0,4

Number of ways to tile a 3 X (n-1) rectangle with 1 X 1 and 2 X 2 square tiles.

Also, number of ways to tile a 2 X (n-1) rectangle with 1 X 2 dominoes and 2 X 2 squares. - Toby Gottfried (toby(AT)gottfriedville.net), Nov 02, 2008.

Also a(n) counts each of the following four things: n-ary quasigroups of order 3 with automorphism group of order 3, n-ary quasigroups of order 3 with automorphism group of order 6, (n-1)-ary quasigroups of order 3 with automorphism group of order 2 and (n-2)-ary quasigroups of order 3. See the McKay-Wanless (2008) paper. - Ian Wanless (ian.wanless(AT)sci.monash.edu.au), Apr 28 2008

Also the number of ways to tie a necktie using n+2 turns. So three turns make an "oriental", four make a "four in hand" and for 5 turns there are 3 methods: "Kelvin", "Nicky" and "Pratt". The formula also arises from a special random walk on a triangular grid with side conditions (see Fink and Mao, 1999). - arne.ring(AT)epost.de, Mar 18 2001

Also the number of compositions of n+1 ending with an odd part (a(2)=3 because 3, 21, 111 are the only compositions of 3 ending with an odd part). Also the number of compositions of n+2 ending with an even part (a(2)=3 because 4, 22, 112 are the only compositions of 4 ending with an even part). - Emeric Deutsch (deutsch(AT)duke.poly.edu), May 08 2001

Arises in study of sorting by merge insertions and in analysis of a method for computing GCDs - see Knuth reference.

Roberto E. Martinez II (remartin(AT)fas.harvard.edu), Jan 07 2002: Number of perfect matchings of a 2 X n grid upon replacing unit squares with tetrahedra (C_4 to K_4):

o----o----o----o...

| \/ | \/ | \/ |

| /\ | /\ | /\ |

o----o----o----o...

Also the numerators of the reduced fractions in the alternating sum 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 + ... - Joshua Zucker (joshua.zucker(AT)stanfordalumni.org), Feb 07 2002

Also, if A(n),B(n),C(n) are the angles of the n-orthic triangle of ABC then A(1) = Pi - 2A, A(n) = s(n)Pi + (-2)^nA where s(n) = (-1)^(n-1) * a(n) [1-orthic triangle = the orthic triangle of ABC, n-orthic triangle = the orthic triangle of the (n-1)-orthic triangle] - Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr), Jun 05 2002

Also the number of words of length n+1 in the two letters s and t that reduce to the identity 1 by using the relations sss=1, tt=1 and stst=1. The generators s and t and the three stated relations generate the group S3. - John W. Layman (layman(AT)math.vt.edu), Jun 14 2002

Sums of pair of consecutive terms give all powers of 2 in increasing order. - Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Aug 15 2002

Excess clockwise moves (over anti-clockwise) needed to move a tower of size n to the clockwise peg is -(-1)^n(2^n - (-1)^n)/3; a(n)=its unsigned version. - Wouter Meeussen (wouter.meeussen(AT)pandora.be), Sep 01 2002

Also the absolute value of the number represented in base -2 by the string of n 1's, the negabinary repunit. The Mersenne numbers (A000225 and its subsequences) are the binary repunits. - Rick L. Shepherd(AT)prodigy.net (rshepherd2(AT)hotmail.com), Sep 16 2002

Note that 3a(n)+(-1)^n=2^n is significant for Pascal' triangle A007318. It arises from a Jacobsthal decomposition of Pascal's triangle illustrated by 1+7+21+35+35+21+7+1 = (7+35+1)+(1+35+7)+(21+21) = 43 + 43 + 42 = 3a(7)-1; 1+8+28+56+70+56+29+8+1 = (1+56+28)+(28+56+1)+(8+70+8) = 85 + 85 + 86 = 3a(8)+1. - Paul Barry (pbarry(AT)wit.ie), Feb 20 2003

Number of positive integers requiring exactly n signed bits in the non-adjacent form representation.

Counts walks between adjacent vertices of a triangle - Paul Barry (pbarry(AT)wit.ie), Nov 17 2003

Comment from Slavik Jablan, Dec 26, 2003: Every amphichiral rational knot written in Conway notation is a palindromic sequence of numbers, not beginning or ending with 1. For example, for 4 <= n <= 12, the amphichiral rational knots are: 2 2, 2 1 1 2, 4 4, 3 1 1 3, 2 2 2 2, 4 1 1 4, 3 1 1 1 1 3, 2 3 3 2, 2 1 2 2 1 2, 2 1 1 1 1 1 1 2, 6 6, 5 1 1 5, 4 2 2 4, 3 3 3 3, 2 4 4 2, 3 2 1 1 2 3, 3 1 2 2 1 3, 2 2 2 2 2 2, 2 2 1 1 1 1 2 2, 2 1 2 1 1 2 1 2, 2 1 1 1 1 1 1 1 1 2. The number of amphichiral knots for n=2k (k=1, 2, 3, ...) we obtain the 0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, ...

a(n+2) counts the binary sequences of total length n made up of codewords from C={0,10,11} - Paul Barry (pbarry(AT)wit.ie), Jan 23 2004

Number of permutations with no fixed points avoiding 231 and 132.

The n-th entry (n>1) of the sequence is equal to the 2,2-entry of the n-th power of the unnormalized 4 by 4 Haar matrix: [1 1 1 0 / 1 1 -1 0 / 1 1 0 1 / 1 1 0 -1]. - Simone Severini (ss54(AT)york.ac.uk), Oct 27 2004

a(n) = number of Motzkin (n+1)-sequences whose flatsteps all occur at level 1 and whose height is <=2. For example, a(4)=5 counts UDUFD, UFDUD, UFFFD, UFUDD, UUDFD. - David Callan (callan(AT)stat.wisc.edu), Dec 09 2004

a(n+1) gives row sums of A059260. - Paul Barry (pbarry(AT)wit.ie), Jan 26 2005

If (m + n) is odd, then 3*(a(m) + a(n)) is always of the form a^2 + 2*b^2, where a and b both equal powers of 2; consequently every factor of (a(m) + a(n)) is always of the form a^2 + 2*b^2. - Matthew Vandermast (ghodges14(AT)comcast.net), Jul 12 2003

Number of "0,0" in f_{n+1}, where f_0 = "1" and f_{n+1} = a sequenece formed by changing all "1"s in f_n to "1,0" and all "0"s in f_n to "0,1" . - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Sep 22 2006

All prime Jacobsthal numbers A049883[n] = {3,5,11,43,683,2731,43691,...} have prime indices except a(4) = 5. All prime Jacobsthal numbers with prime indices (all but a(4) = 5) are of the form (2^p + 1)/3 - the Wagstaff primes A000979[n]. Indices of prime Jacobsthal numbers are listed in A107036[n] = {3,4,5,7,11,13,17,19,23,31,43,61,...}. For n>1 A107036[n] = A000978[n] Numbers n such that (2^n + 1)/3 is prime. - Alexander Adamchuk (alex(AT)kolmogorov.com), Oct 03 2006

Correspondence: a(n)=b(n)*2^(n-1), where b(n) is the sequence of the arithmetic means of previous two terms defined by b(n)=1/2*(b(n-1)+b(n-2)) with initial values b(0)=0, b(1)=1; The g.f. for b(n) is B(x):=x/(1-(x^1+x^2)/2), so the g.f. A(x) for a(n) suffices A(x)=B(2*x)/2. Because b(n) converges to the limit lim (1-x)*B(x)=1/3*(b(0)+2*b(1))=2/3 (for x-->1), it follows that a(n)/2^(n-1) also converges to 2/3 (see also A103770). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Feb 04 2006

Inverse: floor(log_2(a(n))=n-2 for n>=2. Also: log_2(a(n)+a(n-1))=n-1 for n>=1(see also A130249). Characterization: x is a Jacobsthal number if and only if there is a power of 4 (=c) such that x is a root of p(x)=9x(x-c)+(c-1)(2c+1) (see also the indicator sequence A105348). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), May 17 2007

This sequence counts the odd coefficients in the expansion of (1+x+x^2)^(2^n-1), n>=0. - Tewodros Amdeberhan (tewodros(AT)math.mit.edu), Oct 18 2007, Jan 08 2008

2^(n+1) = 2*A005578(n) + 2*a(n) + 2*A000975(n-1); e.g. 2^6 = 64 = 2*A005578(5) + 2*a(5) + 2*A000975(4) = (2*11 + 2*11 + 2*10). Let A005578(n), a(n), A000975(n-1) = triangle (a, b, c). Then ((S-c), (S-b), (S-a)) = (A005578(n-1), a(n-1), A000975(n-2)). Example: (a, b, c) = (11, 11, 10) = (A005578(5), a(5), A000975(4). Then ((S-c), (S-b), (S-a)) = (6, 5, 5) = (A005578(4), a(4), A000975(3)). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 24 2007

Sequence is identical to the absolute values of its inverse binomial transform. A similar result holds for [0,A001045*2^n]. - Paul Curtz (bpcrtz(AT)free.fr), Jan 17 2008

From a(2) on (i.e., 1,3,5,11,21,...) also: least odd number such that the subsets of {a(2),...,a(n)} sum to 2^(n-1) different values, cf. A138000 and A064934. It is interesting to note the pattern of numbers occuring (or not occuring) as such a sum (A003158). - M. F. Hasler (www.univ-ag.fr/~mhasler), Apr 09 2008

a(n) = term (5,1) of n-th power of the 5x5 matrix shown in A121231 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Oct 03 2008]

A147612(a(n)) = 1. [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Nov 08 2008]

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

a(n+1) = Sum(A153778(i): 2^n <= i < 2^(n+1)). [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Jan 01 2009]

Contribution from John Fossaceca (john(AT)fossaceca.net), Jan 31 2009: (Start)

It appears that a(n) is also the number of integers between 2^n and 2^(n+1)

that are divisible by 3 with no remainder (End)

Number of pairs of consecutive odious (or evil) numbers between 2^(n+1) and 2^(n+2), inclusive. [From T. D. Noe (noe(AT)sspectra.com), Feb 05 2009]

Equals eigensequence of triangle A156319 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Feb 07 2009]

Starting with offset 1 = row sums of triangle A156667. [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Feb 12 2009]

A three-dimensional interpretation of a(n+1) is that it gives the number of ways of filling a 2 by 2 by n hole with 1 by 2 by 2 bricks. [From Martin Griffiths (griffm(AT)essex.ac.uk), Mar 28 2009]

Starting with offset 1 = INVERTi transform of A002605: (1, 2, 6, 16, 44,...). [From Gary W. Adamson (qntmpkt(AT)yahoo.com), May 12 2009]

Convolved with (1, 2, 2, 2,...) = A000225: (1, 3, 7, 15, 31,...). [From Gary W. Adamson (qntmpkt(AT)yahoo.com), May 23 2009]

The product of a pair of successive terms is always a trianguler number. - Giuseppe Ottonello, Jun 14 2009

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

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

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

Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.

L. Carlitz, R. Scoville and V. E. Hoggatt, Jr., Representations for a special sequence, Fib. Quart., 10 (1972), 499-518, 550.

D. E. Daykin, D. J. Kleitman and D. B. West, The number of meets between two subsets of a lattice, J. Combin. Theory, A 26 (1979), 135-156.

Th. Fink and Y. Mao. The 85 ways to tie a tie, Fourth Estate, London, 1999; Die 85 Methoden eine Krawatte zu binden. Hoffmann und Kampe, Hamburg, 1999.

International Mathematical Olympiad 2001, Hong Kong Preliminary Selection Contest Problem #16.

Jablan S. and Sazdanovic R., LinKnot: Knot Theory by Computer, World Scientific Press, 2007. See p. 80.

D. E. Knuth, Art of Computer Programming, Vol. 3, Sect. 5.3.1, Eq. 13.

T. Koshy, Fibonacci and Lucas numbers with applications, Wiley, 2001, p. 98.

S. L. Levine, Suppose more rabbits are born, Fib. Quart., 26 (1988), 306-311.

B. D. McKay and I. M. Wanless, A census of small latin hypercubes, SIAM J. Discrete Math. 22, (2008) 719-736.

G. Myerson and A. J. van der Poorten, Some problems concerning recurrence sequences, Amer. Math. Monthly, 102 (1995), 698-705.

S. Roman, Introduction to Coding and Information Theory, Springer Verlag, 1996, 41-42

Two-Year College Math. Jnl., 28 (1997), p. 76.

Robert M. Young, "Excursions in Calculus", MAA, 1992, p. 239

G. B. M. Zerr, Problem 64, American Mathematical Monthly, vol. 3, no. 12, 1896 (p. 311).

T. D. Noe, Table of n, a(n) for n = 0..500

Index entries for sequences related to linear recurrences with constant coefficients

Joerg Arndt, Fxtbook

W. Bosma, Signed bits and fast exponentiation

D. D. Frey and J. A. Sellers, Jacobsthal Numbers and Alternating Sign Matrices, J. Integer Seqs., Vol. 3 (2000), #00.2.3

S. Heubach, Tiling an m X n area with squares of size up to k X k (m <=5), Congressus Numerantium 140 (1999), pp. 43-64.

INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 142

Lee Hae-hwang, Illustration of initial terms in terms of rosemary plants

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

G. Myerson and A. J. van der Poorten, Some problems concerning recurrence sequences, Amer. Math. Monthly, 102 (1995), 698-705.

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.

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

Eric Weisstein's World of Mathematics, Rule 28

Thomas Wieder, HomePage.

Index entries for sequences related to Chebyshev polynomials.

a(n) = 2^(n-1) - a(n-1). a(n) = 2*a(n-1) - (-1)^n = {2^n - (-1)^n}/3.

G.f.: x/(1-x-2*x^2). E.g.f.: (exp(2*x)-exp(-x))/3.

a(2n)=2*a(2n-1)-1 for n>=1, a(2n+1)=2*a(2n)+1 for n>=0. - Lee Hae-hwang (mathmaniac(AT)empal.com), Oct 11 2002; corrected by Mario Catalani (mario.catalani(AT)unito.it), Dec 04 2002

Also a(n) is the coefficient of x^(n-1) in the bivariate Fibonacci polynomials F(n)(x, y)=xF(n-1)(x, y)+yF(n-2)(x, y), with y=2x^2. - Mario Catalani (mario.catalani(AT)unito.it), Dec 04 2002

a(n)=sum{k=1..n, binomial(n, k)(-1)^(n+k)*3^(k-1) }. - Paul Barry (pbarry(AT)wit.ie), Apr 02 2003

The ratios a(n)/2^(n-1) converge to 2/3 and every fraction after 1/2 is the arithmetic mean of the two preceding fractions. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jul 05 2003

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

a(n+1)=sum(k=0, ceil(n/2), 2^k*binomial(n-k, k)) - Benoit Cloitre (benoit7848c(AT)orange.fr), Mar 06 2004

a(2n) = A002450(n) = (4^n - 1)/3; a(2n+1) = A007583(n) = (2^(2n+1) + 1)/3. - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Mar 27 2004

a(n) = round(2^n/3) = (2^n + (-1)^(n-1))/3 so lim n->inf 2^n/a(n) = 3 - Gerald McGarvey (Gerald.McGarvey(AT)comcast.net), Jul 21 2004

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

a(n+1)=sum{k=0..n, binomial(k, n-k)2^(n-k)} - Paul Barry (pbarry(AT)wit.ie), Oct 07 2004

a(n)=sum{k=0..n-1, W(n-k, k)(-1)^(n-k)binomial(2k, k)}, W(n, k) as in A004070. - Paul Barry (pbarry(AT)wit.ie), Dec 17 2004

a(n)=sum{k=0..n, k*binomial(n-1, (n-k)/2)(1+(-1)^(n+k))floor((2k+1)/3)}; a(n+1)=sum{k=0..n, k*binomial(n-1, (n-k)/2)(1+(-1)^(n+k))(A042965(k)+0^k)}; - Paul Barry (pbarry(AT)wit.ie), Jan 17 2005

a(n+1)=ceiling(2^n/3)+floor(2^n/3)=(ceiling(2^n/3))^2-(floor(2^n/3))^2; a(n+1)=A005578(n)+A000975(n-1)=A005578(n)^2-A000975(n-1)^2; - Paul Barry (pbarry(AT)wit.ie), Jan 17 2005

a(n+1)=sum{k=0..n, sum{j=0..n, (-1)^(n-j)*binomial(j, k)}}; - Paul Barry (pbarry(AT)wit.ie), Jan 26 2005

Let M=[1, 1, 0;1, 0, 1;0, 1, 1], then a(n) = (M^n)[2, 1], also matrix characteristic polynomial x^3 - 2*x^2 - x + 2 defines the three step recursion a(0)=0, a(1)=1, a(2)=1, a(n)=2a(n-1)+a(n-2)-2a(n-3) for n>2 - Lambert Klasen (lambert.klasen(AT)gmx.net), Jan 28 2005

a(n)=ceiling(2^(n+1)/3)-ceiling(2^n/3)=A005578(n+1)-A005578(n); - Paul Barry (pbarry(AT)wit.ie), Oct 08 2005

a(n)=floor(2^(n+1)/3)-floor(2^n/3)=A000975(n)-A000975(n-1); - Paul Barry (pbarry(AT)wit.ie), Oct 08 2005

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

a(2n)=Product(d divides n, cyclotomic(d,4))/3. a(2n+1)=Product(d divides 2n+1, cyclotomic(2d,2))/3. - Miklos Kristof (kristmikl(AT)freemail.hu), Mar 07 2007

Further comments and formulae from Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Apr 23 2007: (Start) The a(n) are closely related to nested square roots; this is 2*sin(2^(-n)*pi/2*a(n))=sqr(2-sqr(2-sqr(2-sqr(...sqr(2)))...){n-times the '2', n>=0}.

Also true: 2*cos(2^(-n)*pi*a(n))=sqr(2-sqr(2-sqr(2-sqr(...sqr(2)))...){(n-1)-times the '2', n>=1} as well as

2*sin(2^(-n)*3/2*pi*a(n))=sqr(2+sqr(2+sqr(2+sqr(...sqr(2)))...){n-times the '2', n>=0} and

2*cos(2^(-n)*3*pi*a(n))=-sqr(2+sqr(2+sqr(2+sqr(...sqr(2)))...){(n-1)-times the '2', n>=1}.

a(n)=2^(n+1)/pi*arcsin(b(n+1)/2) where b(n) is defined recursively by b(0)=2, b(n)=sqr(2-b(n-1)).

There is a similar formula regarding the arccos function, this is a(n)=2^n/pi*arccos(b(n)/2).

With respect to the sequence c(n) defined recursively by c(0)=-2, c(n)=sqr(2+c(n-1)) the following fomulas hold true: a(n)=2^n/3*(1-(-1)^n*(1-2/pi*arcsin(c(n+1)/2)); a(n)=2^n/3*(1-(-1)^n*(1-1/pi*arccos(-c(n)/2)). (End)

Sum_{k, 0<=k<=n}A039599(n,k)*a(k)=A049027(n), for n>=1 . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Jun 10 2007

Sum_[k, 0<=k<=n}A039599(n,k)*a(k+1)=A067336(n). - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Jun 10 2007

Row sums of triangle A134317. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Oct 19 2007

Let T = the 3 X 3 matrix [1,1,0; 1,0,1; 0,1,1]. Then T^n * [1,0,0,] = [A005578(n), a(n), A000975(n-1]. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 24 2007

a(n)+a(n+5)=11*2^n. - Paul Curtz (bpcrtz(AT)free.fr), Jan 17 2008

a(n)=sum(K(2, k)*a(n - k),k=1..n), where K(n,k) = k if 0 <= k AND k <= n and K(n,k)=0 else. (When using such a K-coefficient several different arguments to K or several different definitions of K may lead to the same integer sequence. For example, the Fibonacci sequence can be generated in several ways using the K-coefficient.) - Thomas Wieder (thomas.wieder(AT)t-online.de), Jan 13 2008

a(n)+a(n+2k+1)=a(2k+1)*2^n. - Paul Curtz (bpcrtz(AT)free.fr), Feb 12 2008

a(n) = lower left term in the 2 X 2 matrix [0,2; 1,1]^n - Gary W. Adamson (qntmpkt(AT)yahoo.com), Mar 02 2008

a(n+1)=Sum_{k, 0<=k<=n} A109466(n,k)*(-2)^(n-k). [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Oct 26 2008]

For n > 0, a(n) = b(n) - b(n-1), where b(n) is defined by the sequence A000975. [From Kailasam Viswanathan Iyer (kvi(AT)nitt.edu), May 12 2009]

a(n) = sqrt( 8*a(n-1)*a(n-2) + 1 ). E.g. sqrt(3*5*8+1)=11, sqrt(5*11*8+1)=21. - Giuseppe Ottonello, Jun 14 2009

a(2) = 3 because the tiling of the 3x2 rectangle has either only 1 X 1 tiles, or one 2 X 2 tile in one of two positions (together with 2 1 X 1 tiles)

a:=n->sum(binomial(n-k, k)*2^k, k=0..n): - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Sep 30 2006

A001045:=-1/(z+1)/(2*z-1); [S. Plouffe in his 1992 dissertation.]

a := proc(n::integer) # A001045 Jacobsthal sequence: a(n) = a(n-1) + 2a(n-2), with a(0) = 0, a(1) = 1. local k; option remember; if n = 0 then 1 else add(K(2, k)*procname(n - k), k=1..n) end if end proc; K := proc(n::integer, k::integer) local KC; if 0 <= k and k <= n then KC := k else KC := 0 end if; end proc; - Thomas Wieder (thomas.wieder(AT)t-online.de), Jan 13 2008

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+2*a[n-2]od: seq(a[n], n=0..33); # [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Dec 15 2008]

with(combstruct): SeqSeqSeqL := [T, {T=Sequence(S, card > 0), S=Sequence(U, card > 0), U=Sequence(Z, card >1)}, unlabeled]: seq(count(SeqSeqSeqL, size=j), j=1..34); # [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 04 2009]

f[n_] := (2^n - (-1)^n)/3; Table[ f[n], {n, 0, 33}] (from Robert G. Wilson v (rgwv(at)rgwv.com), Dec 05 2005)

Array[(2^# - (-1)^#)/3 &, 33, 0] - Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006

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

(PARI) a(n)=if(n<0, 0, (2^n-(-1)^n)/3)

(PARI) a(n)=if(n==0, 0, if(n==1, 1, if(n==2, 1, 2*a(n-1)+a(n-2)-2*a(n-3)))) for(i=0, 15, print1(a(i), ", ")) M=[1, 1, 0; 1, 0, 1; 0, 1, 1]; for(i=0, 15, print1((M^i)[2, 1], ", ")) (Klasen)

sage: from sage.combinat.sloane_functions import recur_gen2 sage: it = recur_gen2(0, 1, 1, 2) sage: [it.next() for i in range(30)] - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 25 2008

(Other) sage: [lucas_number1(n, 1, -2) for n in xrange(0, 34)] # [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 22 2009]

(Other) sage: [abs(gaussian_binomial(n, 1, -2)) for n in xrange(0, 34)] # [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 28 2009]

Partial sums of this sequence give A000975, where there are additional comments from B. E. Williams and Bill Blewett on the tie problem. Cf. A049883, A026644.

A002487(A001045(n))=A000045(n).

Row sums of A059260. Equals A026644(n) + 1 for n > 1.

a(n)= A073370(n-1, 0), n>=1 (first column of triangle).

Apart from initial term, equals A026644(n+1) + 1.

See also A081857.

Cf. A000978, A000979. Cf. A049883 = primes in this sequence, A107036 = indices of primes, A129738.

Cf. A019322 A066845, A105348, A130249, A130250, 130253, A134317, A005578, A002083, A002487, A113405, A138000, A064934, A003158,

Cf. A147613. [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Nov 08 2008]

A156319 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Feb 07 2009]

A156667 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Feb 12 2009]

A002605 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), May 12 2009]

A000225 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), May 23 2009]

Sequence in context: A122997 A146042 A152046 this_sequence A154917 A167167 A077925

Adjacent sequences: A001042 A001043 A001044 this_sequence A001046 A001047 A001048

nonn,nice,easy,core

N. J. A. Sloane (njas(AT)research.att.com).

More terms from James A. Sellers (sellersj(AT)math.psu.edu), Dec 23 1999. Zerr reference from Len Smiley (smiley(AT)math.uaa.alaska.edu), May 21 2001.

More terms from Simone Severini (ss54(AT)york.ac.uk), Oct 27 2004

Further terms from Lambert Klasen (lambert.klasen(AT)gmx.net), Jan 28 2005