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.

Showing 1-10 of 103 results. Next

A155543 a(n)=2*A081294(n).

Original entry on oeis.org

2, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736, 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664
Offset: 0

Views

Author

Paul Curtz, Jan 24 2009

Keywords

Programs

  • Mathematica
    CoefficientList[Series[2(1-2x)/(1-4x),{x,0,30}],x] (* or *) Join[{2}, NestList[ 4#&,4,30]] (* Harvey P. Dale, Nov 13 2015 *)

Formula

a(n)=A000302(n), n>0.
G.f.: 2*(1-2*x)/(1-4*x). [R. J. Mathar, Jul 23 2009]
Conjecture: a(n)=A090129(2n+2).

Extensions

Edited and extended by R. J. Mathar, Jul 23 2009

A088218 Total number of leaves in all rooted ordered trees with n edges.

Original entry on oeis.org

1, 1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220, 1052049481860, 4116715363800, 16123801841550, 63205303218876, 247959266474052
Offset: 0

Views

Author

Michael Somos, Sep 24 2003

Keywords

Comments

Essentially the same as A001700, which has more information.
Note that the unique rooted tree with no edges has no leaves, so a(0)=1 is by convention. - Michael Somos, Jul 30 2011
Number of ordered partitions of n into n parts, allowing zeros (cf. A097070) is binomial(2*n-1,n) = a(n) = essentially A001700. - Vladeta Jovovic, Sep 15 2004
Hankel transform is A000027; example: Det([1,1,3,10;1,3,10,35;3,10,35,126; 10,35,126,462]) = 4. - Philippe Deléham, Apr 13 2007
a(n) is the number of functions f:[n]->[n] such that for all x,y in [n] if xA045992(n). - Geoffrey Critzer, Apr 02 2009
Hankel transform of the aeration of this sequence is A000027 doubled: 1,1,2,2,3,3,... - Paul Barry, Sep 26 2009
The Fi1 and Fi2 triangle sums of A039599 are given by the terms of this sequence. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, Apr 20 2011
Alternating row sums of Riordan triangle A094527. See the Philippe Deléham formula. - Wolfdieter Lang, Nov 22 2012
(-2)*a(n) is the Z-sequence for the Riordan triangle A110162. For the notion of Z- and A-sequences for Riordan arrays see the W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 22 2012
From Gus Wiseman, Jun 27 2021: (Start)
Also the number of integer compositions of 2n with alternating (or reverse-alternating) sum 0 (ranked by A344619). This is equivalent to Ran Pan's comment at A001700. For example, the a(0) = 1 through a(3) = 10 compositions are:
() (11) (22) (33)
(121) (132)
(1111) (231)
(1122)
(1221)
(2112)
(2211)
(11121)
(12111)
(111111)
For n > 0, a(n) is also the number of integer compositions of 2n with alternating sum 2.
(End)
Number of terms in the expansion of (x_1+x_2+...+x_n)^n. - César Eliud Lozada, Jan 08 2022

Examples

			G.f. = 1 + x + 3*x^2 + 10*x^3 + 35*x^4 + 126*x^5 + 462*x^6 + 1716*x^7 + ...
The five rooted ordered trees with 3 edges have 10 leaves.
..x........................
..o..x.x..x......x.........
..o...o...o.x..x.o..x.x.x..
..r...r....r....r.....r....
		

References

  • L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.

Crossrefs

Same as A001700 modulo initial term and offset.
First differences are A024718.
Main diagonal of A071919 and of A305161.
A signed version is A110556.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A003242 counts anti-run compositions.
A025047 counts wiggly compositions (ascend: A025048, descend: A025049).
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A106356 counts compositions by number of maximal anti-runs.
A124754 gives the alternating sum of standard compositions.
A345197 counts compositions by sum, length, and alternating sum.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0: counted by A088218 (this sequence), ranked by A344619/A344619.
- k = 1: counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2: counted by A088218 (this sequence), ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0: counted by A027306, ranked by A345917/A345918.
- k < 0: counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd: counted by A000302, ranked by A053738/A053738.

Programs

  • Magma
    [Binomial(2*n-1, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2014
  • Maple
    seq(binomial(2*n-1, n),n=0..24); # Peter Luschny, Sep 22 2014
  • Mathematica
    a[ n_] := SeriesCoefficient[(1 - x)^-n, {x, 0, n}];
    c = (1 - (1 - 4 x)^(1/2))/(2 x);CoefficientList[Series[1/(1-(c-1)),{x,0,20}],x] (* Geoffrey Critzer, Dec 02 2010 *)
    Table[Binomial[2 n - 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)
    a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, m! SeriesCoefficient[ (1 + BesselI[0, 2 x]) / 2, {x, 0, m}]]]; (* Michael Somos, Nov 22 2014 *)
  • PARI
    {a(n) = sum( i=0, n, binomial(n+i-2,i))};
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( (1 + 1 / sqrt(1 - 4*x + x * O(x^n))) / 2, n))};
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( 1 / (1 - x + x * O(x^n))^n, n))};
    
  • PARI
    {a(n) = if( n<0, 0, binomial( 2*n - 1, n))};
    
  • PARI
    {a(n) = if( n<1, n==0, polcoeff( subst((1 - x) / (1 - 2*x), x, serreverse( x - x^2 + x * O(x^n))), n))};
    
  • Sage
    def A088218(n):
        return rising_factorial(n,n)/falling_factorial(n,n)
    [A088218(n) for n in (0..24)]  # Peter Luschny, Nov 21 2012
    

Formula

G.f.: (1 + 1 / sqrt(1 - 4*x)) / 2.
a(n) = binomial(2*n - 1, n).
a(n) = (n+1)*A000108(n)/2, n>=1. - B. Dubalski (dubalski(AT)atr.bydgoszcz.pl), Feb 05 2002 (in A060150)
a(n) = (0^n + C(2n, n))/2. - Paul Barry, May 21 2004
a(n) is the coefficient of x^n in 1 / (1 - x)^n and also the sum of the first n coefficients of 1 / (1 - x)^n. Given B(x) with the property that the coefficient of x^n in B(x)^n equals the sum of the first n coefficients of B(x)^n, then B(x) = B(0) / (1 - x).
G.f.: 1 / (2 - C(x)) = (1 - x*C(x))/sqrt(1-4*x) where C(x) is g.f. for Catalan numbers A000108. Second equation added by Wolfdieter Lang, Nov 22 2012.
From Paul Barry, Nov 02 2004: (Start)
a(n) = Sum_{k=0..n} binomial(2*n, k)*cos((n-k)*Pi);
a(n) = Sum_{k=0..n} binomial(n, (n-k)/2)*(1+(-1)^(n-k))*cos(k*Pi/2)/2 (with interpolated zeros);
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*cos((n-2*k)*Pi/2) (with interpolated zeros); (End)
a(n) = A110556(n)*(-1)^n, central terms in triangle A110555. - Reinhard Zumkeller, Jul 27 2005
a(n) = Sum_{0<=k<=n} A094527(n,k)*(-1)^k. - Philippe Deléham, Mar 14 2007
From Paul Barry, Mar 29 2010: (Start)
G.f.: 1/(1-x/(1-2x/(1-(1/2)x/(1-(3/2)x/(1-(2/3)x/(1-(4/3)x/(1-(3/4)x/(1-(5/4)x/(1-... (continued fraction);
E.g.f.: (of aerated sequence) (1 + Bessel_I(0, 2*x))/2. (End)
a(n + 1) = A001700(n). a(n) = A024718(n) - A024718(n - 1).
E.g.f.: E(x) = 1+x/(G(0)-2*x) ; G(k) = (k+1)^2+2*x*(2*k+1)-2*x*(2*k+3)*((k+1)^2)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 21 2011
a(n) = Sum_{k=0..n}(-1)^k*binomial(2*n,n+k). - Mircea Merca, Jan 28 2012
a(n) = rf(n,n)/ff(n,n), where rf is the rising factorial and ff the falling factorial. - Peter Luschny, Nov 21 2012
D-finite with recurrence: n*a(n) +2*(-2*n+1)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012
a(n) = hypergeom([1-n,-n],[1],1). - Peter Luschny, Sep 22 2014
G.f.: 1 + x/W(0), where W(k) = 4*k+1 - (4*k+3)*x/(1 - (4*k+1)*x/(4*k+3 - (4*k+5)*x/(1 - (4*k+3)*x/W(k+1) ))) ; (continued fraction). - Sergei N. Gladkovskii, Nov 13 2014
a(n) = A000984(n) + A001791(n). - Gus Wiseman, Jun 28 2021
E.g.f.: (1 + exp(2*x) * BesselI(0,2*x)) / 2. - Ilya Gutkovskiy, Nov 03 2021
From Amiram Eldar, Mar 12 2023: (Start)
Sum_{n>=0} 1/a(n) = 5/3 + 4*Pi/(9*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 3/5 - 8*log(phi)/(5*sqrt(5)), where phi is the golden ratio (A001622). (End)
a(n) ~ 2^(2*n-1)/sqrt(n*Pi). - Stefano Spezia, Apr 17 2024

A004171 a(n) = 2^(2n+1).

Original entry on oeis.org

2, 8, 32, 128, 512, 2048, 8192, 32768, 131072, 524288, 2097152, 8388608, 33554432, 134217728, 536870912, 2147483648, 8589934592, 34359738368, 137438953472, 549755813888, 2199023255552, 8796093022208, 35184372088832, 140737488355328, 562949953421312
Offset: 0

Views

Author

Keywords

Comments

Same as Pisot sequences E(2, 8), L(2, 8), P(2, 8), T(2, 8). See A008776 for definitions of Pisot sequences.
In the Chebyshev polynomial of degree 2n, a(n) is the coefficient of x^2n. - Benoit Cloitre, Mar 13 2002
1/2 - 1/8 + 1/32 - 1/128 + ... = 2/5. - Gary W. Adamson, Mar 03 2009
From Adi Dani, May 15 2011: (Start)
Number of ways of placing an even number of indistinguishable objects in n + 1 distinguishable boxes with at most 3 objects in box.
Number of compositions of even natural numbers into n + 1 parts less than or equal to 3 (0 is counted as part). (End)
Also the number of maximal cliques in the (n+1)-Sierpinski tetrahedron graph for n > 0. - Eric W. Weisstein, Dec 01 2017
Assuming the Collatz conjecture is true, any starting number eventually leads to a power of 2. A number in this sequence can never be the first power of 2 in a Collatz sequence except of course for the Collatz sequence starting with that number. For example, except for 8, 4, 2, 1, any Collatz sequence that includes 8 must also include 16 (e.g., 5, 16, 8, 4, 2, 1). - Alonso del Arte, Oct 01 2019
First differences of A020988, and thus the "wavelengths" of the local maxima in A020986. See the Brillhart and Morton link, pp. 855-856. - John Keith, Mar 04 2021

Examples

			G.f. = 2 + 8*x + 32*x^2 + 128*x^3 + 512*x^4 + 2048*x^5 + 8192*x^6 + 32768*x^7 + ...
From _Adi Dani_, May 15 2011: (Start)
a(1) = 8 because all compositions of even natural numbers into 2 parts less than or equal to 3 are:
  for 0: (0, 0)
  for 2: (0, 2), (2, 0), (1, 1)
  for 4: (1, 3), (3, 1), (2, 2)
  for 6: (3, 3).
a(2) = 32 because all compositions of even natural numbers into 3 parts less than or equal to 3 are:
  for 0: (0, 0, 0)
  for 2: (0, 0, 2), (0, 2, 0), (2, 0, 0), (0, 1, 1), (1, 0, 1) , (1, 1, 0)
  for 4: (0, 1, 3), (0, 3, 1), (1, 0, 3), (1, 3, 0), (3, 0, 1), (3, 1, 0), (0, 2, 2), (2, 0, 2), (2, 2, 0), (1, 1, 2), (1, 2, 1), (2, 1, 1)
  for 6: (0, 3, 3), (3, 0, 3), (3, 3, 0), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1), (2, 2, 2)
  for 8: (2, 3, 3), (3, 2, 3), (3, 3, 2).
(End)
		

References

  • Adi Dani, Quasicompositions of natural numbers, Proceedings of III congress of mathematicians of Macedonia, Struga Macedonia 29 IX -2 X 2005 pages 225-238.

Crossrefs

Absolute value of A009117. Essentially the same as A081294.
Cf. A132020, A164632. Equals A000980(n) + 2*A181765(n). Cf. A013776.

Programs

Formula

a(n) = 2*4^n.
a(n) = 4*a(n-1).
1 = 1/2 + Sum_{n >= 1} 3/a(n) = 3/6 + 3/8 + 3/32 + 3/128 + 3/512 + 3/2048 + ...; with partial sums: 1/2, 31/32, 127/128, 511/512, 2047/2048, ... - Gary W. Adamson, Jun 16 2003
From Philippe Deléham, Nov 23 2008: (Start)
a(n) = 2*A000302(n).
G.f.: 2/(1-4*x). (End)
a(n) = A081294(n+1) = A028403(n+1) - A000079(n+1) for n >= 1. a(n-1) = A028403(n) - A000079(n). - Jaroslav Krizek, Jul 27 2009
E.g.f.: 2*exp(4*x). - Ilya Gutkovskiy, Nov 01 2016
a(n) = A002063(n)/3 - A000302(n). - Zhandos Mambetaliyev, Nov 19 2016
a(n) = Sum_{k = 0..2*n} (-1)^(k+n)*binomial(4*n + 2, 2*k + 1); a(2*n) = Sum_{k = 0..2*n} binomial(4*n + 2, 2*k + 1) = A013776(n). - Peter Bala, Nov 25 2016
Product_{n>=0} (1 - 1/a(n)) = A132020. - Amiram Eldar, May 08 2023

A007583 a(n) = (2^(2*n + 1) + 1)/3.

Original entry on oeis.org

1, 3, 11, 43, 171, 683, 2731, 10923, 43691, 174763, 699051, 2796203, 11184811, 44739243, 178956971, 715827883, 2863311531, 11453246123, 45812984491, 183251937963, 733007751851, 2932031007403, 11728124029611, 46912496118443, 187649984473771, 750599937895083
Offset: 0

Views

Author

Keywords

Comments

Let u(k), v(k), w(k) be the 3 sequences defined by u(1)=1, v(1)=0, w(1)=0 and u(k+1)=u(k)+v(k)-w(k), v(k+1)=u(k)-v(k)+w(k), w(k+1)=-u(k)+v(k)+w(k); let M(k)=Max(u(k),v(k),w(k)); then a(n)=M(2n)=M(2n-1). - Benoit Cloitre, Mar 25 2002
Also the number of words of length 2n generated by the two letters s and t that reduce to the identity 1 by using the relations ssssss=1, tt=1 and stst=1. The generators s and t along with the three relations generate the dihedral group D6=C2xD3. - Jamaine Paddyfoot (jay_paddyfoot(AT)hotmail.com) and John W. Layman, Jul 08 2002
Binomial transform of A025192. - Paul Barry, Apr 11 2003
Number of walks of length 2n+1 between two adjacent vertices in the cycle graph C_6. Example: a(1)=3 because in the cycle ABCDEF we have three walks of length 3 between A and B: ABAB, ABCB and AFAB. - Emeric Deutsch, Apr 01 2004
Numbers of the form 1 + Sum_{i=1..m} 2^(2*i-1). - Artur Jasinski, Feb 09 2007
Prime numbers of the form 1+Sum[2^(2n-1)] are in A000979. Numbers x such that 1+Sum[2^(2n-1)] is prime for n=1,2,...,x is A127936. - Artur Jasinski, Feb 09 2007
Related to A024493(6n+1), A131708(6n+3), A024495(6n+5). - Paul Curtz, Mar 27 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-6, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^(n-1)*charpoly(A,2). - Milan Janjic, Feb 21 2010
Number of toothpicks in the toothpick structure of A139250 after 2^n stages. - Omar E. Pol, Feb 28 2011
Numbers whose binary representation is "10" repeated (n-1) times with "11" appended on the end, n >= 1. For example 171 = 10101011 (2). - Omar E. Pol, Nov 22 2012
a(n) is the smallest number for which A072219(a(n)) = 2*n+1. - Ramasamy Chandramouli, Dec 22 2012
An Engel expansion of 2 to the base b := 4/3 as defined in A181565, with the associated series expansion 2 = b + b^2/3 + b^3/(3*11) + b^4/(3*11*43) + .... Cf. A007051. - Peter Bala, Oct 29 2013
The positive integer solution (x,y) of 3*x - 2^n*y = 1, n>=0, with smallest x is (a(n/2), 2) if n is even and (a((n-1)/2), 1) if n is odd. - Wolfdieter Lang, Feb 15 2014
The smallest positive number that requires at least n additions and subtractions of powers of 2 to be formed. See Puzzling StackExchange link. - Alexander Cooke Jul 16 2023

References

  • H. W. Gould, Combinatorial Identities, Morgantown, 1972, (1.77), page 10.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Partial sums of A081294.
Cf. location of records in A007302.

Programs

  • GAP
    List([0..25], n-> (2^(2*n+1) + 1)/3); # G. C. Greubel, Dec 25 2019
  • Haskell
    a007583 = (`div` 3) . (+ 1) . a004171
    -- Reinhard Zumkeller, Jan 09 2013
    
  • Magma
    [(2^(2*n+1) + 1)/3: n in [0..30] ]; // Vincenzo Librandi, Apr 28 2011
    
  • Maple
    a[0]:=1:for n from 1 to 50 do a[n]:=4*a[n-1]-1 od: seq(a[n], n=0..23); # Zerinvary Lajos, Feb 22 2008, with correction by K. Spage, Aug 20 2014
    A007583 := proc(n)
        (2^(2*n+1)+1)/3 ;
    end proc: # R. J. Mathar, Feb 19 2015
  • Mathematica
    (* From Michael De Vlieger, Aug 22 2016 *)
    Table[(2^(2n+1) + 1)/3, {n, 0, 23}]
    Table[1 + 2Sum[4^k, {k, 0, n-1}], {n, 0, 23}]
    NestList[4# -1 &, 1, 23]
    Table[Sum[Binomial[n+k, 2k]/2^(k-n), {k, 0, n}], {n, 0, 23}]
    CoefficientList[Series[(1-2x)/(1-5x+4x^2), {x, 0, 23}], x] (* End *)
  • PARI
    a(n)=sum(k=-n\3,n\3,binomial(2*n+1,n+1+3*k))
    
  • PARI
    a=1; for(n=1,23, print1(a,", "); a=bitor(a,3*a)) \\ K. Spage, Aug 20 2014
    
  • PARI
    Vec((1-2*x)/(1-5*x+4*x^2) + O(x^30)) \\ Altug Alkan, Dec 08 2015
    
  • PARI
    apply( {A007583(n)=2<<(2*n)\/3}, [0..25]) \\ M. F. Hasler, Nov 30 2021
    
  • Sage
    [(2^(2*n+1) + 1)/3 for n in (0..25)] # G. C. Greubel, Dec 25 2019
    

Formula

a(n) = 2*A002450(n) + 1.
From Wolfdieter Lang, Apr 24 2001: (Start)
a(n) = Sum_{m = 0..n} A060920(n, m) = A002450(n+1) - 2*A002450(n).
G.f.: (1-2*x)/(1-5*x+4*x^2). (End)
a(n) = Sum_{k = 0..n} binomial(n+k, 2*k)/2^(k - n).
a(n) = 4*a(n-1) - 1, n > 0.
From Paul Barry, Mar 17 2003: (Start)
a(n) = 1 + 2*Sum_{k = 0..n-1} 4^k;
a(n) = A001045(2n+1). (End)
a(n) = A020988(n-1) + 1 = A039301(n+1) - 1 = A083584(n-1) + 2. - Ralf Stephan, Jun 14 2003
a(0) = 1; a(n+1) = a(n) * 4 - 1. - Regis Decamps (decamps(AT)users.sf.net), Feb 04 2004 (correction to lead index by K. Spage, Aug 20 2014)
a(n) = Sum_{i + j + k = n; 0 <= i, j, k <= n} (n+k)!/i!/j!/(2*k)!. - Benoit Cloitre, Mar 25 2004
a(n) = 5*a(n-1) - 4*a(n-2). - Emeric Deutsch, Apr 01 2004
a(n) = 4^n - A001045(2*n). - Paul Barry, Apr 17 2004
a(n) = 2*(A001045(n))^2 + (A001045(n+1))^2. - Paul Barry, Jul 15 2004
a(n) = left and right terms in M^n * [1 1 1] where M = the 3X3 matrix [1 1 1 / 1 3 1 / 1 1 1]. M^n * [1 1 1] = [a(n) A002450(n+1) a(n)] E.g. a(3) = 43 since M^n * [1 1 1] = [43 85 43] = [a(3) A002450(4) a(3)]. - Gary W. Adamson, Dec 18 2004
a(n) = A072197(n) - A020988(n). - Creighton Dement, Dec 31 2004
a(n) = A139250(2^n). - Omar E. Pol, Feb 28 2011
a(n) = A193652(2*n+1). - Reinhard Zumkeller, Aug 08 2011
a(n) = Sum_{k = -floor(n/3)..floor(n/3)} binomial(2*n, n+3*k)/2. - Mircea Merca, Jan 28 2012
a(n) = 2^(2*(n+1)) - A072197(n). - Vladimir Pletser, Apr 12 2014
a(n) == 2*n + 1 (mod 3). Indeed, from Regis Decamps' formula (Feb 04 2004) we have a(i+1) - a(i) == -1 (mod 3), i= 0, 1, ..., n - 1. Summing, we have a(n) - 1 == -n (mod 3), and the formula follows. - Vladimir Shevelev, May 20 2015
For n > 0 a(n) = A133494(0) + 2 * (A133494(n) + Sum_{x = 1..n - 1}Sum_{k = 0..x - 1}(binomial(x - 1, k)*(A133494(k+1) + A133494(n-x+k)))). - J. Conrad, Dec 06 2015
a(n) = Sum_{k = 0..2n} (-2)^k == 1 + Sum_{k = 1..n} 2^(2k-1). - Bob Selcoe, Aug 21 2016
E.g.f.: (1 + 2*exp(3*x))*exp(x)/3. - Ilya Gutkovskiy, Aug 21 2016
A075680(a(n)) = 1, for n > 0. - Ralf Stephan, Jun 17 2025

A046717 a(n) = 2*a(n-1) + 3*a(n-2), a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 5, 13, 41, 121, 365, 1093, 3281, 9841, 29525, 88573, 265721, 797161, 2391485, 7174453, 21523361, 64570081, 193710245, 581130733, 1743392201, 5230176601, 15690529805, 47071589413, 141214768241, 423644304721, 1270932914165, 3812798742493, 11438396227481
Offset: 0

Views

Author

Gervais Deroo and M. Deroo

Keywords

Comments

Form the digraph with matrix A = [0,1,1,1; 1,0,1,1; 1,1,0,1; 1,0,1,1]. Then the sequence 0,1,1,5,... or (3^(n-1)-(-1)^n)/2+0^n/3 with g.f. x(1-x)/(1-2x-3x^2) corresponds to the (1,2) term of A^n. - Paul Barry, Oct 02 2004
3*a(n+1) + a(n) = 4*A060925(n); a(n+1) = A015518(n) + A060925(n); a(n+1) - 6*A015518(n) = (-1)^n. - Creighton Dement, Nov 15 2004
The sequence corresponds to the (1,1) term of the matrix [1,2;2,1]^n. - Simone Severini, Dec 04 2004
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 4 times the bottom to get the new top. The limit of the sequence of fractions is 2. - Cino Hilliard, Sep 25 2005
a(n)^2 + (2*A015518(n))^2 = a(2n). E.g., a(3) = 13, 2*A015518(3) = 14, A046717(6) = 365. 13^2 + 14^2 = 365. - Gary W. Adamson, Jun 17 2006
Equals INVERTi transform of A104934: (1, 2, 8, 28, 100, 356, 1268, ...). - Gary W. Adamson, Jul 21 2010
a(n) is the number of compositions of n when there are 1 type of 1 and 4 types of other natural numbers. - Milan Janjic, Aug 13 2010
An elephant sequence, see A175655. For the central square just one A[5] vector, with decimal value 341, leads to this sequence (without the first leading 1). For the corner squares this vector leads to the companion sequence A015518 (without the leading 0). - Johannes W. Meijer, Aug 15 2010
Pisano period lengths: 1, 1, 2, 1, 4, 2, 6, 4, 2, 4, 10, 2, 6, 6, 4, 8, 16, 2, 18, 4, ... - R. J. Mathar, Aug 10 2012
a(n) is the number of words of length n over a ternary alphabet whose position in the lexicographic order is a multiple of two. - Alois P. Heinz, Apr 13 2022
a(n) is the sum, for k=0..3, of the number of walks of length n between two vertices at distance k of the cube graph. - Miquel A. Fiol, Mar 09 2024

References

  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Crossrefs

The first difference sequence of A015518.
Row sums of triangle A080928.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Cf. A015518.
Cf. A104934. - Gary W. Adamson, Jul 21 2010

Programs

  • Magma
    [n le 2 select 1 else 2*Self(n-1)+3*Self(n-2): n in [1..35]]; // Vincenzo Librandi, Jul 21 2013
    
  • Magma
    [(3^n + (-1)^n)/2: n in [0..30]]; // G. C. Greubel, Jan 07 2018
  • Maple
    a[0]:=1:a[1]:=1:for n from 2 to 50 do a[n]:=2*a[n-1]+3*a[n-2] od: seq(a[n], n=0..33); # Zerinvary Lajos, Dec 14 2008
    seq(denom(((-2)^(2*n)+6^(2*n))/((-2)^n+6^n)),n=0..26)
  • Mathematica
    Table[(3^n + (-1)^n)/2, {n, 0, 30}] (* Artur Jasinski, Dec 10 2006 *)
    CoefficientList[ Series[(1 - x)/(1 - 2x - 3x^2), {x, 0, 30}], x]  (* Robert G. Wilson v, Apr 04 2011 *)
    Table[ MatrixPower[{{1, 2}, {1, 1}}, n][[1, 1]], {n, 0, 30}] (* Robert G. Wilson v, Apr 04 2011 *)
  • PARI
    {a(n) = (3^n+(-1)^n)/2};
    for(n=0,30, print1(a(n), ", ")) /* modified by G. C. Greubel, Jan 07 2018 */
    
  • PARI
    x='x+O('x^30); Vec((1-x)/((1+x)*(1-3*x))) \\ G. C. Greubel, Jan 07 2018
    
  • Sage
    [lucas_number2(n,2,-3)/2 for n in range(0, 27)] # Zerinvary Lajos, Apr 30 2009
    

Formula

G.f.: (1-x)/((1+x)*(1-3*x)).
a(n) = (3^n + (-1)^n)/2.
a(n) = Sum_{k=0..n} binomial(n, 2k)2^(2k). - Paul Barry, Feb 26 2003
Binomial transform of A000302 (powers of 4) with interpolated zeros. Inverse binomial transform of A081294. - Paul Barry, Mar 17 2003
E.g.f.: exp(x)cosh(2x). - Paul Barry, Mar 17 2003
a(n) = ceiling(3^n/4) + floor(3^n/4) = ceiling(3^n/4)^2 - floor(3^n/4)^2. - Paul Barry, Jan 17 2005
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(n,j)C(n-j,k)*(1+(-1)^(j-k))/2. - Paul Barry, May 21 2006
a(n) = Sum_{k=0..n} A098158(n,k)*4^(n-k). - Philippe Deléham, Dec 26 2007
a(n) = (3^n + (-1)^n)/2. - M. F. Hasler, Mar 20 2008
a(n) = 2 A015518(n) + (-1)^n; for n > 0, a(n) = A080925(n). - M. F. Hasler, Mar 20 2008
((1 + sqrt4)^n + (1 - sqrt4)^n)/2. The offset is 0. a(3)=13. - Al Hakanson (hawkuu(AT)gmail.com), Nov 22 2008
If p[1]=1 and p[i]=4 (i > 1), and if A is Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i <= j), A[i,j] = -1, (i = j+1), and A[i,j] = 0 otherwise, then, for n >= 1, a(n) = det A. - Milan Janjic, Apr 29 2010
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(4*k-1)/(x*(4*k+3) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
G.f.: G(0)/2, where G(k) = 1 + (-1)^k/(3^k - 3*9^k*x/(3*3^k*x + (-1)^k/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Oct 17 2013

Extensions

Description corrected by and more terms from Michael Somos

A058622 a(n) = 2^(n-1) - ((1+(-1)^n)/4)*binomial(n, n/2).

Original entry on oeis.org

0, 1, 1, 4, 5, 16, 22, 64, 93, 256, 386, 1024, 1586, 4096, 6476, 16384, 26333, 65536, 106762, 262144, 431910, 1048576, 1744436, 4194304, 7036530, 16777216, 28354132, 67108864, 114159428, 268435456, 459312152, 1073741824, 1846943453
Offset: 0

Views

Author

Yong Kong (ykong(AT)curagen.com), Dec 29 2000

Keywords

Comments

a(n) is the number of n-digit binary sequences that have more 1's than 0's. - Geoffrey Critzer, Jul 16 2009
Maps to the number of walks that end above 0 on the number line with steps being 1 or -1. - Benjamin Phillabaum, Mar 06 2011
Chris Godsil observes that a(n) is the independence number of the (n+1)-folded cube graph; proof is by a Cvetkovic's eigenvalue bound to establish an upper bound and a direct construction of the independent set by looking at vertices at an odd (resp., even) distance from a fixed vertex when n is odd (resp., even). - Stan Wagon, Jan 29 2013
Also the number of subsets of {1,2,...,n} that contain more odd than even numbers. For example, for n=4, a(4)=5 and the 5 subsets are {1}, {3}, {1,3}, {1,2,3}, {1,3,4}. See A014495 when same number of even and odd numbers. - Enrique Navarrete, Feb 10 2018
Also half the number of length-n binary sequences with a different number of zeros than ones. This is also the number of integer compositions of n with nonzero alternating sum, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. Also the number of integer compositions of n+1 with alternating sum <= 0, ranked by A345915 (reverse: A345916). - Gus Wiseman, Jul 19 2021

Examples

			G.f. = x + x^2 + 4*x^3 + 5*x^4 + 16*x^5 + 22*x^6 + 64*x^7 + 93*x^8 + ...
From _Gus Wiseman_, Jul 19 2021: (Start)
The a(1) = 1 through a(5) = 16 compositions with nonzero alternating sum:
  (1)  (2)  (3)      (4)      (5)
            (1,2)    (1,3)    (1,4)
            (2,1)    (3,1)    (2,3)
            (1,1,1)  (1,1,2)  (3,2)
                     (2,1,1)  (4,1)
                              (1,1,3)
                              (1,2,2)
                              (1,3,1)
                              (2,1,2)
                              (2,2,1)
                              (3,1,1)
                              (1,1,1,2)
                              (1,1,2,1)
                              (1,2,1,1)
                              (2,1,1,1)
                              (1,1,1,1,1)
(End)
		

References

  • A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992, Eq. (4.2.1.7)

Crossrefs

The odd bisection is A000302.
The even bisection is A000346.
The following relate to compositions with nonzero alternating sum:
- The complement is counted by A001700 or A138364.
- The version for alternating sum > 0 is A027306.
- The unordered version is A086543 (even bisection: A182616).
- The version for alternating sum < 0 is A294175.
- These compositions are ranked by A345921.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A345197 counts compositions by length and alternating sum.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0: counted by A088218, ranked by A344619/A344619.
- k = 1: counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2: counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k > 0: counted by A027306, ranked by A345917/A345918.
- k < 0: counted by A294175, ranked by A345919/A345920.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd: counted by A000302, ranked by A053738/A053738.

Programs

  • Magma
    [(2^n -(1+(-1)^n)*Binomial(n, Floor(n/2))/2)/2: n in [0..40]]; // G. C. Greubel, Aug 08 2022
    
  • Mathematica
    Table[Sum[Binomial[n, Floor[n/2 + i]], {i, 1, n}], {n, 0, 32}] (* Geoffrey Critzer, Jul 16 2009 *)
    a[n_] := If[n < 0, 0, (2^n - Boole[EvenQ @ n] Binomial[n, Quotient[n, 2]])/2]; (* Michael Somos, Nov 22 2014 *)
    a[n_] := If[n < 0, 0, n! SeriesCoefficient[(Exp[2 x] - BesselI[0, 2 x])/2, {x, 0, n}]]; (* Michael Somos, Nov 22 2014 *)
    Table[2^(n - 1) - (1 + (-1)^n) Binomial[n, n/2]/4, {n, 0, 40}] (* Eric W. Weisstein, Mar 21 2018 *)
    CoefficientList[Series[2 x/((1-2x)(1 + 2x + Sqrt[(1+2x)(1-2x)])), {x, 0, 40}], x] (* Eric W. Weisstein, Mar 21 2018 *)
    ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],ats[#]!=0&]],{n,0,15}] (* Gus Wiseman, Jul 19 2021 *)
  • PARI
    a(n) = 2^(n-1) - ((1+(-1)^n)/4)*binomial(n, n\2); \\ Michel Marcus, Dec 30 2015
    
  • PARI
    my(x='x+O('x^100)); concat(0, Vec(2*x/((1-2*x)*(1+2*x+((1+2*x)*(1-2*x))^(1/2))))) \\ Altug Alkan, Dec 30 2015
    
  • Python
    from math import comb
    def A058622(n): return (1<>1)>>1) if n else 0 # Chai Wah Wu, Aug 25 2025
  • SageMath
    [(2^n - binomial(n, n//2)*((n+1)%2))/2 for n in (0..40)] # G. C. Greubel, Aug 08 2022
    

Formula

a(n) = 2^(n-1) - ((1+(-1)^n)/4)*binomial(n, n/2).
a(n) = Sum_{i=0..floor((n-1)/2)} binomial(n, i).
G.f.: 2*x/((1-2*x)*(1+2*x+((1+2*x)*(1-2*x))^(1/2))). - Vladeta Jovovic, Apr 27 2003
E.g.f: (e^(2x)-I_0(2x))/2 where I_n is the Modified Bessel Function. - Benjamin Phillabaum, Mar 06 2011
Logarithmic derivative of the g.f. of A210736 is a(n+1). - Michael Somos, Nov 22 2014
Even index: a(2n) = 2^(n-1) - A088218(n). Odd index: a(2n+1) = 2^(2n). - Gus Wiseman, Jul 19 2021
D-finite with recurrence n*a(n) +2*(-n+1)*a(n-1) +4*(-n+1)*a(n-2) +8*(n-2)*a(n-3)=0. - R. J. Mathar, Sep 23 2021
a(n) = 2^n-A027306(n). - R. J. Mathar, Sep 23 2021
A027306(n) - a(n) = A126869(n). - R. J. Mathar, Sep 23 2021

A345197 Concatenation of square matrices A(n), each read by rows, where A(n)(k,i) is the number of compositions of n of length k with alternating sum i, where 1 <= k <= n, and i ranges from -n + 2 to n in steps of 2.

Original entry on oeis.org

1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 2, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 2, 3, 0, 0, 2, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 2, 3, 4, 0, 0, 3, 4, 3, 0, 0, 0, 0, 2, 3, 0, 0, 0, 0, 1, 0, 0, 0
Offset: 0

Views

Author

Gus Wiseman, Jul 03 2021

Keywords

Comments

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

Examples

			The matrices for n = 1..7:
  1   0 1   0 0 1   0 0 0 1   0 0 0 0 1   0 0 0 0 0 1   0 0 0 0 0 0 1
      1 0   1 1 0   1 1 1 0   1 1 1 1 0   1 1 1 1 1 0   1 1 1 1 1 1 0
            0 1 0   0 1 2 0   0 1 2 3 0   0 1 2 3 4 0   0 1 2 3 4 5 0
                    0 1 0 0   0 2 2 0 0   0 3 4 3 0 0   0 4 6 6 4 0 0
                              0 0 1 0 0   0 0 2 3 0 0   0 0 3 6 6 0 0
                                          0 0 1 0 0 0   0 0 3 3 0 0 0
                                                        0 0 0 1 0 0 0
Matrix n = 5 counts the following compositions:
           i=-3:        i=-1:          i=1:            i=3:        i=5:
        -----------------------------------------------------------------
   k=1: |    0            0             0               0          (5)
   k=2: |   (14)         (23)          (32)            (41)         0
   k=3: |    0          (131)       (221)(122)   (311)(113)(212)    0
   k=4: |    0       (1211)(1112)  (2111)(1121)         0           0
   k=5: |    0            0          (11111)            0           0
		

Crossrefs

The number of nonzero terms in each matrix appears to be A000096.
The number of zeros in each matrix appears to be A000124.
Row sums and column sums both appear to be A007318 (Pascal's triangle).
The matrix sums are A131577.
Antidiagonal sums appear to be A163493.
The reverse-alternating version is also A345197 (this sequence).
Antidiagonals are A345907.
Traces are A345908.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
A344610 counts partitions by sum and positive reverse-alternating sum.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
Other tetrangles: A318393, A318816, A320808, A321912.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0: counted by A088218, ranked by A344619/A344619.
- k = 1: counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2: counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0: counted by A027306, ranked by A345917/A345918.
- k < 0: counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd: counted by A000302, ranked by A053738/A053738.

Programs

  • Mathematica
    ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
    Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[#]==k&&ats[#]==i&]],{n,0,6},{k,1,n},{i,-n+2,n,2}]

A053754 If k is in the sequence then 2*k and 2*k+1 are not (and 0 is in the sequence); when written in binary k has an even number of bits (0 has 0 digits).

Original entry on oeis.org

0, 2, 3, 8, 9, 10, 11, 12, 13, 14, 15, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148
Offset: 1

Views

Author

Henry Bottomley, Apr 06 2000

Keywords

Comments

Runs of successive terms with same number of bits have length twice powers of 4 (A081294). [Clarified by Michel Marcus, Oct 21 2020]
The sequence A081294 counts compositions of even numbers - Gus Wiseman, Aug 12 2021
A031443 is a subsequence; A179888 is the intersection of this sequence and A032925. - Reinhard Zumkeller, Jul 31 2010
The lower and upper asymptotic densities of this sequence are 1/3 and 2/3, respectively. - Amiram Eldar, Feb 01 2021
From Gus Wiseman, Aug 10 2021: (Start)
Also numbers k such that the k-th composition in standard order (row k of A066099) has even sum. The terms and corresponding compositions begin:
0: () 2: (2) 8: (4)
3: (1,1) 9: (3,1)
10: (2,2)
11: (2,1,1)
12: (1,3)
13: (1,2,1)
14: (1,1,2)
15: (1,1,1,1)
The following pertain to compositions in standard order: A000120, A029837, A070939, A066099, A124767.
(End)

Crossrefs

Positions of even terms in A029837 with offset 0.
The complement (the odd version) is A053738, counted by A000302.
The version for Heinz numbers of partitions is A300061, counted by A058696.

Programs

  • Haskell
    a053754 n = a053754_list !! (n-1)
    a053754_list = 0 : filter (even . a070939) [1..]
    -- Reinhard Zumkeller, Apr 18 2015
    
  • Mathematica
    Select[Range[0, 150], EvenQ @ IntegerLength[#, 2] &] (* Amiram Eldar, Feb 01 2021 *)
  • PARI
    lista(nn) = {my(va = vector(nn)); for (n=2, nn, my(k=va[n-1]+1); while (#select(x->(x==k\2), va), k++); va[n] = k;); va;} \\ Michel Marcus, Oct 20 2020
    
  • PARI
    a(n) = n-1 + (1<Kevin Ryde, Apr 30 2021

Extensions

Offset corrected by Reinhard Zumkeller, Jul 30 2010

A075677 Reduced Collatz function R applied to the odd integers: a(n) = R(2n-1), where R(k) = (3k+1)/2^r, with r as large as possible.

Original entry on oeis.org

1, 5, 1, 11, 7, 17, 5, 23, 13, 29, 1, 35, 19, 41, 11, 47, 25, 53, 7, 59, 31, 65, 17, 71, 37, 77, 5, 83, 43, 89, 23, 95, 49, 101, 13, 107, 55, 113, 29, 119, 61, 125, 1, 131, 67, 137, 35, 143, 73, 149, 19, 155, 79, 161, 41, 167, 85, 173, 11, 179, 91, 185, 47, 191, 97, 197
Offset: 1

Views

Author

T. D. Noe, Sep 25 2002

Keywords

Comments

The even-indexed terms a(2i+2) = 6i+5 = A016969(i), i >= 0 [Comment corrected by Bob Selcoe, Apr 06 2015]. The odd-indexed terms are the same as A067745. Note that this sequence is A016789 with all factors of 2 removed from each term. Also note that a(4i-1) = a(i). No multiple of 3 is in this sequence. See A075680 for the number of iterations of R required to yield 1.
From Bob Selcoe, Apr 06 2015: (Start)
All numbers in this sequence appear infinitely often.
From Eq. 1 and Eq. 2 in Formulas: Eq. 1 is used with 1/3 of the numbers in this sequence, Eq. 2 is used with 2/3 of the numbers.
(End)
Empirical: For arbitrary m, Sum_{n=2..A007583(m)} (a(n) - a(n-1)) = 0. - Fred Daniel Kline, Nov 23 2015
From Wolfdieter Lang, Dec 07 2021: (Start)
Only positive numbers congruent to 1 or 5 modulo 6 appear.
i) For the sequence entry with value A016921(m), for m >= 0, that is, a value from {1, 7, 13, ...}, the indices n are given by the row of array A178415(2*m+1, k), for k >= 1.
ii) For the sequence entry with value A007528(m), for m >= 1, that is, a value from {5, 11, 17, ...}, the indices n are given by the row of array A178415(2*m, k), for k >= 1.
See also the array A347834 with permuted row numbers and columns k >= 0. (End)

Examples

			a(11) = 1 because 21 is the 11th odd number and R(21) = 64/64 = 1.
From _Wolfdieter Lang_, Dec 07 2021: (Start)
i) 1 (mod 6) entry 1 = A016921(0) appears for n = A178415(1, k) = A347834(1, k-1) (the arrays), for k >= 1, that is, for {1, 5, 21, ..} = A002450.
ii) 5 (mod 6) entry 11 = A007528(2) appears for n = A178415(4, k) = A347835(3, k-1) (the arrays), for k >= 1, that is, for {7, 29, 117, ..} = A072261. (End)
		

References

  • Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section E16, pp. 330-336.
  • Victor Klee and Stan Wagon, Old and new unsolved problems in plane geometry and number theory, The Mathematical Association of America, 1991, p. 225, C(2n+1) = a(n+1), n >= 0.
  • Jeffrey C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010; see p. 57, also (90-9), p. 306.

Crossrefs

Cf. A006370, A014682 (for non-reduced Collatz maps), A087230 (A371093), A371094.
Odd bisection of A139391.
Even bisection of A067745, which is also the odd bisection of this sequence.
After the initial 1, the second leftmost column of A256598.
Row 2 of A372283.

Programs

  • Haskell
    a075677 = a000265 . subtract 2 . (* 6) -- Reinhard Zumkeller, Jan 08 2014
    
  • Maple
    f:=proc(n) local t1;
    if n=1 then RETURN(1) else
    t1:=3*n+1;
    while t1 mod 2 = 0 do t1:=t1/2; od;
    RETURN(t1); fi;
    end;
    # N. J. A. Sloane, Jan 21 2011
  • Mathematica
    nextOddK[n_] := Module[{m=3n+1}, While[EvenQ[m], m=m/2]; m]; (* assumes odd n *) Table[nextOddK[n], {n, 1, 200, 2}]
    v[x_] := IntegerExponent[x, 2]; f[x_] := (3*x + 1)/2^v[3*x + 1]; Table[f[2*n - 1], {n, 66}] (* L. Edson Jeffery, May 06 2015 *)
  • PARI
    a(n)=n+=2*n-1;n>>valuation(n,2) \\ Charles R Greathouse IV, Jul 05 2013
    
  • Python
    from sympy import divisors
    def a(n):
        return max(d for d in divisors(n) if d % 2)
    print([a(6*n - 2) for n in range(1, 101)]) # Indranil Ghosh, Apr 15 2017, after formula by Reinhard Zumkeller

Formula

a(n) = A000265(6*n-2) = A000265(3*n-1). - Reinhard Zumkeller, Jan 08 2014
From Bob Selcoe, Apr 05 2015: (Start)
For all n>=1 and for every k, there exists j>=0 dependent upon n and k such that either:
Eq. 1: a(n) = (3n-1)/2^(2j+1) when k = ((4^(j+1)-1)/3) mod 2^(2j+3). Alternatively: a(n) = A016789(n-1)/A081294(j+1) when k = A002450(j+1) mod A081294(j+2). Example: n=51; k=101 == 5 mod 32, j=1. a(51) = 152/8 = 19.
or
Eq. 2: a(n) = (3n-1)/4^j when k = (5*2^(2j+1) - 1)/3 mod 4^(j+1). Alternatively: a(n) = A016789(n-1)/A000302(j) when k = A072197(j) mod A000302(j+1). Example: n=91; k=181 == 53 mod 64, j=2. a(91) = 272/16 = 17.
(End) [Definition corrected by William S. Hilton, Jul 29 2017]
a(n) = a(n + g*2^r) - 6*g, n > -g*2^r. Examples: n=59; a(59)=11, r=5. g=-1: 11 = a(27) = 5 - (-1)*6; g=1: 11 = a(91) = 17 - 1*6; g=2: 11 = a(123) = 23 - 2*6; g=3: 11 = a(155) = 29 - 3*6; etc. - Bob Selcoe, Apr 06 2015
a(n) = a((1 + (3*n - 1)*4^(k-1))/3), k>=1 (cf. A191669). - L. Edson Jeffery, Oct 05 2015
a(n) = a(4n-1). - Bob Selcoe, Aug 03 2017
a(n) = A139391(2n-1). - Antti Karttunen, May 06 2024
Sum_{k=1..n} a(k) ~ n^2. - Amiram Eldar, Aug 26 2024
G.f.: Sum_{k>=1} ((3 + 2*(-1)^k)*x^(3*2^(k - 1) - (-2)^k/3 + 1/3) + (3 - 2*(-1)^k)*x^(2^(k - 1) - (-2)^k/3 + 1/3))/(x^(2^k) - 1)^2. - Miles Wilson, Oct 26 2024

A114121 Expansion of (sqrt(1 - 4*x) + (1 - 2*x))/(2*(1 - 4*x)).

Original entry on oeis.org

1, 2, 7, 26, 99, 382, 1486, 5812, 22819, 89846, 354522, 1401292, 5546382, 21977516, 87167164, 345994216, 1374282019, 5461770406, 21717436834, 86392108636, 343801171354, 1368640564996, 5450095992964, 21708901408216, 86492546019214
Offset: 0

Views

Author

Paul Barry, Feb 13 2006

Keywords

Comments

Second binomial transform of A032443 with interpolated zeros.
a(n) is the total number of lattice points, taken over all Dyck n-paths (A000108), that (i) lie on or above ground level and (ii) lie on or directly below a peak. For example with n = 2, UUDD has 1 peak contributing 3 lattice points--(2, 0), (2, 1) and (2, 2) when the path starts at the origin--and UDUD has 2 peaks, each contributing 2 lattice points and so a(2) = 3 + 4 = 7. - David Callan, Jul 14 2006
Hankel transform is binomial(n + 2, 2). - Paul Barry, Dec 04 2007
Image of (-1)^n under the Riordan array ((1/2)(1/(1 - 4x) + 1/sqrt(1 - 4x)), c(x) - 1), c(x) the g.f. of A000108. - Paul Barry, Jun 15 2008
From Gus Wiseman, Jun 21 2021: (Start)
Also the even bisection of A116406 = number compositions of n with alternating sum >= 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. The a(3) = 26 compositions are:
(6) (33) (114) (1122) (11112) (111111)
(42) (123) (1131) (11121)
(51) (132) (1221) (11211)
(213) (2112) (12111)
(222) (2121) (21111)
(231) (2211)
(312) (3111)
(321)
(411)
(End)

Examples

			G.f. = 1 + 2*x + 7*x^2 + 26*x^3 + 99*x^4 + 382*x^5 + 1486*x^6 + 5812*x^7 + ...
		

Crossrefs

The case of alternating sum = 0 is A001700.
The case of alternating sum < 0 is A008549.
This is the even bisection of A116406.
The restriction to reversed partitions is A344611.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A124754 gives the alternating sum of standard compositions.
A316524 is the alternating sum of the prime indices of n.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.

Programs

  • Maple
    seq(sum(binomial(2*n,2*k+irem(n,2)),k=0..floor((1/2)*n)),n=0..20)
    seq(binomial(2*n-1,n)+4^(n-1)-(1/4)*0^n,n=0..20)
  • Mathematica
    a[ n_] := SeriesCoefficient[((1 + 1/Sqrt[1 - 4 x])/2)^2, {x, 0, n}] (* Michael Somos, Dec 31 2013 *)
    ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],ats[#]>=0&]],{n,0,15,2}] (* Gus Wiseman, Jun 21 2021 *)

Formula

a(n) = Sum_{k=0..n} C(n, k)*2^(n-k-2)*(2^k + C(k, k/2))*(1 + (-1)^k).
a(n) = (A000984(n) + A081294(n))/2.
From Paul Barry, Jun 15 2008: (Start)
G.f.: (1 - 4*x + (1 - 2*x)*sqrt(1 - 4*x))/(2*(1 - 4*x)^(3/2)).
a(n) = Sum_{k=0..n} ( Sum_{j=0..n} C(2*n, n-k-j)*(-1)^j ). (End)
a(n) = Sum_{k=0..n} C(2*n, n-k)*(1 + (-1)^k)/2. - Paul Barry, Aug 06 2009
From Paul Barry, Sep 07 2009: (Start)
a(n) = C(2*n-1, n-1) + (4^n + 3*0^n)/4.
Integral representation a(n) = (1/(2*pi))*(Integral_{x=0..4} x^n/sqrt(x(4 - x))) + (4^n + 0^n)/4. (End)
a(n) = Sum_{k=0..floor(n/2)} C(2*n, 2*k + (n mod 2)). - Mircea Merca, Jun 21 2011
Conjecture: n*a(n) + 2*(3 - 4*n)*a(n-1) + 8*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 07 2012
Conjecture verified using the differential equation (16*x^2-8*x+1)*g'(x) + (8*x-2)*g(x)-2*x=0 satisfied by the G.f. - Robert Israel, Jul 27 2020
a(n) = Sum_{i=0..n} (sum_{j=0..n} binomial(n, i+j)*binomial(n, j-i)). - Yalcin Aktar, Jan 07 2013.
G.f.: (1 + (1 - 4*x)^(-1/2))^2 / 4. Convolution square of A088218. - Michael Somos, Dec 31 2013
0 = (1 + 2*n)*b(n) - (5 + 4*n)*b(n+1) + (4 + 2*n)*b(n+2) if n > 0 where b(n) = a(n) / 4^n. - Michael Somos, Dec 31 2013
0 = b(n+3) * (2*b(n+2) - 7*b(n+1) + 5*b(n)) + b(n+2) * (-b(n+2) + 7*b(n+1) - 7*b(n)) + b(n+1) * (-b(n+1) + 2*b(n)) if n > 0 where b(n) = a(n) / 4^n. - Michael Somos, Dec 31 2013
For n > 0, a(n) = 2^(2n-1) - A008549(n). - Gus Wiseman, Jun 27 2021
a(n) = [x^n] 1/((1-2*x) * (1-x)^(n-1)). - Seiichi Manyama, Apr 10 2024
Showing 1-10 of 103 results. Next