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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A124734 Table with all compositions sorted first by total, then by length and finally lexicographically.

Original entry on oeis.org

1, 2, 1, 1, 3, 1, 2, 2, 1, 1, 1, 1, 4, 1, 3, 2, 2, 3, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 5, 1, 4, 2, 3, 3, 2, 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, 6, 1, 5, 2, 4, 3, 3, 4, 2, 5, 1, 1, 1, 4, 1, 2, 3, 1, 3, 2, 1, 4, 1, 2, 1
Offset: 1

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Author

Keywords

Comments

This is similar to the Abramowitz and Stegun ordering for partitions (see A036036). The standard ordering for compositions is A066099, which is more similar to the Mathematica partition ordering (A080577).
This can be regarded as a table in two ways: with each composition as a row, or with the compositions of each integer as a row. The first way has A124736 as row lengths and A070939 as row sums; the second has A001792 as row lengths and A001788 as row sums.
This sequence includes every finite sequence of positive integers.

Examples

			The table starts:
1
2; 1 1
3; 1 2; 2 1; 1 1 1
4; 1 3; 2 2; 3 1; 1 1 2; 1 2 1; 2 1 1; 1 1 1 1;
		

Crossrefs

Programs

  • Mathematica
    Table[Sort@Flatten[Permutations /@ IntegerPartitions@n, 1], {n, 8}] // Flatten (* Robert Price, Jun 13 2020 *)

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

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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)

A027472 Third convolution of the powers of 3 (A000244).

Original entry on oeis.org

1, 9, 54, 270, 1215, 5103, 20412, 78732, 295245, 1082565, 3897234, 13817466, 48361131, 167403915, 573956280, 1951451352, 6586148313, 22082967873, 73609892910, 244074908070, 805447196631, 2646469360359, 8661172452084, 28242953648100, 91789599356325, 297398301914493, 960825283108362, 3095992578904722
Offset: 3

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Author

Keywords

Comments

Third column of A027465.
With offset = 2, a(n) is the number of length n words on alphabet {u,v,w,z} such that each word contains exactly 2 u's. - Zerinvary Lajos, Dec 29 2007

Crossrefs

Sequences similar to the form q^(n-2)*binomial(n, 2): A000217 (q=1), A001788 (q=2), this sequence (q=3), A038845 (q=4), A081135 (q=5), A081136 (q=6), A027474 (q=7), A081138 (q=8), A081139 (q=9), A081140 (q=10), A081141 (q=11), A081142 (q=12), A027476 (q=15).

Programs

  • Magma
    [3^(n-3)*Binomial(n-1, 2): n in [3..40]]; // G. C. Greubel, May 12 2021
  • Mathematica
    nn=41; Drop[Range[0,nn]!CoefficientList[Series[Exp[x]^3 x^2/2!,{x,0,nn}],x],2] (* Geoffrey Critzer, Oct 03 2013 *)
    LinearRecurrence[{9,-27,27}, {1,9,54}, 40] (* G. C. Greubel, May 12 2021 *)
    Abs[Take[CoefficientList[Series[1/(1+3x^2)^3,{x,0,60}],x],{1,-1,2}]] (* Harvey P. Dale, Mar 03 2022 *)
  • PARI
    a(n)=([0,1,0; 0,0,1; 27,-27,9]^(n-3)*[1;9;54])[1,1] \\ Charles R Greathouse IV, Oct 03 2016
    
  • Sage
    [3^(n-3)*binomial(n-1,2) for n in range(3, 40)] # Zerinvary Lajos, Mar 10 2009
    

Formula

Numerators of sequence a[3,n] in (b^2)[i,j]) where b[i,j] = binomial(i-1, j-1)/2^(i-1) if j <= i, 0 if j > i.
From Wolfdieter Lang: (Start)
a(n) = 3^(n-3)*binomial(n-1, 2).
G.f.: (x/(1-3*x))^3. (Third convolution of A000244, powers of 3.) (End)
a(n) = |A075513(n, 2)|/9, n >= 3.
a(n) = A152818(n-3,2)/2 = A006043(n-3)/2. - Paul Curtz, Jan 07 2009
The sequence 0, 1, 9, 54, ... has e.g.f.: (x + 3*x^2/2)*exp(3*x)/. - Paul Barry, Jul 23 2003
E.g.f.: E(0) where E(k) = 1 + 3*(2*k+3)*x/((2*k+1)^2 - 3*x*(k+2)*(2*k+1)^2/(3*x*(k+2) + 2*(k+1)^2/E(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 23 2012
With offset=2 e.g.f.: x^2*exp(3*x)/2. - Geoffrey Critzer, Oct 03 2013
From Amiram Eldar, Jan 05 2022: (Start)
Sum_{n>=3} 1/a(n) = 6 - 12*log(3/2).
Sum_{n>=3} (-1)^(n+1)/a(n) = 24*log(4/3) - 6. (End)

Extensions

Corrected by T. D. Noe, Nov 07 2006
Better name from Wolfdieter Lang
Terms a(23) onward added by G. C. Greubel, May 12 2021

A128796 a(n) = n*(n-1)*2^n.

Original entry on oeis.org

0, 0, 8, 48, 192, 640, 1920, 5376, 14336, 36864, 92160, 225280, 540672, 1277952, 2981888, 6881280, 15728640, 35651584, 80216064, 179306496, 398458880, 880803840, 1937768448, 4244635648, 9261023232, 20132659200, 43620761600, 94220845056, 202937204736, 435939180544
Offset: 0

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Author

Mohammad K. Azarian, Apr 07 2007

Keywords

Crossrefs

Programs

  • Magma
    [(n^2-n)*2^n: n in [0..30]]; // Vincenzo Librandi, Feb 10 2013
    
  • Mathematica
    CoefficientList[Series[8 x^2/(1 - 2 x)^3, {x, 0, 30}], x] (* Vincenzo Librandi, Feb 10 2013 *)
  • PARI
    a(n)=n*(n-1)<Charles R Greathouse IV, Sep 24 2015

Formula

G.f.: 8*x^2/(1 - 2*x)^3. - Vincenzo Librandi, Feb 10 2013
a(n) = 8*A001788(n-1). - R. J. Mathar, Apr 26 2015
From Amiram Eldar, Jul 11 2020: (Start)
Sum_{n>=2} 1/a(n) = (1 - log(2))/2.
Sum_{n>=2} (-1)^n/a(n) = (3*log(3/2) - 1)/2. (End)
E.g.f.: 4*exp(2*x)*x^2. - Stefano Spezia, Sep 02 2024

A038845 3-fold convolution of A000302 (powers of 4).

Original entry on oeis.org

1, 12, 96, 640, 3840, 21504, 114688, 589824, 2949120, 14417920, 69206016, 327155712, 1526726656, 7046430720, 32212254720, 146028888064, 657129996288, 2937757630464, 13056700579840, 57724360458240, 253987186016256, 1112705767309312, 4855443348258816
Offset: 0

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Author

Keywords

Comments

Also convolution of A002802 with A000984 (central binomial coefficients).
With a different offset, number of n-permutations of 5 objects u, v, w, z, x with repetition allowed, containing exactly two u's. - Zerinvary Lajos, Dec 29 2007
Also convolution of A000302 with A002697, also convolution of A002457 with itself. - Rui Duarte, Oct 08 2011

Crossrefs

Sequences similar to the form q^(n-2)*binomial(n, 2): A000217 (q=1), A001788 (q=2), A027472 (q=3), this sequence (q=4), A081135 (q=5), A081136 (q=6), A027474 (q=7), A081138 (q=8), A081139 (q=9), A081140 (q=10), A081141 (q=11), A081142 (q=12), A027476 (q=15).

Programs

Formula

a(n) = (n+2)*(n+1)*2^(2*n-1).
G.f.: 1/(1-4*x)^3.
a(n) = Sum_{u+v+w+x+y+z=n} f(u)*f(v)*f(w)*f(x)*f(y)*f(z) with f(n)=A000984(n). - Philippe Deléham, Jan 22 2004
a(n) = binomial(n+2,n) * 4^n. - Rui Duarte, Oct 08 2011
E.g.f.: (1 + 8*x + 8*x^2)*exp(4*x). - G. C. Greubel, Jul 20 2019
From Amiram Eldar, Jan 05 2022: (Start)
Sum_{n>=0} 1/a(n) = 8 - 24*log(4/3).
Sum_{n>=0} (-1)^n/a(n) = 40*log(5/4) - 8. (End)

A001789 a(n) = binomial(n,3)*2^(n-3).

Original entry on oeis.org

1, 8, 40, 160, 560, 1792, 5376, 15360, 42240, 112640, 292864, 745472, 1863680, 4587520, 11141120, 26738688, 63504384, 149422080, 348651520, 807403520, 1857028096, 4244635648, 9646899200, 21810380800, 49073356800, 109924319232
Offset: 3

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Author

Keywords

Comments

Number of 3-dimensional cubes in n-dimensional hypercube. - Henry Bottomley, Apr 14 2000
With three leading zeros, this is the second binomial transform of (0,0,0,1,0,0,0,0,...). - Paul Barry, Mar 07 2003
With 3 leading zeros, binomial transform of C(n,3). - Paul Barry, Apr 10 2003
Let M=[1,0,i;0,1,0;i,0,1], i=sqrt(-1). Then 1/det(I-xM)=1/(1-2x)^4. - Paul Barry, Apr 27 2005
If X_1,X_2,...,X_n is a partition of a 2n-set X into 2-blocks then, for n>2, a(n+1) is equal to the number of (n+3)-subsets of X intersecting each X_i (i=1,2,...,n). - Milan Janjic, Jul 21 2007
With offset 0, a(n) is the number of ways to separate [n] into four non-overlapping intervals (allowed to be empty) and then choose a subset from each interval. - Geoffrey Critzer, Feb 07 2009

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.
  • Clifford A. Pickover, The Math Book, From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics, Sterling Publ., NY, 2009, page 282.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • GAP
    List([3..30], n-> Binomial(n,3)*2^(n-3)); # G. C. Greubel, Aug 27 2019
  • Haskell
    a001789 n = a007318 n 3 * 2 ^ (n - 3)
    a001789_list = 1 : zipWith (+) (map (* 2) a001789_list) (drop 2 a001788_list)
    -- Reinhard Zumkeller, Jul 12 2014
    
  • Magma
    [Binomial(n,3)*2^(n-3): n in [3..30]]; // G. C. Greubel, Aug 27 2019
    
  • Maple
    A001789:=1/(2*z-1)**4; # conjectured by Simon Plouffe in his 1992 dissertation
    seq(binomial(n+3,3)*2^n,n=0..25); # Zerinvary Lajos, Jun 03 2008
  • Mathematica
    Table[Binomial[n, 3]*2^(n-3), {n,3,30}] (* Stefan Steinerberger, Apr 18 2006 *)
    LinearRecurrence[{8,-24,32,-16},{1,8,40,160},30] (* Harvey P. Dale, Feb 10 2016 *)
  • PARI
    a(n)=binomial(n,3)<<(n-3) \\ Charles R Greathouse IV, Sep 24 2015
    

Formula

a(n) = 2*a(n-1) + A001788(n-2).
For n>0, a(n+3) = 2*A082138(n) = 8*A080930(n+1).
From Paul Barry, Apr 10 2003: (Start)
G.f. (with three leading zeros): x^3/(1-2*x)^4.
With three leading zeros, a(n) = 8*a(n-1) - 24*a(n-2) + 32*a(n-3) - 16*a(n-4), a(0)=a(1)=a(2)=0, a(3)=1.
E.g.f.: (x^3/3!)*exp(2*x) (with 3 leading zeros). (End)
a(n) = Sum_{i=3..n} binomial(i,3)*binomial(n,i). Example: for n=6, a(6) = 1*20 + 4*15 + 10*6 + 20*1 = 160. - Bruno Berselli, Mar 23 2018
From Amiram Eldar, Jan 06 2022: (Start)
Sum_{n>=3} 1/a(n) = 6*log(2) - 3.
Sum_{n>=3} (-1)^(n+1)/a(n) = 54*log(3/2) - 21. (End)

Extensions

More terms from James Sellers, Apr 15 2000
More terms from Stefan Steinerberger, Apr 18 2006
Formula fixed by Reinhard Zumkeller, Jul 12 2014

A003472 a(n) = 2^(n-4)*C(n,4).

Original entry on oeis.org

1, 10, 60, 280, 1120, 4032, 13440, 42240, 126720, 366080, 1025024, 2795520, 7454720, 19496960, 50135040, 127008768, 317521920, 784465920, 1917583360, 4642570240, 11142168576, 26528972800, 62704844800, 147220070400
Offset: 4

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Author

Keywords

Comments

Number of 4D hypercubes in n-dimensional hypercube. - Henry Bottomley, Apr 14 2000
With four leading zeros, binomial transform of C(n,4). - Paul Barry, Apr 10 2003
If X_1, X_2, ..., X_n is a partition of a 2n-set X into 2-blocks, then, for n>3, a(n) is equal to the number of (n+4)-subsets of X intersecting each X_i (i=1,2,...,n). - Milan Janjic, Jul 21 2007

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.
  • Clifford A. Pickover, The Math Book, From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics, Sterling Publ., NY, 2009, page 282.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n) = A038207(n,4).

Programs

  • GAP
    List([4..30], n-> 2^(n-4)*Binomial(n,4)); # G. C. Greubel, Aug 27 2019
  • Magma
    [2^(n-4)*Binomial(n, 4): n in [4..30]]; // Vincenzo Librandi, Oct 16 2011
    
  • Maple
    A003472:=-1/(2*z-1)**5; # conjectured by Simon Plouffe in his 1992 dissertation
    seq(binomial(n,4)*2^(n-4),n=4..24); # Zerinvary Lajos, Jun 12 2008
  • Mathematica
    Table[2^(n-4) Binomial[n,4],{n,4,50}] (* or *) LinearRecurrence[{10,-40,80,-80,32},{1,10,60,280,1120},50] (* Harvey P. Dale, May 27 2017 *)
  • PARI
    a(n)=binomial(n,4)<<(n-4) \\ Charles R Greathouse IV, May 18 2015
    
  • Sage
    [2^(n-4)*binomial(n,4) for n in (4..30)] # G. C. Greubel, Aug 27 2019
    

Formula

a(n) = 2*a(n-1) + A001789(n-1).
From Paul Barry, Apr 10 2003: (Start)
O.g.f.: x^4/(1-2*x)^5.
E.g.f.: exp(2*x)(x^4/4!) (with 4 leading zeros). (End)
a(n) = Sum_{i=4..n} binomial(i,4)*binomial(n,i). Example: for n=7, a(7) = 1*35 + 5*21 + 15*7 + 35*1 = 280. - Bruno Berselli, Mar 23 2018
From Amiram Eldar, Jan 06 2022: (Start)
Sum_{n>=4} 1/a(n) = 20/3 - 8*log(2).
Sum_{n>=4} (-1)^n/a(n) = 216*log(3/2) - 260/3. (End)

Extensions

More terms from James Sellers, Apr 15 2000

A001815 a(n) = binomial(n,2) * 2^(n-1).

Original entry on oeis.org

0, 0, 2, 12, 48, 160, 480, 1344, 3584, 9216, 23040, 56320, 135168, 319488, 745472, 1720320, 3932160, 8912896, 20054016, 44826624, 99614720, 220200960, 484442112, 1061158912, 2315255808, 5033164800, 10905190400, 23555211264, 50734301184, 108984795136
Offset: 0

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Author

Keywords

Comments

Number of permutations of length n+3 containing 132 and 123 exactly once. Likewise for the pairs (123,213), (231,321), (312,321).
a(n) is the number of ways to assign n distinct contestants to two (not necessarily equal) distinct teams and then choose a captain for each team. - Geoffrey Critzer, Apr 07 2009
Consider all binary words of length n, and assign a weight to each set bit - the leftmost gets a weight of n-1, the rightmost a weight of 0. a(n) gives the sum of the weights of all n-bit words. For example, if n=3, we have 000, 001, 010, 011, 100, 101, 110, 111 with weights of 0, 0, 1, 1, 2, 2, 3, 3, giving a sum of 12.
a(n) is the number of North-East paths from (0,0) to (n+2,n+2) that have exactly one east step below y = x-1 and exactly one east step above y = x+1. This is related to the paired pattern P_1 and P_2. More details can be found in Pan and Remmel's link. - Ran Pan, Feb 03 2016
a(n) is the number of diagonals of length sqrt(2) in an n-dimensional hypercube (same as diagonals of its two-dimensional faces). - Stanislav Sykora, Oct 23 2016
a(n) is the number of ways to select a team from n players with at least two players, two of whom are the captain and the goalkeeper. - Wojciech Raszka, Apr 10 2019
a(n) is the sum of N_0*N_1 for all binary strings of length n, where N_0 and N_1 are the number of 0's and 1's in the string, respectively. For example, if n=3, we have 000, 001, 010, 011, 100, 101, 110, 111 with products 0, 2, 2, 2, 2, 2, 2, 0, giving a sum of 12. - Sigurd Kittilsen and Jens Otten, Sep 17 2020

Examples

			G.f. = 2*x^2 + 12*x^3 + 48*x^4 + 160*x^5 + 480*x^6 + 1344*x^7 + 3584*x^8 + ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 801.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    [Binomial(n,2)*2^(n-1): n in [0..30]]; // Vincenzo Librandi, Mar 14 2014
  • Maple
    A001815 := proc(n)
        2^(n-2)*n*(n-1) ;
    end proc: # R. J. Mathar, Mar 12 2014
  • Mathematica
    Table[Binomial[n, 2]*2^(n-1), {n, 0, 28}] (* Arkadiusz Wesolowski, Dec 21 2011 *)
    CoefficientList[Series[2 x^2/(1 - 2 x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, Mar 14 2014 *)
    LinearRecurrence[{6,-12,8},{0,0,2},30] (* Harvey P. Dale, May 19 2018 *)
  • PARI
    a(n)=binomial(n,2)<<(n-1) \\ Charles R Greathouse IV, Dec 21 2011
    
  • PARI
    my(x='x+O('x^100)); concat([0, 0], Vec(2*x^2/(1-2*x)^3)) \\ Altug Alkan, Nov 01 2015
    
  • Sage
    [lucas_number1(n, 2, 0)*binomial(n,2) for n in range(0, 29)] # Zerinvary Lajos, Mar 10 2009
    

Formula

G.f.: 2*x^2/(1 - 2*x)^3. [Simon Plouffe in his 1992 dissertation]
a(n) = A090802(n, 2).
a(n) = Sum_{i=0..n} i*(n-i)*binomial(n, i). - Benoit Cloitre, Nov 11 2004
a(n) = Sum_{k=0..n} k*2^(k-1). - Zerinvary Lajos, Oct 09 2006
a(n) = Sum_{j=0..n} binomial(n-1,j)*n*j. - Zerinvary Lajos, Oct 19 2006
E.g.f.: x^2*exp(2*x). - Geoffrey Critzer, Apr 07 2009
a(n) = 2^(n-2)*n*(n-1). - Tobias Friedrich (tfried(AT)mpi-inf.mpg.de), Jun 18 2009
a(n) = 2*a(n-1) + n*2^n.
For n > 0, a(n) = 2*A001788(n-1). - Stanislav Sykora, Oct 23 2016
a(n) = a(1-n) * 2^(2*n-1) for all n in Z. - Michael Somos, Oct 25 2016
a(n) = Sum_{k=0..n-1} Sum_{i=0..n-1} k * binomial(n-1,i). - Wesley Ivan Hurt, Sep 20 2017
From Amiram Eldar, Jan 07 2022: (Start)
Sum_{n>=0} 1/a(n) = 2*(1 - log(2)).
Sum_{n>=0} (-1)^n/a(n) = 6*log(3/2) - 2. (End)

A081139 9th binomial transform of (0,0,1,0,0,0,...).

Original entry on oeis.org

0, 0, 1, 27, 486, 7290, 98415, 1240029, 14880348, 172186884, 1937102445, 21308126895, 230127770466, 2447722649502, 25701087819771, 266895911974545, 2745215094595320, 28001193964872264, 283512088894331673
Offset: 0

Views

Author

Paul Barry, Mar 08 2003

Keywords

Comments

Starting at 1, the three-fold convolution of A001019 (powers of 9).

Crossrefs

Sequences similar to the form q^(n-2)*binomial(n, 2): A000217 (q=1), A001788 (q=2), A027472 (q=3), A038845 (q=4), A081135 (q=5), A081136 (q=6), A027474 (q=7), A081138 (q=8), this sequence (q=9), A081140 (q=10), A081141 (q=11), A081142 (q=12), A027476 (q=15).
Cf. A001019.

Programs

  • Magma
    [9^n* Binomial(n+2, 2): n in [-2..20]]; // Vincenzo Librandi, Oct 16 2011
  • Mathematica
    LinearRecurrence[{27,-243,729},{0,0,1},30] (* Harvey P. Dale, Jan 30 2018 *)

Formula

a(n) = 27*a(n-1) - 243*a(n-2) + 729*a(n-3), a(0)=a(1)=0, a(2)=1.
a(n) = 9^(n-2)*binomial(n, 2).
G.f.: x^2/(1-9*x)^3.
E.g.f.: (x^2/2)*exp(9*x). - G. C. Greubel, May 13 2021
From Amiram Eldar, Jan 06 2022: (Start)
Sum_{n>=2} 1/a(n) = 18 - 144*log(9/8).
Sum_{n>=2} (-1)^n/a(n) = 180*log(10/9) - 18. (End)

A091894 Triangle read by rows: T(n,k) is the number of Dyck paths of semilength n, having k ddu's [here u = (1,1) and d = (1,-1)].

Original entry on oeis.org

1, 1, 2, 4, 1, 8, 6, 16, 24, 2, 32, 80, 20, 64, 240, 120, 5, 128, 672, 560, 70, 256, 1792, 2240, 560, 14, 512, 4608, 8064, 3360, 252, 1024, 11520, 26880, 16800, 2520, 42, 2048, 28160, 84480, 73920, 18480, 924, 4096, 67584, 253440, 295680, 110880, 11088, 132
Offset: 0

Views

Author

Emeric Deutsch, Mar 10 2004

Keywords

Comments

Number of Dyck paths of semilength n, having k uu's with midpoint at even height. Example: T(4,1) = 6 because we have u(uu)duddd, u(uu)udddd, udu(uu)ddd, u(uu)dddud, u(uu)ddudd and uud(uu)ddd [here u = (1,1), d = (1,-1) and the uu's with midpoint at even height are shown between parentheses]. Row sums are the Catalan numbers (A000108). T(2n+1,n) = A000108(n) (the Catalan numbers). Sum_{k>=0} k*T(n,k) = binomial(2n-2,n-3) = A002694(n-1).
Sometimes called the Touchard distribution (after Touchard's Catalan number identity). T(n,k) = number of full binary trees on 2n edges with k deep interior vertices (deep interior means you have to traverse at least 2 edges to reach a leaf) = number of binary trees on n-1 edges with k vertices having a full complement of 2 children. - David Callan, Jul 19 2004
From David Callan, Oct 25 2004: (Start)
T(n,k) = number of ordered trees on n edges with k prolific edges. A prolific edge is one whose child vertex has at least two children. For example with n=3, drawing ordered trees down from the root, /|\ has no prolific edge and the only tree with one prolific edge has the shape of an inverted Y, so T(3,1)=1.
Proof: Consider the following bijection, recorded by Emeric Deutsch, from ordered trees on n edges to Dyck n-paths. For a given ordered tree, traverse the tree in preorder (walk-around from root order). To each node of outdegree r there correspond r upsteps followed by 1 downstep; nothing corresponds to the last leaf. This bijection sends prolific edges to noninitial ascents of length >=2, that is, to DUU's. Then reverse the resulting Dyck n-path so that prolific edges correspond to DDU's. (End)
T(n,k) is the number of Łukasiewicz paths of length n having k fall steps (1,-1) that start at an even level. A Łukasiewicz path of length n is a path in the first quadrant from (0,0) to (n,0) using rise steps (1,k) for any positive integer k, level steps (1,0) and fall steps (1,-1) (see R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge Univ. Press, Cambridge, 1999, p. 223, Exercise 6.19w; the integers are the slopes of the steps). Example: T(3,1) = 1 because we have U(2)(D)D, where U(2) = (1,2), D = (1,-1) and the fall steps that start at an even level are shown between parentheses. Row n contains ceiling(n/2) terms (n >= 1). - Emeric Deutsch, Jan 06 2005
Number of binary trees with n-1 edges and k+1 leaves (a binary tree is a rooted tree in which each vertex has at most two children and each child of a vertex is designated as its left or right child). - Emeric Deutsch, Jul 31 2006
Number of full binary trees with 2n edges and k+1 vertices both children of which are leaves (n >= 1; a full binary tree is a rooted tree in which each vertex has either 0 or two children). - Emeric Deutsch, Dec 26 2006
Number of ordered trees with n edges and k jumps. In the preorder traversal of an ordered tree, any transition from a node at a deeper level to a node on a strictly higher level is called a jump. - Emeric Deutsch, Jan 18 2007
It is remarkable that we can generate the coefficients of the right hand columns of triangle A175136 with the aid of the coefficients in the rows of the triangle given above. See A175136 for more information. - Johannes W. Meijer, May 06 2011
The antidiagonal sums equal A152225. - Johannes W. Meijer, Sep 13 2012
This array also counts 231-avoiding permutations according to the number of peaks, i.e., positions w[i-1] < w[i] > w[i+1]. For example, 123, 213, 312, and 321 have no peaks, while 132 has one peak. Note also T(n,k) = 2^(n - 1 - 2*k)*A055151(n-1,k). - Kyle Petersen, Aug 02 2013

Examples

			T(4,1) = 6 because we have uduu(ddu)d, uu(ddu)dud, uuu(ddu)dd, uu(ddu)udd, uudu(ddu)d and uuud(ddu)d [here u = (1,1), d = (1,-1) and the ddu's are shown between parentheses].
Triangle begins:
    1;
    1;
    2;
    4,    1;
    8,    6;
   16,   24,    2;
   32,   80,   20;
   64,  240,  120,   5;
  128,  672,  560,  70;
  256, 1792, 2240, 560, 14;
  ...
		

References

  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Section 4.3.

Crossrefs

The first few columns equal A011782, A001788, 2*A003472, 5*A002409, 14*A140325 and 42*A172242. - Johannes W. Meijer, Sep 13 2012

Programs

  • GAP
    T:=Concatenation([1],Flat(List([1..13],n->List([0..Int((n-1)/2)],k->2^(n-2*k-1)*Binomial(n-1,2*k)*Binomial(2*k,k)/(k+1))))); # Muniru A Asiru, Nov 29 2018
    
  • Maple
    a := proc(n,k) if n=0 and k=0 then 1 elif n=0 then 0 else 2^(n-2*k-1)*binomial(n-1,2*k)*binomial(2*k,k)/(k+1) fi end: 1,seq(seq(a(n,k),k=0..(n-1)/2),n=1..15);
  • Mathematica
    A091894[n_] := Prepend[Table[ CoefficientList[ 2^i (1 - z)^((2 i + 3)/2) Hypergeometric2F1[(i + 3)/2, (i + 4)/2, 2, z], z], {i, 0, n}], {1}] (* computes a table of the first n rows. Stumbled accidentally on it. Perhaps someone can find a relationship here? Thies Heidecke (theidecke(AT)astrophysik.uni-kiel.de), Sep 23 2008 *)
    Join[{1},Select[Flatten[Table[2^(n-2k-1) Binomial[n-1,2k] Binomial[2k,k]/ (k+1), {n,20},{k,0,n}]],#!=0&]] (* Harvey P. Dale, Mar 05 2012 *)
    p[n_] := 2^n Hypergeometric2F1[(1 - n)/2, -n/2, 2, x]; Flatten[Join[{{1}}, Table[CoefficientList[p[n], x], {n, 0, 12}]]] (* Peter Luschny, Jan 23 2018 *)
  • PARI
    {T(n, k) = if( n<1, n==0 && k==0, polcoeff( polcoeff( serreverse( x / (1 + 2*x*y + x^2) + x * O(x^n)), n), n-1 - 2*k))} /* Michael Somos, Sep 25 2006 */
    
  • Sage
    [1] + [[2^(n-2*k-1)*binomial(n-1,2*k)*binomial(2*k,k)/(k+1) for k in (0..floor((n-1)/2))] for n in (1..12)] # G. C. Greubel, Nov 30 2018

Formula

T(n,k) = 2^(n - 2*k - 1)*binomial(n-1,2*k)*binomial(2*k,k)/(k + 1), T(0,0) = 1, for 0 <= k <= floor((n-1)/2).
G.f.: G = G(t,z) satisfies: t*z*G^2 - (1 - 2*z + 2*t*z)*G + 1 - z + t*z = 0.
With first row zero, the o.g.f. is g(x,t) = (1 - 2*x - sqrt((1 - 2*x)^2 - 4*t*x^2)) / (2*t*x) with the inverse ginv(x,t) = x / (1 + 2*x + t*x^2), an o.g.f. for shifted A207538 and A133156 mod signs, so A134264 and A125181 can be used to interpret the polynomials of this entry. Cf. A097610. - Tom Copeland, Feb 08 2016
If we delete the first 1 from the data these are the coefficients of the polynomials p(n) = 2^n*hypergeom([(1 - n)/2, - n/2], [2], x). - Peter Luschny, Jan 23 2018
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