A010502 -id:A010502 - cp's OEIS frontend

A248274 Egyptian fraction representation of sqrt(48) (A010502) using a greedy function.

6, 2, 3, 11, 253, 121406, 26366884520, 849309519459745289215, 1275274072254463235178765644812100983170793, 6840927687383150447046638299623716679409134458251745775753167712124043749551513939746

Offset: 0

Robert G. Wilson v, Oct 04 2014

nonn

Egyptian fraction representations of the square roots: A006487, A224231, A248235-A248322.

Egyptian fraction representations of the cube roots: A129702, A132480-A132574.

Egyptian[nbr_] := Block[{lst = {IntegerPart[nbr]}, cons = N[ FractionalPart[ nbr], 2^20], denom, iter = 8}, While[ iter > 0, denom = Ceiling[ 1/cons]; AppendTo[ lst, denom]; cons -= 1/denom; iter--]; lst]; Egyptian[ Sqrt[ 48]]

A010527 Decimal expansion of sqrt(3)/2.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

This is the ratio of the height of an equilateral triangle to its base.
Essentially the same sequence arises from decimal expansion of square root of 75, which is 8.6602540378443864676372317...
Also the real part of i^(1/3), the cubic root of i. - Stanislav Sykora, Apr 25 2012
Gilbert & Pollak conjectured that this is the Steiner ratio rho_2, the least upper bound of the ratio of the length of the Steiner minimal tree to the length of the minimal tree in dimension 2. (See Ivanov & Tuzhilin for the status of this conjecture as of 2012.) - Charles R Greathouse IV, Dec 11 2012
Surface area of a regular icosahedron with unit edge is 5*sqrt(3), i.e., 10 times this constant. - Stanislav Sykora, Nov 29 2013
Circumscribed sphere radius for a cube with unit edges. - Stanislav Sykora, Feb 10 2014
Also the ratio between the height and the pitch, used in the Unified Thread Standard (UTS). - Enrique Pérez Herrero, Nov 13 2014
Area of a 30-60-90 triangle with shortest side equal to 1. - Wesley Ivan Hurt, Apr 09 2016
If a, b, c are the sides of a triangle ABC and h_a, h_b, h_c the corresponding altitudes, then (h_a+h_b+h_c) / (a+b+c) <= sqrt(3)/2; equality is obtained only when the triangle is equilateral (see Mitrinovic reference). - Bernard Schott, Sep 26 2022

			0.86602540378443864676372317...

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Sections 8.2, 8.3 and 8.6, pp. 484, 489, and 504.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §12.4 Theorems and Formulas (Solid Geometry), pp. 450-451.
D. S. Mitrinovic, E. S. Barnes, D. C. B. Marsh, and J. R. M. Radok, Elementary Inequalities, Tutorial Text 1 (1964), P. Noordhoff LTD, Groningen, problem 6.8, page 114.

Harry J. Smith, Table of n, a(n) for n = 0..20000
E. N. Gilbert and H. O. Pollak, Steiner minimal trees, SIAM J. Appl. Math. 16, (1968), pp. 1-29.
A. O. Ivanov and A. A. Tuzhilin, The Steiner ratio Gilbert-Pollak conjecture is still open, Algorithmica 62:1-2 (2012), pp. 630-632.
Matt Parker, The mystery of 0.866025403784438646763723170752936183471402626905190314027903489, Stand-up Maths, YouTube video, Feb 14 2024.
Simon Plouffe, Plouffe's Inverter, sqrt(3)/2 to 10000 digits.
Simon Plouffe, Sqrt(3)/2 to 5000 digits.
Eric Weisstein's World of Mathematics, Lebesgue Minimal Problem.
Wikipedia, Icosahedron.
Wikipedia, Platonic solid.
Wikipedia, Unified Thread Standard.
Index entries for algebraic numbers, degree 2.

Cf. A010153.
Cf. Platonic solids surfaces: A002194 (tetrahedron), A010469 (octahedron), A131595 (dodecahedron).
Cf. Platonic solids circumradii: A010503 (octahedron), A019881 (icosahedron), A179296 (dodecahedron), A187110 (tetrahedron).
Cf. A126664 (continued fraction), A144535/A144536 (convergents).
Cf. A002194, A010502, A020821, A104956, A152623 (other geometric inequalities).

SetDefaultRealField(RealField(100)); Sqrt(3)/2; // G. C. Greubel, Nov 02 2018

Digits:=100: evalf(sqrt(3)/2); # Wesley Ivan Hurt, Apr 09 2016

RealDigits[Sqrt[3]/2, 10, 200][[1]] (* Vladimir Joseph Stephan Orlovsky, Feb 21 2011 *)

default(realprecision, 20080); x=10*(sqrt(3)/2); for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b010527.txt", n, " ", d));  \\ Harry J. Smith, Jun 02 2009

sqrt(3)/2 \\ Michel Marcus, Apr 10 2016

Equals cos(30 degrees). - Kausthub Gudipati, Aug 15 2011
Equals A002194/2. - Stanislav Sykora, Nov 30 2013
From Amiram Eldar, Jun 29 2020: (Start)
Equals sin(Pi/3) = cos(Pi/6).
Equals Integral_{x=0..Pi/3} cos(x) dx. (End)
Equals 1/(10*A020832). - Bernard Schott, Sep 29 2022
Equals x^(x^(x^...)) where x = (3/4)^(1/sqrt(3)) (infinite power tower). - Michal Paulovic, Jun 25 2023
Equals 2F1(-1/4,1/4 ; 1/2 ; 3/4) . - R. J. Mathar, Aug 31 2025

Last term corrected and more terms added by Harry J. Smith, Jun 02 2009

A001108 a(n)-th triangular number is a square: a(n+1) = 6*a(n) - a(n-1) + 2, with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 8, 49, 288, 1681, 9800, 57121, 332928, 1940449, 11309768, 65918161, 384199200, 2239277041, 13051463048, 76069501249, 443365544448, 2584123765441, 15061377048200, 87784138523761, 511643454094368, 2982076586042449, 17380816062160328, 101302819786919521

Offset: 0

N. J. A. Sloane

b(0)=0, c(0)=1, b(i+1)=b(i)+c(i), c(i+1)=b(i+1)+b(i); then a(i) (the number in the sequence) is 2b(i)^2 if i is even, c(i)^2 if i is odd and b(n)=A000129(n) and c(n)=A001333(n). - Darin Stephenson (stephenson(AT)cs.hope.edu) and Alan Koch
For n > 1 gives solutions to A007913(2x) = A007913(x+1). - Benoit Cloitre, Apr 07 2002
If (X,X+1,Z) is a Pythagorean triple, then Z-X-1 and Z+X are in the sequence.
For n >= 2, a(n) gives exactly the positive integers m such that 1,2,...,m has a perfect median. The sequence of associated perfect medians is A001109. Let a_1,...,a_m be an (ordered) sequence of real numbers, then a term a_k is a perfect median if Sum_{j=1..k-1} a_j = Sum_{j=k+1..m} a_j. See Puzzle 1 in MSRI Emissary, Fall 2005. - Asher Auel, Jan 12 2006
This is the r=8 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found.
Also, 1^3 + 2^3 + 3^3 + ... + a(n)^3 = k(n)^4 where k(n) is A001109. - Anton Vrba (antonvrba(AT)yahoo.com), Nov 18 2006
If T_x = y^2 is a triangular number which is also a square, the least number which is both triangular and square and greater than T_x is T_(3*x + 4*y + 1) = (2*x + 3*y + 1)^2 (W. Sierpiński 1961). - Richard Choulet, Apr 28 2009
If (a,b) is a solution of the Diophantine equation 0 + 1 + 2 + ... + x = y^2, then a or (a+1) is a perfect square. If (a,b) is a solution of the Diophantine equation 0 + 1 + 2 + ... + x = y^2, then a or a/8 is a perfect square. If (a,b) and (c,d) are two consecutive solutions of the Diophantine equation 0 + 1 + 2 + ... + x = y^2 with a < c, then a+b = c-d and ((d+b)^2, d^2-b^2) is a solution, too. If (a,b), (c,d) and (e,f) are three consecutive solutions of the Diophantine equation 0 + 1 + 2 + ... + x = y^2 with a < c < e, then (8*d^2, d*(f-b)) is a solution, too. - Mohamed Bouhamida, Aug 29 2009
If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation 0 + 1 + 2 + ... + x = y^2 with p < r, then r = 3p + 4q + 1 and s = 2p + 3q + 1. - Mohamed Bouhamida, Sep 02 2009
Also numbers k such that (ceiling(sqrt(k*(k+1)/2)))^2 - k*(k+1)/2 = 0. - Ctibor O. Zizka, Nov 10 2009
From Lekraj Beedassy, Mar 04 2011: (Start)
Let x=a(n) be the index of the associated triangular number T_x=1+2+3+...+x and y=A001109(n) be the base of the associated perfect square S_y=y^2. Now using the identity S_y = T_y + T_{y-1}, the defining T_x = S_y may be rewritten as T_y = T_x - T_{y-1}, or 1+2+3+...+y = y+(y+1)+...+x. This solves the Strand Magazine House Number problem mentioned in A001109 in references from Poo-Sung Park and John C. Butcher. In a variant of the problem, solving the equation 1+3+5+...+(2*x+1) = (2*x+1)+(2*x+3)+...+(2*y-1) implies S_(x+1) = S_y - S_x, i.e., with (x,x+1,y) forming a Pythagorean triple, the solutions are given by pairs of x=A001652(n), y=A001653(n). (End)
If P = 8*n +- 1 is a prime, then P divides a((P-1)/2); e.g., 7 divides a(3) and 41 divides a(20). Also, if P = 8*n +- 3 is prime, then 4*P divides (a((P-1)/2) + a((P+1)/2) + 3). - Kenneth J Ramsey, Mar 05 2012
Starting at a(2), a(n) gives all the dimensions of Euclidean k-space in which the ratio of outer to inner Soddy hyperspheres' radii for k+1 identical kissing hyperspheres is rational. The formula for this ratio is (1+3k+2*sqrt(2k*(k+1)))/(k-1) where k is the dimension. So for a(3) = 49, the ratio is 6 in the 49th dimension. See comment for A010502. - Frank M Jackson, Feb 09 2013
Conjecture: For n>1 a(n) is the index of the first occurrence of -n in sequence A123737. - Vaclav Kotesovec, Jun 02 2015
For n=2*k, k>0, a(n) is divisible by 8 (deficient), so since all proper divisors of deficient numbers are deficient, then a(n) is deficient. For n=2*k+1, k>0, a(n) is odd.  If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. sigma(a(5)) = 1723 < 3362 = 2*a(5). In either case, a(n) is deficient. - Muniru A Asiru, Apr 14 2016
The squares of NSW numbers (A008843) interleaved with twice squares from A084703, where A008843(n) = A002315(n)^2 and A084703(n) = A001542(n)^2. Conjecture: Also numbers n such that sigma(n) = A000203(n) and sigma(n-th triangular number) = A074285(n) are both odd numbers. - Jaroslav Krizek, Aug 05 2016
For n > 0, numbers for which the number of odd divisors of both n and of n + 1 is odd. - Gionata Neri, Apr 30 2018
a(n) will be solutions to some (A000217(k) + A000217(k+1))/2. - Art Baker, Jul 16 2019
For n >= 2, a(n) is the base for which A058331(A001109(n)) is a length-3 repunit. Example: for n=2, A001109(2)=6 and A058331(6)=73 and 73 in base a(2)=8 is 111. See Grantham and Graves. - Michel Marcus, Sep 11 2020

			a(1) = ((3 + 2*sqrt(2)) + (3 - 2*sqrt(2)) - 2) / 4 = (3 + 3 - 2) / 4 = 4 / 4 = 1;
a(2) = ((3 + 2*sqrt(2))^2 + (3 - 2*sqrt(2))^2 - 2) / 4 = (9 + 4*sqrt(2) + 8 + 9 - 4*sqrt(2) + 8 - 2) / 4 = (18 + 16 - 2) / 4 = (34 - 2) / 4 = 32 / 4 = 8, etc.

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 193.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 204.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 10.
M. S. Klamkin, "International Mathematical Olympiads 1978-1985," (Supplementary problem N.T.6)
W. Sierpiński, Pythagorean triangles, Dover Publications, Inc., Mineola, NY, 2003, pp. 21-22 MR2002669
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 257-258.

Indranil Ghosh, Table of n, a(n) for n = 0..1304 (terms 0..200 from T. D. Noe)
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
M. A. Asiru, All square chiliagonal numbers, Int J Math Edu Sci Technol, 47:7(2016), 1123-1134.
Elwyn Berlekamp and Joe P. Buhler, Puzzle Column, Emissary, MSRI Newsletter, Fall 2005. Problem 1, (6 MB).
Henk Bruin and Robbert Fokkink, On the records and zeros of a deterministic random walk, arXiv:2503.11734 [math.DS], 2025. See p. 5.
Zongyun Chen, Steven J. Miller, and Chenghan Wu, Geometric Proof of the Irrationality of Square-Roots for Select Integers, arXiv:2410.14434 [math.HO], 2024. See pp. 10-11.
Leonhard Euler, De solutione problematum diophanteorum per numeros integros, Par. 19.
H. G. Forder, A Simple Proof of a Result on Diophantine Approximation, Math. Gaz., 47 (1963), 237-238.
Jon Grantham and Hester Graves, The abc Conjecture Implies That Only Finitely Many Cullen Numbers Are Repunits, arXiv:2009.04052 [math.NT], 2020.
D. B. Hayes, Calculemus!, American Scientist, 96 (Sep-Oct 2008), 362-366.
Olcay Karaatli and Refik Keskin, On some diophantine equations related to square triangular and balancing numbers, J. Alg. Number Theory 4 (2) (2010) 71-89.
Refik Keskin and Olcay Karaatli, Some New Properties of Balancing Numbers and Square Triangular Numbers, Journal of Integer Sequences, Vol. 15 (2012), Article #12.1.4.
P. Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
MSRI newsletter, Emissary, Fall 2005.
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2020.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
B. Polster and M. Ross, Marching in squares, arXiv preprint arXiv:1503.04658 [math.HO], 2015.
K. Ramsey, Generalized Proof re Square Triangular Numbers.
K. Ramsey, Generalized Proof re Square Triangular Numbers, digest of 2 messages in Triangular_and_Fibonacci_Numbers Yahoo group, May 27, 2005 - Oct 10, 2011.
D. L. Vestal, Review of "Pythagorean Triangles" (Chapter 4) by W. Sierpiński.
Eric Weisstein's World of Mathematics, Square Triangular Number.
Eric Weisstein's World of Mathematics, Triangular Number.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).
Index entries for two-way infinite sequences

Cf. A000217, A000290, A001109, A001110, A007913, A000203, A084301, A001652, A072221.
Partial sums of A002315. A000129, A005319.
a(n) = A115598(n), n > 0. - Hermann Stamm-Wilbrandt, Jul 27 2014

a001108 n = a001108_list !! n
a001108_list = 0 : 1 : map (+ 2)
   (zipWith (-) (map (* 6) (tail a001108_list)) a001108_list)
-- Reinhard Zumkeller, Jan 10 2012

m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(1+x)/((1-x)*(1-6*x+x^2)))); // G. C. Greubel, Jul 15 2018

A001108:=-(1+z)/(z-1)/(z**2-6*z+1); # Simon Plouffe in his 1992 dissertation, without the leading 0

Table[(1/2)(-1 + Sqrt[1 + Expand[8(((3 + 2Sqrt[2])^n - (3 - 2Sqrt[2])^n)/(4Sqrt[2]))^2]]), {n, 0, 100}] (* Artur Jasinski, Dec 10 2006 *)
Transpose[NestList[{#[[2]],#[[3]],6#[[3]]-#[[2]]+2}&,{0,1,8},20]][[1]] (* Harvey P. Dale, Sep 04 2011 *)
LinearRecurrence[{7, -7, 1}, {0, 1, 8}, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)

PARI
```
a(n)=(real((3+quadgen(32))^n)-1)/2
```

a(n)=(subst(poltchebi(abs(n)),x,3)-1)/2

a(n)=if(n<0,a(-n),(polsym(1-6*x+x^2,n)[n+1]-2)/4)

x='x+O('x^99); concat(0, Vec(x*(1+x)/((1-x)*(1-6*x+x^2)))) \\ Altug Alkan, May 01 2018

a(0) = 0, a(n+1) = 3*a(n) + 1 + 2*sqrt(2*a(n)*(a(n)+1)). - Jim Nastos, Jun 18 2002
a(n) = floor( (1/4) * (3+2*sqrt(2))^n ). - Benoit Cloitre, Sep 04 2002
a(n) = A001653(k)*A001653(k+n) - A001652(k)*A001652(k+n) - A046090(k)*A046090(k+n). - Charlie Marion, Jul 01 2003
a(n) = A001652(n-1) + A001653(n-1) = A001653(n) - A046090(n) = (A001541(n)-1)/2 = a(-n). - Michael Somos, Mar 03 2004
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3). - Antonio Alberto Olivares, Oct 23 2003
a(n) = Sum_{r=1..n} 2^(r-1)*binomial(2n, 2r). - Lekraj Beedassy, Aug 21 2004
If n > 1, then both A000203(n) and A000203(n+1) are odd numbers: n is either a square or twice a square. - Labos Elemer, Aug 23 2004
a(n) = (T(n, 3)-1)/2 with Chebyshev's polynomials of the first kind evaluated at x=3: T(n, 3) = A001541(n). - Wolfdieter Lang, Oct 18 2004
G.f.: x*(1+x)/((1-x)*(1-6*x+x^2)). Binet form: a(n) = ((3+2*sqrt(2))^n + (3-2*sqrt(2))^n - 2)/4. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 26 2002
a(n) = floor(sqrt(2*A001110(n))) = floor(A001109(n)*sqrt(2)) = 2*(A000129(n)^2) - (n mod 2) = A001333(n)^2 - 1 + (n mod 2). - Henry Bottomley, Apr 19 2000, corrected by Eric Rowland, Jun 23 2017
A072221(n) = 3*a(n) + 1. - David Scheers, Dec 25 2006
A028982(a(n)) + 1 = A028982(a(n) + 1). - Juri-Stepan Gerasimov, Mar 28 2011
a(n+1)^2 + a(n)^2 + 1 = 6*a(n+1)*a(n) + 2*a(n+1) + 2*a(n). - Charlie Marion, Sep 28 2011
a(n) = 2*A001653(m)*A053141(n-m-1) + A002315(m)*A046090(n-m-1) + a(m) with m < n; otherwise, a(n) = 2*A001653(m)*A053141(m-n) - A002315(m)*A001652(m-n) + a(m). See Link to Generalized Proof re Square Triangular Numbers. - Kenneth J Ramsey, Oct 13 2011
a(n) = A048739(2n-2), n > 0. - Richard R. Forberg, Aug 31 2013
From Peter Bala, Jan 28 2014: (Start)
A divisibility sequence: that is, a(n) divides a(n*m) for all n and m. Case P1 = 8, P2 = 12, Q = 1 of the 3-parameter family of linear divisibility sequences found by Williams and Guy.
a(2*n+1) = A002315(n)^2 = Sum_{k = 0..4*n + 1} Pell(n), where Pell(n) = A000129(n).
a(2*n) = (1/2)*A005319(n)^2 = 8*A001109(n)^2.
(2,1) entry of the 2 X 2 matrix T(n,M), where M = [0, -3; 1, 4] and T(n,x) is the Chebyshev polynomial of the first kind. (End)
E.g.f.: exp(x)*(exp(2*x)*cosh(2*sqrt(2)*x) - 1)/2. - Stefano Spezia, Oct 25 2024

More terms from Larry Reeves (larryr(AT)acm.org), Apr 19 2000

More terms from Lekraj Beedassy, Aug 21 2004

A103974 Smaller sides (a) in (a,a,a+1)-integer triangle with integer area.

Original entry on oeis.org

1, 5, 65, 901, 12545, 174725, 2433601, 33895685, 472105985, 6575588101, 91586127425, 1275630195845, 17767236614401, 247465682405765, 3446752317066305, 48007066756522501, 668652182274248705

Offset: 1

Zak Seidov, Feb 23 2005

Corresponding areas are: 0, 12, 1848, 351780, 68149872, 13219419708, 2564481115560 (see A104009).
What is the next term? Is the sequence finite? The possible last two digits of "a" are (it may help in searching for more terms): {01, 05, 09, 15, 19, 25, 29, 33, 35, 39, 45, 49, 51, 55, 59, 65, 69, 75, 79, 83, 85, 89, 95, 99}.
Equivalently, positive integers a such that 3/16*a^4 + 1/4*a^3 - 1/8*a^2 - 1/4*a - 1/16 is a square (A000290), a direct result of Heron's formula. Conjecture: lim_{n->oo} a(n+1)/a(n) = 7 + 4*sqrt(3) (= 7 + A010502). - Rick L. Shepherd, Sep 04 2005
Values x^2 + y^2, where the pair (x, y) solves for x^2 - 3y^2=1, i.e., a(n)= (A001075(n))^2 + (A001353(n))^2 = A055793(n) + A098301(n). - Lekraj Beedassy, Jul 13 2006
Floretion Algebra Multiplication Program, FAMP Code: 1lestes[ 3'i - 2'j + 'k + 3i' - 2j' + k' - 4'ii' - 3'jj' + 4'kk' - 'ij' - 'ji' + 3'jk' + 3'kj' + 4e ]

Christian Aebi, and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Project Euler, Problem 94: Almost Equilateral Triangles.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A011922, A007655, A001353, A102341, A103975, A016064, A011945, A010502 (4*sqrt(3)), A000290 (square numbers), A350916.

A:=rsolve({-A(n+3)+15*A(n+2)-15*A(n+1)+A(n), A(0) = 1, A(1) = 5, A(2)=65}, A(n), makeproc); # Mihailovs

f[n_] := Simplify[((2 + Sqrt[3])^(2n) + (2 - Sqrt[3])^(2n) + 1)/3]; Table[ f[n], {n, 0, 16}] (* Or *)
a[1] = 1; a[2] = 5; a[3] = 65; a[n_] := a[n] = 15a[n - 1] - 15a[n - 2] + a[n - 3]; Table[ a[n], {n, 17}] (* Or *)
CoefficientList[ Series[(1 - 10x + 5x^2)/(1 - 15x + 15x^2 - x^3), {x, 0, 16}], x] (* Or *)
Range[0, 16]! CoefficientList[ Simplify[ Series[(E^x + E^((7 + 4Sqrt[3])x) + E^((7 - 4Sqrt[3])x))/3, {x, 0, 16}]], x] (* Robert G. Wilson v, Mar 24 2005 *)

for(a=1,10^6, b=a; c=a+1; s=(a+b+c)/2; if(issquare(s*(s-a)*(s-b)*(s-c)), print1(a,","))) /* Uses Heron's formula */ \\ Rick L. Shepherd, Sep 04 2005

Composite of comments from Alec Mihailovs (alec(AT)mihailovs.com) and David Terr, Mar 07 2005: (Start)
"a(n)^2 = A011922(n)^2 + (4*A007655(n))^2, so that A011922(n) = 1/2 base of triangles, A007655(n) = 1/4 height of triangles (conjectured by Paul Hanna).
Area is (a+1)/4*sqrt((3*a+1)*(a-1)). If a is even, the numerator is odd and the area is not an integer. That means a=2*k-1. In this case, Area=k*sqrt((3*k-1)*(k-1)).
Solving equation (3*k-1)*(k-1)=y^2, we get k=(2+sqrt(1+3*y^2))/3. That means that 1+3*y^2=x^2 with integer x and y. This is a Pell equation, all solutions of which have the form x=((2+sqrt(3))^n+(2-sqrt(3))^n)/2, y=((2+sqrt(3))^n-(2-sqrt(3))^n)/(2*sqrt(3)). Therefore k=(x+2)/3 is an integer only for even n. Then a=2*k-1=(2*x+1)/3 with even n. Q.E.D.
a(n)=(1/3)*((2+sqrt(3))^(2*n-2)+(2-sqrt(3))^(2*n-2)+1).
Recurrence: a(n+3)=15*a(n+2)-15*a(n+1)+a(n), a(0)=1, a(1)=5, a(2)=65.
G.f.: x*(1-10*x+5*x^2)/(1-15*x+15*x^2-x^3).
E.g.f.: 1/3*(exp(x)+exp((7+4*sqrt(3))*x)+exp((7-4*sqrt(3))*x)).
a(n) = 4U(n)^2 + 1, where U(1) = 0, U(2)=1 and U(n+1) = 4U(n) - U(n-1) for n>1. (U(n), V(n)) is the n-th solution to Pell's equation 3U(n)^2 + 1 = V(n)^2. (U(n) is the sequence A001353.)" (End)
a(n+1) = A098301(n+1) + A055793(n+2) - Creighton Dement, Apr 18 2005
a(n) = floor((7+4*sqrt(3))*a(n-1))-4, n>=3. - Rick L. Shepherd, Sep 04 2005
a(n)= [1+14*A007655(n+2)-194*A007655(n+1)]/3. - R. J. Mathar, Nov 16 2007
For n>=3, a(n) = 14*a(n-1) - a(n-2) - 4. It is one of 10 second-order linear recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916. - Max Alekseyev, Jan 22 2022

More terms from Creighton Dement, Apr 18 2005

Edited by Max Alekseyev, Jan 22 2022

A178789 a(n) = 4^(n-1) + 2: Number of acute angles after n iterations of the Koch snowflake construction.

Original entry on oeis.org

3, 6, 18, 66, 258, 1026, 4098, 16386, 65538, 262146, 1048578, 4194306, 16777218, 67108866, 268435458, 1073741826, 4294967298, 17179869186, 68719476738, 274877906946, 1099511627778, 4398046511106, 17592186044418, 70368744177666

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jun 14 2010

Starting from an equilateral triangle, at each step each straight segment is replaced by a "/\" shape of four segments of equal length, with the acute angle in the middle pointing to the exterior. The sequence counts the angles which are (i.e., already were) at both extremities, plus the one newly created acute angle in the middle of each former segment. At step n, there are 3*4^(n-1) straight segments, therefore a(n+1) = a(n) + 3*4^(n-1). - M. F. Hasler, Dec 17 2013

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Mario Raso, Integer Sequences in Cryptography: A New Generalized Family and its Application, Ph. D. Thesis, Sapienza University of Rome (Italy 2025). See p. 113.
Larry Riddle, Koch Curve.
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A002001, A000302, A010502, A047849, A164346.

[2^(2*(n-1)) + 2: n in [1..30]]; // Vincenzo Librandi, Feb 02 2013

A178789:=n->2+4^(n-1); seq(A178789(n), n=1..30); # Wesley Ivan Hurt, Dec 17 2013

a=b=3;lst={a};Do[a=a+b;b*=4;AppendTo[lst,a],{n,40}];lst
Flatten[Table[2^(2*(n-1)) + 2, {n, 1, 50}]](* or *)   CoefficientList[Series[(3 - 9*x)/(1 - 5*x + 4*x^2),{x, 0, 100}], x] (* Vincenzo Librandi, Feb 02 2013 *)

A178789=n->4^(n-1)+2  \\ - M. F. Hasler, Dec 17 2013

G.f.: 3*x*(1 - 3*x)/(1 - 5*x + 4*x^2).
a(n) = 3 * A047849(n-1).
a(n) = 2^(2*(n-1)) + 2. - Vincenzo Librandi, Feb 02 2013
a(n+1) = a(n) + 3*4^(n-1) = a(n) + A002001(n) for n > 0. - M. F. Hasler, Dec 17 2013
a(n) = 2 + A000302(n-1). - Omar E. Pol, Dec 18 2013

A040041 Continued fraction for sqrt(48).

Original entry on oeis.org

6, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1, 12, 1

Offset: 0

N. J. A. Sloane

			6.9282032302755091741097853... = 6 + 1/(1 + 1/(12 + 1/(1 + 1/(12 + ...)))). - _Harry J. Smith_, Jun 06 2009

James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 276.

Harry J. Smith, Table of n, a(n) for n = 0..20000
G. Xiao, Contfrac.
Index entries for continued fractions for constants.
Index entries for linear recurrences with constant coefficients, signature (0,1).

Cf. A010502 (decimal expansion).

Digits := 100: convert(evalf(sqrt(N)),confrac,90,'cvgts'):

ContinuedFraction[Sqrt[48],300] (* Vladimir Joseph Stephan Orlovsky, Mar 07 2011 *)

{ allocatemem(932245000); default(realprecision, 24000); x=contfrac(sqrt(48)); for (n=0, 20000, write("b040041.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 06 2009

From Amiram Eldar, Nov 12 2023: (Start)
Multiplicative with a(2^e) = 12, and a(p^e) = 1 for an odd prime p.
Dirichlet g.f.: zeta(s) * (1 + 11/2^s). (End)
G.f.: (6 + x + 6*x^2)/(1 - x^2). - Stefano Spezia, Jul 27 2025

A041083 Denominators of continued fraction convergents to sqrt(48).

Original entry on oeis.org

1, 1, 13, 14, 181, 195, 2521, 2716, 35113, 37829, 489061, 526890, 6811741, 7338631, 94875313, 102213944, 1321442641, 1423656585, 18405321661, 19828978246, 256353060613, 276182038859, 3570537526921, 3846719565780, 49731172316281, 53577891882061

Offset: 0

N. J. A. Sloane

The following remarks assume an offset of 1. This is the sequence of Lehmer numbers U_n(sqrt(R),Q) for the parameters R = 12 and Q = -1; it is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for all positive integers n and m. Consequently, this is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, May 28 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Eric W. Weisstein, MathWorld: Lehmer Number
Index entries for linear recurrences with constant coefficients, signature (0,14,0,-1).

Cf. A010502, A041082, A002530.

Table[Denominator[FromContinuedFraction[ContinuedFraction[Sqrt[48], n]]], {n, 1, 50}] (* Vladimir Joseph Stephan Orlovsky, Jun 23 2011 *)
Denominator[Convergents[Sqrt[48], 40]] (* Vincenzo Librandi, Oct 24 2013 *)
LinearRecurrence[{0,14,0,-1},{1,1,13,14},30] (* Harvey P. Dale, Mar 15 2015 *)

From Colin Barker, Jul 15 2012: (Start)
a(n) = 14*a(n-2) - a(n-4).
G.f.: (1+x-x^2)/((1-4*x+x^2)*(1+4*x+x^2)). (End)
From Peter Bala, May 28 2014: (Start)
The following remarks assume an offset of 1.
Let alpha = sqrt(3) + 2 and beta = sqrt(3) - 2 be the roots of the equation x^2 - sqrt(12)*x - 1 = 0. Then a(n) = (alpha^n - beta^n)/(alpha - beta) for n odd, while a(n) = (alpha^n - beta^n)/(alpha^2 - beta^2) for n even.
a(n) = Product_{k = 1..floor((n-1)/2)} ( 12 + 4*cos^2(k*Pi/n) ).
Recurrence equations: a(0) = 0, a(1) = 1 and for n >= 1, a(2*n) = a(2*n - 1) + a(2*n - 2) and a(2*n + 1) = 12*a(2*n) + a(2*n - 1). (End)

A354129 Decimal expansion of 7 + 4*sqrt(3).

Original entry on oeis.org

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

Offset: 2

Stefano Spezia, May 18 2022

The largest root of x^2 - 14*x + 1 = 0.

Apart from leading digits the same as A010502. - R. J. Mathar, May 24 2022

			13.92820323027550917410978536602...

Nicholas Owad and Anastasiia Tsvietkova, Random meander model for links, arXiv:2205.03451 [math.GT], 2022, p. 13.

Cf. A002194, A010502, A019973 (square root), A354128 (multiplicative inverse).

Mathematica
```
First[RealDigits[N[7+4Sqrt[3],92]]]
```

Equals (1 + sqrt(3))^4 / 4. - Vaclav Kotesovec, May 18 2022
Equals (2 + sqrt(3))^2 = A019973^2. - Jianing Song, May 27 2022
Equals exp(arccosh(7)). - Amiram Eldar, Jul 06 2023

A386740 Decimal expansion of the volume of an augmented sphenocorona with unit edges.

Original entry on oeis.org

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

Offset: 1

Paolo Xausa, Aug 01 2025

The augmented sphenocorona is Johnson solid J_87.

			1.7510539003719224011954274331734292512511481527...

Paolo Xausa, Table of n, a(n) for n = 1..10000
Wikipedia, Augmented sphenocorona.

Cf. A010502 (surface area - 1).

Cf. A010507, A020775, A115754, A386739, A386741.

First[RealDigits[Sqrt[1 + 3*Sqrt[3/2] + Sqrt[13 + Sqrt[54]]]/2 + 1/Sqrt[18], 10, 100]] (* or *)
First[RealDigits[PolyhedronData["J87", "Volume"], 10, 100]]

Equals sqrt(1 + 3*sqrt(3/2) + sqrt(13 + 3*sqrt(6)))/2 + 1/(3*sqrt(2)) = sqrt(1 + 3*A115754  + sqrt(13 + A010507))/2 + A020775.
Equals A386739 + A020775.
Equals the largest real root of 45137758519296*x^16 - 110336743047168*x^14 - 191069246324736*x^12 + 209269081571328*x^10 + 364547659290624*x^8 - 58793017190400*x^6 + 3306865979520*x^4 - 1275399855936*x^2 + 1439671249.

A041082 Numerators of continued fraction convergents to sqrt(48).

Original entry on oeis.org

6, 7, 90, 97, 1254, 1351, 17466, 18817, 243270, 262087, 3388314, 3650401, 47193126, 50843527, 657315450, 708158977, 9155223174, 9863382151, 127515808986, 137379191137, 1776066102630, 1913445293767

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (0,14,0,-1).

Cf. A010502, A041083.

Numerator[Convergents[Sqrt[48], 30]] (* Vincenzo Librandi, Oct 25 2013 *)
LinearRecurrence[{0,14,0,-1},{6,7,90,97},30] (* Harvey P. Dale, Aug 30 2020 *)

a(n) = 14*a(n-2)-a(n-4). G.f.: (6+7*x+6*x^2-x^3)/((1-4*x+x^2)*(1+4*x+x^2)). [Colin Barker, Jul 15 2012]

cp's OEIS Frontend

A248274 Egyptian fraction representation of sqrt(48) (A010502) using a greedy function.

Views

Author

Keywords

Crossrefs

Programs

Mathematica

A010527 Decimal expansion of sqrt(3)/2.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Formula

Extensions

A001108 a(n)-th triangular number is a square: a(n+1) = 6*a(n) - a(n-1) + 2, with a(0) = 0, a(1) = 1.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

PARI

PARI

PARI

Formula

Extensions

A103974 Smaller sides (a) in (a,a,a+1)-integer triangle with integer area.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A178789 a(n) = 4^(n-1) + 2: Number of acute angles after n iterations of the Koch snowflake construction.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A040041 Continued fraction for sqrt(48).

Views

Author

Keywords

Examples

References

Links

Crossrefs

Programs