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Numbers k such that k^2 is centered hexagonal.
(Formerly M4915 N2108)
48

%I M4915 N2108 #147 Sep 18 2024 12:46:12

%S 1,13,181,2521,35113,489061,6811741,94875313,1321442641,18405321661,

%T 256353060613,3570537526921,49731172316281,692665874901013,

%U 9647591076297901,134373609193269601,1871582937629476513,26067787517619401581,363077442309042145621

%N Numbers k such that k^2 is centered hexagonal.

%C Chebyshev T-sequence with Diophantine property. - _Wolfdieter Lang_, Nov 29 2002

%C a(n) = L(n,14), where L is defined as in A108299; see also A028230 for L(n,-14). - _Reinhard Zumkeller_, Jun 01 2005

%C Numbers x satisfying x^2 + y^3 = (y+1)^3. Corresponding y given by A001921(n)={A028230(n)-1}/2. - _Lekraj Beedassy_, Jul 21 2006

%C 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

%C Also numbers n such that RootMeanSquare(1,3,...,2*n-1) is an integer. - _Ctibor O. Zizka_, Sep 04 2008

%C a(n), with n>1, is the length of the cevian of equilateral triangle whose side length is the term b(n) of the sequence A028230. This cevian divides the side (2*x+1) of the triangle in two integer segments x and x+1. - _Giacomo Fecondo_, Oct 09 2010

%C 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

%C 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

%C Numbers n such that there exists positive x with x^2 + x + 1 = 3n^2. - _Jeffrey Shallit_, Dec 11 2017

%C Given by the denominators of the continued fractions [1,(1,2)^i,3,(1,2)^{i-1},1]. - _Jeffrey Shallit_, Dec 11 2017

%C 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

%D E.-A. Majol, Note #2228, L'Intermédiaire des Mathématiciens, 9 (1902), pp. 183-185. - _N. J. A. Sloane_, Mar 03 2022

%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H G. C. Greubel, <a href="/A001570/b001570.txt">Table of n, a(n) for n = 1..870</a> (terms 1..101 from T. D. Noe)

%H Alex Fink, Richard K. Guy, and Mark Krusemeyer, <a href="https://doi.org/10.11575/cdm.v3i2.61940">Partitions with parts occurring at most thrice</a>, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.

%H G. Julia, <a href="https://gjmaths.pagesperso-orange.fr/contenu/trcvenon.pdf">Triangles dont un angle mesure 120 degrés</a>, Problème Capes, part C (in French).

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H J.-C. Novelli and J.-Y. Thibon, <a href="http://arxiv.org/abs/1403.5962">Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions</a>, arXiv preprint arXiv:1403.5962 [math.CO], 2014.

%H Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.

%H Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992.

%H Sociedad Magic Penny Patagonia, <a href="http://www.magicpenny.org/engteorema.htm">Leonardo en Patagonia</a>

%H V. Thebault, <a href="http://www.jstor.org/stable/2305033">Consecutive cubes with difference a square</a>, Amer. Math. Monthly, 56 (1949), 174-175.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HexNumber.html">Hex Number</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Beal%27s_conjecture">Beal's conjecture</a>

%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>

%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (14,-1).

%F a(n) = ((2 + sqrt(3))^(2*n - 1) + (2 - sqrt(3))^(2*n - 1)) / 4. - _Michael Somos_, Feb 15 2011

%F G.f.: x * (1 - x) / (1 -14*x + x^2). - _Michael Somos_, Feb 15 2011

%F 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

%F 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. See A049310 and A053120. S(-1, x)=0, S(n, 14)=A007655(n+1) and T(n, 2)=A001075(n). - _Wolfdieter Lang_, Nov 29 2002

%F a(n) = A001075(n)*A001075(n+1) - 1 and thus (a(n)+1)^6 has divisors A001075(n)^6 and A001075(n+1)^6 congruent to -1 modulo a(n) (cf. A350916). - _Max Alekseyev_, Jan 23 2022

%F 4*a(n)^2 - 3*b(n)^2 = 1 with b(n)=A028230(n+1), n>=0.

%F a(n)*a(n+3) = 168 + a(n+1)*a(n+2). - _Ralf Stephan_, May 29 2004

%F a(n) = 14*a(n-1) - a(n-2), a(0) = a(1) = 1. a(1 - n) = a(n) (compare A122571).

%F a(n) = 12*A076139(n) + 1 = 4*A076140(n) + 1. - _Alexander Adamchuk_, Apr 06 2007

%F 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

%F 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

%F a(n) = sqrt((3*A028230(n+1)^2 + 1)/4).

%F a(n) = A098301(n+1) - A001353(n)*A001835(n).

%F a(n) = A000217(A001571(n-1)) + A000217(A133161(n)), n>=1. - _Ivan N. Ianakiev_, Sep 24 2013

%F a(n)^2 = A001922(n-1)^3 - A001921(n-1)^3, for n >= 1. - _Bernard Schott_, Nov 20 2022

%F 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

%F a(n) = A003154(A101265(n)). - _Andrea Pinos_, Dec 19 2022

%e G.f. = x + 13*x^2 + 181*x^3 + 2521*x^4 + 35113*x^5 + 489061*x^6 + 6811741*x^7 + ...

%p A001570:=-(-1+z)/(1-14*z+z**2); # _Simon Plouffe_ in his 1992 dissertation.

%t NestList[3 + 7*#1 + 4*Sqrt[1 + 3*#1 + 3*#1^2] &, 0, 24] (* _Zak Seidov_, May 06 2007 *)

%t 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 *)

%t a[c_, n_] := Module[{},

%t p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];

%t d := Denominator[Convergents[Sqrt[c], n p]];

%t t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];

%t Return[t];

%t ] (* Complement of A041017 *)

%t a[12, 20] (* _Gerry Martens_, Jun 07 2015 *)

%t LinearRecurrence[{14, -1}, {1, 13}, 19] (* _Jean-François Alcover_, Sep 26 2017 *)

%t CoefficientList[Series[x (1-x)/(1-14x+x^2),{x,0,20}],x] (* _Harvey P. Dale_, Sep 18 2024 *)

%o (PARI) {a(n) = real( (2 + quadgen( 12)) ^ (2*n - 1)) / 2}; /* _Michael Somos_, Feb 15 2011 */

%o (Magma) [((2 + Sqrt(3))^(2*n - 1) + (2 - Sqrt(3))^(2*n - 1))/4: n in [1..50]]; // _G. C. Greubel_, Nov 04 2017

%Y Bisection of A003500/4. Cf. A006051, A001921, A001922.

%Y One half of odd part of bisection of A001075. First differences of A007655.

%Y Cf. A077417 with companion A077416.

%Y Row 14 of array A094954.

%Y Cf. A076139, A076140, A102871.

%Y A122571 is another version of the same sequence.

%Y Row 2 of array A188646.

%Y Cf. similar sequences listed in A238379.

%Y Cf. A028231, which gives the corresponding values of x in 3n^2 = x^2 + x + 1.

%Y Similar sequences of the type cosh((2*m+1)*arccosh(k))/k are listed in A302329. This is the case k=2.

%K nonn,easy,nice

%O 1,2

%A _N. J. A. Sloane_