0,3

a(n) = A001045(n)+(-1)^n = A000079(n)-2*A001045(n). - Paul Barry (pbarry(AT)wit.ie), Feb 20 2003

Conjecture: a(n) = the number of fractions in the infinite Farey row of 2^n terms with even denominators. Compare the Salamin & Gosper item in the Beeler et al. link. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Oct 27 2003

Counts closed walks starting and ending at the same vertex of a triangle. 3a(n)=P(C_n,3) chromatic polynomial for 3 colors on cyclic graph C_n. A078008(n)+2*A001045(n)=2^n provides decomposition of Pascal's triangle. - Paul Barry (pbarry(AT)wit.ie), Nov 17 2003

Permutations with one fixed point avoiding 123 and 132.

a(n) = A014113(n-1) for n>0; a(n) = A052953(n-1) - 2*(n mod 2) = sum of n-th row of the triangle in A108561. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Jun 10 2005

General form: iterate k=2^n-k. See also A001045. [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Dec 11 2008]

Contribution from Gary W. Adamson (qntmpkt(AT)yahoo.com), May 25 2009: (Start)

Signed (A151575) [1, 0, 2, -2, 6, -10, 22, -42, 86,...] = an operator for toothpick

sequences. The signed sequence (A151575) convolved with A151548 = toothpick sequence

A139250. The signed sequence (A151575) convolved with A151555 = toothpick sequence A153006. (End)

Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.

Leonhard Euler, Introductio in analysin infinitorum (1748), section 216.

I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, John Wiley and Sons, N.Y., 1983, ex. 1.1.10a.

Beeler, M., Gosper, R. W., & Schroeppel, R. C., R. HAKMEM. MIT AI Memo 239, Feb. 29, 1972. (Item #54 by Salamin & Gosper)

T. Mansour and A. Robertson, Refined restricted permutations....

Index entries for sequences related to linear recurrences with constant coefficients

Index entries for sequences related to Chebyshev polynomials.

Euler expands(1-x)/(1-x-2*x^2) into an infinite series and finds that the coefficient of the n-th term is (2^n + (-1)^n 2)/3. Section 226 shows that Euler could have easily found the recursion relation: a(n) = a(n-1) + 2a(n-2) with a(0) = 1 and a(1) = 0. - V. Frederick Rickey (fred-rickey(AT)usma.edu), Feb 10 2006. [Typos corrected by Jaume Oliver i Lafont, Jun 01 2009]

a(n)=sum_{k=0..floor(n, 3)} binomial(n, f(n)+3k) where f(n)=(0, 2, 1, 0, 2, 1, ...)=A080424(n). - Paul Barry (pbarry(AT)wit.ie), Feb 20 2003

E.g.f. (exp(2x)+2exp(-x))/3. - Paul Barry (pbarry(AT)wit.ie), Apr 20 2003

a(n)=(1/3)(2^n+2(-1)^n) - Mario Catalani (mario.catalani(AT)unito.it), Aug 29 2003

a(n)=T(n, i/(2sqrt(2)))(-i*sqrt(2)^n-U(n-1, i/(2sqrt(2)))(-i*sqrt(2))^(n-1)/2 - Paul Barry (pbarry(AT)wit.ie), Nov 17 2003

a(0)=1, a(n)=2a(n-1)+2(-1)^n, n>0; a(n)=sum{k=0..n, (-1)^k(2^(n-k-1)+0^(n-k)/2)}. - Paul Barry (pbarry(AT)wit.ie), Jul 30 2004

A137208(n+1)-2*A137208(n)=a(n) signed. - Paul Curtz (bpcrtz(AT)free.fr), Aug 03 2008

a(n) = A001045(n+1)-A001045(n) - Paul Curtz (bpcrtz(AT)free.fr), Feb 09 2009

k=0; lst={1, k}; Do[k=2^n-k; AppendTo[lst, k], {n, 1, 5!}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Dec 11 2008]

Cf. A001045 [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Dec 11 2008]

See A151575 for a signed version.

Bisections: A047849, A020988. [From R. J. Mathar (mathar(AT)strw.leidenunvi.nl), Feb 25 2009]

Cf. A151548, A139250, A151555, A153006 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), May 25 2009]

Sequence in context: A167399 A019310 A014113 this_sequence A151575 A076907 A153897

Adjacent sequences: A078005 A078006 A078007 this_sequence A078009 A078010 A078011

nonn

N. J. A. Sloane (njas(AT)research.att.com), Nov 17 2002