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A001570 Numbers n such that n^2 is centered hexagonal.
(Formerly M4915 N2108)
42
1, 13, 181, 2521, 35113, 489061, 6811741, 94875313, 1321442641, 18405321661, 256353060613, 3570537526921, 49731172316281, 692665874901013, 9647591076297901, 134373609193269601, 1871582937629476513, 26067787517619401581, 363077442309042145621 (list; graph; refs; listen; history; text; internal format)
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 in A108299; see also A028230 for L(n,-14). - Reinhard Zumkeller, Jun 01 2005

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

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 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

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

REFERENCES

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

T. D. Noe, Table of n, a(n) for n=1..101

Tanya Khovanova, Recursive Sequences

J.-C. Novelli, 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.

Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 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 sequences related to Chebyshev polynomials.

Index entries for two-way infinite sequences

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. 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

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) (compare A122571).

a(n) = 12*A076139(n) + 1 = 4*A076140(n) + 1. - Alexander Adamchuk, Apr 06 2007

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

a(n) = A000217(A001571(n-1)) + A000217(A133161(n)), n>=1. - Ivan N. Ianakiev, Sep 24 2013

EXAMPLE

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 Plouffe in his 1992 dissertation.

MATHEMATICA

NestList[3 + 7*#1 + 4*Sqrt[1 + 3*#1 + 3*#1^2] &, 0, 24] (* Zak Seidov, May 06 2007 *)

q=6; s=0; lst={}; Do[s+=n; If[Sqrt[q*s+1]==Floor[Sqrt[q*s+1]], AppendTo[lst, Sqrt[q*s+1]]], {n, 0, 9!}]; lst (* Vladimir Joseph Stephan Orlovsky, Apr 02 2009 *)

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 of A041017 *)

a[12, 20] (* Gerry Martens, Jun 07 2015 *)

PROG

(PARI) {a(n) = real( (2 + quadgen( 12)) ^ (2*n - 1)) / 2} /* Michael Somos, Feb 15 2011 */

(PARI) {a(n) = n = abs( 2*n - 1); round( 2 ^ (n-2) * prod( k=1, n, 2 - sin(2 * Pi * k / n)))} /* Michael Somos, Feb 15 2011 */

CROSSREFS

Bisection of A003500/4. Cf. A006051, A001922.

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

Cf. A077417 with companion A077416.

Row 14 of array A094954.

Cf. A076139, A076140, A102871.

A122571 is another version of the same sequence.

Row 2 of array A188646.

Cf. similar sequences listed in A238379.

Sequence in context: A083576 A189432 A122571 * A239902 A020544 A009015

Adjacent sequences:  A001567 A001568 A001569 * A001571 A001572 A001573

KEYWORD

nonn,easy,nice

AUTHOR

N. J. A. Sloane.

STATUS

approved

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Last modified July 24 18:02 EDT 2017. Contains 289776 sequences.