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 12 results. Next

A057711 a(0)=0, a(1)=1, a(n) = n*2^(n-2) for n >= 2.

Original entry on oeis.org

0, 1, 2, 6, 16, 40, 96, 224, 512, 1152, 2560, 5632, 12288, 26624, 57344, 122880, 262144, 557056, 1179648, 2490368, 5242880, 11010048, 23068672, 48234496, 100663296, 209715200, 436207616, 905969664, 1879048192, 3892314112, 8053063680
Offset: 0

Views

Author

Bernhard Wolf (wolf(AT)cs.tu-berlin.de), Oct 24 2000

Keywords

Comments

Number of states in the planning domain FERRY, when n-3 cars are at one of two shores while the (n-2)nd car may be on the ferry or at one of the shores.
If the ferry could board any number of cars (instead of only one), the number of states would form the Pisot sequence P(2,6) (A008776). In addition, if k shores existed, the sequence would form the Pisot sequence P(k,k(k+1)). This corresponds to the BRIEFCASE planning domain.
a(i) is the number of occurrences of the number 1 in all palindromic compositions of n = 2*(i+1). - Silvia Heubach (sheubac(AT)calstatela.edu), Jan 10 2003. E.g., there are 5 palindromic compositions of 6, namely 111111 11211 2112 1221 141, containing a total of 16 1's.
Number of occurrences of 00's in all circular binary words of length n. Example: a(3)=6 because in the circular binary words 000, 001, 010, 011, 100, 101, 110 and 111 we have a total of 3+1+1+0+1+0+0+0=6 occurrences of 00. a(n) = Sum_{k=0..n} k*A119458(n,k). - Emeric Deutsch, May 20 2006
a(n) is the number of permutations on [n] for which the entries of each left factor form a circular subinterval of [n]. A subset I of [n] forms a circular subinterval of [n] if it is an ordinary interval [a,b] or has the form [1,a]-union-[b,n] for 1 <= a < b <= n. For example, (5,4,2) is a left factor of the permutation (5,4,2,1,3) which does not form a circular subinterval of [5] and a(4)=16 counts all 24 permutations of [4] except the eight whose first two entries are 1,3 (in either order) or 2,4. - David Callan, Mar 30 2007
a(n) is the total number of runs in all Boolean (n-1)-strings. For example, the 8 Boolean 3-strings, 000, 001, 010, 011, 100, 101, 110, 111 have 1, 2, 3, 2, 2, 3, 2, 1 runs respectively. - David Callan, Jul 22 2008
From Gary W. Adamson, Jul 31 2010: (Start)
Starting with "1" = (1, 2, 4, 8, ...) convolved with (1, 0, 2, 4, 8, ...).
Example: a(6) = 96 = (32, 16, 8, 4, 2, 1) dot (1, 0, 2, 4, 8, 16) = (32 + 0 + 16 + 16 + 16, + 16) = 32 + 4*16 (End)
An elephant sequence, see A175654. For the corner squares 24 A[5] vectors, with decimal values between 27 and 432, lead to this sequence (without the leading 0). For the central square these vectors lead to the companion sequence A087447 (without the first leading 1). - Johannes W. Meijer, Aug 15 2010
Starting with 1 = (1, 1, 2, 4, 8, 16, ...) convolved with (1, 1, 3, 7, 15, 31, ...). - Gary W. Adamson, Oct 26 2010
a(n) is the number of ways to draw simple polygonal chains for n vertices lying on a circle. - Anton Zakharov, Dec 31 2016
Also the number of edges, maximal cliques, and maximum cliques in the n-folded cube graph for n > 3. - Eric W. Weisstein, Dec 01 2017 and Mar 21 2018
Number of pairs of compositions of n corresponding to a seaweed algebra of index n-2 for n > 2. - Nick Mayers, Jun 25 2018
Starting with 1, 2, 6, 16, ..., number of permutations of length n>0 avoiding the partially ordered pattern (POP) {1>2, 1>3} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the second and third elements. - Sergey Kitaev, Dec 08 2020

Examples

			a(1)=6 because the palindromic compositions of n=4 are 4, 1+2+1, 1+1+1+1 and 2+2 and they contain 6 ones. - Silvia Heubach (sheubac(AT)calstatela.edu), Jan 10 2003
		

Crossrefs

Pisot sequence P(2, 6) (A008776), Pisot sequence P(k, k(k+1))
Cf. A119458.

Programs

  • Magma
    [Ceiling(n*2^(n-2)) : n in [0..40]]; // Vincenzo Librandi, Sep 22 2011
    
  • Mathematica
    Join[{0, 1}, Table[n 2^(n - 2), {n, 2, 30}]] (* Eric W. Weisstein, Dec 01 2017 *)
    Join[{0, 1}, LinearRecurrence[{4, -4}, {2, 6}, 20]] (* Eric W. Weisstein, Dec 01 2017 *)
    CoefficientList[Series[x (1 - 2 x + 2 x^2)/(1 - 2 x)^2, {x, 0, 20}], x] (* Eric W. Weisstein, Dec 01 2017 *)
  • PARI
    a(n)=ceil(n*2^(n-2)) \\ Charles R Greathouse IV, Oct 31 2011
    
  • PARI
    x='x+O('x^50); concat(0, Vec(x*(1-2*x+2*x^2)/(1-2*x)^2)) \\ Altug Alkan, Nov 01 2015

Formula

a(n) = ceiling(n*2^(n-2)).
Binomial transform of (0, 1, 0, 3, 0, 5, 0, 7, ...).
From Paul Barry, Apr 06 2003: (Start)
a(0)=0, a(n) = n*(0^(n-1) + 2^(n-1))/2, n > 0.
a(n) = Sum_{k=0..n} binomial(n, 2k+1)*(2k+1).
E.g.f.: x*exp(x)*cosh(x). (End)
The sequence 1, 1, 6, 16, ... is the binomial transform of A016813 with interpolated zeros. - Paul Barry, Jul 25 2003
For n > 1, a(n) = Sum_{k=0..n} (k-n/2)^2 C(n, k). (n+1)*a(n) = A001788(n). - Mario Catalani (mario.catalani(AT)unito.it), Nov 26 2003
From Paul Barry, May 07 2004: (Start)
a(n) = n*2^(n-2) - Sum_{k=0..n} binomial(n, k)*k*(-1)^k.
G.f.: x*(1-2*x+2*x^2)/(1-2*x)^2. (End)
a(n+1) = ceiling(binomial(n+1,1)*2^(n-1)). - Zerinvary Lajos, Nov 01 2006
a(n+1) = Sum_{k=0..n} A196389(n,k)*2^k. - Philippe Deléham, Oct 31 2011
a(0)=0, a(1)=1, a(2)=2, a(3)=6, a(n+1) = 4*a(n)-4*a(n-1) for n >= 3. - Philippe Deléham, Feb 20 2013
a(n) = A002064(n-1) - A002064(n-2), for n >= 2. - Ivan N. Ianakiev, Dec 29 2013
From Amiram Eldar, Aug 05 2020: (Start)
Sum_{n>=1} 1/a(n) = 4*log(2) - 1.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(3/2) - 1. (End)

A175655 Eight bishops and one elephant on a 3 X 3 chessboard. G.f.: (1+x-5*x^2)/(1-3*x-x^2+6*x^3).

Original entry on oeis.org

1, 4, 8, 22, 50, 124, 290, 694, 1628, 3838, 8978, 21004, 48962, 114022, 265004, 615262, 1426658, 3305212, 7650722, 17697430, 40911740, 94528318, 218312114, 503994220, 1163124866, 2683496134, 6189647948, 14273690782
Offset: 0

Views

Author

Johannes W. Meijer, Aug 06 2010, Aug 10 2010

Keywords

Comments

a(n) represents the number of n-move routes of a fairy chess piece starting in the central square (m = 5) on a 3 X 3 chessboard. This fairy chess piece behaves like a bishop on the eight side and corner squares but on the central square the bishop turns into a raging elephant, see A175654.
For the central square the 512 elephants lead to 46 different elephant sequences, see the cross-references for examples.
The sequence above corresponds to 16 A[5] vectors with decimal values 71, 77, 101, 197, 263, 269, 293, 323, 326, 329, 332, 353, 356, 389, 449 and 452. These vectors lead for the side squares to A000079 and for the corner squares to A175654.

Crossrefs

Cf. Elephant sequences central square [decimal value A[5]]: A000007 [0], A000012 [16], A000045 [1], A011782 [2], A000079 [3], A003945 [42], A099036 [11], A175656 [7], A105476 [69], A168604 [26], A045891 [19], A078057 [21], A151821 [170], A175657 [43], 4*A172481 [15; n>=-1], A175655 [71, this sequence], 4*A026597 [325; n>=-1], A033484 [58], A087447 [27], A175658 [23], A026150 [85], A175661 [171], A036563 [186], A098156 [59], A046717 [341], 2*A001792 [187; n>=1 with a(0)=1], A175659 [343].

Programs

  • Magma
    I:=[1, 4, 8]; [n le 3 select I[n] else 3*Self(n-1)+Self(n-2)-6*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jul 21 2013
    
  • Maple
    with(LinearAlgebra): nmax:=27; m:=5; A[5]:= [0,0,1,0,0,0,1,1,1]: A:=Matrix([[0,0,0,0,1,0,0,0,1], [0,0,0,1,0,1,0,0,0], [0,0,0,0,1,0,1,0,0], [0,1,0,0,0,0,0,1,0], A[5], [0,1,0,0,0,0,0,1,0], [0,0,1,0,1,0,0,0,0], [0,0,0,1,0,1,0,0,0], [1,0,0,0,1,0,0,0,0]]): for n from 0 to nmax do B(n):=A^n: a(n):= add(B(n)[m,k],k=1..9): od: seq(a(n), n=0..nmax);
  • Mathematica
    CoefficientList[Series[(1 + x - 5 x^2) / (1 - 3 x - x^2 + 6 x^3), {x, 0, 40}], x] (* Vincenzo Librandi, Jul 21 2013 *)
    LinearRecurrence[{3,1,-6},{1,4,8},40] (* Harvey P. Dale, Dec 25 2024 *)
  • PARI
    a(n)=([0,1,0; 0,0,1; -6,1,3]^n*[1;4;8])[1,1] \\ Charles R Greathouse IV, Oct 03 2016

Formula

G.f.: (1+x-5*x^2)/(1-3*x-x^2+6*x^3).
a(n) = 3*a(n-1) + a(n-2) - 6*a(n-3) with a(0)=1, a(1)=4 and a(2)=8.
a(n) = ((10+8*A)*A^(-n-1) + (10+8*B)*B^(-n-1))/13 - 2^n with A = (-1+sqrt(13))/6 and B = (-1-sqrt(13))/6.
Limit_{k->oo} a(n+k)/a(k) = (-1)^(n)*2*A000244(n)/(A075118(n)-A006130(n-1)*sqrt(13)).
E.g.f.: 2*exp(x/2)*(13*cosh(sqrt(13)*x/2) + 5*sqrt(13)*sinh(sqrt(13)*x/2))/13 - cosh(2*x) - sinh(2*x). - Stefano Spezia, Jan 31 2023

A282011 Number T(n,k) of k-element subsets of [n] having an even sum; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 2, 4, 6, 3, 0, 1, 3, 6, 10, 9, 3, 0, 1, 3, 9, 19, 19, 9, 3, 1, 1, 4, 12, 28, 38, 28, 12, 4, 1, 1, 4, 16, 44, 66, 60, 40, 20, 5, 0, 1, 5, 20, 60, 110, 126, 100, 60, 25, 5, 0, 1, 5, 25, 85, 170, 226, 226, 170, 85, 25, 5, 1, 1, 6, 30, 110, 255, 396, 452, 396, 255, 110, 30, 6, 1
Offset: 0

Views

Author

Alois P. Heinz, Feb 04 2017

Keywords

Comments

Row n is symmetric if and only if n mod 4 in {0,3} (or if T(n,n) = 1).

Examples

			T(5,0) = 1: {}.
T(5,1) = 2: {2}, {4}.
T(5,2) = 4: {1,3}, {1,5}, {2,4}, {3,5}.
T(5,3) = 6: {1,2,3}, {1,2,5}, {1,3,4}, {1,4,5}, {2,3,5}, {3,4,5}.
T(5,4) = 3: {1,2,3,4}, {1,2,4,5}, {2,3,4,5}.
T(5,5) = 0.
T(7,7) = 1: {1,2,3,4,5,6,7}.
Triangle T(n,k) begins:
  1;
  1, 0;
  1, 1,  0;
  1, 1,  1,   1;
  1, 2,  2,   2,   1;
  1, 2,  4,   6,   3,   0;
  1, 3,  6,  10,   9,   3,   0;
  1, 3,  9,  19,  19,   9,   3,   1;
  1, 4, 12,  28,  38,  28,  12,   4,   1;
  1, 4, 16,  44,  66,  60,  40,  20,   5,   0;
  1, 5, 20,  60, 110, 126, 100,  60,  25,   5,  0;
  1, 5, 25,  85, 170, 226, 226, 170,  85,  25,  5, 1;
  1, 6, 30, 110, 255, 396, 452, 396, 255, 110, 30, 6, 1;
		

Crossrefs

Columns k=0..10 give (offsets may differ): A000012, A004526, A002620, A005993, A005994, A032092, A032093, A018211, A018212, A282077, A282078.
Row sums give A011782.
Main diagonal gives A133872(n+1).
Lower diagonals T(n+j,n) for j=1..10 give: A004525(n+1), A282079, A228705, A282080, A282081, A282082, A282083, A282084, A282085, A282086.
T(2n,n) gives A119358.

Programs

  • Maple
    b:= proc(n, s) option remember; expand(
          `if`(n=0, s, b(n-1, s)+x*b(n-1, irem(s+n, 2))))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n, 1)):
    seq(T(n), n=0..16);
  • Mathematica
    Flatten[Table[Sum[Binomial[Ceiling[n/2],2j]Binomial[Floor[n/2],k-2j],{j,0,Floor[(n+1)/4]}],{n,0,10},{k,0,n}]] (* Indranil Ghosh, Feb 26 2017 *)
  • PARI
    a(n,k)=sum(j=0,floor((n+1)/4),binomial(ceil(n/2),2*j)*binomial(floor(n/2),k-2*j));
    tabl(nn)={for(n=0,nn,for(k=0,n,print1(a(n,k),", "););print(););} \\ Indranil Ghosh, Feb 26 2017

Formula

T(n,k) = Sum_{j=0..floor((n+1)/4)} C(ceiling(n/2),2*j) * C(floor(n/2),k-2*j).
T(n,k) = A007318(n,k) - A159916(n,k).
Sum_{k=0..n} k * T(n,k) = A057711(n-1) for n>0.
Sum_{k=0..n} (k+1) * T(n,k) = A087447(n) + [n=2].

A131047 (1/2) * (A007318 - A007318^(-1)).

Original entry on oeis.org

1, 0, 2, 1, 0, 3, 0, 4, 0, 4, 1, 0, 10, 0, 5, 0, 6, 0, 20, 0, 6, 1, 0, 21, 0, 35, 0, 7, 0, 8, 0, 56, 0, 56, 0, 8, 1, 0, 36, 0, 126, 0, 84, 0, 9
Offset: 1

Views

Author

Gary W. Adamson, Jun 12 2007

Keywords

Comments

Row sums = (1, 2, 4, 8, ...). A131047 * (1,2,3, ...) = A087447 starting (1, 4, 10, 24, 56, ...). A generalized set of analogous triangles: (1/(Q+1)) * (P^Q - 1/P), Q an integer, generates triangles with row sums = powers of (Q+1). Cf. A131048, A131049, A131050, A131051 for triangles having Q = 2,3,4 and 5, respectively.
A007318, Pascal's triangle, = this triangle + A119467, since one triangle = the zeros or masks of the other. - Gary W. Adamson, Jun 12 2007

Examples

			First few rows of the triangle:
  1;
  0, 2;
  1, 0,  3;
  0, 4,  0,  4;
  1, 0, 10,  0,  5;
  0, 6,  0, 20,  0, 6;
  1, 0, 21,  0, 35, 0, 7;
  ...
		

Crossrefs

Formula

Let A007318 (Pascal's triangle) = P, then A131047 = (1/2) * (P - 1/P); deleting the right border of zeros.

A127984 a(n) = (n/3 + 7/9)*2^(n - 1) + (-1)^n/9.

Original entry on oeis.org

1, 3, 7, 17, 39, 89, 199, 441, 967, 2105, 4551, 9785, 20935, 44601, 94663, 200249, 422343, 888377, 1864135, 3903033, 8155591, 17010233, 35418567, 73633337, 152859079, 316902969, 656175559, 1357090361, 2803659207, 5786275385, 11930464711, 24576757305
Offset: 1

Views

Author

Artur Jasinski, Feb 09 2007

Keywords

Comments

a(n) is the number of runs of strictly increasing parts in all compositions of n. a(3) = 7: (1)(1)(1), (12), (2)(1), (3). - Alois P. Heinz, Apr 30 2017
From Hugo Pfoertner, Feb 19 2020: (Start)
a(n)/2^(n-2) apparently is the expected number of flips of a fair coin to completion of a game where the player advances by 1 for heads and by 2 for tails, starting at position 0 and repeating to flip until the target n+1 is exactly reached. If the position n (1 below the target) is reached, the player stays at this position and continues to flip the coin and count the flips until he can advance by 1.
The expected number of flips for targets 1, 2, 3,... , found by inversion of the corresponding Markov matrices, is 2, 2, 3, 7/2, 17/4, 39/8, 89/16, 199/32, 441/64, ...
Target 1 needs an expected number of 2 flips and would require a(0) = 1/2.
n=1, target n+1 = 2: 1 / 2^(1-2) = 2;
n=2, target n+1 = 3: 3 / 2^(2-2) = 3;
n=3, target n+1 = 4: 7 / 2^(3-2) = 7/2.
(End)

Crossrefs

Programs

  • Magma
    [(n/3+7/9)*2^(n-1)+(-1)^n/9: n in [1..35]]; // Vincenzo Librandi, Jun 15 2017
  • Maple
    A127984:=n->(n/3 + 7/9)*2^(n - 1) + (-1)^n/9; seq(A127984(n), n=1..50); # Wesley Ivan Hurt, Mar 14 2014
  • Mathematica
    Table[(n/3 + 7/9)2^(n - 1) + (-1)^n/9, {n, 50}] (* Artur Jasinski *)
    CoefficientList[Series[(1 - 2 x^2) / ((-1 + 2 x)^2 (1 + x)), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 15 2017 *)

Formula

a(n) = (n/3 + 7/9)*2^(n - 1) + (-1)^n/9.
From R. J. Mathar, Apr 04 2008: (Start)
O.g.f.: -x*(-1+2x^2)/((-1+2x)^2*(1+x)).
a(n) = 3*a(n-1) - 4*a(n-3). (End)
a(n) + a(n+1) = A087447(n+1). - R. J. Mathar, Feb 21 2009
A172481(n) = a(n) + 2^(n-1). Application: Problem 11623, AMM 119 (2012) 161. - Stephen J. Herschkorn, Feb 11 2012
From Wolfdieter Lang, Jun 14 2017: (Start)
a(n) = f(n+1)*2^(n-1), where f(n) is a rational Fibonacci type sequence based on fuse(a,b) = (a+b+1)/2 with f(0) = 0, f(1) = 1 and f(n) = fuse(f(n-1),f(n-2)), for n >= 2. For fuse(a,b) see the Jeff Erickson link under A188545. Proof: f(n) = (3*n+4 - (-1)^n/2^(n-2))/9, n >= 0, by induction.
a(n) = a(n-1) + a(n-2) + 2^(n-2), n >= 1, with inputs a(-1) = 0, a(0) = 1/2.
(End)
E.g.f.: (2*exp(-x) + exp(2*x)*(7 + 6*x) - 9)/18. - Stefano Spezia, Feb 19 2020

A129953 First differences of A129952.

Original entry on oeis.org

0, 1, 4, 10, 24, 56, 128, 288, 640, 1408, 3072, 6656, 14336, 30720, 65536, 139264, 294912, 622592, 1310720, 2752512, 5767168, 12058624, 25165824, 52428800, 109051904, 226492416, 469762048, 973078528, 2013265920, 4160749568
Offset: 0

Views

Author

Paul Curtz, Jun 10 2007

Keywords

Crossrefs

Programs

  • Magma
    m:=16; S:=&cat[ [ 1, 2*i ]: i in [0..m] ]; T:=[ &+[ Binomial(j-1, k-1)*S[k]: k in [1..j] ]: j in [1..2*m] ]; [ T[n+1]-T[n]: n in[1..2*m-1] ]; // Klaus Brockhaus, Jun 17 2007
    
  • PARI
    {m=29; print1(0, ",", 1, ","); for(n=2, m, print1((n+2)*2^(n-2), ","))} \\ Klaus Brockhaus, Jun 17 2007
    
  • Python
    def A129953(n): return n+2<1 else n # Chai Wah Wu, Oct 03 2024

Formula

a(n) = A129952(n+1) - A129952(n).
a(n) = A087447(n) for n > 0.
a(0) = 0, a(1) = 1; for n > 1, a(n) = (n+2)*2^(n-2).
G.f.: x*(1-2*x^2)/(1-2*x)^2.

Extensions

Edited and extended by Klaus Brockhaus, Jun 17 2007

A306145 Expansion of (1/(1 - x)) * Sum_{k>=0} x^(2*k+1) / Product_{j=1..2*k+1} (1 - x^j).

Original entry on oeis.org

0, 1, 2, 4, 6, 10, 15, 23, 33, 49, 69, 98, 135, 187, 253, 343, 456, 607, 797, 1045, 1355, 1755, 2252, 2884, 3666, 4651, 5863, 7375, 9226, 11517, 14310, 17741, 21904, 26988, 33130, 40586, 49558, 60394, 73383, 88996, 107642, 129958, 156519, 188178, 225734, 270335, 323078, 385494
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 19 2018

Keywords

Comments

Partial sums of A027193.
From Gus Wiseman, Jun 23 2021: (Start)
Also the number of even-length integer partitions of 2n+1 with exactly one odd part. For example, the a(1) = 1 through a(5) = 10 partitions are:
(2,1) (3,2) (4,3) (5,4) (6,5)
(4,1) (5,2) (6,3) (7,4)
(6,1) (7,2) (8,3)
(2,2,2,1) (8,1) (9,2)
(3,2,2,2) (10,1)
(4,2,2,1) (4,3,2,2)
(4,4,2,1)
(5,2,2,2)
(6,2,2,1)
(2,2,2,2,2,1)
Also partitions of 2n+1 with even greatest part and alternating sum 1.
(End)

Crossrefs

First differences are A027193.
The ordered version appears to be A087447 modulo initial terms.
The version for odd instead of even-length partitions is A304620.
The case of strict partitions is A318156.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A027187 counts partitions of even length, with strict case A067661.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A344610 counts partitions by sum and positive reverse-alternating sum.

Programs

  • Mathematica
    nmax = 47; CoefficientList[Series[1/(1 - x) Sum[x^(2 k + 1)/Product[(1 - x^j), {j, 1, 2 k + 1}], {k, 0, nmax}], {x, 0, nmax}], x]
    nmax = 47; CoefficientList[Series[(1 - EllipticTheta[4, 0, x])/(2 (1 - x) QPochhammer[x]), {x, 0, nmax}], x]
    Table[Length[Select[IntegerPartitions[n],EvenQ[Length[#]]&&Count[#,?OddQ]==1&]],{n,1,30,2}] (* _Gus Wiseman, Jun 23 2021 *)

Formula

a(n) = A000070(n) - A304620(n).
a(n) ~ exp(Pi*sqrt(2*n/3)) / (2^(5/2)*Pi*sqrt(n)). - Vaclav Kotesovec, Aug 20 2018

A318156 Expansion of (1/(1 - x)) * Sum_{k>=1} x^(k*(2*k-1)) / Product_{j=1..2*k-1} (1 - x^j).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 7, 9, 12, 16, 21, 27, 35, 44, 55, 69, 85, 104, 127, 154, 186, 224, 268, 320, 381, 452, 534, 630, 741, 869, 1017, 1187, 1382, 1606, 1862, 2155, 2489, 2869, 3301, 3792, 4349, 4979, 5692, 6497, 7405, 8429, 9581, 10876, 12331, 13963, 15792, 17840, 20131, 22691
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 19 2018

Keywords

Comments

Partial sums of A067659.

Examples

			From _Gus Wiseman_, Jul 18 2021: (Start)
Also the number of strict integer partitions of 2n+1 of even length with exactly one odd part. For example, the a(1) = 1 through a(8) = 12 partitions are:
  (2,1)  (3,2)  (4,3)  (5,4)  (6,5)   (7,6)      (8,7)      (9,8)
         (4,1)  (5,2)  (6,3)  (7,4)   (8,5)      (9,6)      (10,7)
                (6,1)  (7,2)  (8,3)   (9,4)      (10,5)     (11,6)
                       (8,1)  (9,2)   (10,3)     (11,4)     (12,5)
                              (10,1)  (11,2)     (12,3)     (13,4)
                                      (12,1)     (13,2)     (14,3)
                                      (6,4,2,1)  (14,1)     (15,2)
                                                 (6,4,3,2)  (16,1)
                                                 (8,4,2,1)  (6,5,4,2)
                                                            (8,4,3,2)
                                                            (8,6,2,1)
                                                            (10,4,2,1)
Also the number of integer partitions of 2n+1 covering an initial interval and having even maximum and alternating sum 1.
(End)
		

Crossrefs

Partial sums of A067659.
The following relate to strict integer partitions of 2n+1 of even length with exactly one odd part.
- Allowing any length gives A036469.
- The non-strict version is A306145.
- The version for odd length is A318155 (non-strict: A304620).
- Allowing any number of odd parts gives A343942 (odd bisection of A067661).
A000041 counts partitions.
A027187 counts partitions of even length (strict: A067661).
A078408 counts strict partitions of 2n+1 (odd bisection of A000009).
A103919 counts partitions by sum and alternating sum (reverse: A344612).

Programs

  • Maple
    b:= proc(n, i, t) option remember; `if`(n>i*(i+1)/2, 0,
          `if`(n=0, t, add(b(n-i*j, i-1, abs(t-j)), j=0..min(n/i, 1))))
        end:
    a:= proc(n) option remember; b(n$2, 0)+`if`(n>0, a(n-1), 0) end:
    seq(a(n), n=0..60);
  • Mathematica
    nmax = 53; CoefficientList[Series[1/(1 - x) Sum[x^(k (2 k - 1))/Product[(1 - x^j), {j, 1, 2 k - 1}], {k, 1, nmax}], {x, 0, nmax}], x]
    nmax = 53; CoefficientList[Series[(QPochhammer[-x, x] - QPochhammer[x])/(2 (1 - x)), {x, 0, nmax}], x]
    Table[Length[Select[IntegerPartitions[2n+1],UnsameQ@@#&&EvenQ[Length[#]]&&Count[#,?OddQ]==1&]],{n,0,15}] (* _Gus Wiseman, Jul 18 2021 *)

Formula

a(n) = A036469(n) - A318155(n).
a(n) = A318155(n) - A078616(n).
a(n) ~ exp(Pi*sqrt(n/3)) * 3^(1/4) / (4*Pi*n^(1/4)). - Vaclav Kotesovec, Aug 20 2018

A087448 3^(n-1)(n+3)/2-(n-1)/2.

Original entry on oeis.org

1, 2, 7, 26, 93, 322, 1091, 3642, 12025, 39362, 127935, 413338, 1328597, 4251522, 13551739, 43046714, 136314609, 430467202, 1355971703, 4261625370, 13366006861, 41841412802, 130754415027, 407953774906, 1270932914153
Offset: 0

Views

Author

Paul Barry, Sep 05 2003

Keywords

Comments

Binomial transform of A087447. Second binomial transform of 1,0,3,0,5,0,7,... (A005408).

Formula

G.f.: (1-2*x)*(1-4*x+5*x^2)/((1-x)^2*(1-3*x)^2). [Colin Barker, Mar 18 2012]

A131052 A131047 * A000012.

Original entry on oeis.org

1, 2, 2, 4, 3, 3, 8, 8, 4, 4, 16, 15, 15, 5, 5, 32, 32, 26, 26, 6, 6, 64, 63, 63, 42, 42, 7, 7, 128, 128, 120, 120, 64, 64, 8, 8
Offset: 1

Views

Author

Gary W. Adamson, Jun 12 2007

Keywords

Comments

Row sums = A087447: (1, 4, 10, 24, 56, ...). A131053 = A000012 * A131047.

Examples

			First few rows of the triangle:
   1;
   2,  2;
   4,  3,  3;
   8,  8,  4,  4;
  16, 15, 15,  5,  5;
  32, 32, 26, 26,  6,  6;
  64, 63, 63, 42, 42,  7,  7;
  ...
		

Crossrefs

Formula

A131047 * A000012 as infinite lower triangular matrices.
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