1,2

Chebyshev T-sequence with diophantine property.

a(n) = L(n,14), where L is defined as in A108299; see also A028230 for L(n,-14). - Reinhard Zumkeller (reinhard.zumkeller(AT)lhsystems.com), 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 (blekraj(AT)yahoo.com), 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 (alex(AT)kolmogorov.com), Apr 06 2007

V. Thebault, Consecutive cubes with difference a square, Amer. Math. Monthly, 56 (1949), 174-175.

T. D. Noe, Table of n, a(n) for n=0..100

Tanya Khovanova, Recursive Sequences

S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures}, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

Sociedad Magic Penny Patagonia, Leonardo en Patagonia

Eric Weisstein's World of Mathematics, Hex Number

Index entries for sequences related to Chebyshev polynomials.

a(n) = (1/4)*((2+sqrt(3))^(2*n+1)+(2-sqrt(3))^(2*n+1)).

G.f.: (1-x)/(1-14*x+x^2).

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).

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). - R. Stephan, May 29 2004

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

a(n) = 12*A076139(n) + 1 = 4*A076140(n) + 1. - Alexander Adamchuk (alex(AT)kolmogorov.com), 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 (zakseidov(AT)yahoo.com), 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

A001570:=-(-1+z)/(1-14*z+z**2); [S. Plouffe in his 1992 dissertation.]

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

(PARI) a(n)=real((2+quadgen(12))^(2*n+1))/2 /* Michael Somos Apr 30 2005 */

(PARI) a(n)= n=abs(1+2*n); round(2^(n-2)*prod(k=1, n, 2-sin(2*Pi*k/n)))

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

One half of odd part of bisection of A001075.

Cf. A077417 with companion A077416.

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

Row 14 of array A094954.

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

Cf. A076139, A076140, A102871.

A122571 is another version of the same sequence.

Adjacent sequences: A001567 A001568 A001569 this_sequence A001571 A001572 A001573

Sequence in context: A127390 A083576 A122571 this_sequence A020544 A009015 A067385

nonn,easy,nice

njas

Chebyshev comments from Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Nov 29 2002