OFFSET
1,2
COMMENTS
Chebyshev T-sequence with Diophantine property. -Wolfdieter Lang,Nov 29 2002
a(n) = L(n,14), where L is defined as inA108299;see alsoA028230for L(n,-14). -Reinhard Zumkeller,Jun 01 2005
Numbers x satisfying x^2 + y^3 = (y+1)^3. Corresponding y given byA001921(n)={A028230(n)-1}/2. -Lekraj Beedassy,Jul 21 2006
Mod[ a(n), 12 ] = 1. (a(n) - 1)/12 =A076139(n) = Triangular numbers that are one-third of another triangular number. (a(n) - 1)/4 =A076140(n) = Triangular numbers T(k) that are three times another triangular number. -Alexander Adamchuk,Apr 06 2007
Also numbers n such that RootMeanSquare(1,3,...,2*n-1) is an integer. -Ctibor O. Zizka,Sep 04 2008
a(n), with n>1, is the length of the cevian of equilateral triangle whose side length is the term b(n) of the sequenceA028230.This cevian divides the side (2*x+1) of the triangle in two integer segments x and x+1. -Giacomo Fecondo,Oct 09 2010
For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(12)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. -John M. Campbell,Jul 08 2011
Beal's conjecture would imply that set intersection of this sequence with the perfect powers (A001597) equals {1}. In other words, existence of a nontrivial perfect power in this sequence would disprove Beal's conjecture. -Max Alekseyev,Mar 15 2015
Numbers n such that there exists positive x with x^2 + x + 1 = 3n^2. -Jeffrey Shallit,Dec 11 2017
Given by the denominators of the continued fractions [1,(1,2)^i,3,(1,2)^{i-1},1]. -Jeffrey Shallit,Dec 11 2017
A near-isosceles integer-sided triangle with an angle of 2*Pi/3 is a triangle whose sides (a, a+1, c) satisfy Diophantine equation (a+1)^3 - a^3 = c^2. For n >= 2, the largest side c is given by a(n) while smallest and middle sides (a, a+1) = (A001921(n-1),A001922(n-1)) (see Julia link). -Bernard Schott,Nov 20 2022
REFERENCES
E.-A. Majol, Note #2228, L'Intermédiaire des Mathématiciens, 9 (1902), pp. 183-185. -N. J. A. Sloane,Mar 03 2022
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).
LINKS
G. C. Greubel,Table of n, a(n) for n = 1..870(terms 1..101 from T. D. Noe)
Alex Fink, Richard K. Guy, and Mark Krusemeyer,Partitions with parts occurring at most thrice,Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
G. Julia,Triangles dont un angle mesure 120 degrés,Problème Capes, part C (in French).
Tanya Khovanova,Recursive Sequences
J.-C. Novelli and J.-Y. Thibon,Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions,arXiv preprint arXiv:1403.5962 [math.CO], 2014.
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.
Sociedad Magic Penny Patagonia,Leonardo en Patagonia
V. Thebault,Consecutive cubes with difference a square,Amer. Math. Monthly, 56 (1949), 174-175.
Eric Weisstein's World of Mathematics,Hex Number
Wikipedia,Beal's conjecture
Index entries for linear recurrences with constant coefficients,signature (14,-1).
FORMULA
a(n) = ((2 + sqrt(3))^(2*n - 1) + (2 - sqrt(3))^(2*n - 1)) / 4. -Michael Somos,Feb 15 2011
G.f.: x * (1 - x) / (1 -14*x + x^2). -Michael Somos,Feb 15 2011
Let q(n, x) = Sum_{i=0, n} x^(n-i)*binomial(2*n-i, i) then a(n) = q(n, 12). -Benoit Cloitre,Dec 10, 2002
a(n) = S(n, 14) - S(n-1, 14) = T(2*n+1, 2)/2 with S(n, x):= U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. SeeA049310andA053120.S(-1, x)=0, S(n, 14)=A007655(n+1) and T(n, 2)=A001075(n). -Wolfdieter Lang,Nov 29 2002
a(n) =A001075(n)*A001075(n+1) - 1 and thus (a(n)+1)^6 has divisorsA001075(n)^6 andA001075(n+1)^6 congruent to -1 modulo a(n) (cf.A350916). -Max Alekseyev,Jan 23 2022
4*a(n)^2 - 3*b(n)^2 = 1 with b(n)=A028230(n+1), n>=0.
a(n)*a(n+3) = 168 + a(n+1)*a(n+2). -Ralf Stephan,May 29 2004
a(n) = 14*a(n-1) - a(n-2), a(0) = a(1) = 1. a(1 - n) = a(n) (compareA122571).
a(n) = (1/12)*((7-4*sqrt(3))^n*(3-2*sqrt(3))+(3+2*sqrt(3))*(7+4*sqrt(3))^n -6). -Zak Seidov,May 06 2007
a(n) =A102871(n)^2+(A102871(n)-1)^2; sum of consecutive squares. E.g. a(4)=36^2+35^2. - Mason Withers (mwithers(AT)semprautilities.com), Jan 26 2008
a(n) = sqrt((3*A028230(n+1)^2 + 1)/4).
a(n) = 2^(2*n-3)*Product_{k=1..2*n-1} (2 - sin(2*Pi*k/(2*n-1))).Michael Somos,Dec 18 2022
EXAMPLE
G.f. = x + 13*x^2 + 181*x^3 + 2521*x^4 + 35113*x^5 + 489061*x^6 + 6811741*x^7 +...
MAPLE
A001570:=-(-1+z)/(1-14*z+z**2); #Simon Plouffein his 1992 dissertation.
MATHEMATICA
NestList[3 + 7*#1 + 4*Sqrt[1 + 3*#1 + 3*#1^2] &, 0, 24] (*Zak Seidov,May 06 2007 *)
f[n_]:= Simplify[(2 + Sqrt@3)^(2 n - 1) + (2 - Sqrt@3)^(2 n - 1)]/4; Array[f, 19] (*Robert G. Wilson v,Oct 28 2010 *)
a[c_, n_]:= Module[{},
p:= Length[ContinuedFraction[ Sqrt[ c]][[2]]];
d:= Denominator[Convergents[Sqrt[c], n p]];
t:= Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
Return[t];
] (* Complement ofA041017*)
a[12, 20] (*Gerry Martens,Jun 07 2015 *)
LinearRecurrence[{14, -1}, {1, 13}, 19] (*Jean-François Alcover,Sep 26 2017 *)
CoefficientList[Series[x (1-x)/(1-14x+x^2), {x, 0, 20}], x] (*Harvey P. Dale,Sep 18 2024 *)
PROG
(PARI) {a(n) = real( (2 + quadgen( 12)) ^ (2*n - 1)) / 2}; /*Michael Somos,Feb 15 2011 */
(Magma) [((2 + Sqrt(3))^(2*n - 1) + (2 - Sqrt(3))^(2*n - 1))/4: n in [1..50]]; //G. C. Greubel,Nov 04 2017
CROSSREFS
Row 14 of arrayA094954.
A122571is another version of the same sequence.
Row 2 of arrayA188646.
Cf. similar sequences listed inA238379.
Cf.A028231,which gives the corresponding values of x in 3n^2 = x^2 + x + 1.
Similar sequences of the type cosh((2*m+1)*arccosh(k))/k are listed inA302329.This is the case k=2.
KEYWORD
nonn,easy,nice
AUTHOR
STATUS
approved