Reinhard Zumkeller, Apr 27 2014

seq(6*n^2,n=0..44); # Nathaniel Johnston, Jun 26 2011

6 Range[44]^2 (* Michael De Vlieger, Apr 02 2017 *)
LinearRecurrence[{3,-3,1},{0,6,24},50] (* Harvey P. Dale, Jul 03 2017 *)

vector(100,n,6*(n-1)^2) \\ Derek Orr, Mar 11 2015

a(n) = A000290(n)*6. - Omar E. Pol, Dec 11 2008
a(n) = A001105(n)*3 = A033428(n)*2. - Omar E. Pol, Dec 13 2008
a(n) = 12*n + a(n-1) - 6, with a(0)=0. - Vincenzo Librandi, Aug 05 2010
G.f.: 6*x*(1+x)/(1-x)^3. - Colin Barker, Feb 14 2012
For n > 0: a(n) = A005897(n) - 2. - Reinhard Zumkeller, Apr 27 2014
a(n) = 3*floor(1/(1-cos(1/n))) = floor(1/(1-n*sin(1/n))) for n > 0. - Clark Kimberling, Oct 08 2014
a(n) = t(4*n) - 4*t(n), where t(i) = i*(i+k)/2 for any k. Special case (k=1): a(n) = A000217(4*n) - 4*A000217(n). - Bruno Berselli, Aug 31 2017
From Amiram Eldar, Feb 03 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/36.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/72 (A086729).
Product_{n>=1} (1 + 1/a(n)) = sqrt(6)*sinh(Pi/sqrt(6))/Pi.
Product_{n>=1} (1 - 1/a(n)) = sqrt(6)*sin(Pi/sqrt(6))/Pi. (End)
E.g.f.: 6*exp(x)*x*(1 + x). - Stefano Spezia, Aug 19 2022

More terms from Larry Reeves (larryr(AT)acm.org), Nov 08 2001

A029549 a(n + 3) = 35a(n + 2) - 35a(n + 1) + a(n), with a(0) = 0, a(1) = 6, a(2) = 210.

Original entry on oeis.org

0, 6, 210, 7140, 242556, 8239770, 279909630, 9508687656, 323015470680, 10973017315470, 372759573255306, 12662852473364940, 430164224521152660, 14612920781245825506, 496409142337836914550, 16863297918705209269200, 572855720093639278238256

Offset: 0

Don N. Page

Triangular numbers that are twice other triangular numbers. - Don N. Page
Triangular numbers that are also pronic numbers. These will be shown to have a Pythagorean connection in a paper in preparation. - Stuart M. Ellerstein (ellerstein(AT)aol.com), Mar 09 2002
In other words, triangular numbers which are products of two consecutive numbers. E.g., a(2) = 210: 210 is a triangular number which is the product of two consecutive numbers: 14 * 15. - Shyam Sunder Gupta, Oct 26 2002
Coefficients of the series giving the best rational approximations to sqrt(8). The partial sums of the series 3 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(8) = 2 sqrt(2), which constitute every second convergent of the continued fraction. The corresponding continued fractions are [2; 1, 4, 1], [2; 1, 4, 1, 4, 1], [2; 1, 4, 1, 4, 1, 4, 1], [2; 1, 4, 1, 4, 1, 4, 1, 4, 1] and so forth. - Gene Ward Smith, Sep 30 2006
This sequence satisfy the same recurrence as A165518. - Ant King, Dec 13 2010
Intersection of A000217 and A002378.
This is the sequence of areas, x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n) = x(n) + 1,z(n)) with x(0) = 0, y(0) = 1, z(0) = 1, a(0) = 0 and x(1) = 3, y(1) = 4, z(1) = 5, a(1) = 6. - George F. Johnson, Aug 20 2012

Reinhard Zumkeller, Table of n, a(n) for n = 0..100
H. J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
Shyam Sunder Gupta Fascinating Triangular Numbers
Roger B. Nelson, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2103.03019 [math.GM], 2021.
Vladimir Pletser, Searching for multiple of triangular numbers being triangular numbers, 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2021.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A123478, A123479, A123480, A123482, A075528, A082405 (first differences).
Cf. A000129, A001108, A002203, A009111, A245031.
Cf. A000217, A001109, A001652, A002315, A002378, A011900, A029546, A046090, A046729, A053141, A084159, A096979, A157259, A165518.

List([0..20], n-> (Lucas(2,-1, 4*n+2)[2] -6)/32 ); # G. C. Greubel, Jan 13 2020

a029549 n = a029549_list !! n
a029549_list = [0,6,210] ++
   zipWith (+) a029549_list
               (map (* 35) $ tail delta)
   where delta = zipWith (-) (tail a029549_list) a029549_list
-- Reinhard Zumkeller, Sep 19 2011

(makelist(binom(n,2),n,1,999999),intersection(%%,2*%%)) /* Bill Gosper, Feb 07 2010 */

R:=PowerSeriesRing(Integers(), 25); [0] cat Coefficients(R!(6/(1-35*x+35*x^2-x^3))); // G. C. Greubel, Jul 15 2018

A029549 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[0,6]) ;
    else
        34*procname(n-1)-procname(n-2)+6 ;
    end if;
end proc: # R. J. Mathar, Feb 05 2016

Table[Floor[(Sqrt[2] + 1)^(4n + 2)/32], {n, 0, 20} ] (* Original program from author, corrected by Ray Chandler, Jul 09 2015 *)
CoefficientList[Series[6/(1 - 35x + 35x^2 - x^3), {x, 0, 14}], x]
Intersection[#, 2#] &@ Table[Binomial[n, 2], {n, 999999}] (* Bill Gosper, Feb 07 2010 *)
LinearRecurrence[{35, -35, 1}, {0, 6, 210}, 20] (* Harvey P. Dale, Jun 06 2011 *)
(LucasL[4Range[20] - 2, 2] -6)/32 (* G. C. Greubel, Jan 13 2020 *)

concat(0,Vec(6/(1-35*x+35*x^2-x^3)+O(x^25))) \\ Charles R Greathouse IV, Jun 13 2013

[(lucas_number2(4*n+2, 2, -1) -6)/32 for n in (0..20)] # G. C. Greubel, Jan 13 2020

val triNums = (0 to 39999).map(n => (n * n + n)/2)
triNums.filter( % 2 == 0).filter(n => (triNums.contains(n/2))) // _Alonso del Arte, Jan 12 2020

G.f.: 6*x/(1 - 35*x + 35*x^2 - x^3) = 6*x /( (1-x)*(1 - 34*x + x^2) ).
a(n) = 6*A029546(n-1) = 2*A075528(n).
a(n) = -3/16 + ((3+2*sqrt(2))/32) *(17 + 12*sqrt(2))^n + ((3-2*sqrt(2))/32) *(17 - 12*sqrt(2))^n. - Gene Ward Smith, Sep 30 2006
From Bill Gosper, Feb 07 2010: (Start)
a(n) = (cosh((4*n + 2)*log(1 + sqrt(2))) - 3)/16.
a(n) = binomial(A001652(n) + 1, 2) = 2*binomial(A053141(n) + 1, 2). (End)
a(n) = binomial(A046090(n), 2) = A000217(A001652(n)). - Mitch Harris, Apr 19 2007, R. J. Mathar, Jun 26 2009
a(n) = ceiling((3 + 2*sqrt(2))^(2n + 1) - 6)/32 = floor((1/32) (1+sqrt(2))^(4n+2)). - Ant King, Dec 13 2010
Sum_{n >= 1} 1/a(n) = 3 - 2*sqrt(2) = A157259 - 4. - Ant King, Dec 13 2010
a(n) = a(n - 1) + A001109(2n). - Charlie Marion, Feb 10 2011
a(n+2) = 34*a(n + 1) - a(n) + 6. - Charlie Marion, Feb 11 2011
From George F. Johnson, Aug 20 2012: (Start)
a(n) = ((3 + 2*sqrt(2))^(2*n + 1) + (3 - 2*sqrt(2))^(2*n + 1) - 6)/32.
8*a(n) + 1 = (A002315(n))^2, 4*a(n) + 1 = (A000129(2*n + 1))^2, 32*a(n)^2 + 12*a(n) + 1 are perfect squares.
a(n + 1) = 17*a(n) + 3 + 3*sqrt((8*a(n) + 1)*(4*a(n) + 1)).
a(n - 1) = 17*a(n) + 3 - 3*sqrt((8*a(n) + 1)*(4*a(n) + 1)).
a(n - 1)*a(n + 1) = a(n)*(a(n) - 6), a(n) = A096979(2*n).
a(n) = (1/2)*A084159(n)*A046729(n) = (1/2)*A001652(n)*A046090(n).
Limit_{n->infinity} a(n)/a(n - 1) = 17 + 12*sqrt(2).
Limit_{n->infinity} a(n)/a(n - 2) = (17 + 12*sqrt(2))^2 = 577 + 408*sqrt(2).
Limit_{n->infinity} a(n)/a(n - r) = (17 + 12*sqrt(2))^r.
Limit_{n->infinity} a(n - r)/a(n) = (17 + 12*sqrt(2))^(-r) = (17 - 12*sqrt(2))^r. (End)
a(n) = 3 * T( b(n) ) + (2*b(n) + 1)*sqrt( T( b(n) ) ) where b(n) = A001108(n) (indices of the square triangular numbers), T(n) = A000217(n) (the n-th triangular number). - Dimitri Papadopoulos, Jul 07 2017
a(n) = (Pell(2*n + 1)^2 - 1)/4 = (Q(4*n + 2) - 6)/32, where Q(n) are the Pell-Lucas numbers (A002203). - G. C. Greubel, Jan 13 2020
a(n) = A002378(A011900(n)-1) = A002378(A053141(n)). - Pontus von Brömssen, Sep 11 2024

Additional comments from Christian G. Bower, Sep 19 2002; T. D. Noe, Nov 07 2006; and others

Edited by N. J. A. Sloane, Apr 18 2007, following suggestions from Andrew S. Plewe and Tanya Khovanova

A009112 Areas of Pythagorean triangles: numbers which can be the area of a right triangle with integer sides.

Original entry on oeis.org

6, 24, 30, 54, 60, 84, 96, 120, 150, 180, 210, 216, 240, 270, 294, 330, 336, 384, 480, 486, 504, 540, 546, 600, 630, 720, 726, 750, 756, 840, 864, 924, 960, 990, 1014, 1080, 1176, 1224, 1320, 1344, 1350, 1386, 1470, 1500, 1536, 1560, 1620, 1710, 1716, 1734, 1890

Offset: 1

David W. Wilson

Number of terms < 10^k for increasing values of k: 1, 7, 34, 150, 636, 2536, 9757, 35987, 125350, 407538, ..., .
All terms are divisible by 6.
Also positive integers m with four (or more) different divisors (p, q, r, s) such that m = p*q = r*s and s = p+q+r. - Jose Aranda, Jun 28 2023

			30 belongs to the sequence as the area of the triangle (5,12,13) is 30.
6 is in the sequence because it is the area of the 3-4-5 triangle.

Robert Israel, Table of n, a(n) for n = 1..10000
Ron Knott, Pythagorean Triangles
B. Miller, Nasty Numbers, The Mathematics Teacher 73 (1980), page 649.
Supriya Mohanty and S. P. Mohanty, Pythagorean Numbers, Fibonacci Quarterly 28 (1990), 31-42.

Union of A009111, A009127, A024365, A177021.
A073120 is a subsequence.
See A256418 for the numbers 4*a(n).

N:= 10^4: # to get all entries <= N
A:= {}:
for t from 1 to floor(sqrt(2*N)) do
   F:= select(f -> f[2]::odd,ifactors(2*t)[2]);
   d:= mul(f[1],f=F);
   for e from ceil(sqrt(t/d)) do
     s:= d*e^2;
     r:= sqrt(2*t*s);
     a:= (r+s)*(r+t)/2;
     if a > N then break fi;
     A:= A union {a};
   od
od:
A;
# if using Maple 11 or earlier, uncomment the next line
# sort(convert(A,list)); # Robert Israel, Apr 06 2015

lst = {}; Do[ If[ IntegerQ[c = Sqrt[a^2 + b^2]], AppendTo[lst, a*b/2]; lst = Union@ lst], {a, 4, 180}, {b, a - 1, Floor[ Sqrt[a]], -1}]; Take[lst, 51] (* Vladimir Joseph Stephan Orlovsky, Nov 23 2010 *)
g@A_ := With[{div = Divisors@(2*A)}, AnyTrue[Sqrt@(Plus@@({#, 2*A/#}^2))& /@Take[div, Floor[(Length@div)/2]],IntegerQ]];
Select[Range@5000, g@# &] (* Hans Rudolf Widmer, Sep 25 2023 *)

is_A009112(n)={ my(N=1+#n=divisors(2*n)); for( i=1, N\2, issquare(n[i]^2+n[N-i]^2) & return(1)) } \\ M. F. Hasler, Dec 09 2010

is_A009112 = lambda n: any(is_square(a**2+(2*n/a)**2) for a in divisors(2*n)) # D. S. McNeil, Dec 09 2010

A024365 Areas of right triangles with coprime integer sides.

Original entry on oeis.org

6, 30, 60, 84, 180, 210, 330, 504, 546, 630, 840, 924, 990, 1224, 1320, 1386, 1560, 1710, 1716, 2310, 2340, 2574, 2730, 3036, 3570, 3900, 4080, 4290, 4620, 4914, 5016, 5610, 5814, 6090, 6630, 7140, 7440, 7854, 7956, 7980, 8970, 8976, 9690, 10374

Offset: 1

David W. Wilson

Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, (A, B) = 1, A <= B); sequence gives areas A*B/2.
By Theorem 2 of Mohanty and Mohanty, all these numbers are primitive Pythagorean. - T. D. Noe, Sep 24 2013
This sequence also gives Fibonacci's congruous numbers (without multiplicity, in increasing order) divided by 4. See A258150. - Wolfdieter Lang, Jun 14 2015
The same as A024406 with duplicates removed. All terms are multiples of 6, cf. A258151. - M. F. Hasler, Jan 20 2019

			6 is in the sequence because it is the area of the 3-4-5 triangle.
a(7) = 210 corresponds to the two primitive Pythagorean triangles (21, 20, 29) and (35, 12, 37). See A024406. - _Wolfdieter Lang_, Jun 14 2015

T. D. Noe, Table of n, a(n) for n = 1..10000 (corrected by _Giovanni Resta_, Jan 21 2019)
Supriya Mohanty and S. P. Mohanty, Pythagorean Numbers, Fibonacci Quarterly 28 (1990), 31-42.

Cf. A009111, A009112, A024406 (with multiplicity), A258150, A024407, A258151 (terms divided by 6).

Subsequence of A073120 and A147778.

nn = 22; (* nn must be even *) t = Union[Flatten[Table[If[GCD[u, v] == 1 && Mod[u, 2] + Mod[v, 2] == 1, u v (u^2 - v^2), 0], {u, nn}, {v, u - 1}]]]; Select[Rest[t], # < nn (nn^2 - 1) &] (* T. D. Noe, Sep 19 2013 *)

select( {is_A024365(n)=my(N=1+#n=divisors(2*n)); for(i=1, N\2, gcd(n[i], n[N-i])==1 && issquare(n[i]^2+n[N-i]^2) && return(n[i]))}, [1..10^4]) \\ is_A024365 returns the smaller leg if n is a term, else 0. - M. F. Hasler, Jun 06 2024

Positive integers of the form u*v*(u^2 - v^2) where 2uv and u^2 - v^2 are coprime or, alternatively, where u, v are coprime and one of them is even.

a(n) = 6*A258151(n). - M. F. Hasler, Jan 20 2019

Additional comments James R. Buddenhagen, Aug 10 2008 and from Max Alekseyev, Nov 12 2008

Edited by N. J. A. Sloane, Nov 20 2008 at the suggestion of R. J. Mathar

A078522 Numbers k such that (k+1)(2k+1) is a perfect square.

Original entry on oeis.org

0, 24, 840, 28560, 970224, 32959080, 1119638520, 38034750624, 1292061882720, 43892069261880, 1491038293021224, 50651409893459760, 1720656898084610640, 58451683124983302024, 1985636569351347658200, 67453191674820837076800, 2291422880374557112953024

Offset: 1

Joseph L. Pe, Jan 07 2003

Equivalently, both k+1 and 2*k+1 are perfect squares.
The square roots of (k+1)*(2*k+1) are in A046176.
Also numbers k such that 3*A000217(k) + A000217(k+1) is a perfect square. - Bruno Berselli, Nov 17 2016
From Sergey Pavlov, Mar 14 2017: (Start)
The sequence of areas k(n)*q(n)/2, of the ordered Pythagorean triples (k(n), q(n) = k(n) + 2, c(n)) with k(1)=0, q(1)=2, c(1)=0, a(1)=0, and k(2)=6, q(2)=8, c(2)=10, a(2)=24 (conjectured).
Conjecture: let f(n) be a sequence of form x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n) = x(n) + v, z(n)) with x(1)=0, y(1)=v, z(1)=0, f(1)=0, where v is an even number. Then there exists such subset p(i) that p(1) = 0, p(2) = 24*(v/2)^2, for any i > 2, p(i) = 34*p(i-1) - p(i-2) + 24*(v/2)^2, and any p(i) is a term of the above sequence f(n) (see also the first formula by Benoit Cloitre in the Formula section).
(End)

Colin Barker, Table of n, a(n) for n = 1..650
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000217, A001542, A001653, A002315, A008844, A009111, A029546, A029549, A046176.

Cf. A278310: numbers m such that T(m) + 3*T(m+1) is a square.

a:=[0,24];; for n in [3..20] do a[n]:=34*a[n-1]-a[n-2]+24; od; a; # G. C. Greubel, Jan 13 2020

I:=[0,24]; [n le 2 select I[n] else 34*Self(n-1) - Self(n-2) + 24: n in [1..20]]; // Marius A. Burtea, Sep 15 2019

seq(coeff(series(24*x^2/((1-x)*(1-34*x+x^2)), x, n+1), x, n), n = 1..20); # G. C. Greubel, Jan 13 2020

RecurrenceTable[{a[1]==0, a[2]==24, a[n]==34a[n-1] -a[n-2] +24}, a[n], {n,20}]
Drop[CoefficientList[Series[24*x^2/((1-x)*(1-34*x+x^2)), {x,0,20}], x], 1] (* Indranil Ghosh, Mar 15 2017 *)
Table[3*(ChebyshevT[n, 17] -16*ChebyshevU[n-1, 17] -1)/4, {n,20}] (* G. C. Greubel, Jan 13 2020 *)

concat(0, Vec(24*x^2/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Nov 21 2016

def A078522_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 24*x^2/((1-x)*(1-34*x+x^2)) ).list()
a=A078522_list(20); a[1:] # G. C. Greubel, Jan 13 2020

From Benoit Cloitre, Jan 19 2003: (Start)
a(1) = 0, a(2) = 24; for n > 2, a(n) = 34*a(n-1) - a(n-2) + 24.
a(n) = floor(A*B^n), where A = (3 + 2*sqrt(2))/8 and B = 17 + 12*sqrt(2).
a(n) = A008844(n) - 1. (End)
From R. J. Mathar, Sep 21 2011: (Start)
G.f.: 24*x^2/( (1-x)*(1-34*x+x^2) ).
a(n) = 24*A029546(n-2). (End)
a(n) = (A001653(n)^2 - 1)/2 = A002315(n-1)^2 - 1. - Tomohiro Yamada, Sep 15 2019
a(n) = (3/4)*(ChebyshevT(n, 17) - 16*Chebyshev(n-1, 17) - 1). - G. C. Greubel, Jan 13 2020
From Amiram Eldar, Dec 02 2024: (Start)
a(n) = A001542(n-1)*A001542(n).
Sum_{n>=2} 1/a(n) = (3 - 2*sqrt(2))/4. (End)

Edited by Bruno Berselli, Nov 17 2016

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

Henry Bottomley, Aug 01 2000

nonn

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)

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

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.

T. D. Noe, Table of n, a(n) for n = 1..10000
Andrew Bremner and Richard K. Guy, A Dozen Difficult Diophantine Dilemmas, The American Mathematical Monthly, Vol. 95, No. 1 (1988), pp. 31-36.
Ernest J. Eckert, Problem 994, Crux Mathematicorum, Vol. 10, No. 10 (1984), p. 318, entire issue.
Richard K. Guy, Comment to Problem 994, Crux Mathematicorum, Vol. 12, No. 5 (1986), p. 109, entire issue.
Henry Plane, Calcule-moi un parallélépipède..., AMPEP, PLOT No. 22 (2002), pp. 22-23.
Giovanni Resta, Saint-Exupery numbers.
Antoine de Saint Exupéry, Le Problème du Pharaon, Succession Saint Exupéry - d'Agay, 2018.
Eric Weisstein's World of Mathematics, Pythagorean Triple.

Cf. A009004, A009012, A009111, A020886, A057097, A057098, A057099, A057100.

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 *)

a(n) = 60*A057097(n) = A057098(n)*A057099(n)*A057100(n).

A055193 Smallest number that is the area of n distinct Pythagorean triangles.

Original entry on oeis.org

6, 210, 840, 341880, 71831760, 64648584000, 2216650756320, 22861058133513600

Offset: 1

David W. Wilson, Jun 30 2000

a(9) <= 14456267383718160000 the triangles are 3188885700-9066657600-9611101500, 8199991800-3525922400-8925917000, 2333331000-12391098720-12608876280, 29569667400-977776800-29585829000, 11569432875-2499045120-11836258005, 2179485000-13265764512-13443610488, 8493324840-3404148000-9150125160, 1027776750-28131143040-28149911790, 313939080-92096004000-92096539080. - Felipe Villaseñor, Oct 29 2024

			a(5) = 71831760 is area of 5 Pythagorean triangles: 2415-59488-59537, 2640-54418-54482, 5070-28336-28786, 7280-19734-21034, 10010-14352-17498
From _Sture Sjöstedt_, Jun 09 2017: (Start)
The area of 7280-19734-21034 is (2*13)^2*the area of 280-759-809.
The area of 10010-14352-17498 is (2*13)^2*the area of 385-552-673.
These triangles have the same area as the triangles I get by solving p^2-p*q+q^2=r^2. r=169, p=15, q=176, (q-p)=161 Area=r*p*q*(q-p)
q=176 and r=169 gives 2415-59488-59537;
r=169 and q-p=161 gives 2640-54418-54482;
r=169 and p=15 gives 5070-28336-28786. (End)

Cf. A009111, A093536.

Edited by N. J. A. Sloane, Sep 15 2008 at the suggestion of R. J. Mathar

a(8) added by Duncan Moore, Mar 10 2017

A009127 Area of more than one Pythagorean triangle.

Original entry on oeis.org

210, 840, 1320, 1890, 2730, 3360, 4914, 5250, 5280, 7560, 7980, 10290, 10920, 11880, 13440, 17010, 18480, 19656, 21000, 21120, 24570, 25410, 29400, 30240, 30600, 31920, 32130, 33000, 34650, 35490, 41160, 41580, 43680, 44226, 45144, 47250, 47520, 50490

Offset: 1

David W. Wilson

nonn

Duplicate numbers in A009111. The number 210 appears twice; 840 appears three times; 341880 appears four times. See A055193 for the sequence - T. D. Noe, Oct 14 2013

Giovanni Resta, Table of n, a(n) for n = 1..10000

Cf. A009111.

A177063 Number of Pythagorean triangles with area n.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 1

M. F. Hasler, Dec 09 2010

nonn

The first term > 1 is a(210) = 2, cf. A009127, A055193 and A024407. Up to there the sequence coincides with the characteristic function of A009112. The triangles are not necessarily primitive.

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries related to Pythagorean triples.

Cf. A009111, A009127, A177021.

a(n)=my(N=1+#n=divisors(2*n));sum(i=1,N\2,issquare(n[i]^2+n[N-i]^2));

Secondary offset added by Antti Karttunen, Nov 24 2017

A228875 Pairs of Pythagorean numbers differing by 6.

Original entry on oeis.org

24, 30, 54, 60, 210, 216, 330, 336, 480, 486, 540, 546, 720, 726, 750, 756, 1344, 1350, 1710, 1716, 2160, 2166, 8664, 8670, 8970, 8976, 10080, 10086, 10290, 10296, 12144, 12150, 15600, 15606, 18144, 18150, 24570, 24576, 28560, 28566, 30240, 30246, 34650, 34656

Offset: 1

T. D. Noe, Sep 24 2013

nonn

These are pairs of terms of A009111 differing by 6 -- the least amount possible. Mohanty and Mohanty call these twin Pythagorean numbers. In Theorem 9, they prove that there are an infinite number of these numbers.

Supriya Mohanty and S. P. Mohanty, Pythagorean Numbers, Fibonacci Quarterly 28 (1990), 31-42.

Cf. A009111 (Pythagorean numbers).

Cf. A228876, A228877 (upper and lower Pythagorean twins).

cp's OEIS Frontend

A033581 a(n) = 6*n^2.

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A029549 a(n + 3) = 35*a(n + 2) - 35*a(n + 1) + a(n), with a(0) = 0, a(1) = 6, a(2) = 210.

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A009112 Areas of Pythagorean triangles: numbers which can be the area of a right triangle with integer sides.

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A024365 Areas of right triangles with coprime integer sides.

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A078522 Numbers k such that (k+1)*(2*k+1) is a perfect square.

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

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A029549 a(n + 3) = 35a(n + 2) - 35a(n + 1) + a(n), with a(0) = 0, a(1) = 6, a(2) = 210.

A078522 Numbers k such that (k+1)(2k+1) is a perfect square.