cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A027383 a(2*n) = 3*2^n - 2; a(2*n+1) = 2^(n+2) - 2.

Original entry on oeis.org

1, 2, 4, 6, 10, 14, 22, 30, 46, 62, 94, 126, 190, 254, 382, 510, 766, 1022, 1534, 2046, 3070, 4094, 6142, 8190, 12286, 16382, 24574, 32766, 49150, 65534, 98302, 131070, 196606, 262142, 393214, 524286, 786430, 1048574, 1572862, 2097150, 3145726, 4194302, 6291454
Offset: 0

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Author

Keywords

Comments

Number of balanced strings of length n: let d(S) = #(1's) - #(0's), # == count in S, then S is balanced if every substring T of S has -2 <= d(T) <= 2.
Number of "fold lines" seen when a rectangular piece of paper is folded n+1 times along alternate orthogonal directions and then unfolded. - Quim Castellsaguer (qcastell(AT)pie.xtec.es), Dec 30 1999
Also the number of binary strings with the property that, when scanning from left to right, once the first 1 is seen in position j, there must be a 1 in positions j+2, j+4, ... until the end of the string. (Positions j+1, j+3, ... can be occupied by 0 or 1.) - Jeffrey Shallit, Sep 02 2002
a(n-1) is also the Moore lower bound on the order of a (3,n)-cage. - Eric W. Weisstein, May 20 2003 and Jason Kimberley, Oct 30 2011
Partial sums of A016116. - Hieronymus Fischer, Sep 15 2007
Equals row sums of triangle A152201. - Gary W. Adamson, Nov 29 2008
From John P. McSorley, Sep 28 2010: (Start)
a(n) = DPE(n+1) is the total number of k-double-palindromes of n up to cyclic equivalence. See sequence A180918 for the definitions of a k-double-palindrome of n and of cyclic equivalence. Sequence A180918 is the 'DPE(n,k)' triangle read by rows where DPE(n,k) is the number of k-double-palindromes of n up to cyclic equivalence. For example, we have a(4) = DPE(5) = DPE(5,1) + DPE(5,2) + DPE(5,3) + DPE(5,4) + DPE(5,5) = 0 + 2 + 2 + 1 + 1 = 6.
The 6 double-palindromes of 5 up to cyclic equivalence are 14, 23, 113, 122, 1112, 11111. They come from cyclic equivalence classes {14,41}, {23,32}, {113,311,131}, {122,212,221}, {1112,2111,1211,1121}, and {11111}. Hence a(n)=DPE(n+1) is the total number of cyclic equivalence classes of n containing at least one double-palindrome.
(End)
From Herbert Eberle, Oct 02 2015: (Start)
For n > 0, there is a red-black tree of height n with a(n-1) internal nodes and none with less.
In order a red-black tree of given height has minimal number of nodes, it has exactly 1 path with strictly alternating red and black nodes. All nodes outside this height defining path are black.
Consider:
mrbt5 R
/ \
/ \
/ \
/ B
/ / \
mrbt4 B / B
/ \ B E E
/ B E E
mrbt3 R E E
/ \
/ B
mrbt2 B E E
/ E
mrbt1 R
E E
(Red nodes shown as R, blacks as B, externals as E.)
Red-black trees mrbt1, mrbt2, mrbt3, mrbt4, mrbt5 of respective heights h = 1, 2, 3, 4, 5; all minimal in the number of internal nodes, namely 1, 2, 4, 6, 10.
Recursion (let n = h-1): a(-1) = 0, a(n) = a(n-1) + 2^floor(n/2), n >= 0.
(End)
Also the number of strings of length n with the digits 1 and 2 with the property that the sum of the digits of all substrings of uneven length is not divisible by 3. An example with length 8 is 21221121. - Herbert Kociemba, Apr 29 2017
a(n-2) is the number of achiral n-bead necklaces or bracelets using exactly two colors. For n=4, the four arrangements are AAAB, AABB, ABAB, and ABBB. - Robert A. Russell, Sep 26 2018
Partial sums of powers of 2 repeated 2 times, like A200672 where is 3 times. - Yuchun Ji, Nov 16 2018
Also the number of binary words of length n with cuts-resistance <= 2, where, for the operation of shortening all runs by one, cuts-resistance is the number of applications required to reach an empty word. Explicitly, these are words whose sequence of run-lengths, all of which are 1 or 2, has no odd-length run of 1's sandwiched between two 2's. - Gus Wiseman, Nov 28 2019
Also the number of up-down paths with n steps such that the height difference between the highest and lowest points is at most 2. - Jeremy Dover, Jun 17 2020
Also the number of non-singleton integer compositions of n + 2 with no odd part other than the first or last. Including singletons gives A052955. This is an unsorted (or ordered) version of A351003. The version without even (instead of odd) interior parts is A001911, complement A232580. Note that A000045(n-1) counts compositions without odd parts, with non-singleton case A077896, and A052952/A074331 count non-singleton compositions without even parts. Also the number of compositions y of n + 1 such that y_i = y_{i+1} for all even i. - Gus Wiseman, Feb 19 2022

Examples

			After 3 folds one sees 4 fold lines.
Example: a(3) = 6 because the strings 001, 010, 100, 011, 101, 110 have the property.
Binary: 1, 10, 100, 110, 1010, 1110, 10110, 11110, 101110, 111110, 1011110, 1111110, 10111110, 11111110, 101111110, 111111110, 1011111110, 1111111110, 10111111110, ... - _Jason Kimberley_, Nov 02 2011
Example: Partial sums of powers of 2 repeated 2 times:
a(3) = 1+1+2 = 4;
a(4) = 1+1+2+2 = 6;
a(5) = 1+1+2+2+4 = 10.
_Yuchun Ji_, Nov 16 2018
		

References

  • John P. McSorley: Counting k-compositions of n with palindromic and related structures. Preprint, 2010. [John P. McSorley, Sep 28 2010]

Crossrefs

Moore lower bound on the order of a (k,g) cage: A198300 (square); rows: A000027 (k=2), this sequence (k=3), A062318 (k=4), A061547 (k=5), A198306 (k=6), A198307 (k=7), A198308 (k=8), A198309 (k=9), A198310 (k=10), A094626 (k=11); columns: A020725 (g=3), A005843 (g=4), A002522 (g=5), A051890 (g=6), A188377 (g=7). - Jason Kimberley, Oct 30 2011
Cf. A000066 (actual order of a (3,g)-cage).
Bisections are A033484 (even) and A000918 (odd).
a(n) = A305540(n+2,2), the second column of the triangle.
Numbers whose binary expansion is a balanced word are A330029.
Binary words counted by cuts-resistance are A319421 or A329860.
The complementary compositions are counted by A274230(n-1) + 1, with bisections A060867 (even) and A134057 (odd).
Cf. A000346, A000984, A001405, A001700, A011782 (compositions).
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A029744 = {s(n), n>=1}, the numbers 2^k and 3*2^k, as the parent: A029744 (s(n)); A052955 (s(n)-1), A027383 (s(n)-2), A354788 (s(n)-3), A347789 (s(n)-4), A209721 (s(n)+1), A209722 (s(n)+2), A343177 (s(n)+3), A209723 (s(n)+4); A060482, A136252 (minor differences from A354788 at the start); A354785 (3*s(n)), A354789 (3*s(n)-7). The first differences of A029744 are 1,1,1,2,2,4,4,8,8,... which essentially matches eight sequences: A016116, A060546, A117575, A131572, A152166, A158780, A163403, A320770. The bisections of A029744 are A000079 and A007283. - N. J. A. Sloane, Jul 14 2022

Programs

  • Haskell
    import Data.List (transpose)
    a027383 n = a027383_list !! n
    a027383_list = concat $ transpose [a033484_list, drop 2 a000918_list]
    -- Reinhard Zumkeller, Jun 17 2015
    
  • Magma
    [2^Floor((n+2)/2)+2^Floor((n+1)/2)-2: n in [0..50]]; // Vincenzo Librandi, Aug 16 2011
    
  • Maple
    a[0]:=0:a[1]:=1:for n from 2 to 100 do a[n]:=2*a[n-2]+2 od: seq(a[n], n=1..41); # Zerinvary Lajos, Mar 16 2008
  • Mathematica
    a[n_?EvenQ] := 3*2^(n/2)-2; a[n_?OddQ] := 2^(2+(n-1)/2)-2; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Oct 21 2011, after Quim Castellsaguer *)
    LinearRecurrence[{1, 2, -2}, {1, 2, 4}, 41] (* Robert G. Wilson v, Oct 06 2014 *)
    Table[Length[Select[Tuples[{0,1},n],And[Max@@Length/@Split[#]<=2,!MatchQ[Length/@Split[#],{_,2,ins:1..,2,_}/;OddQ[Plus[ins]]]]&]],{n,0,15}] (* Gus Wiseman, Nov 28 2019 *)
  • PARI
    a(n)=2^(n\2+1)+2^((n+1)\2)-2 \\ Charles R Greathouse IV, Oct 21 2011
    
  • Python
    def a(n): return 2**((n+2)//2) + 2**((n+1)//2) - 2
    print([a(n) for n in range(43)]) # Michael S. Branicky, Feb 19 2022

Formula

a(0)=1, a(1)=2; thereafter a(n+2) = 2*a(n) + 2.
a(2n) = 3*2^n - 2 = A033484(n);
a(2n-1) = 2^(n+1) - 2 = A000918(n+1).
G.f.: (1 + x)/((1 - x)*(1 - 2*x^2)). - David Callan, Jul 22 2008
a(n) = Sum_{k=0..n} 2^min(k, n-k).
a(n) = 2^floor((n+2)/2) + 2^floor((n+1)/2) - 2. - Quim Castellsaguer (qcastell(AT)pie.xtec.es)
a(n) = 2^(n/2)*(3 + 2*sqrt(2) + (3-2*sqrt(2))*(-1)^n)/2 - 2. - Paul Barry, Apr 23 2004
a(n) = A132340(A052955(n)). - Reinhard Zumkeller, Aug 20 2007
a(n) = A052955(n+1) - 1. - Hieronymus Fischer, Sep 15 2007
a(n) = A132666(a(n+1)) - 1. - Hieronymus Fischer, Sep 15 2007
a(n) = A132666(a(n-1)+1) for n > 0. - Hieronymus Fischer, Sep 15 2007
A132666(a(n)) = a(n-1) + 1 for n > 0. - Hieronymus Fischer, Sep 15 2007
G.f.: (1 + x)/((1 - x)*(1 - 2*x^2)). - David Callan, Jul 22 2008
a(n) = 2*( (a(n-2)+1) mod (a(n-1)+1) ), n > 1. - Pierre Charland, Dec 12 2010
a(n) = A136252(n-1) + 1, for n > 0. - Jason Kimberley, Nov 01 2011
G.f.: (1+x*R(0))/(1-x), where R(k) = 1 + 2*x/( 1 - x/(x + 1/R(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 16 2013
a(n) = 2^((2*n + 3*(1-(-1)^n))/4)*3^((1+(-1)^n)/2) - 2. - Luce ETIENNE, Sep 01 2014
a(n) = a(n-1) + 2^floor((n-1)/2) for n>0, a(0)=1. - Yuchun Ji, Nov 23 2018
E.g.f.: 3*cosh(sqrt(2)*x) - 2*cosh(x) + 2*sqrt(2)*sinh(sqrt(2)*x) - 2*sinh(x). - Stefano Spezia, Apr 06 2022

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Mar 24 2000
Replaced definition with a simpler one. - N. J. A. Sloane, Jul 09 2022

A029744 Numbers of the form 2^n or 3*2^n.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 128, 192, 256, 384, 512, 768, 1024, 1536, 2048, 3072, 4096, 6144, 8192, 12288, 16384, 24576, 32768, 49152, 65536, 98304, 131072, 196608, 262144, 393216, 524288, 786432, 1048576, 1572864, 2097152, 3145728, 4194304
Offset: 1

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Keywords

Comments

This entry is a list, and so has offset 1. WARNING: However, in this entry several comments, formulas and programs seem to refer to the original version of this sequence which had offset 0. - M. F. Hasler, Oct 06 2014
Number of necklaces with n-1 beads and two colors that are the same when turned over and hence have reflection symmetry. [edited by Herbert Kociemba, Nov 24 2016]
The subset {a(1),...,a(2k)} contains all proper divisors of 3*2^k. - Ralf Stephan, Jun 02 2003
Let k = any nonnegative integer and j = 0 or 1. Then n+1 = 2k + 3j and a(n) = 2^k*3^j. - Andras Erszegi (erszegi.andras(AT)chello.hu), Jul 30 2005
Smallest number having no fewer prime factors than any predecessor, a(0)=1; A110654(n) = A001222(a(n)); complement of A116451. - Reinhard Zumkeller, Feb 16 2006
A093873(a(n)) = 1. - Reinhard Zumkeller, Oct 13 2006
a(n) = a(n-1) + a(n-2) - gcd(a(n-1), a(n-2)), n >= 3, a(1)=2, a(2)=3. - Ctibor O. Zizka, Jun 06 2009
Where records occur in A048985: A193652(n) = A048985(a(n)) and A193652(n) < A048985(m) for m < a(n). - Reinhard Zumkeller, Aug 08 2011
A002348(a(n)) = A000079(n-3) for n > 2. - Reinhard Zumkeller, Mar 18 2012
Without initial 1, third row in array A228405. - Richard R. Forberg, Sep 06 2013
Also positions of records in A048673. A246360 gives the record values. - Antti Karttunen, Sep 23 2014
Known in numerical mathematics as "Bulirsch sequence", used in various extrapolation methods for step size control. - Peter Luschny, Oct 30 2019
For n > 1, squares of the terms can be expressed as the sum of two powers of two: 2^x + 2^y. - Karl-Heinz Hofmann, Sep 08 2022

Crossrefs

Cf. A056493, A038754, A063759. Union of A000079 and A007283.
First differences are in A016116(n-1).
Row sums of the triangle in sequence A119963. - John P. McSorley, Aug 31 2010
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A029744 = {s(n), n>=1}, the numbers 2^k and 3*2^k, as the parent. There may be minor differences from (s(n)) at the start, and a shift of indices. A029744 (s(n)); A052955 (s(n)-1), A027383 (s(n)-2), A354788 (s(n)-3), A060482 (s(n)-3); A136252 (s(n)-3); A347789 (s(n)-4), A209721 (s(n)+1), A209722 (s(n)+2), A343177 (s(n)+3), A209723 (s(n)+4); A354785 (3*s(n)), A061776 (3*s(n)-6); A354789 (3*s(n)-7). The first differences of A029744 are 1,1,1,2,2,4,4,8,8,... which essentially matches eight sequences: A016116, A060546, A117575, A131572, A152166, A158780, A163403, A320770. The bisections of A029744 are A000079 and A007283. - N. J. A. Sloane, Jul 14 2022

Programs

  • Haskell
    a029744 n = a029744_list !! (n-1)
    a029744_list = 1 : iterate
       (\x -> if x `mod` 3 == 0 then 4 * x `div` 3 else 3 * x `div` 2) 2
    -- Reinhard Zumkeller, Mar 18 2012
    
  • Maple
    1,seq(op([2^i,3*2^(i-1)]),i=1..100); # Robert Israel, Sep 23 2014
  • Mathematica
    CoefficientList[Series[(-x^2 - 2*x - 1)/(2*x^2 - 1), {x, 0, 200}], x] (* Vladimir Joseph Stephan Orlovsky, Jun 10 2011 *)
    Function[w, DeleteCases[Union@ Flatten@ w, k_ /; k > Max@ First@ w]]@ TensorProduct[{1, 3}, 2^Range[0, 22]] (* Michael De Vlieger, Nov 24 2016 *)
    LinearRecurrence[{0,2},{1,2,3},50] (* Harvey P. Dale, Jul 04 2017 *)
  • PARI
    a(n)=if(n%2,3/2,2)<<((n-1)\2)\1
    
  • Python
    def A029744(n):
        if n == 1: return 1
        elif n % 2 == 0: return 2**(n//2)
        else: return 3 * 2**((n-3)//2) # Karl-Heinz Hofmann, Sep 08 2022
  • Scheme
    (define (A029744 n) (cond ((<= n 1) n) ((even? n) (expt 2 (/ n 2))) (else (* 3 (expt 2 (/ (- n 3) 2)))))) ;; Antti Karttunen, Sep 23 2014
    

Formula

a(n) = 2*A000029(n) - A000031(n).
For n > 2, a(n) = 2*a(n - 2); for n > 3, a(n) = a(n - 1)*a(n - 2)/a(n - 3). G.f.: (1 + x)^2/(1 - 2*x^2). - Henry Bottomley, Jul 15 2001, corrected May 04 2007
a(0)=1, a(1)=1 and a(n) = a(n-2) * ( floor(a(n-1)/a(n-2)) + 1 ). - Benoit Cloitre, Aug 13 2002
(3/4 + sqrt(1/2))*sqrt(2)^n + (3/4 - sqrt(1/2))*(-sqrt(2))^n. a(0)=1, a(2n) = a(n-1)*a(n), a(2n+1) = a(n) + 2^floor((n-1)/2). - Ralf Stephan, Apr 16 2003 [Seems to refer to the original version with offset=0. - M. F. Hasler, Oct 06 2014]
Binomial transform is A048739. - Paul Barry, Apr 23 2004
E.g.f.: (cosh(x/sqrt(2)) + sqrt(2)sinh(x/sqrt(2)))^2.
a(1) = 1; a(n+1) = a(n) + A000010(a(n)). - Stefan Steinerberger, Dec 20 2007
u(2)=1, v(2)=1, u(n)=2*v(n-1), v(n)=u(n-1), a(n)=u(n)+v(n). - Jaume Oliver Lafont, May 21 2008
For n => 3, a(n) = sqrt(2*a(n-1)^2 + (-2)^(n-3)). - Richard R. Forberg, Aug 20 2013
a(n) = A064216(A246360(n)). - Antti Karttunen, Sep 23 2014
a(n) = sqrt((17 - (-1)^n)*2^(n-4)) for n >= 2. - Anton Zakharov, Jul 24 2016
Sum_{n>=1} 1/a(n) = 8/3. - Amiram Eldar, Nov 12 2020
a(n) = 2^(n/2) if n is even. a(n) = 3 * 2^((n-3)/2) if n is odd and for n>1. - Karl-Heinz Hofmann, Sep 08 2022

Extensions

Corrected and extended by Joe Keane (jgk(AT)jgk.org), Feb 20 2000

A005418 Number of (n-1)-bead black-white reversible strings; also binary grids; also row sums of Losanitsch's triangle A034851; also number of caterpillar graphs on n+2 vertices.

Original entry on oeis.org

1, 2, 3, 6, 10, 20, 36, 72, 136, 272, 528, 1056, 2080, 4160, 8256, 16512, 32896, 65792, 131328, 262656, 524800, 1049600, 2098176, 4196352, 8390656, 16781312, 33558528, 67117056, 134225920, 268451840, 536887296, 1073774592, 2147516416, 4295032832
Offset: 1

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Author

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Comments

Equivalently, walks on triangle, visiting n+2 vertices, so length n+1, n "corners"; the symmetry group is S3, reversing a walk does not count as different. Walks are not self-avoiding. - Colin Mallows
Slavik V. Jablan observes that this is also the number of rational knots and links with n+2 crossings (cf. A018240). See reference. [Corrected by Andrey Zabolotskiy, Jun 18 2020]
Number of bit strings of length (n-1), not counting strings which are the end-for-end reversal or the 0-for-1 reversal of each other as different. - Carl Witty (cwitty(AT)newtonlabs.com), Oct 27 2001
The formula given in page 1095 of the Balasubramanian reference can be used to derive this sequence. - Parthasarathy Nambi, May 14 2007
Also number of compositions of n up to direction, where a composition is considered equivalent to its reversal, see example. - Franklin T. Adams-Watters, Oct 24 2009
Number of normally non-isomorphic realizations of the associahedron of type I starting with dimension 2 in Ceballos et al. - Tom Copeland, Oct 19 2011
Number of fibonacenes with n+2 hexagons. See the Balaban and the Dobrynin references. - Emeric Deutsch, Apr 21 2013
From the point of view of binary grids, it is a (1,n)-rectangular grid. A225826 to A225834 are the numbers of binary pattern classes in the (m,n)-rectangular grid, 1 < m < 11. - Yosu Yurramendi, May 19 2013
Number of n-vertex difference graphs (bipartite 2K_2-free graphs) [Peled & Sun, Thm. 9]. - Falk Hüffner, Jan 10 2016
The offset should be 0, since the first row of A034851 is row 0. The name would then be: "Number of n bead...". - Daniel Forgues, Jul 26 2018
a(n) is the number of non-isomorphic generalized rigid ladders with n cells. A generalized rigid ladder with n cells is a graph with vertex set is the union of {u_0, u_1, ..., u_n} and {v_0, v_1, ..., v_n}, and for every 0 <= i <= n-1, the edges are of the form {u_i,u_i+1}, {v_i, v_i+1}, {u_i,v_i} and either {u_i,v_i+1} or {u_i+1,v_i}. - Christian Barrientos, Jul 29 2018
Also number of non-isomorphic stairs with n+1 cells. A stair is a snake polyomino allowing only two directions for adjacent cells: east and north. - Christian Barrientos and Sarah Minion, Jul 29 2018
From Robert A. Russell, Oct 28 2018: (Start)
There are two different unoriented row colorings using two colors that give us very similar results here, a difference of one in the offset. In an unoriented row, chiral pairs are counted as one.
a(n) is the number of color patterns (set partitions) of an unoriented row of length n using two or fewer colors (subsets). Two color patterns are equivalent if the colors are permutable.
a(n+1) is the number of ways to color an unoriented row of length n using two noninterchangeable colors (one need not use both colors).
See the examples below of these two different colorings. (End)
Also arises from the enumeration of types of based polyhedra with exactly two triangular faces [Rademacher]. - N. J. A. Sloane, Apr 24 2020
a(n) is the number of (unlabeled) 2-paths with n+4 vertices. (A 2-path with order n at least 4 can be constructed from a 3-clique by iteratively adding a new 2-leaf (vertex of degree 2) adjacent to an existing 2-clique containing an existing 2-leaf.) - Allan Bickle, Apr 05 2022
a(n) is the number of caterpillars with a perfect matching and order 2n+2. - Christian Barrientos, Sep 12 2023
a(n) is also the number of distinct planar embeddings of the (n+2)-centipede graph (up to at least n=8 and likely for all larger n). - Eric W. Weisstein, May 21 2024
a(n) is also the number of distinct planar embeddings of the 2 X (n+2) grid graph i.e., the (n+2)-ladder graph. - Eric W. Weisstein, May 21 2024
Dimension of the homogeneous component of degree n of the free Jordan algebra on two generators (or, in this case, the free special Jordan algebra on two generators). It follows from (Shirshov 1956, Cohn 1959). - Vladimir Dotsenko, Mar 29 2025

Examples

			a(5) = 10 because there are 16 compositions of 5 (shown as <vectors>) but only 10 equivalence classes (shown as {sets}): {<5>}, {<4,1>,<1,4>}, {<3,2>,<2,3>}, {<3,1,1>,<1,1,3>}, {<1,3,1>},{<2,2,1>,<1,2,2>}, {<2,1,2>}, {<2,1,1,1>,<1,1,1,2>}, {<1,2,1,1>,<1,1,2,1>}, {<1,1,1,1,1>}. - _Geoffrey Critzer_, Nov 02 2012
G.f. = x + 2*x^2 + 3*x^3 + 6*x^4 + 10*x^5 + 20*x^6 + 36*x^7 + 72*x^8 + ... - _Michael Somos_, Jun 24 2018
From _Robert A. Russell_, Oct 28 2018: (Start)
For a(5)=10, the 4 achiral patterns (set partitions) are AAAAA, AABAA, ABABA, and ABBBA. The 6 chiral pairs are AAAAB-ABBBB, AAABA-ABAAA, AAABB-AABBB, AABAB-ABABB, AABBA-ABBAA, and ABAAB-ABBAB. The colors are permutable.
For n=4 and a(n+1)=10, the 4 achiral colorings are AAAA, ABBA, BAAB, and BBBB. The 6 achiral pairs are AAAB-BAAA, AABA-ABAA, AABB-BBAA, ABAB-BABA, ABBB-BBBA, and BABB-BBAB. The colors are not permutable. (End)
		

References

  • K. Balasubramanian, "Combinatorial Enumeration of Chemical Isomers", Indian J. Chem., (1978) vol. 16B, pp. 1094-1096. See page 1095.
  • Wayne M. Dymacek, Steinhaus graphs. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 399--412, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561065 (81f:05120)
  • Jablan S. and Sazdanovic R., LinKnot: Knot Theory by Computer, World Scientific Press, 2007.
  • Joseph S. Madachy: Madachy's Mathematical Recreations. New York: Dover Publications, Inc., 1979, p. 46 (first publ. by Charles Scribner's Sons, New York, 1966, under the title: Mathematics on Vacation)
  • M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2.]
  • C. A. Pickover, Keys to Infinity, Wiley 1995, p. 75.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Column 2 of A320750 (set partitions).
Cf. A131577 (oriented), A122746(n-3) (chiral), A016116 (achiral), for set partitions with up to two subsets.
Column 2 of A277504, offset by one (colors not permutable).
Cf. A000079 (oriented), A122746(n-2) (chiral), and A060546 (achiral), for a(n+1).

Programs

  • Haskell
    a005418 n = sum $ a034851_row (n - 1) -- Reinhard Zumkeller, Jan 14 2012
    
  • Maple
    A005418 := n->2^(n-2)+2^(floor(n/2)-1): seq(A005418(n), n=1..34);
  • Mathematica
    LinearRecurrence[{2,2,-4}, {1,2,3}, 40] (* or *) Table[2^(n-2)+2^(Floor[n/2]-1), {n,40}] (* Harvey P. Dale, Jan 18 2012 *)
  • PARI
    A005418(n)= 2^(n-2) + 2^(n\2-1); \\ Joerg Arndt, Sep 16 2013
    
  • Python
    def A005418(n): return 1 if n == 1 else 2**((m:= n//2)-1)*(2**(n-m-1)+1) # Chai Wah Wu, Feb 03 2022

Formula

a(n) = 2^(n-2) + 2^(floor(n/2) - 1).
G.f.: -x*(-1 + 3*x^2) / ( (2*x - 1)*(2*x^2 - 1) ). - Simon Plouffe in his 1992 dissertation
G.f.: x*(1+2*x)*(1-3*x^2)/((1-4*x^2)*(1-2*x^2)), not reduced. - Wolfdieter Lang, May 08 2001
a(n) = 6*a(n - 2) - 8*a(n - 4). a(2*n) = A063376(n - 1) = 2*a(2*n - 1); a(2*n + 1) = A007582(n). - Henry Bottomley, Jul 14 2001
a(n+2) = 2*a(n+1) - A077957(n) with a(1) = 1, a(2) = 2. - Yosu Yurramendi, Oct 24 2008
a(n) = 2*a(n-1) + 2*a(n-2) - 4*a(n-3). - Jaume Oliver Lafont, Dec 05 2008
Union of A007582 and A161168. Union of A007582 and A063376. - Jaroslav Krizek, Aug 14 2009
G.f.: G(0); G(k) = 1 + 2*x/(1 - x*(1+2^(k+1))/(x*(1+2^(k+1)) + (1+2^k)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 12 2011
a(2*n) = 2*a(2*n-1) and a(2*n+1) = a(2*n) + 4^(n-1) with a(1) = 1. - Johannes W. Meijer, Aug 26 2013
From Robert A. Russell, Oct 28 2018: (Start)
a(n) = (A131577(n) + A016116(n)) / 2 = A131577(n) - A122746(n-3) = A122746(n-3) + A016116(n), for set partitions with up to two subsets.
a(n+1) = (A000079(n) + A060546(n)) / 2 = A000079(n) - A122746(n-2) = A122746(n-2) + A060546(n), for two colors that do not permute.
a(n) = Sum_{j=0..k} (S2(n,j) + Ach(n,j)) / 2, where k=2 is the maximum number of colors, S2(n,k) is the Stirling subset number A008277, and Ach(n,k) = [n>=0 & n<2 & n==k] + [n>1]*(k*Ach(n-2,k) + Ach(n-2,k-1) + Ach(n-2,k-2)).
a(n+1) = (k^n + k^ceiling(n/2)) / 2, where k=2 is number of colors we can use. (End)
E.g.f.: (cosh(2*x) + 2*cosh(sqrt(2)*x) + sinh(2*x) + sqrt(2)*sinh(sqrt(2)*x) - 3)/4. - Stefano Spezia, Jun 01 2022

A065941 T(n,k) = binomial(n-floor((k+1)/2), floor(k/2)). Triangle read by rows, for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 4, 3, 3, 1, 1, 1, 5, 4, 6, 3, 1, 1, 1, 6, 5, 10, 6, 4, 1, 1, 1, 7, 6, 15, 10, 10, 4, 1, 1, 1, 8, 7, 21, 15, 20, 10, 5, 1, 1, 1, 9, 8, 28, 21, 35, 20, 15, 5, 1, 1, 1, 10, 9, 36, 28, 56, 35, 35, 15, 6, 1, 1, 1, 11, 10, 45, 36, 84, 56, 70, 35, 21, 6, 1
Offset: 0

Views

Author

Len Smiley, Nov 29 2001

Keywords

Comments

Also the q-Stirling2 numbers at q = -1. - Peter Luschny, Mar 09 2020
Row sums give the Fibonacci sequence. So do the alternating row sums.
Triangle of coefficients of polynomials defined by p(-1,x) = p(0,x) = 1, p(n, x) = x*p(n-1, x) + p(n-2, x), for n >= 1. - Benoit Cloitre, May 08 2005 [rewritten with correct offset. - Wolfdieter Lang, Feb 18 2020]
Another version of triangle in A103631. - Philippe Deléham, Jan 01 2009
The T(n,k) coefficients appear in appendix 2 of Parks's remarkable article "A new proof of the Routh-Hurwitz stability criterion using the second method of Liapunov" if we assume that the b(n) coefficients are all equal to 1 and ignore the first column. The complete version of this triangle including the first column is A103631. - Johannes W. Meijer, Aug 11 2011
Signed ++--++..., the roots are chaotic using f(x) --> x^2 - 2 with cycle lengths shown in A003558 by n-th rows. Example: given row 3, x^3 + x^2 - 2x - 1; the roots are (a = 1.24697, ...; b = -0.445041, ...; c = -1.802937, ...). Then (say using seed b with x^2 - 2) we obtain the trajectory -0.445041, ... -> -1.80193, ... -> 1.24697, ...; matching the entry "3" in A003558(3). - Gary W. Adamson, Sep 06 2011
From Gary W. Adamson, Aug 25 2019: (Start)
Roots to the polynomials and terms in A003558 can all be obtained from the numbers below using a doubling series mod N procedure as follows: (more than one row may result). Any row ends when the trajectory produces a term already used. Then try the next higher odd term not used as the leftmost term, then repeat.
For example, for N = 11, we get: (1, 2, 4, 3, 5), showing that when confronted with two choices after the 4: (8 and -3), pick the smaller (abs) term, = 3. Then for the next row pick 7 (not used) and repeat the algorithm; succeeding only if the trajectory produces new terms. But 7 is also (-4) mod 11 and 4 was used. Therefore what I call the "r-t table" (for roots trajectory) has only one row: (1, 2, 4, 3, 5). Conjecture: The numbers of terms in the first row is equal to A003558 corresponding to N, i.e., 5 in this case with period 2.
Now for the roots to the polynomials. Pick N = 7. The polynomial is x^3 - x^2 - 2x + 1 = 0, with roots 1.8019..., -1.2469... and 0.445... corresponding to 2*cos(j*Pi/N), N = 7, and j = (1, 2, and 3). The terms (1, 2, 3) are the r-t terms for N = 7. For 11, the r-t terms are (1, 2, 4, 3, 5). This implies that given any roots of the corresponding polynomial, they are cyclic using f(x) --> x^2 - 2 with cycle lengths shown in A003558. The terms thus generated are 2*cos(j*Pi), with j = (1, 2, 4, 3, 5). Check: Begin with 2*j*Pi/N, with j = 1 (1.9189...). The other trajectory terms are: --> 1.6825..., --> 0.83083..., -1.3097...; 545...; (a 5 period and cyclic since we can begin with any of the constants). The r-t table for odd N begins as follows:
3...............1
5...............1, 2
7...............1, 2, 3
9...............1, 2, 4
...............3 (singleton terms reduce to "1") (9 has two rows)
11...............1, 2, 4, 3, 5
13...............1, 2, 4, 5, 3, 6
15...............1, 2, 4, 7
................3, 6 (dividing through by the gcd gives (1, 2))
................5. (singleton terms reduce to "1")
The result is that 15 has 3 factors (since 3 rows), and the values of those factors are the previous terms "N", corresponding to the r-t terms in each row. Thus, the first row is new, the second (1, 2), corresponds to N = 5, and the "1" in row 3 corresponds to N = 3. The factors are those values apart from 15 and 1. Note that all of the unreduced r-t terms in all rows for N form a complete set of the terms 1 through (N-1)/2 without duplication. (End)
From Gary W. Adamson, Sep 30 2019: (Start)
The 3 factors of the 7th degree polynomial for 15: (x^7 - x^6 - 6x^5 + 5x^4 + 10x^3 - 6x^2 - 4x + 1) can be determined by getting the roots for 2*cos(j*Pi/1), j = (1, 2, 4, 7) and finding the corresponding polynomial, which is x^4 + x^3 - 4x^2 - 4x + 1. This is the minimal polynomial for N = 15 as shown in Table 2, p. 46 of (Lang). The degree of this polynomial is 4, corresponding to the entry in A003558 for 15, = 4. The trajectories (3, 6) and (5) are j values for 2*cos(j*Pi/15) which are roots to x^2 - x - 1 (relating to the pentagon), and (x - 1), relating to the triangle. (End)
From Gary W. Adamson, Aug 21 2019: (Start)
Matrices M of the form: (1's in the main diagonal, -1's in the subdiagonal, and the rest zeros) are chaotic if we replace (f(x) --> x^2 - 2) with f(x) --> M^2 - 2I, where I is the Identity matrix [1, 0, 0; 0, 1, 0; 0, 0, 1]. For example, with the 3 X 3 matrix M: [0, 0, 1; 0, 1, -1; 1, -1, 0]; the f(x) trajectory is:
....M^2 - 2I: [-1, -1, 0; -1, 0, -1; 0, -1, 0], then for the latter,
....M^2 - 2I: [0, 1, 1; 1, 0, 0; 1, 0, -1]. The cycle ends with period 3 since the next matrix is (-1) * the seed matrix. As in the case with f(x) --> x^2 - 2, the eigenvalues of the 3 chaotic matrices are (abs) 1.24697, 0.44504... and 1.80193, ... Also, the characteristic equations of the 3 matrices are the same as or variants of row 4 of the triangle below: (x^3 + x - 2x - 1) with different signs. (End)
Received from Herb Conn, Jan 2004: (Start)
Let x = 2*cos(2A) (A = Angle); then
sin(A)/sin A = 1
sin(3A)/sin A = x + 1
sin(5A)/sin A = x^2 + x - 1
sin(7A)/sin A = x^3 + x - 2x - 1
sin(9A)/sin A = x^4 + x^3 - 3x^2 - 2x + 1
... (signed ++--++...). (End)
Or Pascal's triangle (A007318) with duplicated diagonals. Also triangle of coefficients of polynomials defined by P_0(x) = 1 and for n>=1, P_n(x) = F_n(x) + F_(n+1)(x), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} C(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 12 2012
The matrix inverse is given by
1;
1, 1;
0, -1, 1;
0, 1, -2, 1;
0, 0, 1, -2, 1;
0, 0, -1, 3, -3, 1;
0, 0, 0, -1, 3, -3, 1;
0, 0, 0, 1, -4, 6, -4, 1;
0, 0, 0, 0, 1, -4, 6, -4, 1;
... apart from signs the same as A124645. - R. J. Mathar, Mar 12 2013

Examples

			Triangle T(n, k) begins:
n\k 0  1  2  3   4   5  6   7  8  9 ...
---------------------------------------
[0] 1,
[1] 1, 1,
[2] 1, 1, 1,
[3] 1, 1, 2, 1,
[4] 1, 1, 3, 2,  1,
[5] 1, 1, 4, 3,  3,  1,
[6] 1, 1, 5, 4,  6,  3,  1,
[7] 1, 1, 6, 5, 10,  6,  4,  1,
[8] 1, 1, 7, 6, 15, 10, 10,  4,  1,
[9] 1, 1, 8, 7, 21, 15, 20, 10,  5, 1,
---------------------------------------
From _Gary W. Adamson_, Oct 23 2019: (Start)
Consider the roots of the polynomials corresponding to odd N such that for N=7 the polynomial is (x^3 + x^2 - 2x - 1) and the roots (a, b, c) are (-1.8019377..., 1.247697..., and -0.445041...). The discriminant of a polynomial derived from the roots is the square of the product of successive differences: ((a-b), (b-c), (c-a))^2 in this case, resulting in 49, matching the method derived from the coefficients of a cubic. For our purposes we use the product of the differences, not the square, resulting in (3.048...) * (1.69202...) * (1.35689...) = 7.0. Conjecture: for all polynomials in the set, the product of the differences of the roots = the corresponding N. For N = 7, we get x^3 - 7x + 7. It appears that for all prime N's, these resulting companion polynomials are monic (left coefficient is 1), and all other coefficients are N or multiples thereof, with the rightmost term = N. The companion polynomials for the first few primes are:
  N =  5:  x^2 - 5;
  N =  7:  x^3 - 7x + 7;
  N = 11:  x^5 - 11x^3 + 11x^2 + 11x - 11;
  N = 13:  x^6 - 13x^4 + 13x^3 + 26x^2 - 39x + 13;
  N = 17:  x^8 - 17x^6 + 17x^5 + 68x^4 - 119x^3 + 17x^2 + 51x - 17;
  N = 19:  x^9 - 19x^7 + 19x^6 + 95x^5 - 171x^4 - 19x^3 + 190x^2 - 114x + 19. (End)
		

Crossrefs

Cf. A065942 (central stalk sequence), A000045 (row sums), A108299.
Reflected version of A046854.
Some triangle sums (see A180662): A000045 (Fi1), A016116 (Kn21), A000295 (Kn23), A094967 (Fi2), A000931 (Ca2), A001519 (Gi3), A000930 (Ze3).

Programs

  • Haskell
    a065941 n k = a065941_tabl !! n !! k
    a065941_row n = a065941_tabl !! n
    a065941_tabl = iterate (\row ->
       zipWith (+) ([0] ++ row) (zipWith (*) (row ++ [0]) a059841_list)) [1]
    -- Reinhard Zumkeller, May 07 2012
    
  • Magma
    [Binomial(n - Floor((k+1)/2), Floor(k/2)): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 10 2019
    
  • Maple
    A065941 := proc(n,k): binomial(n-floor((k+1)/2),floor(k/2)) end: seq(seq(A065941(n,k), k=0..n), n=0..15); # Johannes W. Meijer, Aug 11 2011
    A065941 := proc(n,k) option remember: local j: if k=0 then 1 elif k=1 then 1: elif k>=2 then add(procname(j,k-2), j=k-2..n-2) fi: end: seq(seq(A065941(n,k), k=0..n), n=0..15);  # Johannes W. Meijer, Aug 11 2011
    # The function qStirling2 is defined in A333143.
    seq(print(seq(qStirling2(n, k, -1), k=0..n)), n=0..9);
    # Peter Luschny, Mar 09 2020
  • Mathematica
    Flatten[Table[Binomial[n-Floor[(k+1)/2],Floor[k/2]],{n,0,15},{k,0,n}]] (* Harvey P. Dale, Dec 11 2011 *)
  • PARI
    T065941(n, k) = binomial(n-(k+1)\2, k\2); \\ Michel Marcus, Apr 28 2014
    
  • Sage
    [[binomial(n - floor((k+1)/2), floor(k/2)) for k in (0..n)] for n in (0..15)] # G. C. Greubel, Jul 10 2019

Formula

T(n, k) = binomial(n-floor((k+1)/2), floor(k/2)).
As a square array read by antidiagonals, this is given by T1(n, k) = binomial(floor(n/2) + k, k). - Paul Barry, Mar 11 2003
Triangle is a reflection of that in A066170 (absolute values). - Gary W. Adamson, Feb 16 2004
Recurrences: T(k, 0) = 1, T(k, n) = T(k-1, n) + T(k-2, n-2), or T(k, n) = T(k-1, n) + T(k-1, n-1) if n even, T(k-1, n-1) if n odd. - Ralf Stephan, May 17 2004
G.f.: sum[n, sum[k, T(k, n)x^ky^n]] = (1+xy)/(1-y-x^2y^2). sum[n>=0, T(k, n)y^n] = y^k/(1-y)^[k/2]. - Ralf Stephan, May 17 2004
T(n, k) = A108299(n, k)*A087960(k) = abs(A108299(n, k)). - Reinhard Zumkeller, Jun 01 2005
From Johannes W. Meijer, Aug 11 2011: (Start)
T(n,k) = A046854(n, n-k) = abs(A066170(n, n-k)).
T(n+k, n-k) = A109223(n,k).
T(n, k) = sum(T(j, k-2), j=k-2..n-2), 2 <= k <= n, n>=2;
T(n, 0) =1, T(n+1, 1) = 1, n >= 0. (End)
For n > 1: T(n, k) = T(n-2, k) + T(n-1, k), 1 < k < n. - Reinhard Zumkeller, Apr 24 2013

A052955 a(2n) = 2*2^n - 1, a(2n+1) = 3*2^n - 1.

Original entry on oeis.org

1, 2, 3, 5, 7, 11, 15, 23, 31, 47, 63, 95, 127, 191, 255, 383, 511, 767, 1023, 1535, 2047, 3071, 4095, 6143, 8191, 12287, 16383, 24575, 32767, 49151, 65535, 98303, 131071, 196607, 262143, 393215, 524287, 786431, 1048575, 1572863, 2097151, 3145727
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

a(n) is the least k such that A056792(k) = n.
One quarter of the number of positive integer (n+2) X (n+2) arrays with every 2 X 2 subblock summing to 1. - R. H. Hardin, Sep 29 2008
Number of length n+1 left factors of Dyck paths having no DUU's (here U=(1,1) and D=(1,-1)). Example: a(4)=7 because we have UDUDU, UUDDU, UUDUD, UUUDD, UUUDU, UUUUD, and UUUUU (the paths UDUUD, UDUUU, and UUDUU do not qualify).
Number of binary palindromes < 2^n (see A006995). - Hieronymus Fischer, Feb 03 2012
Partial sums of A016116 (omitting the initial term). - Hieronymus Fischer, Feb 18 2012
a(n - 1), n > 1, is the number of maximal subsemigroups of the monoid of order-preserving or -reversing partial injective mappings on a set with n elements. - Wilf A. Wilson, Jul 21 2017
Number of monomials of the algebraic normal form of the Boolean function representing the n-th bit of the product 3x in terms of the bits of x. - Sebastiano Vigna, Oct 04 2020

Examples

			G.f. = 1 + 2*x + 3*x^2 + 5*x^3 + 7*x^4 + 11*x^5 + 15*x^6 + 23*x^7 + ... - _Michael Somos_, Jun 24 2018
		

Crossrefs

Cf. A000225 for even terms, A055010 for odd terms. See also A056792.
Essentially 1 more than A027383, 2 more than A060482. [Comment corrected by Klaus Brockhaus, Aug 09 2009]
Union of A000225 & A055010.
For partial sums see A027383.
See A016116 for the first differences.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A029744 = {s(n), n>=1}, the numbers 2^k and 3*2^k, as the parent: A029744 (s(n)); A052955 (s(n)-1), A027383 (s(n)-2), A354788 (s(n)-3), A347789 (s(n)-4), A209721 (s(n)+1), A209722 (s(n)+2), A343177 (s(n)+3), A209723 (s(n)+4); A060482, A136252 (minor differences from A354788 at the start); A354785 (3*s(n)), A354789 (3*s(n)-7). The first differences of A029744 are 1,1,1,2,2,4,4,8,8,... which essentially matches eight sequences: A016116, A060546, A117575, A131572, A152166, A158780, A163403, A320770. The bisections of A029744 are A000079 and A007283. - N. J. A. Sloane, Jul 14 2022

Programs

  • GAP
    List([0..45], n-> ((5-(-1)^n)/2)*2^((2*n-1+(-1)^n)/4)-1); # G. C. Greubel, Oct 22 2019
    
  • Haskell
    a052955 n = a052955_list !! n
    a052955_list = 1 : 2 : map ((+ 1) . (* 2)) a052955_list
    -- Reinhard Zumkeller, Feb 22 2012
    
  • Magma
    [((5-(-1)^n)/2)*2^((2*n-1+(-1)^n)/4)-1: n in [0..45]]; // G. C. Greubel, Oct 22 2019
    
  • Maple
    spec := [S,{S=Prod(Sequence(Prod(Union(Z,Z),Z)),Union(Sequence(Z),Z))}, unlabeled ]: seq(combstruct[count ](spec,size=n), n=0..20);
    a[0]:=0:a[1]:=1:for n from 2 to 100 do a[n]:=2*a[n-2]+2 od: seq(a[n]/2, n=2..43); # Zerinvary Lajos, Mar 16 2008
  • Mathematica
    a[n_]:= If[EvenQ[n], 2^(n/2+1) -1, 3*2^((n-1)/2) -1]; Table[a[n], {n, 0, 41}] (* Robert G. Wilson v, Jun 05 2004 *)
    a[0]=1; a[1]=2; a[n_]:= a[n]= 2 a[n-2] +1; Array[a, 42, 0]
    a[n_]:= (2 + Mod[n, 2]) 2^Quotient[n, 2] - 1; (* Michael Somos, Jun 24 2018 *)
  • PARI
    a(n)=(2+n%2)<<(n\2)-1 \\ Charles R Greathouse IV, Jun 19 2011
    
  • PARI
    {a(n) = (n%2 + 2) * 2^(n\2) - 1}; /* Michael Somos, Jun 24 2018 */
    
  • Perl
    # command line argument tells how high to take n
    # Beyond a(38) = 786431 you may need a special code to handle large integers
      $lim = shift;
      sub show{};
    $n = $incr = $P = 1;
    show($n, $incr, $P);
    $incr = 1;
    for $n (2..$lim) {
        $P += $incr;
        show($n, $P, $incr, $P);
        $incr *=2 if ($n % 2); # double the increment after an odd n
    }
    sub show {
        my($n, $P) = @_;
        printf("%4d\t%16g\n", $n, $P);
    }
    # Mark A. Mandel (thnidu aT  g ma(il) doT c0m), Dec 29 2010
    
  • Python
    def A052955(n): return ((2|n&1)<<(n>>1))-1 # Chai Wah Wu, Jul 13 2023
  • Sage
    [((5-(-1)^n)/2)*2^((2*n-1+(-1)^n)/4)-1 for n in (0..45)] # G. C. Greubel, Oct 22 2019
    

Formula

a(0)=1, a(1)=2; thereafter a(n) = 2*a(n-2) + 1, n >= 2.
G.f.: (1 + x - x^2)/((1 - x)*(1 - 2*x^2)).
a(n) = -1 + Sum_{alpha = RootOf(-1 + 2*Z^2)} (1/4) * (3 + 4*alpha) * alpha^(-1-n). (That is, the sum is indexed by the roots of the polynomial -1 + 2*Z^2.)
a(n) = 2^(n/2) * (3*sqrt(2)/4 + 1 - (3*sqrt(2)/4 - 1) * (-1)^n) - 1. - Paul Barry, May 23 2004
a(n) = 1 + Sum_{k=0..n-1} A016116(k). - Robert G. Wilson v, Jun 05 2004
A132340(a(n)) = A027383(n). - Reinhard Zumkeller, Aug 20 2007
From Hieronymus Fischer, Sep 15 2007: (Start)
a(n) = A027383(n-1) + 1 for n>0.
a(n) = A132666(a(n+1)-1).
a(n) = A132666(a(n-1)) + 1 for n>0.
A132666(a(n)) = a(n+1) - 1. (End)
a(n) = A027383(n+1)/2. - Zerinvary Lajos, Mar 16 2008
a(n) = (5 - (-1)^n)/2*2^floor(n/2) - 1. - Hieronymus Fischer, Feb 03 2012
a(2n+1) = (a(2*n) + a(2*n+2))/2. Combined with a(n) = 2*a(n-2) + 1, n >= 2 and a(0) = 1, this specifies the sequence. - Richard R. Forberg, Nov 30 2013
a(n) = ((5 - (-1)^n)/2)*2^((2*n - 1 + (-1)^n)/4) - 1. - Luce ETIENNE, Sep 20 2014
a(n) = -(2^(n+1)) * A107659(-3-n) for all n in Z. - Michael Somos, Jun 24 2018
E.g.f.: (1/4)*exp(-sqrt(2)*x)*(4 - 3*sqrt(2) + (4 + 3*sqrt(2))*exp(2*sqrt(2)*x) - 4*exp(x + sqrt(2)*x)). - Stefano Spezia, Oct 22 2019
A term k appears in this sequence <=> 4 does not divide binomial(k, j) for any j in 0..k. - Peter Luschny, Jun 28 2025

Extensions

Formula and more terms from Henry Bottomley, May 03 2000
Additional comments from Robert G. Wilson v, Jan 29 2001
Minor edits from N. J. A. Sloane, Jul 09 2022

A077957 Powers of 2 alternating with zeros.

Original entry on oeis.org

1, 0, 2, 0, 4, 0, 8, 0, 16, 0, 32, 0, 64, 0, 128, 0, 256, 0, 512, 0, 1024, 0, 2048, 0, 4096, 0, 8192, 0, 16384, 0, 32768, 0, 65536, 0, 131072, 0, 262144, 0, 524288, 0, 1048576, 0, 2097152, 0, 4194304, 0, 8388608, 0, 16777216, 0, 33554432, 0, 67108864, 0, 134217728, 0, 268435456
Offset: 0

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Author

N. J. A. Sloane, Nov 17 2002

Keywords

Comments

Normally sequences like this are not included, since with the alternating 0's deleted it is already in the database.
Inverse binomial transform of A001333. - Paul Barry, Feb 25 2003
"Sloping binary representation" of powers of 2 (A000079), slope=-1 (see A037095 and A102370). - Philippe Deléham, Jan 04 2008
0,1,0,2,0,4,0,8,0,16,... is the inverse binomial transform of A000129 (Pell numbers). - Philippe Deléham, Oct 28 2008
Number of maximal self-avoiding walks from the NW to SW corners of a 3 X n grid.
Row sums of the triangle in A204293. - Reinhard Zumkeller, Jan 14 2012
Pisano period lengths: 1, 1, 4, 1, 8, 4, 6, 1, 12, 8, 20, 4, 24, 6, 8, 1, 16, 12, 36, 8, ... . - R. J. Mathar, Aug 10 2012
This sequence occurs in the length L(n) = sqrt(2)^n of Lévy's C-curve at the n-th iteration step. Therefore, L(n) is the Q(sqrt(2)) integer a(n) + a(n-1)*sqrt(2), with a(-1) = 0. For a variant of this C-curve see A251732 and A251733. - Wolfdieter Lang, Dec 08 2014
a(n) counts walks (closed) on the graph G(1-vertex,2-loop,2-loop). Equivalently the middle entry (2,2) of A^n where the adjacency matrix of digraph is A=(0,1,0;1,0,1;0,1,0). - David Neil McGrath, Dec 19 2014
a(n-2) is the number of compositions of n into even parts. For example, there are 4 compositions of 6 into even parts: (6), (222), (42), and (24). - David Neil McGrath, Dec 19 2014
Also the number of alternately constant compositions of n + 2, ranked by A351010. The alternately strict version gives A000213. The unordered version is A035363, ranked by A000290, strict A035457. - Gus Wiseman, Feb 19 2022
a(n) counts degree n fixed points of GF(2)[x]'s automorphisms. Proof: given a field k, k[x]'s automorphisms are determined by k's automorphisms and invertible affine maps x -> ax + b. GF(2) is rigid and has only one unit so its only nontrivial automorphism is x -> x + 1. For n = 0 we have 1 fixed point, the constant polynomial 1. (Taking the convention that 0 is not a degree 0 polynomial.) For n = 1 we have 0 fixed points as x -> x + 1 -> x are the only degree 1 polynomials. Note that if f(x) is a fixed point, then f(x) + 1 is also a fixed point. Given f(x) a degree n fixed point, we can assume WLOG x | f(x). Applying the automorphism, we then have x + 1 | f(x). Now note that f(x) / (x^2 + x) must be a fixed point, so any fixed point of degree n must either be of the form g(x) * (x^2 + x) or g(x) * (x^2 + x) + 1 for a unique degree n - 2 fixed point g(x). Therefore we have the recurrence relation a(n) = 2 * a(n - 2) as desired. - Keith J. Bauer, Mar 19 2024

Crossrefs

Column k=3 of A219946. - Alois P. Heinz, Dec 01 2012
Cf. A016116 (powers repeated).

Programs

  • GAP
    Flat(List([0..30],n->[2^n,0])); # Muniru A Asiru, Aug 05 2018
  • Haskell
    a077957 = sum . a204293_row  -- Reinhard Zumkeller, Jan 14 2012
    
  • Magma
    &cat [[2^n,0]: n in [0..20]]; // Vincenzo Librandi, Apr 03 2018
    
  • Maple
    seq(op([2^n,0]),n=0..100); # Robert Israel, Dec 23 2014
  • Mathematica
    a077957[n_] := Riffle[Table[2^i, {i, 0, n - 1}], Table[0, {n}]]; a077957[29] (* Michael De Vlieger, Dec 22 2014 *)
    CoefficientList[Series[1/(1 - 2*x^2), {x,0,50}], x] (* G. C. Greubel, Apr 12 2017 *)
    LinearRecurrence[{0, 2}, {1, 0}, 54] (* Robert G. Wilson v, Jul 23 2018 *)
    Riffle[2^Range[0,30],0,{2,-1,2}] (* Harvey P. Dale, Jan 06 2022 *)
  • PARI
    a(n)=if(n<0||n%2, 0, 2^(n/2))
    
  • Sage
    def A077957():
        x, y = -1, 1
        while True:
            yield -x
            x, y = x + y, x - y
    a = A077957(); [next(a) for i in range(40)]  # Peter Luschny, Jul 11 2013
    

Formula

G.f.: 1/(1-2*x^2).
E.g.f.: cosh(x*sqrt(2)).
a(n) = (1 - n mod 2) * 2^floor(n/2).
a(n) = sqrt(2)^n*(1+(-1)^n)/2. - Paul Barry, May 13 2003
a(n) = 2*a(n-2) with a(0)=1, a(1)=0. - Jim Singh, Jul 12 2018

A046854 Triangle read by rows: T(n, k) = binomial(floor((n+k)/2), k), n >= k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 3, 3, 4, 1, 1, 1, 3, 6, 4, 5, 1, 1, 1, 4, 6, 10, 5, 6, 1, 1, 1, 4, 10, 10, 15, 6, 7, 1, 1, 1, 5, 10, 20, 15, 21, 7, 8, 1, 1, 1, 5, 15, 20, 35, 21, 28, 8, 9, 1, 1, 1, 6, 15, 35, 35, 56, 28, 36, 9, 10, 1, 1, 1, 6, 21, 35, 70, 56, 84, 36, 45, 10, 11, 1, 1
Offset: 0

Views

Author

Keywords

Comments

Row sums are Fibonacci(n+2). Diagonal sums are A016116. - Paul Barry, Jul 07 2004
Riordan array (1/(1-x), x/(1-x^2)). Matrix inverse is A106180. - Paul Barry, Apr 24 2005
As an infinite lower triangular matrix * [1,2,3,...] = A055244. - Gary W. Adamson, Dec 23 2008
From Emeric Deutsch, Jun 18 2010: (Start)
T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n} of size k, starting with an odd number (Terquem's problem, see the Riordan reference, p. 17). Example: T(8,5)=6 because we have 12345, 12347, 12367, 12567, 14567, and 34567.
T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n,n+1} of size k, starting with an even number. Example: T(7,4)=5 because we have 2345, 2347, 2367, 2567, and 4567. (End)
From L. Edson Jeffery, Mar 01 2011: (Start)
This triangle can be constructed as follows. Interlace two copies of the table of binomial coefficients to get the preliminary table
1
1
1 1
1 1
1 2 1
1 2 1
1 3 3 1
1 3 3 1
...,
then shift each entire r-th column up r rows, r=0,1,2,.... Also, a signed version of this sequence (A187660 in tabular form) begins with
1;
1, -1;
1, -1, -1;
1, -2, -1, 1;
1, -2, -3, 1, 1;
...
(compare with A066170, A130777). Let T(N,k) denote the k-th entry in row N of the signed table. Then, for N>1, row N gives the coefficients of the characteristic function p_N(x) = Sum_{k=0..N} T(N,k)*x^(N-k) = 0 of the N X N matrix U_N=[(0 ... 0 1);(0 ... 0 1 1);...;(0 1 ... 1);(1 ... 1)]. Now let Q_r(t) be a polynomial with recurrence relation Q_r(t)=t*Q_(r-1)(t)-Q_(r-2)(t) (r>1), with Q_0(t)=1 and Q_1(t)=t. Then p_N(x)=0 has solutions Q_(N-1)(phi_j), where phi_j=2*(-1)^(j-1)*cos(j*Pi/(2*N+1)), j=1,2,...,N.
For example, row N=3 is {1,-2,-1,1}, giving the coefficients of the characteristic function p_3(x) = x^3-2*x^2-x+1 = 0 for the 3 X 3 matrix U_3=[(0 0 1);(0 1 1);(1 1 1)], with eigenvalues Q_2(phi_j)=[2*(-1)^(j-1)*cos(j*Pi/7)]^2-1, j=1,2,3. (End)
Given the signed polynomials (+--++--,...) of the triangle, the largest root of the n-th row polynomial is the longest (2n+1) regular polygon diagonal length, with edge = 1. Example: the largest root to x^3 - 2x^2 - x + 1 = 0 is 2.24697...; the longest heptagon diagonal, sin(3*Pi/7)/sin(Pi/7). - Gary W. Adamson, Sep 06 2011
Given the signed polynomials from Gary W. Adamson's comment, the largest root of the n-th polynomial also equals the length from the center to a corner (vertex) of a regular 2*(2*n+1)-sided polygon with side (edge) length = 1. - L. Edson Jeffery, Jan 01 2012
Put f(x,0) = 1 and f(x,n) = x + 1/f(x,n-1). Then f(x,n) = u(x,n)/v(x,n), where u(x,n) and v(x,n) are polynomials. The flattened triangles of coefficients of u and v are both essentially A046854, as indicated by the Mathematica program headed "Polynomials". - Clark Kimberling, Oct 12 2014
From Jeremy Dover, Jun 07 2016: (Start)
T(n,k) is the number of binary strings of length n+1 starting with 0 that have exactly k pairs of consecutive 0's and no pairs of consecutive 1's.
T(n,k) is the number of binary strings of length n+2 starting with 1 that have exactly k pairs of consecutive 0's and no pairs of consecutive 1's. (End)

Examples

			Triangle begins:
  1;
  1 1;
  1 1 1;
  1 2 1 1;
  1 2 3 1 1;
  1 3 3 4 1 1;
  ...
		

References

  • J. Riordan, An Introduction to Combinatorial Analysis, Princeton University Press, 1978. [Emeric Deutsch, Jun 18 2010]

Crossrefs

Reflected version of A065941, which is considered the main entry. A deficient version is in A030111.
Cf. A055244. - Gary W. Adamson, Dec 23 2008

Programs

  • GAP
    Flat(List([0..16], n-> List([0..n], k-> Binomial(Int((n+k)/2), k) ))); # G. C. Greubel, Jul 13 2019
  • Haskell
    a046854 n k = a046854_tabl !! n !! k
    a046854_row n = a046854_tabl !! n
    a046854_tabl = [1] : f [1] [1,1] where
       f us vs = vs : f vs  (zipWith (+) (us ++ [0,0]) ([0] ++ vs))
    -- Reinhard Zumkeller, Apr 24 2013
    
  • Magma
    [Binomial(Floor((n+k)/2), k): k in [0..n], n in [0..16]]; // G. C. Greubel, Jul 13 2019
    
  • Maple
    A046854:= proc(n,k): binomial(floor(n/2+k/2), k) end: seq(seq(A046854(n,k),k=0..n),n=0..16); # Nathaniel Johnston, Jun 30 2011
  • Mathematica
    Table[Binomial[Floor[(n+k)/2], k], {n,0,16}, {k,0,n}]//Flatten
    (* next program: Polynomials *)
    z = 12; f[x_, n_] := x + 1/f[x, n - 1]; f[x_, 1] = 1;
    t = Table[Factor[f[x, n]], {n, 1, z}]
    u = Flatten[CoefficientList[Numerator[t], x]] (* this sequence *)
    v = Flatten[CoefficientList[Denominator[t], x]]
    (* Clark Kimberling, Oct 13 2014 *)
  • PARI
    T(n,k) = binomial((n+k)\2, k); \\ G. C. Greubel, Jul 13 2019
    
  • Sage
    [[binomial(floor((n+k)/2), k) for k in (0..n)] for n in (0..16)] # G. C. Greubel, Jul 13 2019
    

Formula

T(n,k) = binomial(floor((n+k)/2), k).
G.f.: (1+x)/(1-x*y-x^2). - Ralf Stephan, Feb 13 2005
Triangle = A097806 * A168561, as infinite lower triangular matrices. - Gary W. Adamson, Oct 28 2007
T(n,k) = A065941(n,n-k) = abs(A130777(n,k)) = abs(A066170(n,k)) = abs(A187660(n,k)). - Johannes W. Meijer, Aug 08 2011
For n > 1: T(n, k) = T(n-1, k-1) + T(n-2, k), 0 < k < n-1. - Reinhard Zumkeller, Apr 24 2013
T(n,k) = A168561(n,k) + A168561(n-1,k). - R. J. Mathar, Feb 10 2024

A155585 a(n) = 2^n*E(n, 1) where E(n, x) are the Euler polynomials.

Original entry on oeis.org

1, 1, 0, -2, 0, 16, 0, -272, 0, 7936, 0, -353792, 0, 22368256, 0, -1903757312, 0, 209865342976, 0, -29088885112832, 0, 4951498053124096, 0, -1015423886506852352, 0, 246921480190207983616, 0, -70251601603943959887872, 0, 23119184187809597841473536, 0
Offset: 0

Views

Author

Paul D. Hanna, Jan 24 2009

Keywords

Comments

Previous name was: a(n) = Sum_{k=0..n-1} (-1)^(k)*C(n-1,k)*a(n-1-k)*a(k) for n>0 with a(0)=1.
Factorials have a similar recurrence: f(n) = Sum_{k=0..n-1} C(n-1,k)*f(n-1-k)*f(k), n > 0.
Related to A102573: letting T(q,r) be the coefficient of n^(r+1) in the polynomial 2^(q-n)/n times Sum_{k=0..n} binomial(n,k)*k^q, then A155585(x) = Sum_{k=0..x-1} T(x,k)*(-1)^k. See Mathematica code below. - John M. Campbell, Nov 16 2011
For the difference table and the relation to the Seidel triangle see A239005. - Paul Curtz, Mar 06 2014
From Tom Copeland, Sep 29 2015: (Start)
Let z(t) = 2/(e^(2t)+1) = 1 + tanh(-t) = e.g.f.(-t) for this sequence = 1 - t + 2*t^3/3! - 16*t^5/5! + ... .
dlog(z(t))/dt = -z(-t), so the raising operators that generate Appell polynomials associated with this sequence, A081733, and its reciprocal, A119468, contain z(-d/dx) = e.g.f.(d/dx) as the differential operator component.
dz(t)/dt = z*(z-2), so the assorted relations to a Ricatti equation, the Eulerian numbers A008292, and the Bernoulli numbers in the Rzadkowski link hold.
From Michael Somos's formula below (drawing on the Edwards link), y(t,1)=1 and x(t,1) = (1-e^(2t))/(1+e^(2t)), giving z(t) = 1 + x(t,1). Compare this to the formulas in my list in A008292 (Sep 14 2014) with a=1 and b=-1,
A) A(t,1,-1) = A(t) = -x(t,1) = (e^(2t)-1)/(1+e^(2t)) = tanh(t) = t + -2*t^3/3! + 16*t^5/5! + -272*t^7/7! + ... = e.g.f.(t) - 1 (see A000182 and A000111)
B) Ainv(t) = log((1+t)/(1-t))/2 = tanh^(-1)(t) = t + t^3/3 + t^5/5 + ..., the compositional inverse of A(t)
C) dA/dt = (1-A^2), relating A(t) to a Weierstrass elliptic function
D) ((1-t^2)d/dt)^n t evaluated at t=0, a generator for the sequence A(t)
F) FGL(x,y)= (x+y)/(1+xy) = A(Ainv(x) + Ainv(y)), a related formal group law corresponding to the Lorentz FGL (Lorentz transformation--addition of parallel velocities in special relativity) and the Atiyah-Singer signature and the elliptic curve (1-t^2)*s = t^3 in Tate coordinates according to the Lenart and Zainoulline link and the Buchstaber and Bunkova link (pp. 35-37) in A008292.
A133437 maps the reciprocal odd natural numbers through the refined faces of associahedra to a(n).
A145271 links the differential relations to the geometry of flow maps, vector fields, and thereby formal group laws. See Mathworld for links of tanh to other geometries and statistics.
Since the a(n) are related to normalized values of the Bernoulli numbers and the Riemann zeta and Dirichlet eta functions, there are links to Witten's work on volumes of manifolds in two-dimensional quantum gauge theories and the Kervaire-Milnor formula for homotopy groups of hyperspheres (see my link below).
See A101343, A111593 and A059419 for this and the related generator (1 + t^2) d/dt and associated polynomials. (End)
With the exception of the first term (1), entries are the alternating sums of the rows of the Eulerian triangle, A008292. - Gregory Gerard Wojnar, Sep 29 2018

Examples

			E.g.f.: 1 + x - 2*x^3/3! + 16*x^5/5! - 272*x^7/7! + 7936*x^9/9! -+ ... = exp(x)/cosh(x).
O.g.f.: 1 + x - 2*x^3 + 16*x^5 - 272*x^7 + 7936*x^9 - 353792*x^11 +- ...
O.g.f.: 1 + x/(1+2*x) + 2!*x^2/((1+2*x)*(1+4*x)) + 3!*x^3/((1+2*x)*(1+4*x)*(1+6*x)) + ...
		

Crossrefs

Equals row sums of A119879. - Johannes W. Meijer, Apr 20 2011
(-1)^n*a(n) are the alternating row sums of A123125. - Wolfdieter Lang, Jul 12 2017

Programs

  • Maple
    A155585 := n -> 2^n*euler(n, 1): # Peter Luschny, Jan 26 2009
    a := proc(n) option remember; `if`(n::even, 0^n, -(-1)^n - add((-1)^k*binomial(n,k) *a(n-k), k = 1..n-1)) end: # Peter Luschny, Jun 01 2016
    # Or via the recurrence of the Fubini polynomials:
    F := proc(n) option remember; if n = 0 then return 1 fi;
    expand(add(binomial(n, k)*F(n-k)*x, k = 1..n)) end:
    a := n -> (-2)^n*subs(x = -1/2, F(n)):
    seq(a(n), n = 0..30); # Peter Luschny, May 21 2021
  • Mathematica
    a[m_] := Sum[(-2)^(m - k) k! StirlingS2[m, k], {k, 0, m}] (* Peter Luschny, Apr 29 2009 *)
    poly[q_] :=  2^(q-n)/n*FunctionExpand[Sum[Binomial[n, k]*k^q, {k, 0, n}]]; T[q_, r_] :=  First[Take[CoefficientList[poly[q], n], {r+1, r+1}]]; Table[Sum[T[x, k]*(-1)^k, {k, 0, x-1}], {x, 1, 16}] (* John M. Campbell, Nov 16 2011 *)
    f[n_] := (-1)^n 2^(n+1) PolyLog[-n, -1]; f[0] = -f[0]; Array[f, 27, 0] (* Robert G. Wilson v, Jun 28 2012 *)
  • PARI
    a(n)=if(n==0,1,sum(k=0,n-1,(-1)^(k)*binomial(n-1,k)*a(n-1-k)*a(k)))
    
  • PARI
    a(n)=local(X=x+x*O(x^n));n!*polcoeff(exp(X)/cosh(X),n)
    
  • PARI
    a(n)=polcoeff(sum(m=0,n,m!*x^m/prod(k=1,m,1+2*k*x+x*O(x^n))),n) \\ Paul D. Hanna, Jul 20 2011
    
  • PARI
    {a(n) = local(A); if( n<0, 0, A = x * O(x^n); n! * polcoeff( 1 + sinh(x + A) / cosh(x + A), n))} /* Michael Somos, Jan 16 2012 */
    
  • PARI
    a(n)=local(A=1+x);for(i=1,n,A=sum(k=0,n,intformal(subst(A,x,-x)+x*O(x^n))^k/k!));n!*polcoeff(A,n)
    for(n=0,30,print1(a(n),", ")) \\ Paul D. Hanna, Nov 25 2013
    
  • Python
    from sympy import bernoulli
    def A155585(n): return (((2<<(m:=n+1))-2)*bernoulli(m)<>1) if n&1 else (0 if n else 1) # Chai Wah Wu, Apr 14 2023
  • Sage
    def A155585(n) :
        if n == 0 : return 1
        return add(add((-1)^(j+1)*binomial(n+1,k-j)*j^n for j in (0..k)) for k in (1..n))
    [A155585(n) for n in (0..26)] # Peter Luschny, Jul 23 2012
    
  • Sage
    def A155585_list(n): # Akiyama-Tanigawa algorithm
        A = [0]*(n+1); R = []
        for m in range(n+1) :
            d = divmod(m+3, 4)
            A[m] = 0 if d[1] == 0 else (-1)^d[0]/2^(m//2)
            for j in range(m, 0, -1) :
                A[j - 1] = j * (A[j - 1] - A[j])
            R.append(A[0])
        return R
    A155585_list(30) # Peter Luschny, Mar 09 2014
    

Formula

E.g.f.: exp(x)*sech(x) = exp(x)/cosh(x). (See A009006.) - Paul Barry, Mar 15 2006
Sequence of absolute values is A009006 (e.g.f. 1+tan(x)).
O.g.f.: Sum_{n>=0} n! * x^n / Product_{k=1..n} (1 + 2*k*x). - Paul D. Hanna, Jul 20 2011
a(n) = 2^n*E_{n}(1) where E_{n}(x) are the Euler polynomials. - Peter Luschny, Jan 26 2009
a(n) = EL_{n}(-1) where EL_{n}(x) are the Eulerian polynomials. - Peter Luschny, Aug 03 2010
a(n+1) = (4^n-2^n)*B_n(1)/n, where B_{n}(x) are the Bernoulli polynomials (B_n(1) = B_n for n <> 1). - Peter Luschny, Apr 22 2009
G.f.: 1/(1-x+x^2/(1-x+4*x^2/(1-x+9*x^2/(1-x+16*x^2/(1-...))))) (continued fraction). - Paul Barry, Mar 30 2010
G.f.: -log(x/(exp(x)-1))/x = Sum_{n>=0} a(n)*x^n/(2^(n+1)*(2^(n+1)-1)*n!). - Vladimir Kruchinin, Nov 05 2011
E.g.f.: exp(x)/cosh(x) = 2/(1+exp(-2*x)) = 2/(G(0) + 1); G(k) = 1 - 2*x/(2*k + 1 - x*(2*k+1)/(x - (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 10 2011
E.g.f. is x(t,1) + y(t,1) where x(t,a) and y(t,a) satisfy y(t,a)^2 = (a^2 - x(t,a)^2) / (1 - a^2 * x(t,a)^2) and dx(t,a) / dt = y(t,a) * (1 - a * x(t,a)^2) and are the elliptic functions of Edwards. - Michael Somos, Jan 16 2012
E.g.f.: 1/(1 - x/(1+x/(1 - x/(3+x/(1 - x/(5+x/(1 - x/(7+x/(1 - x/(9+x/(1 +...))))))))))), a continued fraction. - Paul D. Hanna, Feb 11 2012
E.g.f. satisfies: A(x) = Sum_{n>=0} Integral( A(-x) dx )^n / n!. - Paul D. Hanna, Nov 25 2013
a(n) = -2^(n+1)*Li_{-n}(-1). - Peter Luschny, Jun 28 2012
a(n) = Sum_{k=1..n} Sum_{j=0..k} (-1)^(j+1)*binomial(n+1,k-j)*j^n for n > 0. - Peter Luschny, Jul 23 2012
From Sergei N. Gladkovskii, Oct 25 2012 to Dec 16 2013: (Start)
Continued fractions:
G.f.: 1 + x/T(0) where T(k) = 1 + (k+1)*(k+2)*x^2/T(k+1).
E.g.f.: exp(x)/cosh(x) = 1 + x/S(0) where S(k) = (2*k+1) + x^2/S(k+1).
E.g.f.: 1 + x/(U(0)+x) where U(k) = 4*k+1 - x/(1 + x/(4*k+3 - x/(1 + x/U(k+1)))).
E.g.f.: 1 + tanh(x) = 4*x/(G(0)+2*x) where G(k) = 1 - (k+1)/(1 - 2*x/(2*x + (k+1)^2/G(k+1)));
G.f.: 1 + x/G(0) where G(k) = 1 + 2*x^2*(2*k+1)^2 - x^4*(2*k+1)*(2*k+2)^2*(2*k+3)/G(k+1) (due to Stieltjes).
E.g.f.: 1 + x/(G(0) + x) where G(k) = 1 - 2*x/(1 + (k+1)/G(k+1)).
G.f.: 2 - 1/Q(0) where Q(k) = 1 + x*(k+1)/( 1 - x*(k+1)/Q(k+1)).
G.f.: 2 - 1/Q(0) where Q(k) = 1 + x*k^2 + x/(1 - x*(k+1)^2/Q(k+1)).
G.f.: 1/Q(0) where Q(k) = 1 - 2*x + x*(k+1)/(1-x*(k+1)/Q(k+1)).
G.f.: 1/Q(0) where Q(k) = 1 - x*(k+1)/(1 + x*(k+1)/Q(k+1)).
E.g.f.: 1 + x*Q(0) where Q(k) = 1 - x^2/( x^2 + (2*k+1)*(2*k+3)/Q(k+1)).
G.f.: 2 - T(0)/(1+x) where T(k) = 1 - x^2*(k+1)^2/(x^2*(k+1)^2 + (1+x)^2/T(k+1)).
E.g.f.: 1/(x - Q(0)) where Q(k) = 4*k^2 - 1 + 2*x + x^2*(2*k-1)*(2*k+3)/Q(k+1). (End)
G.f.: 1 / (1 - b(1)*x / (1 - b(2)*x / (1 - b(3)*x / ... ))) where b = A001057. - Michael Somos, Jan 03 2013
From Paul Curtz, Mar 06 2014: (Start)
a(2n) = A000007(n).
a(2n+1) = (-1)^n*A000182(n+1).
a(n) is the binomial transform of A122045(n).
a(n) is the row sum of A081658. For fractional Euler numbers see A238800.
a(n) + A122045(n) = 2, 1, -1, -2, 5, 16, ... = -A163982(n).
a(n) - A122045(n) = -A163747(n).
a(n) is the Akiyama-Tanigawa transform applied to 1, 0, -1/2, -1/2, -1/4, 0, ... = A046978(n+3)/A016116(n). (End)
a(n) = 2^(2*n+1)*(zeta(-n,1/2) - zeta(-n, 1)), where zeta(a, z) is the generalized Riemann zeta function. - Peter Luschny, Mar 11 2015
a(n) = 2^(n + 1)*(2^(n + 1) - 1)*Bernoulli(n + 1, 1)/(n + 1). (From Bill Gosper, Oct 28 2015) - N. J. A. Sloane, Oct 28 2015 [See the above comment from Peter Luschny, Apr 22 2009.]
a(n) = -(n mod 2)*((-1)^n + Sum_{k=1..n-1} (-1)^k*C(n,k)*a(n-k)) for n >= 1. - Peter Luschny, Jun 01 2016
a(n) = (-2)^n*F_{n}(-1/2), where F_{n}(x) is the Fubini polynomial. - Peter Luschny, May 21 2021

Extensions

New name from Peter Luschny, Mar 12 2015

A009545 Expansion of e.g.f. sin(x)*exp(x).

Original entry on oeis.org

0, 1, 2, 2, 0, -4, -8, -8, 0, 16, 32, 32, 0, -64, -128, -128, 0, 256, 512, 512, 0, -1024, -2048, -2048, 0, 4096, 8192, 8192, 0, -16384, -32768, -32768, 0, 65536, 131072, 131072, 0, -262144, -524288, -524288, 0, 1048576, 2097152, 2097152, 0, -4194304, -8388608, -8388608, 0, 16777216, 33554432
Offset: 0

Views

Author

Keywords

Comments

Also first of the two associated sequences a(n) and b(n) built from a(0)=0 and b(0)=1 with the formulas a(n) = a(n-1) + b(n-1) and b(n) = -a(n-1) + b(n-1). The initial terms of the second sequence b(n) are 1, 1, 0, -2, -4, -4, 0, 8, 16, 16, 0, -32, -64, -64, 0, 128, 256, ... The points Mn(a(n)+b(n)*I) of the complex plane are located on the spiral logarithmic rho = 2*(1/2)^(2*theta)/Pi) and on the straight lines drawn from the origin with slopes: infinity, 1/2, 0, -1/2. - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 30 2007
A000225: (1, 3, 7, 15, 31, ...) = 2^n - 1 = INVERT transform of A009545 starting (1, 2, 2, 0, -4, -8, ...). (Cf. comments in A144081). - Gary W. Adamson, Sep 10 2008
Pisano period lengths: 1, 1, 8, 1, 4, 8, 24, 1, 24, 4, 40, 8, 12, 24, 8, 1, 16, 24, 72, 4, ... - R. J. Mathar, Aug 10 2012
The variant 0, 1, -2, 2, 0, -4, 8, -8, 0, 16, -32, 32, 0, -64, (with different signs) is the Lucas U(-2,2) sequence. - R. J. Mathar, Jan 08 2013
(1+i)^n = A146559(n) + a(n)*i where i = sqrt(-1). - Philippe Deléham, Feb 13 2013
This is the Lucas U(2,2) sequence. - Raphie Frank, Nov 28 2015
{A146559, A009545} are the difference analogs of {cos(x),sin(x)} (cf. [Shevelev] link). - Vladimir Shevelev, Jun 08 2017

Crossrefs

Cf. A009116. For minor variants of this sequence see A108520, A084102, A099087.
a(2*n) = A056594(n)*2^n, n >= 1, a(2*n+1) = A057077(n)*2^n.
This is the next term in the sequence A015518, A002605, A000129, A000079, A001477.
Cf. A000225, A144081. - Gary W. Adamson, Sep 10 2008
Cf. A146559.

Programs

  • Magma
    I:=[0,1,2,2]; [n le 4 select I[n] else -4*Self(n-4): n in [1..60]]; // Vincenzo Librandi, Nov 29 2015
    
  • Maple
    t1 := sum(n*x^n, n=0..100): F := series(t1/(1+x*t1), x, 100): for i from 0 to 50 do printf(`%d, `, coeff(F, x, i)) od: # Zerinvary Lajos, Mar 22 2009
    G(x):=exp(x)*sin(x): f[0]:=G(x): for n from 1 to 54 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..50 ); # Zerinvary Lajos, Apr 05 2009
    A009545 := n -> `if`(n<2, n, 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], 2)):
    seq(simplify(A009545(n)), n=0..50); # Peter Luschny, Dec 17 2015
  • Mathematica
    nn=104; Range[0,nn-1]! CoefficientList[Series[Sin[x]Exp[x], {x,0,nn}], x] (* T. D. Noe, May 26 2007 *)
    Join[{a=0,b=1},Table[c=2*b-2*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 17 2011 *)
    f[n_] := (1 + I)^(n - 2) + (1 - I)^(n - 2); Array[f, 51, 0] (* Robert G. Wilson v, May 30 2011 *)
    LinearRecurrence[{2,-2},{0,1},110] (* Harvey P. Dale, Oct 13 2011 *)
  • PARI
    x='x+O('x^66); Vec(serlaplace(exp(x)*sin(x))) /* Joerg Arndt, Apr 24 2011 */
    
  • PARI
    x='x+O('x^100); concat(0, Vec(x/(1-2*x+2*x^2))) \\ Altug Alkan, Dec 04 2015
    
  • Python
    def A009545(n): return ((0, 1, 2, 2)[n&3]<<((n>>1)&-2))*(-1 if n&4 else 1) # Chai Wah Wu, Feb 16 2024
  • Sage
    [lucas_number1(n,2,2) for n in range(0, 51)] # Zerinvary Lajos, Apr 23 2009
    
  • Sage
    def A146559():
        x, y = 0, -1
        while True:
            yield x
            x, y = x - y, x + y
    a = A146559(); [next(a) for i in range(40)]  # Peter Luschny, Jul 11 2013
    

Formula

a(0)=0; a(1)=1; a(2)=2; a(3)=2; a(n) = -4*a(n-4), n>3. - Larry Reeves (larryr(AT)acm.org), Aug 24 2000
Imaginary part of (1+i)^n. - Marc LeBrun
G.f.: x/(1 - 2*x + 2*x^2).
E.g.f.: sin(x)*exp(x).
a(n) = S(n-1, sqrt(2))*(sqrt(2))^(n-1) with S(n, x)= U(n, x/2) Chebyshev's polynomials of the 2nd kind, Cf. A049310, S(-1, x) := 0.
a(n) = ((1+i)^n - (1-i)^n)/(2*i) = 2*a(n-1) - 2*a(n-2) (with a(0)=0 and a(1)=1). - Henry Bottomley, May 10 2001
a(n) = (1+i)^(n-2) + (1-i)^(n-2). - Benoit Cloitre, Oct 28 2002
a(n) = Sum_{k=0..n-1} (-1)^floor(k/2)*binomial(n-1, k). - Benoit Cloitre, Jan 31 2003
a(n) = 2^(n/2)sin(Pi*n/4). - Paul Barry, Sep 17 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k+1)*(-1)^k. - Paul Barry, Sep 20 2003
a(n+1) = Sum_{k=0..n} 2^k*A109466(n,k). - Philippe Deléham, Nov 13 2006
a(n) = 2*((1/2)^(2*theta(n)/Pi))*cos(theta(n)) where theta(4*p+1) = p*Pi + Pi/2, theta(4*p+2) = p*Pi + Pi/4, theta(4*p+3) = p*Pi - Pi/4, theta(4*p+4) = p*Pi - Pi/2, or a(0)=0, a(1)=1, a(2)=2, a(3)=2, and for n>3 a(n)=-4*a(n-4). Same formulas for the second sequence replacing cosines with sines. For example: a(0) = 0, b(0) = 1; a(1) = 0+1 = 1, b(1) = -0+1 = 1; a(2) = 1+1 = 2, b(2) = -1+1 = 0; a(3) = 2+0 = 2, b(3) = -2+0 = -2. - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 30 2007
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3), n > 3, which implies the sequence is identical to its fourth differences. Binomial transform of 0, 1, 0, -1. - Paul Curtz, Dec 21 2007
Logarithm g.f. arctan(x/(1-x)) = Sum_{n>0} a(n)/n*x^n. - Vladimir Kruchinin, Aug 11 2010
a(n) = A046978(n) * A016116(n). - Paul Curtz, Apr 24 2011
E.g.f.: exp(x) * sin(x) = x + x^2/(G(0)-x); G(k) = 2k + 1 + x - x*(2k+1)/(4k+3+x+x^2*(4k+3)/( (2k+2)*(4k+5) - x^2 - x*(2k+2)*(4k+5)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 15 2011
a(n) = Im( (1+i)^n ) where i=sqrt(-1). - Stanislav Sykora, Jun 11 2012
G.f.: x*U(0) where U(k) = 1 + x*(k+3) - x*(k+1)/U(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Oct 10 2012
G.f.: G(0)*x/(2*(1-x)), where G(k) = 1 + 1/(1 - x*(k+1)/(x*(k+2) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 25 2013
G.f.: x + x^2*W(0), where W(k) = 1 + 1/(1 - x*(k+1)/( x*(k+2) + 1/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 28 2013
G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(4*k+2 - 2*x)/( x*(4*k+4 - 2*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 06 2013
a(n) = (A^n - B^n)/(A - B), where A = 1 + i and B = 1 - i; A and B are solutions of x^2 - 2*x + 2 = 0. - Raphie Frank, Nov 28 2015
a(n) = 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], 2) for n >= 2. - Peter Luschny, Dec 17 2015
a(k+m) = a(k)*A146559(m) + a(m)*A146559(k). - Vladimir Shevelev, Jun 08 2017

Extensions

Extended with signs by Olivier Gérard, Mar 15 1997
More terms from Larry Reeves (larryr(AT)acm.org), Aug 24 2000
Definition corrected by Joerg Arndt, Apr 24 2011

A060546 a(n) = 2^ceiling(n/2).

Original entry on oeis.org

1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 32, 64, 64, 128, 128, 256, 256, 512, 512, 1024, 1024, 2048, 2048, 4096, 4096, 8192, 8192, 16384, 16384, 32768, 32768, 65536, 65536, 131072, 131072, 262144, 262144, 524288, 524288, 1048576, 1048576, 2097152, 2097152
Offset: 0

Views

Author

André Barbé (Andre.Barbe(AT)esat.kuleuven.ac.be), Apr 03 2001

Keywords

Comments

a(n) is also the number of median-reflective (palindrome) symmetric patterns in a top-down equilateral triangular arrangement of closely packed black and white cells satisfying the local matching rule of Pascal's triangle modulo 2, where n is the number of cells in each edge of the arrangement. The matching rule is such that any elementary top-down triangle of three neighboring cells in the arrangement contains either one or three white cells.
The number of possibilities for an n-game (sub)set of tennis with neither player gaining a 2-game advantage. (Motivated by the marathon Isner-Mahut match at Wimbledon, 2010.) - Barry Cipra, Jun 28 2010
Number of achiral rows of n colors using up to two colors. For a(3)=4, the rows are AAA, ABA, BAB, and BBB. - Robert A. Russell, Nov 07 2018
Also the number of walks of length n on the graph x--y--z starting at y. - Sean A. Irvine, May 30 2025

Crossrefs

Column k=2 of A321391.
Cf. A000079 (oriented), A005418(n+1) (unoriented), A122746(n-2) (chiral).
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A029744 = {s(n), n>=1}, the numbers 2^k and 3*2^k, as the parent: A029744 (s(n)); A052955 (s(n)-1), A027383 (s(n)-2), A354788 (s(n)-3), A347789 (s(n)-4), A209721 (s(n)+1), A209722 (s(n)+2), A343177 (s(n)+3), A209723 (s(n)+4); A060482, A136252 (minor differences from A354788 at the start); A354785 (3*s(n)), A354789 (3*s(n)-7). The first differences of A029744 are 1,1,1,2,2,4,4,8,8,... which essentially matches eight sequences: A016116, A060546, A117575, A131572, A152166, A158780, A163403, A320770. The bisections of A029744 are A000079 and A007283. - N. J. A. Sloane, Jul 14 2022

Programs

  • Magma
    [2^Ceiling(n/2): n in [0..50]]; // G. C. Greubel, Nov 07 2018
  • Maple
    for n from 0 to 100 do printf(`%d,`,2^ceil(n/2)) od:
  • Mathematica
    2^Ceiling[Range[0,50]/2] (* or *) Riffle[2^Range[0, 25], 2^Range[25]] (* Harvey P. Dale, Mar 05 2013 *)
    LinearRecurrence[{0, 2}, {1, 2}, 40] (* Robert A. Russell, Nov 07 2018 *)
  • PARI
    a(n) = { 2^ceil(n/2) } \\ Harry J. Smith, Jul 06 2009
    

Formula

a(n) = 2^ceiling(n/2).
a(n) = A016116(n+1) for n >= 1.
a(n) = 2^A008619(n-1) for n >= 1.
G.f.: (1 + 2*x) / (1 - 2*x^2). - Ralf Stephan, Jul 15 2013
E.g.f.: cosh(sqrt(2)*x) + sqrt(2)*sinh(sqrt(2)*x). - Stefano Spezia, Feb 02 2023

Extensions

More terms from James Sellers, Apr 04 2001
a(0)=1 prepended by Robert A. Russell, Nov 07 2018
Edited by N. J. A. Sloane, Nov 10 2018
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