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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A000346 a(n) = 2^(2*n+1) - binomial(2*n+1, n+1).

Original entry on oeis.org

1, 5, 22, 93, 386, 1586, 6476, 26333, 106762, 431910, 1744436, 7036530, 28354132, 114159428, 459312152, 1846943453, 7423131482, 29822170718, 119766321572, 480832549478, 1929894318332, 7744043540348, 31067656725032, 124613686513778, 499744650202436
Offset: 0

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Keywords

Comments

Also a(n) = 2nd elementary symmetric function of binomial(n,0), binomial(n,1), ..., binomial(n,n).
Also a(n) = one half the sum of the heights, over all Dyck (n+2)-paths, of the vertices that are at even height and terminate an upstep. For example with n=1, these vertices are indicated by asterisks in the 5 Dyck 3-paths: UU*UDDD, UU*DU*DD, UDUU*DD, UDUDUD, UU*DDUD, yielding a(1)=(2+4+2+0+2)/2=5. - David Callan, Jul 14 2006
Hankel transform is (-1)^n*(2n+1); the Hankel transform of sum(k=0..n, C(2*n,k) ) - C(2n,n) is (-1)^n*n. - Paul Barry, Jan 21 2007
Row sums of the Riordan matrix (1/(1-4x),(1-sqrt(1-4x))/2) (A187926). - Emanuele Munarini, Mar 16 2011
From Gus Wiseman, Jul 19 2021: (Start)
For n > 0, a(n-1) is also the number of integer compositions of 2n with nonzero alternating sum, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. These compositions are ranked by A053754 /\ A345921. For example, the a(3-1) = 22 compositions of 6 are:
(6) (1,5) (1,1,4) (1,1,1,3) (1,1,1,1,2)
(2,4) (1,2,3) (1,1,3,1) (1,1,2,1,1)
(4,2) (1,4,1) (1,2,1,2) (2,1,1,1,1)
(5,1) (2,1,3) (1,3,1,1)
(2,2,2) (2,1,2,1)
(3,1,2) (3,1,1,1)
(3,2,1)
(4,1,1)
(End)

Examples

			G.f. = 1 + 5*x + 22*x^2 + 93*x^3 + 386*x^4 + 1586*x^5 + 6476*x^6 + ...
		

References

  • T. Myers and L. Shapiro, Some applications of the sequence 1, 5, 22, 93, 386, ... to Dyck paths and ordered trees, Congressus Numerant., 204 (2010), 93-104.
  • D. Phulara and L. W. Shapiro, Descendants in ordered trees with a marked vertex, Congressus Numerantium, 205 (2011), 121-128.
  • 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

Cf. A000108, A014137, A014318. A column of A058893. Subdiagonal of A053979.
Bisection of A058622 and (possibly) A007008.
Even bisection of A294175 (without the first two terms).
The following relate to compositions of 2n with alternating sum k.
- The k > 0 case is counted by A000302.
- The k <= 0 case is counted by A000302.
- The k != 0 case is counted by A000346 (this sequence).
- The k = 0 case is counted by A001700 or A088218.
- The k < 0 case is counted by A008549.
- The k >= 0 case is counted by A114121.
A011782 counts compositions.
A086543 counts partitions with nonzero alternating sum (bisection: A182616).
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.

Programs

  • Magma
    [2^(2*n+1) - Binomial(2*n+1,n+1): n in [0..30]]; // Vincenzo Librandi, Jun 07 2011
  • Maple
    seq(2^(2*n+1)-binomial(2*n,n)*(2*n+1)/(n+1), n=0..12); # Emanuele Munarini, Mar 16 2011
  • Mathematica
    Table[2^(2n+1)-Binomial[2n,n](2n+1)/(n+1),{n,0,20}] (* Emanuele Munarini, Mar 16 2011 *)
    a[ n_] := If[ n<-4, 0, (4^(n + 1) - Binomial[2 n + 2, n + 1]) / 2]; (* Michael Somos, Jan 25 2014 *)
  • Maxima
    makelist(2^(2*n+1)-binomial(2*n,n)*(2*n+1)/(n+1),n,0,12); /* Emanuele Munarini, Mar 16 2011 */
    
  • PARI
    {a(n) = if( n<-4, 0, n++; (2^(2*n) - binomial(2*n, n)) / 2)}; /* Michael Somos, Jan 25 2014 */
    

Formula

G.f.: c(x)/(1-4x), c(x) = g.f. of Catalan numbers.
Convolution of Catalan numbers and powers of 4.
Also one half of convolution of central binomial coeffs. A000984(n), n=0, 1, 2, ... with shifted central binomial coeffs. A000984(n), n=1, 2, 3, ...
a(n) = A045621(2n+1) = (1/2)*A068551(n+1).
a(n) = Sum_{k=0..n} A000984(k)*A001700(n-k). - Philippe Deléham, Jan 22 2004
a(n) = Sum_{k=0..n+1} binomial(n+k, k-1)2^(n-k+1). - Paul Barry, Nov 13 2004
a(n) = Sum_{i=0..n} binomial(2n+2, i). See A008949. - Ed Catmur (ed(AT)catmur.co.uk), Dec 09 2006
a(n) = Sum_{k=0..n+1, C(2n+2,k)} - C(2n+2,n+1). - Paul Barry, Jan 21 2007
Logarithm g.f. log(1/(2-C(x)))=sum(n>0, a(n)/n*x^n), C(x)=(1-sqrt(1-4*x))/2x (A000108). - Vladimir Kruchinin, Aug 10 2010
D-finite with recurrence: (n+3) a(n+2) - 2(4n+9) a(n+1) + 8(2n+3) a(n) = 0. - Emanuele Munarini, Mar 16 2011
E.g.f.:exp(2*x)*(2*exp(2*x) - BesselI(0,2*x) - BesselI(1,2*x)).
This is the first derivative of exp(2*x)*(exp(2*x) - BesselI(0,2*x))/2. See the e.g.f. of A032443 (which has a plus sign) and the remarks given there. - Wolfdieter Lang, Jan 16 2012
a(n) = Sum_{0<=iMircea Merca, Apr 05 2012
A000346 = A004171 - A001700. See A032443 for the sum. - M. F. Hasler, Jan 02 2014
0 = a(n) * (256*a(n+1) - 224*a(n+2) + 40*a(n+3)) + a(n+1) * (-32*a(n+1) + 56*a(n+2) - 14*a(n+3)) + a(n+2) * (-2*a(n+2) + a(n+3)) if n>-5. - Michael Somos, Jan 25 2014
REVERT transform is [1,-5,28,-168,1056,...] = alternating signed version of A069731. - Michael Somos, Jan 25 2014
Convolution square is A038806. - Michael Somos, Jan 25 2014
BINOMIAL transform of A055217(n-1) is a(n-1). - Michael Somos, Jan 25 2014
(n+1) * a(n) = A033504(n). - Michael Somos, Jan 25 2014
Recurrence: (n+1)*a(n) = 512*(2*n-7)*a(n-5) + 256*(13-5*n)*a(n-4) + 64*(10*n-17)*a(n-3) + 32*(4-5*n)*a(n-2) + 2*(10*n+1)*a(n-1), n>=5. - Fung Lam, Mar 21 2014
Asymptotic approximation: a(n) ~ 2^(2n+1)*(1-1/sqrt(n*Pi)). - Fung Lam, Mar 21 2014
a(n) = Sum_{m = n+2..2*(n+1)} binomial(2*(n+1), m), n >= 0. - Wolfdieter Lang, May 22 2015
a(n) = A000302(n) + A008549(n). - Gus Wiseman, Jul 19 2021
a(n) = Sum_{j=1..n+1} Sum_{k=1..j} 2^(j-k)*binomial(n+k-1, n). - Fabio Visonà, May 04 2022
a(n) = (1/2)*(-1)^n*binomial(-(n+1), n+2)*hypergeom([1, 2*n + 3], [n + 3], 1/2). - Peter Luschny, Nov 29 2023

Extensions

Corrected by Christian G. Bower

A002054 Binomial coefficient C(2n+1, n-1).

Original entry on oeis.org

1, 5, 21, 84, 330, 1287, 5005, 19448, 75582, 293930, 1144066, 4457400, 17383860, 67863915, 265182525, 1037158320, 4059928950, 15905368710, 62359143990, 244662670200, 960566918220, 3773655750150, 14833897694226, 58343356817424, 229591913401900
Offset: 1

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Comments

a(n) = number of permutations in S_{n+2} containing exactly one 312 pattern. E.g., S_3 has a_1 = 1 permutations containing exactly one 312 pattern, and S_4 has a_2 = 5 permutations containing exactly one 312 pattern, namely 1423, 2413, 3124, 3142, and 4231. This comment is also true if 312 is replaced by any of 132, 213, or 231 (but not 123 or 321, for which see A003517). [Comment revised by N. J. A. Sloane, Nov 26 2022]
Number of valleys in all Dyck paths of semilength n+1. Example: a(2)=5 because UD*UD*UD, UD*UUDD, UUDD*UD, UUD*UDD, UUUDDD, where U=(1,1), D=(1,-1) and the valleys are shown by *. - Emeric Deutsch, Dec 05 2003
Number of UU's (double rises) in all Dyck paths of semilength n+1. Example: a(2)=5 because UDUDUD, UDU*UDD, U*UDDUD, U*UDUDD, U*U*UDDD, the double rises being shown by *. - Emeric Deutsch, Dec 05 2003
Number of peaks at level higher than one (high peaks) in all Dyck paths of semilength n+1. Example: a(2)=5 because UDUDUD, UDUU*DD, UU*DDUD, UU*DU*DD, UUU*DDD, the high peaks being shown by *. - Emeric Deutsch, Dec 05 2003
Number of diagonal dissections of a convex (n+3)-gon into n regions. Number of standard tableaux of shape (n,n,1) (see Stanley reference). - Emeric Deutsch, May 20 2004
Number of dissections of a convex (n+3)-gon by noncrossing diagonals into several regions, exactly n-1 of which are triangular. Example: a(2)=5 because the convex pentagon ABCDE is dissected by any of the diagonals AC, BD, CE, DA, EB into regions containing exactly 1 triangle. - Emeric Deutsch, May 31 2004
Number of jumps in all full binary trees with n+1 internal nodes. In the preorder traversal of a full binary 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
a(n) is the total number of nonempty Dyck subpaths in all Dyck paths (A000108) of semilength n. For example, the Dyck path UUDUUDDD has Dyck subpaths stretching over positions 1-8 (the entire path), 2-3, 2-7, 4-7, 5-6 and so contributes 5 to a(4). - David Callan, Jul 25 2008
a(n+1) is the total number of ascents in the set of all n-permutations avoiding the pattern 132. For example, a(2) = 5 because there are 5 ascents in the set 123, 213, 231, 312, 321. - Cheyne Homberger, Oct 25 2013
Number of increasing tableaux of shape (n+1,n+1) with largest entry 2n+1. An increasing tableau is a semistandard tableau with strictly increasing rows and columns, and set of entries an initial segment of the positive integers. Example: a(2) = 5 counts the five tableaux (124)(235), (123)(245), (124)(345), (134)(245), (123)(245). - Oliver Pechenik, May 02 2014
a(n) is the number of noncrossing partitions of 2n+1 into n-1 blocks of size 2 and 1 block of size 3. - Oliver Pechenik, May 02 2014
Number of paths in the half-plane x>=0, from (0,0) to (2n+1,3), and consisting of steps U=(1,1) and D=(1,-1). For example, for n=2, we have the 5 paths: UUUUD, UUUDU, UUDUU, UDUUU, DUUUU. - José Luis Ramírez Ramírez, Apr 19 2015
From Gus Wiseman, Aug 20 2021: (Start)
Also the number of binary numbers with 2n+2 digits and with two more 0's than 1's. For example, the a(2) = 5 binary numbers are: 100001, 100010, 100100, 101000, 110000, with decimal values 33, 34, 36, 40, 48. Allowing first digit 0 gives A001791, ranked by A345910/A345912.
Also the number of integer compositions of 2n+2 with alternating sum -2, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(3) = 21 compositions are:
(35) (152) (1124) (11141) (111113)
(251) (1223) (12131) (111212)
(1322) (13121) (111311)
(1421) (14111) (121112)
(2114) (121211)
(2213) (131111)
(2312)
(2411)
The following pertain to these compositions:
- The unordered version is A344741.
- Ranked by A345924 (reverse: A345923).
- A345197 counts compositions by length and alternating sum.
- A345925 ranks compositions with alternating sum 2 (reverse: A345922).
(End)

Examples

			G.f. = x + 5*x^2 + 21*x^3 + 84*x^4 + 330*x^5 + 1287*x^6 + 5005*x^7 + ...
		

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. 828.
  • George Grätzer, General Lattice Theory. Birkhauser, Basel, 1998, 2nd edition, p. 474, line -3.
  • 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.
  • 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

Diagonal 4 of triangle A100257. Also a diagonal of A033282.
Equals (1/2) A024483(n+2). Bisection of A037951 and A037955.
Cf. A001263.
Column k=1 of A263771.
Counts terms of A031445 with 2n+2 digits in binary.
Cf. binomial(2*n+m, n): A000984 (m = 0), A001700 (m = 1), A001791 (m = 2), A002694 (m = 4), A003516 (m = 5), A002696 (m = 6), A030053 - A030056, A004310 - A004318.

Programs

  • GAP
    List([1..25],n->Binomial(2*n+1,n-1)); # Muniru A Asiru, Aug 09 2018
    
  • Magma
    [Binomial(2*n+1, n-1): n in [1..30]]; // Vincenzo Librandi, Apr 20 2015
    
  • Maple
    with(combstruct): seq((count(Composition(2*n+2), size=n)), n=1..24); # Zerinvary Lajos, May 03 2007
  • Mathematica
    CoefficientList[Series[8/(((Sqrt[1-4x] +1)^3)*Sqrt[1-4x]), {x,0,22}], x] (* Robert G. Wilson v, Aug 08 2011 *)
    a[ n_]:= Binomial[2 n + 1, n - 1]; (* Michael Somos, Apr 25 2014 *)
  • PARI
    {a(n) = binomial( 2*n+1, n-1)};
    
  • Python
    from _future_ import division
    A002054_list, b = [], 1
    for n in range(1,10**3):
        A002054_list.append(b)
        b = b*(2*n+2)*(2*n+3)//(n*(n+3)) # Chai Wah Wu, Jan 26 2016
    
  • Sage
    [binomial(2*n+1, n-1) for n in (1..25)] # G. C. Greubel, Mar 22 2019

Formula

a(n) = Sum_{j=0..n-1} binomial(2*j, j) * binomial(2*n - 2*j, n-j-1)/(j+1). - Yong Kong (ykong(AT)curagen.com), Dec 26 2000
G.f.: z*C^4/(2-C), where C=[1-sqrt(1-4z)]/(2z) is the Catalan function. - Emeric Deutsch, Jul 05 2003
From Wolfdieter Lang, Jan 09 2004: (Start)
a(n) = binomial(2*n+1, n-1) = n*C(n+1)/2, C(n)=A000108(n) (Catalan).
G.f.: (1 - 2*x - (1-3*x)*c(x))/(x*(1-4*x)) with g.f. c(x) of A000108. (End)
G.f.: z*C(z)^3/(1-2*z*C(z)), where C(z) is the g.f. of Catalan numbers. - José Luis Ramírez Ramírez, Apr 19 2015
G.f.: 2F1(5/2, 2; 4; 4*x). - R. J. Mathar, Aug 09 2015
D-finite with recurrence: a(n+1) = a(n)*(2*n+3)*(2*n+2)/(n*(n+3)). - Chai Wah Wu, Jan 26 2016
From Ilya Gutkovskiy, Aug 30 2016: (Start)
E.g.f.: (BesselI(0,2*x) + (1 - 1/x)*BesselI(1,2*x))*exp(2*x).
a(n) ~ 2^(2*n+1)/sqrt(Pi*n). (End)
a(n) = (1/(n+1))*Sum_{i=0..n-1} (n+1-i)*binomial(2n+2,i), n >= 1. - Taras Goy, Aug 09 2018
G.f.: (x - 1 + (1 - 3*x)/sqrt(1 - 4*x))/(2*x^2). - Michael Somos, Jul 28 2021
From Amiram Eldar, Jan 24 2022: (Start)
Sum_{n>=1} 1/a(n) = 5/3 - 2*Pi/(9*sqrt(3)).
Sum_{n>=1} (-1)^(n+1)/a(n) = 52*log(phi)/(5*sqrt(5)) - 7/5, where phi is the golden ratio (A001622). (End)
a(n) = A001405(2*n+1) - A000108(n+1), n >= 1 (from Eremin link, page 7). - Gennady Eremin, Sep 05 2023
G.f.: x/(1 - 4*x)^2 * c(-x/(1 - 4*x))^3, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. - Peter Bala, Feb 03 2024
From Peter Bala, Oct 13 2024: (Start)
a(n) = Integral_{x = 0..4} x^n * w(x) dx, where the weight function w(x) = 1/(2*Pi) * sqrt(x)*(x - 3)/sqrt(4 - x) (see Penson).
G.f. x*/sqrt(1 - 4*x) * c(x)^3. (End)

A002893 a(n) = Sum_{k=0..n} binomial(n,k)^2 * binomial(2*k,k).

Original entry on oeis.org

1, 3, 15, 93, 639, 4653, 35169, 272835, 2157759, 17319837, 140668065, 1153462995, 9533639025, 79326566595, 663835030335, 5582724468093, 47152425626559, 399769750195965, 3400775573443089, 29016970072920387, 248256043372999089
Offset: 0

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Comments

This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
a(n) is the 2n-th moment of the distance from the origin of a 3-step random walk in the plane. - Peter M. W. Gill (peter.gill(AT)nott.ac.uk), Feb 27 2004
a(n) is the number of Abelian squares of length 2n over a 3-letter alphabet. - Jeffrey Shallit, Aug 17 2010
Consider 2D simple random walk on honeycomb lattice. a(n) gives number of paths of length 2n ending at origin. - Sergey Perepechko, Feb 16 2011
Row sums of A318397 the square of A008459. - Peter Bala, Mar 05 2013
Conjecture: For each n=1,2,3,... the polynomial g_n(x) = Sum_{k=0..n} binomial(n,k)^2*binomial(2k,k)*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013
This is one of the Apery-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
a(n) is the sum of the squares of the coefficients of (x + y + z)^n. - Michael Somos, Aug 25 2018
a(n) is the constant term in the expansion of (1 + (1 + x) * (1 + y) + (1 + 1/x) * (1 + 1/y))^n. - Seiichi Manyama, Oct 28 2019

Examples

			G.f.: A(x) = 1 + 3*x + 15*x^2 + 93*x^3 + 639*x^4 + 4653*x^5 + 35169*x^6 + ...
G.f.: A(x) = 1/(1-3*x) + 6*x^2*(1-x)/(1-3*x)^4 + 90*x^4*(1-x)^2/(1-3*x)^7 + 1680*x^6*(1-x)^3/(1-3*x)^10 + 34650*x^8*(1-x)^4/(1-3*x)^13 + ... - _Paul D. Hanna_, Feb 26 2012
		

References

  • Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
  • 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

The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.

Programs

  • Magma
    [&+[Binomial(n, k)^2 * Binomial(2*k, k): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, Aug 26 2018
    
  • Maple
    series(1/GaussAGM(sqrt((1-3*x)*(1+x)^3), sqrt((1+3*x)*(1-x)^3)), x=0, 42) # Gheorghe Coserea, Aug 17 2016
    A002893 := n -> hypergeom([1/2, -n, -n], [1, 1], 4):
    seq(simplify(A002893(n)), n=0..20); # Peter Luschny, May 23 2017
  • Mathematica
    Table[Sum[Binomial[n,k]^2 Binomial[2k,k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Aug 19 2011 *)
    a[ n_] := If[ n < 0, 0, HypergeometricPFQ[ {1/2, -n, -n}, {1, 1}, 4]]; (* Michael Somos, Oct 16 2013 *)
    a[n_] := SeriesCoefficient[BesselI[0, 2*Sqrt[x]]^3, {x, 0, n}]*n!^2; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Dec 30 2013 *)
    a[ n_] := If[ n < 0, 0, Block[ {x, y, z},  Expand[(x + y + z)^n] /. {t_Integer -> t^2, x -> 1, y -> 1, z -> 1}]]; (* Michael Somos, Aug 25 2018 *)
  • PARI
    {a(n) = if( n<0, 0, n!^2 * polcoeff( besseli(0, 2*x + O(x^(2*n+1)))^3, 2*n))};
    
  • PARI
    {a(n) = sum(k=0, n, binomial(n, k)^2 * binomial(2*k, k))}; /* Michael Somos, Jul 25 2007 */
    
  • PARI
    {a(n)=polcoeff(sum(m=0,n, (3*m)!/m!^3 * x^(2*m)*(1-x)^m / (1-3*x+x*O(x^n))^(3*m+1)),n)} \\ Paul D. Hanna, Feb 26 2012
    
  • PARI
    N = 42; x='x + O('x^N); v = Vec(1/agm(sqrt((1-3*x)*(1+x)^3), sqrt((1+3*x)*(1-x)^3))); vector((#v+1)\2, k, v[2*k-1])  \\ Gheorghe Coserea, Aug 17 2016
    
  • SageMath
    def A002893(n): return simplify(hypergeometric([1/2,-n,-n], [1,1], 4))
    [A002893(n) for n in range(31)] # G. C. Greubel, Jan 21 2023

Formula

a(n) = Sum_{m=0..n} binomial(n, m) * A000172(m). [Barrucand]
D-finite with recurrence: (n+1)^2 a(n+1) = (10*n^2+10*n+3) * a(n) - 9*n^2 * a(n-1). - Matthijs Coster, Apr 28 2004
Sum_{n>=0} a(n)*x^n/n!^2 = BesselI(0, 2*sqrt(x))^3. - Vladeta Jovovic, Mar 11 2003
a(n) = Sum_{p+q+r=n} (n!/(p!*q!*r!))^2 with p, q, r >= 0. - Michael Somos, Jul 25 2007
a(n) = 3*A087457(n) for n>0. - Philippe Deléham, Sep 14 2008
a(n) = hypergeom([1/2, -n, -n], [1, 1], 4). - Mark van Hoeij, Jun 02 2010
G.f.: 2*sqrt(2)/Pi/sqrt(1-6*z-3*z^2+sqrt((1-z)^3*(1-9*z))) * EllipticK(8*z^(3/2)/(1-6*z-3*z^2+sqrt((1-z)^3*(1-9*z)))). - Sergey Perepechko, Feb 16 2011
G.f.: Sum_{n>=0} (3*n)!/n!^3 * x^(2*n)*(1-x)^n / (1-3*x)^(3*n+1). - Paul D. Hanna, Feb 26 2012
Asymptotic: a(n) ~ 3^(2*n+3/2)/(4*Pi*n). - Vaclav Kotesovec, Sep 11 2012
G.f.: 1/(1-3*x)*(1-6*x^2*(1-x)/(Q(0)+6*x^2*(1-x))), where Q(k) = (54*x^3 - 54*x^2 + 9*x -1)*k^2 + (81*x^3 - 81*x^2 + 18*x -2)*k + 33*x^3 - 33*x^2 +9*x - 1 - 3*x^2*(1-x)*(1-3*x)^3*(k+1)^2*(3*k+4)*(3*k+5)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Jul 16 2013
G.f.: G(0)/(2*(1-9*x)^(2/3)), where G(k) = 1 + 1/(1 - 3*(3*k+1)^2*x*(1-x)^2/(3*(3*k+1)^2*x*(1-x)^2 - (k+1)^2*(1-9*x)^2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 31 2013
a(n) = [x^(2n)] 1/agm(sqrt((1-3*x)*(1+x)^3), sqrt((1+3*x)*(1-x)^3)). - Gheorghe Coserea, Aug 17 2016
0 = +a(n)*(+a(n+1)*(+729*a(n+2) -1539*a(n+3) +243*a(n+4)) +a(n+2)*(-567*a(n+2) +1665*a(n+3) -297*a(n+4)) +a(n+3)*(-117*a(n+3) +27*a(n+4))) +a(n+1)*(+a(n+1)*(-324*a(n+2) +720*a(n+3) -117*a(n+4)) +a(n+2)*(+315*a(n+2) -1000*a(n+3) +185*a(n+4)) +a(n+3)*(+80*a(n+3) -19*a(n+4))) +a(n+2)*(+a(n+2)*(-9*a(n+2) +35*a(n+3) -7*a(n+4)) +a(n+3)*(-4*a(n+3) +a(n+4))) for all n in Z. - Michael Somos, Oct 30 2017
G.f. y=A(x) satisfies: 0 = x*(x - 1)*(9*x - 1)*y'' + (27*x^2 - 20*x + 1)*y' + 3*(3*x - 1)*y. - Gheorghe Coserea, Jul 01 2018
Sum_{k>=0} binomial(2*k,k) * a(k) / 6^(2*k) = A086231 = (sqrt(3)-1) * (Gamma(1/24) * Gamma(11/24))^2 / (32*Pi^3). - Vaclav Kotesovec, Apr 23 2023
From Bradley Klee, Jun 05 2023: (Start)
The g.f. T(x) obeys a period-annihilating ODE:
0=3*(-1 + 3*x)*T(x) + (1 - 20*x + 27*x^2)*T'(x) + x*(-1 + x)*(-1 + 9*x)*T''(x).
The periods ODE can be derived from the following Weierstrass data:
g2 = (3/64)*(1 + 3*x)*(1 - 15*x + 75*x^2 + 3*x^3);
g3 = -(1/512)*(-1 + 6*x + 3*x^2)*(1 - 12*x + 30*x^2 - 540*x^3 + 9*x^4);
which determine an elliptic surface with four singular fibers. (End)
a(n) = Sum_{k = 0..n} binomial(n, k)^2 * binomial(3*k, 2*n) (Almkvist, p. 16). - Peter Bala, May 22 2025

A289780 p-INVERT of the positive integers (A000027), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 4, 14, 47, 156, 517, 1714, 5684, 18851, 62520, 207349, 687676, 2280686, 7563923, 25085844, 83197513, 275925586, 915110636, 3034975799, 10065534960, 33382471801, 110713382644, 367182309614, 1217764693607, 4038731742156, 13394504020957, 44423039068114
Offset: 0

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Author

Clark Kimberling, Aug 10 2017

Keywords

Comments

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x).
Taking p(S) = 1 - S gives the INVERT transform of s, so that p-INVERT is a generalization of the INVERT transform (e.g., A033453).
Guide to p-INVERT sequences using p(S) = 1 - S - S^2:
t(A000012) = t(1,1,1,1,1,1,1,...) = A001906
t(A000290) = t(1,4,9,16,25,36,...) = A289779
t(A000027) = t(1,2,3,4,5,6,7,8,...) = A289780
t(A000045) = t(1,2,3,5,8,13,21,...) = A289781
t(A000032) = t(2,1,3,4,7,11,14,...) = A289782
t(A000244) = t(1,3,9,27,81,243,...) = A289783
t(A000302) = t(1,4,16,64,256,...) = A289784
t(A000351) = t(1,5,25,125,625,...) = A289785
t(A005408) = t(1,3,5,7,9,11,13,...) = A289786
t(A005843) = t(2,4,6,8,10,12,14,...) = A289787
t(A016777) = t(1,4,7,10,13,16,...) = A289789
t(A016789) = t(2,5,8,11,14,17,...) = A289790
t(A008585) = t(3,6,9,12,15,18,...) = A289795
t(A000217) = t(1,3,6,10,15,21,...) = A289797
t(A000225) = t(1,3,7,15,31,63,...) = A289798
t(A000578) = t(1,8,27,64,625,...) = A289799
t(A000984) = t(1,2,6,20,70,252,...) = A289800
t(A000292) = t(1,4,10,20,35,56,...) = A289801
t(A002620) = t(1,2,4,6,9,12,16,...) = A289802
t(A001906) = t(1,3,8,21,55,144,...) = A289803
t(A001519) = t(1,1,2,5,13,34,...) = A289804
t(A103889) = t(2,1,4,3,6,5,8,7,,...) = A289805
t(A008619) = t(1,1,2,2,3,3,4,4,...) = A289806
t(A080513) = t(1,2,2,3,3,4,4,5,...) = A289807
t(A133622) = t(1,2,1,3,1,4,1,5,...) = A289809
t(A000108) = t(1,1,2,5,14,42,...) = A081696
t(A081696) = t(1,1,3,9,29,97,...) = A289810
t(A027656) = t(1,0,2,0,3,0,4,0,5...) = A289843
t(A175676) = t(1,0,0,2,0,0,3,0,...) = A289844
t(A079977) = t(1,0,1,0,2,0,3,...) = A289845
t(A059841) = t(1,0,1,0,1,0,1,...) = A289846
t(A000040) = t(2,3,5,7,11,13,...) = A289847
t(A008578) = t(1,2,3,5,7,11,13,...) = A289828
t(A000142) = t(1!, 2!, 3!, 4!, ...) = A289924
t(A000201) = t(1,3,4,6,8,9,11,...) = A289925
t(A001950) = t(2,5,7,10,13,15,...) = A289926
t(A014217) = t(1,2,4,6,11,17,29,...) = A289927
t(A000045*) = t(0,1,1,2,3,5,...) = A289975 (* indicates prepended 0's)
t(A000045*) = t(0,0,1,1,2,3,5,...) = A289976
t(A000045*) = t(0,0,0,1,1,2,3,5,...) = A289977
t(A290990*) = t(0,1,2,3,4,5,...) = A290990
t(A290990*) = t(0,0,1,2,3,4,5,...) = A290991
t(A290990*) = t(0,0,01,2,3,4,5,...) = A290992

Examples

			Example 1:  s = (1,2,3,4,5,6,...) = A000027 and p(S) = 1 - S.
S(x) = x + 2x^2 + 3x^3 + 4x^4 + ...
p(S(x)) = 1 - (x + 2x^2 + 3x^3 + 4x^4 + ... )
- p(0) + 1/p(S(x)) = -1 + 1 + x + 3x^2 + 8x^3 + 21x^4 + ...
T(x) = 1 + 3x + 8x^2 + 21x^3 + ...
t(s) = (1,3,8,21,...) = A001906.
***
Example 2:  s = (1,2,3,4,5,6,...) = A000027 and p(S) = 1 - S - S^2.
S(x) =  x + 2x^2 + 3x^3 + 4x^4 + ...
p(S(x)) = 1 - ( x + 2x^2 + 3x^3 + 4x^4 + ...) - ( x + 2x^2 + 3x^3 + 4x^4 + ...)^2
- p(0) + 1/p(S(x)) = -1 + 1 + x + 4x^2 + 14x^3 + 47x^4 + ...
T(x) = 1 + 4x + 14x^2 + 47x^3 + ...
t(s) = (1,4,14,47,...) = A289780.
		

Crossrefs

Cf. A000027.

Programs

  • GAP
    P:=[1,4,14,47];; for n in [5..10^2] do P[n]:=5*P[n-1]-7*P[n-2]+5*P[n-3]-P[n-4]; od; P; # Muniru A Asiru, Sep 03 2017
  • Mathematica
    z = 60; s = x/(1 - x)^2; p = 1 - s - s^2;
    Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
    Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289780 *)
  • PARI
    x='x+O('x^99); Vec((1-x+x^2)/(1-5*x+7*x^2-5*x^3+x^4)) \\ Altug Alkan, Aug 13 2017
    

Formula

G.f.: (1 - x + x^2)/(1 - 5 x + 7 x^2 - 5 x^3 + x^4).
a(n) = 5*a(n-1) - 7*a(n-2) + 5*a(n-3) - a(n-4).

A001790 Numerators in expansion of 1/sqrt(1-x).

Original entry on oeis.org

1, 1, 3, 5, 35, 63, 231, 429, 6435, 12155, 46189, 88179, 676039, 1300075, 5014575, 9694845, 300540195, 583401555, 2268783825, 4418157975, 34461632205, 67282234305, 263012370465, 514589420475, 8061900920775, 15801325804719
Offset: 0

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Author

Keywords

Comments

Also numerator of e(n-1,n-1) (see Maple line).
Leading coefficient of normalized Legendre polynomial.
Common denominator of expansions of powers of x in terms of Legendre polynomials P_n(x).
Also the numerator of binomial(2*n,n)/2^n. - T. D. Noe, Nov 29 2005
This sequence gives the numerators of the Maclaurin series of the Lorentz factor (see Wikipedia link) of 1/sqrt(1-b^2) = dt/dtau where b=u/c is the velocity in terms of the speed of light c, u is the velocity as observed in the reference frame where time t is measured and tau is the proper time. - Stephen Crowley, Apr 03 2007
Truncations of rational expressions like those given by the numerator operator are artifacts in integer formulas and have many disadvantages. A pure integer formula follows. Let n$ denote the swinging factorial and sigma(n) = number of '1's in the base-2 representation of floor(n/2). Then a(n) = (2*n)$ / sigma(2*n) = A056040(2*n) / A060632(2*n+1). Simply said: this sequence is the odd part of the swinging factorial at even indices. - Peter Luschny, Aug 01 2009
It appears that a(n) = A060818(n)*A001147(n)/A000142(n). - James R. Buddenhagen, Jan 20 2010
The convolution of sequence binomial(2*n,n)/4^n with itself is the constant sequence with all terms = 1.
a(n) equals the denominator of Hypergeometric2F1[1/2, n, 1 + n, -1] (see Mathematica code below). - John M. Campbell, Jul 04 2011
a(n) = numerator of (1/Pi)*Integral_{x=-oo..+oo} 1/(x^2-2*x+2)^n dx. - Leonid Bedratyuk, Nov 17 2012
a(n) = numerator of the mean value of cos(x)^(2*n) from x = 0 to 2*Pi. - Jean-François Alcover, Mar 21 2013
Constant terms for normalized Legendre polynomials. - Tom Copeland, Feb 04 2016
From Ralf Steiner, Apr 07 2017: (Start)
By analytic continuation to the entire complex plane there exist regularized values for divergent sums:
a(n)/A060818(n) = (-2)^n*sqrt(Pi)/(Gamma(1/2 - n)*Gamma(1 + n)).
Sum_{k>=0} a(k)/A060818(k) = -i.
Sum_{k>=0} (-1)^k*a(k)/A060818(k) = 1/sqrt(3).
Sum_{k>=0} (-1)^(k+1)*a(k)/A060818(k) = -1/sqrt(3).
a(n)/A046161(n) = (-1)^n*sqrt(Pi)/(Gamma(1/2 - n)*Gamma(1 + n)).
Sum_{k>=0} (-1)^k*a(k)/A046161(k) = 1/sqrt(2).
Sum_{k>=0} (-1)^(k+1)*a(k)/A046161(k) = -1/sqrt(2). (End)
a(n) = numerator of (1/Pi)*Integral_{x=-oo..+oo} 1/(x^2+1)^n dx. (n=1 is the Cauchy distribution.) - Harry Garst, May 26 2017
Let R(n, d) = (Product_{j prime to d} Pochhammer(j / d, n)) / n!. Then the numerators of R(n, 2) give this sequence and the denominators are A046161. For d = 3 see A273194/A344402. - Peter Luschny, May 20 2021
Using WolframAlpha, it appears a(n) gives the numerator in the residues of f(z) = 2z choose z at odd negative half integers. E.g., the residues of f(z) at z = -1/2, -3/2, -5/2 are 1/(2*Pi), 1/(16*Pi), and 3/(256*Pi) respectively. - Nicholas Juricic, Mar 31 2022
a(n) is the numerator of (1/Pi) * Integral_{x=-oo..+oo} sech(x)^(2*n+1) dx. The corresponding denominator is A046161. - Mohammed Yaseen, Jul 29 2023
a(n) is the numerator of (1/Pi) * Integral_{x=0..Pi/2} sin(x)^(2*n) dx. The corresponding denominator is A101926(n). - Mohammed Yaseen, Sep 19 2023

Examples

			1, 1, 3/2, 5/2, 35/8, 63/8, 231/16, 429/16, 6435/128, 12155/128, 46189/256, ...
binomial(2*n,n)/4^n => 1, 1/2, 3/8, 5/16, 35/128, 63/256, 231/1024, 429/2048, 6435/32768, ...
		

References

  • P. J. Davis, Interpolation and Approximation, Dover Publications, 1975, p. 372.
  • W. Feller, An Introduction to Probability Theory and Its Applications, Vol. 1, 2nd ed. New York: Wiley, 1968; Chap. III, Eq. 4.1.
  • 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).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 6, equation 6:14:6 at page 51.
  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 102.

Crossrefs

Cf. A060818 (denominator of binomial(2*n,n)/2^n), A061549 (denominators).
Cf. A123854 (denominators).
Cf. A161198 (triangle of coefficients for (1-x)^((-1-2*n)/2)).
Cf. A163590 (odd part of the swinging factorial).
Cf. A001405.
First column and diagonal 1 of triangle A100258.
Bisection of A036069.
Bisections give A061548 and A063079.
Inverse Moebius transform of A180403/A046161.
Numerators of [x^n]( (1-x)^(p/2) ): A161202 (p=5), A161200 (p=3), A002596 (p=1), this sequence (p=-1), A001803 (p=-3), A161199 (p=-5), A161201 (p=-7).

Programs

  • Magma
    A001790:= func< n | Numerator((n+1)*Catalan(n)/4^n) >;
    [A001790(n): n in [0..40]]; // G. C. Greubel, Sep 23 2024
  • Maple
    e := proc(l,m) local k; add(2^(k-2*m)*binomial(2*m-2*k,m-k)*binomial(m+k,m)*binomial(k,l),k=l..m); end;
    # From Peter Luschny, Aug 01 2009: (Start)
    swing := proc(n) option remember; if n = 0 then 1 elif irem(n, 2) = 1 then swing(n-1)*n else 4*swing(n-1)/n fi end:
    sigma := n -> 2^(add(i,i=convert(iquo(n,2),base,2))):
    a := n -> swing(2*n)/sigma(2*n); # (End)
    A001790 := proc(n) binomial(2*n, n)/4^n ; numer(%) ; end proc : # R. J. Mathar, Jan 18 2013
  • Mathematica
    Numerator[ CoefficientList[ Series[1/Sqrt[(1 - x)], {x, 0, 25}], x]]
    Table[Denominator[Hypergeometric2F1[1/2, n, 1 + n, -1]], {n, 0, 34}]   (* John M. Campbell, Jul 04 2011 *)
    Numerator[Table[(-2)^n*Sqrt[Pi]/(Gamma[1/2 - n]*Gamma[1 + n]),{n,0,20}]] (* Ralf Steiner, Apr 07 2017 *)
    Numerator[Table[Binomial[2n,n]/2^n, {n, 0, 25}]] (* Vaclav Kotesovec, Apr 07 2017 *)
    Table[Numerator@LegendreP[2 n, 0]*(-1)^n, {n, 0, 25}] (* Andres Cicuttin, Jan 22 2018 *)
    A = {1}; Do[A = Append[A, 2^IntegerExponent[n, 2]*(2*n - 1)*A[[n]]/n], {n, 1, 25}]; Print[A] (* John Lawrence, Jul 17 2020 *)
  • PARI
    {a(n) = if( n<0, 0, polcoeff( pollegendre(n), n) * 2^valuation((n\2*2)!, 2))};
    
  • PARI
    a(n)=binomial(2*n,n)>>hammingweight(n); \\ Gleb Koloskov, Sep 26 2021
    
  • Sage
    # uses[A000120]
    @CachedFunction
    def swing(n):
        if n == 0: return 1
        return swing(n-1)*n if is_odd(n) else 4*swing(n-1)/n
    A001790 = lambda n: swing(2*n)/2^A000120(2*n)
    [A001790(n) for n in (0..25)]  # Peter Luschny, Nov 19 2012
    

Formula

a(n) = numerator( binomial(2*n,n)/4^n ) (cf. A046161).
a(n) = A000984(n)/A001316(n) where A001316(n) is the highest power of 2 dividing C(2*n, n) = A000984(n). - Benoit Cloitre, Jan 27 2002
a(n) = denominator of (2^n/binomial(2*n,n)). - Artur Jasinski, Nov 26 2011
a(n) = numerator(L(n)), with rational L(n):=binomial(2*n,n)/2^n. L(n) is the leading coefficient of the Legendre polynomial P_n(x).
L(n) = (2*n-1)!!/n! with the double factorials (2*n-1)!! = A001147(n), n >= 0.
Numerator in (1-2t)^(-1/2) = 1 + t + (3/2)t^2 + (5/2)t^3 + (35/8)t^4 + (63/8)t^5 + (231/16)t^6 + (429/16)t^7 + ... = 1 + t + 3*t^2/2! + 15*t^3/3! + 105*t^4/4! + 945*t^5/5! + ... = e.g.f. for double factorials A001147 (cf. A094638). - Tom Copeland, Dec 04 2013
From Ralf Steiner, Apr 08 2017: (Start)
a(n)/A061549(n) = (-1/4)^n*sqrt(Pi)/(Gamma(1/2 - n)*Gamma(1 + n)).
Sum_{k>=0} a(k)/A061549(k) = 2/sqrt(3).
Sum_{k>=0} (-1)^k*a(k)/A061549(k) = 2/sqrt(5).
Sum_{k>=0} (-1)^(k+1)*a(k)/A061549(k) = -2/sqrt(5).
a(n)/A123854(n) = (-1/2)^n*sqrt(Pi)/(gamma(1/2 - n)*gamma(1 + n)).
Sum_{k>=0} a(k)/A123854(k) = sqrt(2).
Sum_{k>=0} (-1)^k*a(k)/A123854(k) = sqrt(2/3).
Sum_{k>=0} (-1)^(k+1)*a(k)/A123854(k) = -sqrt(2/3). (End)
a(n) = 2^A007814(n)*(2*n-1)*a(n-1)/n. - John Lawrence, Jul 17 2020
Sum_{k>=0} A086117(k+3)/a(k+2) = Pi. - Antonio Graciá Llorente, Aug 31 2024
a(n) = A001803(n)/(2*n+1). - G. C. Greubel, Sep 23 2024

A006134 a(n) = Sum_{k=0..n} binomial(2*k,k).

Original entry on oeis.org

1, 3, 9, 29, 99, 351, 1275, 4707, 17577, 66197, 250953, 956385, 3660541, 14061141, 54177741, 209295261, 810375651, 3143981871, 12219117171, 47564380971, 185410909791, 723668784231, 2827767747951, 11061198475551, 43308802158651, 169719408596403, 665637941544507
Offset: 0

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Author

Keywords

Comments

The expression a(n) = B^n*Sum_{ k=0..n } binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5). - N. J. A. Sloane, Jan 21 2009
T(n+1,1) from table A045912 of characteristic polynomial of negative Pascal matrix. - Michael Somos, Jul 24 2002
p divides a((p-3)/2) for p=11, 13, 23, 37, 47, 59, 61, 71, 73, 83, 97, 107, 109, 131, 157, 167, ...: A097933. Also primes congruent to {1, 2, 3, 11} mod 12 or primes p such that 3 is a square mod p (excluding 2 and 3) A038874. - Alexander Adamchuk, Jul 05 2006
Partial sums of the even central binomial coefficients. For p prime >=5, a(p-1) = 1 or -1 (mod p) according as p = 1 or -1 (mod 3) (see Pan and Sun link). - David Callan, Nov 29 2007
First column of triangle A187887. - Michel Marcus, Jun 23 2013
From Gus Wiseman, Apr 20 2023: (Start)
Also the number of nonempty subsets of {1,...,2n+1} with median n+1, where the median of a multiset is either the middle part (for odd length), or the average of the two middle parts (for even length). The odd/even-length cases are A000984 and A006134(n-1). For example, the a(0) = 1 through a(2) = 9 subsets are:
{1} {2} {3}
{1,3} {1,5}
{1,2,3} {2,4}
{1,3,4}
{1,3,5}
{2,3,4}
{2,3,5}
{1,2,4,5}
{1,2,3,4,5}
Alternatively, a(n-1) is the number of nonempty subsets of {1,...,2n-1} with median n.
(End)

Examples

			1 + 3*x + 9*x^2 + 29*x^3 + 99*x^4 + 351*x^5 + 1275*x^6 + 4707*x^7 + 17577*x^8 + ...
		

References

  • Marko Petkovsek, Herbert Wilf and Doron Zeilberger, A=B, A K Peters, 1996, p. 22.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A000984 (first differences), A097933, A038874, A132310.
Equals A066796 + 1.
Odd bisection of A100066.
Row sums of A361654 (also column k = 2).
A007318 counts subsets by length, A231147 by median, A013580 by integer median.
A359893 and A359901 count partitions by median.

Programs

  • MATLAB
    n=10; x=pascal(n); trace(x)
    
  • Magma
    &cat[ [&+[ Binomial(2*k, k): k in [0..n]]]: n in [0..30]]; // Vincenzo Librandi, Aug 13 2015
  • Maple
    A006134 := proc(n) sum(binomial(2*k,k),k=0..n); end;
    a := n -> -binomial(2*(n+1),n+1)*hypergeom([1,n+3/2],[n+2], 4) - I/sqrt(3):
    seq(simplify(a(n)), n=0..24); # Peter Luschny, Oct 29 2015
    # third program:
    A006134 := series(exp(2*x)*BesselI(0, 2*x) + exp(x)*int(BesselI(0, 2*x)*exp(x), x), x = 0, 25):
    seq(n!*coeff(A006134, x, n), n=0..24); # Mélika Tebni, Feb 27 2024
  • Mathematica
    Table[Sum[((2k)!/(k!)^2),{k,0,n}], {n,0,50}] (* Alexander Adamchuk, Jul 05 2006 *)
    a[ n_] := (4/3) Binomial[ 2 n, n] Hypergeometric2F1[ 1/2, 1, -n + 1/2, -1/3] (* Michael Somos, Jun 20 2012 *)
    Accumulate[Table[Binomial[2n,n],{n,0,30}]] (* Harvey P. Dale, Jan 11 2015 *)
    CoefficientList[Series[1/((1 - x) Sqrt[1 - 4 x]), {x, 0, 33}], x] (* Vincenzo Librandi, Aug 13 2015 *)
  • Maxima
    makelist(sum(binomial(2*k,k),k,0,n),n,0,12); /* Emanuele Munarini, Mar 15 2011 */
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( charpoly( matrix( n+1, n+1, i, j, -binomial( i+j-2, i-1))), 1))} \\ Michael Somos, Jul 10 2002
    
  • PARI
    {a(n)=binomial(2*n,n)*sum(k=0,2*n,(-1)^k*polcoeff((1+x+x^2)^n,k)/binomial(2*n,k))} \\ Paul D. Hanna, Aug 21 2007
    
  • PARI
    my(x='x+O('x^100)); Vec(1/((1-x)*sqrt(1-4*x))) \\ Altug Alkan, Oct 29 2015
    

Formula

From Alexander Adamchuk, Jul 05 2006: (Start)
a(n) = Sum_{k=0..n} (2k)!/(k!)^2.
a(n) = A066796(n) + 1, n>0. (End)
G.f.: 1/((1-x)*sqrt(1-4*x)).
D-finite with recurrence: (n+2)*a(n+2) - (5*n+8)*a(n+1) + 2*(2*n+3)*a(n) = 0. - Emanuele Munarini, Mar 15 2011
a(n) = C(2n,n) * Sum_{k=0..2n} (-1)^k*trinomial(n,k)/C(2n,k) where trinomial(n,k) = [x^k] (1 + x + x^2)^n. E.g. a(2) = C(4,2)*(1/1 - 2/4 + 3/6 - 2/4 + 1/1) = 6*(3/2) = 9 ; a(3) = C(6,3)*(1/1 - 3/6 + 6/15 - 7/20 + 6/15 - 3/6 + 1/1) = 20*(29/20) = 29. - Paul D. Hanna, Aug 21 2007
From Alzhekeyev Ascar M, Jan 19 2012: (Start)
a(n) = Sum_{ k=0..n } b(k)*binomial(n+k,k), where b(k)=0 for n-k == 2 (mod 3), b(k)=1 for n-k == 0 or 1 (mod 6), and b(k)=-1 for n-k== 3 or 4 (mod 6).
a(n) = Sum_{ k=0..n-1 } c(k)*binomial(2n,k) + binomial(2n,n), where c(k)=0 for n-k == 0 (mod 3), c(k)=1 for n-k== 1 (mod 3), and c(k)=-1 for n-k==2 (mod 3). (End)
a(n) ~ 2^(2*n+2)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Nov 06 2012
G.f.: G(0)/2/(1-x), where G(k)= 1 + 1/(1 - 2*x*(2*k+1)/(2*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
G.f.: G(0)/(1-x), where G(k)= 1 + 4*x*(4*k+1)/( (4*k+2) - x*(4*k+2)*(4*k+3)/(x*(4*k+3) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 26 2013
a(n) = Sum_{k = 0..n} binomial(n+1,k+1)*A002426(k). - Peter Bala, Oct 29 2015
a(n) = -binomial(2*(n+1),n+1)*hypergeom([1,n+3/2],[n+2], 4) - i/sqrt(3). - Peter Luschny, Oct 29 2015
a(n) = binomial(2*n, n)*hypergeom([1,-n], [1/2-n], 1/4). - Peter Luschny, Mar 16 2016
From Gus Wiseman, Apr 20 2023: (Start)
a(n+1) - a(n) = A000984(n).
a(n) = A013580(2n+1,n+1) (conjectured).
a(n) = 2*A024718(n) - 1.
a(n) = A100066(2n+1).
a(n) = A231147(2n+1,n+1) (conjectured). (End)
a(n) = Sum_{k=0..floor(n/3)} 3^(n-3*k) * binomial(n-k,2*k) * binomial(2*k,k) (Sawhney, 2017). - Amiram Eldar, Feb 24 2024
From Mélika Tebni, Feb 27 2024: (Start)
Limit_{n -> oo} a(n) / A281593(n) = 2.
E.g.f.: exp(2*x)*BesselI(0,2*x) + exp(x)*integral( BesselI(0,2*x)*exp(x) ) dx. (End)
a(n) = [(x*y)^n] 1/((1 - (x + y))*(1 - x*y)). - Stefano Spezia, Feb 16 2025
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*binomial(2*n+1-k, n-2*k). - Michael Weselcouch, Jun 17 2025
a(n) = binomial(1+2*n, n)*hypergeom([1, (1-n)/2, -n/2], [-1-2*n, 2+n], 4). - Stefano Spezia, Jun 18 2025

Extensions

Simpler definition from Alexander Adamchuk, Jul 05 2006

A268555 Diagonal of the rational function of six variables 1/((1 - w - u v - u v w) * (1 - z - x y)).

Original entry on oeis.org

1, 6, 78, 1260, 22470, 424116, 8305836, 166929048, 3419932230, 71109813060, 1496053026468, 31777397077608, 680354749147164, 14664155597771400, 317877850826299800, 6924815555276838960, 151505459922479997510, 3327336781596164286180
Offset: 0

Views

Author

N. J. A. Sloane, Feb 29 2016

Keywords

Comments

Also diagonal of rational function R(x,y,z) = 1 /(1 - x - y - z - x*y + x*z).
Annihilating differential operator: x*(16*x^2-24*x+1)*(d/dx)^2 + (48*x^2-48*x+1)*(d/dx) + 12*x-6.

Examples

			G.f. = 1 + 6*x + 78*x^2 + 1260*x^3 + 22470*x^4 + 424116*x^5 + 8305836*x^6 + ...
		

Crossrefs

Programs

  • GAP
    List([0..20],n->Binomial(2*n,n)*Sum([0..n],k->Binomial(n,k)*Binomial(n+k,k))); # Muniru A Asiru, Mar 19 2018
  • Maple
    A268555 := proc(n)
        1/(1-w-u*v-u*v*w)/(1-z-x*y) ;
        coeftayl(%,x=0,n) ;
        coeftayl(%,y=0,n) ;
        coeftayl(%,z=0,n) ;
        coeftayl(%,u=0,n) ;
        coeftayl(%,v=0,n) ;
        coeftayl(%,w=0,n) ;
    end proc:
    seq(A268555(n),n=0..40) ; # R. J. Mathar, Mar 10 2016
    seq(binomial(2*n,n)*add(binomial(n,k)*binomial(n+k,k), k = 0..n), n = 0..20); # Peter Bala, Mar 19 2018
  • Mathematica
    sc = SeriesCoefficient;
    a[n_] := 1/(1-w-u*v-u*v*w)/(1-z-x*y) // sc[#, {x, 0, n}]& // sc[#, {y, 0, n}]& // sc[#, {z, 0, n}]& // sc[#, {u, 0, n}]& // sc[#, {v, 0, n}]& // sc[#, {w, 0, n}]&;
    Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Nov 14 2017 *)
    a[n_] := Product[Hypergeometric2F1[-n, -n, 1, i], {i, 1, 2}];
    Table[a[n], {n, 0, 17}]  (* Peter Luschny, Mar 19 2018 *)
  • PARI
    \\ system("wget http://www.jjj.de/pari/hypergeom.gpi");
    read("hypergeom.gpi");
    N = 18; x = 'x + O('x^N);
    Vec(hypergeom_sym([1/12, 5/12],[1],6912*x^4*(1-24*x+16*x^2)/(1-24*x+48*x^2)^3, N)/(1-24*x+48*x^2)^(1/4)) \\ Gheorghe Coserea, Jul 05 2016
    
  • PARI
    {a(n) = if( n<1, n==0, my(A = vector(n+1)); A[1] = 1; A[2] = 6; for(k=2, n, A[k+1] = (6*(2*k-1)^2*A[k] - 4*(2*k-1)*(2*k-3)*A[k-1]) / k^2); A[n+1])}; /* Michael Somos, Jan 22 2017 */
    
  • PARI
    diag(expr, N=22, var=variables(expr)) = {
      my(a = vector(N));
      for (k = 1, #var, expr = taylor(expr, var[#var - k + 1], N));
      for (n = 1, N, a[n] = expr;
        for (k = 1, #var, a[n] = polcoeff(a[n], n-1)));
      return(a);
    };
    diag(1/(1 - x - y - z - x*y + x*z), 18)
    \\ test: diag(1/(1-x-y-z-x*y+x*z)) == diag(1/((1-w-u*v-u*v*w)*(1-z-x*y)))
    \\ Gheorghe Coserea, Jun 15 2018
    

Formula

D-finite with recurrence: n^2*a(n) -6*(2*n-1)^2*a(n-1) +4*(2*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Mar 10 2016
a(n) ~ sqrt(4+3*sqrt(2)) * 2^(2*n-3/2) * (1+sqrt(2))^(2*n) / (Pi*n). - Vaclav Kotesovec, Jul 01 2016
G.f.: hypergeom([1/12, 5/12],[1],6912*x^4*(1-24*x+16*x^2)/(1-24*x+48*x^2)^3)/(1-24*x+48*x^2)^(1/4).
0 = x*(16*x^2-24*x+1)*y'' + (48*x^2-48*x+1)*y' + (12*x-6)*y, where y is g.f.
a(n) = A000984(n)*A001850(n) = C(2*n,n)*Sum_{k = 0..n} C(n,k)*C(n+k,k). - Peter Bala, Mar 19 2018
a(n) = [(x*y)^n] (1 + x + y + 2*x*y)^(2*n), so the sequence is the diagonal of the three variable rational function 1/(1 - u*(1 + x + y + 2*x*y)^2). - Peter Bala, Oct 30 2024

A008459 Square the entries of Pascal's triangle.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 9, 9, 1, 1, 16, 36, 16, 1, 1, 25, 100, 100, 25, 1, 1, 36, 225, 400, 225, 36, 1, 1, 49, 441, 1225, 1225, 441, 49, 1, 1, 64, 784, 3136, 4900, 3136, 784, 64, 1, 1, 81, 1296, 7056, 15876, 15876, 7056, 1296, 81, 1, 1, 100, 2025, 14400, 44100, 63504, 44100, 14400, 2025, 100, 1
Offset: 0

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Author

Keywords

Comments

Number of lattice paths from (0, 0) to (n, n) with steps (1, 0) and (0, 1), having k right turns. - Emeric Deutsch, Nov 23 2003
Product of A007318 and A105868. - Paul Barry, Nov 15 2005
Number of partitions that fit in an n X n box with Durfee square k. - Franklin T. Adams-Watters, Feb 20 2006
From Peter Bala, Oct 23 2008: (Start)
Narayana numbers of type B. Row n of this triangle is the h-vector of the simplicial complex dual to an associahedron of type B_n (a cyclohedron) [Fomin & Reading, p. 60]. See A063007 for the corresponding f-vectors for associahedra of type B_n. See A001263 for the h-vectors for associahedra of type A_n. The Hilbert transform of this triangular array is A108625 (see A145905 for the definition of this term).
Let A_n be the root lattice generated as a monoid by {e_i - e_j: 0 <= i, j <= n + 1}. Let P(A_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the h-vectors of a unimodular triangulation of P(A_n) [Ardila et al.]. A063007 is the corresponding array of f-vectors for these type A_n polytopes. See A086645 for the array of h-vectors for type C_n polytopes and A108558 for the array of h-vectors associated with type D_n polytopes.
(End)
The n-th row consists of the coefficients of the polynomial P_n(t) = Integral_{s = 0..2*Pi} (1 + t^2 - 2*t*cos(s))^n/Pi/2 ds. For example, when n = 3, we get P_3(t) = t^6 + 9*t^4 + 9*t^2 + 1; the coefficients are 1, 9, 9, 1. - Theodore Kolokolnikov, Oct 26 2010
Let E(y) = Sum_{n >= 0} y^n/n!^2 = BesselJ(0, 2*sqrt(-y)). Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence n!^2 as defined in Wang and Wang. - Peter Bala, Jul 24 2013
From Colin Defant, Sep 16 2018: (Start)
Let s denote West's stack-sorting map. T(n,k) is the number of permutations pi of [n+1] with k descents such that s(pi) avoids the patterns 132, 231, and 321. T(n,k) is also the number of permutations pi of [n+1] with k descents such that s(pi) avoids the patterns 132, 312, and 321.
T(n,k) is the number of permutations of [n+1] with k descents that avoid the patterns 1342, 3142, 3412, and 3421. (End)
The number of convex polyominoes whose smallest bounding rectangle has size (k+1)*(n+1-k) and which contain the lower left corner of the bounding rectangle (directed convex polyominoes). - Günter Rote, Feb 27 2019
Let P be the poset [n] X [n] ordered by the product order. T(n,k) is the number of antichains in P containing exactly k elements. Cf. A063746. - Geoffrey Critzer, Mar 28 2020

Examples

			Pascal's triangle begins
  1
  1  1
  1  2   1
  1  3   3   1
  1  4   6   4   1
  1  5  10  10   5   1
  1  6  15  20  15   6   1
  1  7  21  35  35  21   7   1
...
so the present triangle begins
  1
  1   1
  1   4    1
  1   9    9     1
  1  16   36    16     1
  1  25  100   100    25    1
  1  36  225   400   225   36   1
  1  49  441  1225  1225  441  49   1
...
		

References

  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Chapter 12.
  • J. Riordan, An introduction to combinatorial analysis, Dover Publications, Mineola, NY, 2002, page 191, Problem 15. MR1949650
  • P. G. Tait, On the Linear Differential Equation of the Second Order, Proceedings of the Royal Society of Edinburgh, 9 (1876), 93-98 (see p. 97) [From Tom Copeland, Sep 09 2010, vol number corrected Sep 10 2010]

Crossrefs

Row sums are in A000984. Columns 0-3 are A000012, A000290, A000537, A001249.
Family of polynomials (see A062145): this sequence (c=1), A132813 (c=2), A062196 (c=3), A062145 (c=4), A062264 (c=5), A062190 (c=6).
Cf. A007318, A055133, A116647, A001263, A086645, A063007, A108558, A108625 (Hilbert transform), A145903, A181543, A086645 (logarithmic derivative), A105868 (inverse binomial transform), A093118.

Programs

  • GAP
    Flat(List([0..10],n->List([0..n],k->Binomial(n,k)^2))); # Muniru A Asiru, Mar 30 2018
    
  • Magma
    /* As triangle */ [[Binomial(n, k)^2: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Dec 15 2016
    
  • Maple
    seq(seq(binomial(n, k)^2, k=0..n), n=0..10);
  • Mathematica
    Table[Binomial[n, k]^2, {n, 0, 11}, {k, 0, n}]//Flatten (* Alonso del Arte, Dec 08 2013 *)
  • Maxima
    create_list(binomial(n,k)^2,n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • Maxima
    T(n,k):=if n=k then 1 else if k=0 then 1 else T(n-1,k)*(n+k)/(n-k)+T(n-1,k-1); /* Vladimir Kruchinin, Oct 18 2014 */
    
  • Maxima
    A(x,y):=1/sqrt(1-2*x-2*x*y+x^2-2*x^2*y+x^2*y^2);
    taylor(x*A(x,y)+x*y*A(x,y)+sqrt(1+4*x^2*y*A(x,y)^2),x,0,7,y,0,7); /* Vladimir Kruchinin, Oct 23 2020 */
    
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, binomial(n, k)^2)}; /* Michael Somos, May 03 2004 */
    
  • PARI
    {T(n,k)=polcoeff(polcoeff(sum(m=0,n,(2*m)!/m!^2*x^(2*m)*y^m/(1-x-x*y+x*O(x^n))^(2*m+1)),n,x),k,y)} \\ Paul D. Hanna, Oct 31 2010
    
  • Python
    def A008459(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2))**2 # Chai Wah Wu, Nov 12 2024

Formula

T(n,k) = A007318(n,k)^2. - Sean A. Irvine, Mar 29 2018
E.g.f.: exp((1+y)*x)*BesselI(0, 2*sqrt(y)*x). - Vladeta Jovovic, Nov 17 2003
G.f.: 1/sqrt(1-2*x-2*x*y+x^2-2*x^2*y+x^2*y^2); g.f. for row n: (1-t)^n P_n[(1+t)/(1-t)] where the P_n's are the Legendre polynomials. - Emeric Deutsch, Nov 23 2003 [The original version of the bivariate g.f. has been modified with the roles of x and y interchanged so that now x corresponds to n and y to k. - Petros Hadjicostas, Oct 22 2017]
G.f. for column k is Sum_{j = 0..k} C(k, j)^2*x^(k+j)/(1 - x)^(2*k+1). - Paul Barry, Nov 15 2005
Column k has g.f. (x^k)*Legendre_P(k, (1+x)/(1-x))/(1 - x)^(k+1) = (x^k)*Sum_{j = 0..k} C(k, j)^2*x^j/(1 - x)^(2*k+1). - Paul Barry, Nov 19 2005
Let E be the operator D*x*D, where D denotes the derivative operator d/dx. Then (1/n!^2) * E^n(1/(1 - x)) = (row n generating polynomial)/(1 - x)^(2*n+1) = Sum_{k >= 0} binomial(n+k, k)^2*x^k. For example, when n = 3 we have (1/3!)^2*E^3(1/(1 - x)) = (1 + 9*x + 9*x^2 + x^3)/(1 - x)^7 = (1/3!)^2 * Sum_{k >= 0} ((k+1)*(k+2)*(k+3))^2*x^k. - Peter Bala, Oct 23 2008
G.f.: A(x, y) = Sum_{n >= 0} (2*n)!/n!^2 * x^(2*n)*y^n/(1 - x - x*y)^(2*n+1). - Paul D. Hanna, Oct 31 2010
From Peter Bala, Jul 24 2013: (Start)
Let E(y) = Sum_{n >= 0} y^n/n!^2 = BesselJ(0, 2*sqrt(-y)). Generating function: E(y)*E(x*y) = 1 + (1 + x)*y + (1 + 4*x + x^2)*y^2/2!^2 + (1 + 9*x + 9*x^2 + x^3)*y^3/3!^2 + .... Cf. the unsigned version of A021009 with generating function exp(y)*E(x*y).
The n-th power of this array has the generating function E(y)^n*E(x*y). In particular, the matrix inverse A055133 has the generating function E(x*y)/E(y). (End)
T(n,k) = T(n-1,k)*(n+k)/(n-k) + T(n-1,k-1), T(n,0) = T(n,n) = 1. - Vladimir Kruchinin, Oct 18 2014
Observe that the recurrence T(n,k) = T(n-1,k)*(n+k)/(n-k) - T(n-1,k-1), for n >= 2 and 1 <= k < n, with boundary conditions T(n,0) = T(n,n) = 1 gives Pascal's triangle A007318. - Peter Bala, Dec 21 2014
n-th row polynomial R(n, x) = [z^n] (1 + (1 + x)*z + x*z^2)^n. Note that 1/n*[z^(n-1)] (1 + (1 + x)*z + x*z^2)^n gives the row polynomials of A001263. - Peter Bala, Jun 24 2015
Binomial transform of A105868. If G(x,t) = 1/sqrt(1 - 2*(1 + t)*x + (1 - t)^2*x^2) denotes the o.g.f. of this array then 1 + x*d/dx log(G(x,t)) = 1 + (1 + t)*x + (1 + 6*t + t^2)*x^2 + ... is the o.g.f. for A086645. - Peter Bala, Sep 06 2015
T(n,k) = Sum_{i=0..n} C(n-i,k)*C(n,i)*C(n+i,i)*(-1)^(n-i-k). - Vladimir Kruchinin, Jan 14 2018
G.f. satisfies A(x,y) = x*A(x,y)+x*y*A(x,y)+sqrt(1+4*x^2*y*A(x,y)^2). - Vladimir Kruchinin, Oct 23 2020
G.f. satisfies the differential equation y * d^2(A(x,y))/dy^2 - x^2 * d^2(x*A(x,y))/dx^2 + 2*x^2* A(x,y)^3 = 0. - Sergii Voloshyn, Mar 07 2025
T(n,k) = Sum_{i=0..n} C(2*n+1,i)*C(n+k-i,n)^2*(-1)^i. - Natalia L. Skirrow, Apr 14 2025

A027306 a(n) = 2^(n-1) + ((1 + (-1)^n)/4)*binomial(n, n/2).

Original entry on oeis.org

1, 1, 3, 4, 11, 16, 42, 64, 163, 256, 638, 1024, 2510, 4096, 9908, 16384, 39203, 65536, 155382, 262144, 616666, 1048576, 2449868, 4194304, 9740686, 16777216, 38754732, 67108864, 154276028, 268435456, 614429672, 1073741824, 2448023843
Offset: 0

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Author

Keywords

Comments

Inverse binomial transform of A027914. Hankel transform (see A001906 for definition) is {1, 2, 3, 4, ..., n, ...}. - Philippe Deléham, Jul 21 2005
Number of walks of length n on a line that starts at the origin and ends at or above 0. - Benjamin Phillabaum, Mar 05 2011
Number of binary integers (i.e., with a leading 1 bit) of length n+1 which have a majority of 1-bits. E.g., for n+1=4: (1011, 1101, 1110, 1111) a(3)=4. - Toby Gottfried, Dec 11 2011
Number of distinct symmetric staircase walks connecting opposite corners of a square grid of side n > 1. - Christian Barrientos, Nov 25 2018
From Gus Wiseman, Aug 20 2021: (Start)
Also the number of integer compositions of n + 1 with alternating sum > 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. These compositions are ranked by A345917. For example, the a(0) = 1 through a(4) = 11 compositions are:
(1) (2) (3) (4) (5)
(21) (31) (32)
(111) (112) (41)
(211) (113)
(122)
(212)
(221)
(311)
(1121)
(2111)
(11111)
The following relate to these compositions:
- The unordered version is A027193.
- The complement is counted by A058622.
- The reverse unordered version is A086543.
- The version for alternating sum >= 0 is A116406.
- The version for alternating sum < 0 is A294175.
- Ranked by A345917. (End)
The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 07 2022

Examples

			From _Gus Wiseman_, Aug 20 2021: (Start)
The a(0) = 1 through a(4) = 11 binary numbers with a majority of 1-bits (Gottfried's comment) are:
  1   11   101   1011   10011
           110   1101   10101
           111   1110   10110
                 1111   10111
                        11001
                        11010
                        11011
                        11100
                        11101
                        11110
                        11111
The version allowing an initial zero is A058622.
(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. (4.2.1.6)

Crossrefs

a(n) = Sum{(k+1)T(n, m-k)}, 0<=k<=[ (n+1)/2 ], T given by A008315.
Column k=2 of A226873. - Alois P. Heinz, Jun 21 2013
The even bisection is A000302.
The odd bisection appears to be A032443.

Programs

  • GAP
    List([0..35],n->Sum([0..Int(n/2)],k->Binomial(n,k))); # Muniru A Asiru, Nov 27 2018
  • Haskell
    a027306 n = a008949 n (n `div` 2)  -- Reinhard Zumkeller, Nov 14 2014
    
  • Magma
    [2^(n-1)+(1+(-1)^n)/4*Binomial(n, n div 2): n in [0..40]]; // Vincenzo Librandi, Jun 19 2016
    
  • Maple
    a:= proc(n) add(binomial(n, j), j=0..n/2) end:
    seq(a(n), n=0..32); # Zerinvary Lajos, Mar 29 2009
  • Mathematica
    Table[Sum[Binomial[n, k], {k, 0, Floor[n/2]}], {n, 1, 35}]
    (* Second program: *)
    a[0] = a[1] = 1; a[2] = 3; a[n_] := a[n] = (2(n-1)(2a[n-2] + a[n-1]) - 8(n-2) a[n-3])/n; Array[a, 33, 0] (* Jean-François Alcover, Sep 04 2016 *)
  • PARI
    a(n)=if(n<0,0,(2^n+if(n%2,0,binomial(n, n/2)))/2)
    

Formula

a(n) = Sum_{k=0..floor(n/2)} binomial(n,k).
Odd terms are 2^(n-1). Also a(2n) - 2^(2n-1) is given by A001700. a(n) = 2^n + (n mod 2)*binomial(n, (n-1)/2).
E.g.f.: (exp(2x) + I_0(2x))/2.
O.g.f.: 2*x/(1-2*x)/(1+2*x-((1+2*x)*(1-2*x))^(1/2)). - Vladeta Jovovic, Apr 27 2003
a(n) = A008949(n, floor(n/2)); a(n) + a(n-1) = A248574(n), n > 0. - Reinhard Zumkeller, Nov 14 2014
From Peter Bala, Jul 21 2015: (Start)
a(n) = [x^n]( 2*x - 1/(1 - x) )^n.
O.g.f.: (1/2)*( 1/sqrt(1 - 4*x^2) + 1/(1 - 2*x) ).
Inverse binomial transform is (-1)^n*A246437(n).
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 2*x^2 + 3*x^3 + 6*x^4 + 10*x^5 + ... is the o.g.f. for A001405. (End)
a(n) = Sum_{k=1..floor((n+1)/2)} binomial(n-1,(2n+1-(-1)^n)/4 -k). - Anthony Browne, Jun 18 2016
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, Aug 09 2017

Extensions

Better description from Robert G. Wilson v, Aug 30 2000 and from Yong Kong (ykong(AT)curagen.com), Dec 28 2000

A005810 a(n) = binomial(4n,n).

Original entry on oeis.org

1, 4, 28, 220, 1820, 15504, 134596, 1184040, 10518300, 94143280, 847660528, 7669339132, 69668534468, 635013559600, 5804731963800, 53194089192720, 488526937079580, 4495151581425648, 41432089765583440, 382460951663844400
Offset: 0

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Comments

Start off with 0 balls in a box. Find the number of ways you can throw 3 balls back out. Then continue to throw 4 balls into the box after each stage. (I.e., the first stage is 0. Then at the next stage there are 4 ways to throw 3 balls back out.) - Ruppi Rana (ruppirana007(AT)hotmail.com), Mar 03 2004
Central coefficients of A094527. - Paul Barry, Mar 08 2011
This is the case m = 2n in Catalan's formula (2m)!*(2n)!/(m!*(m+n)!*n!) - see Umberto Scarpis in References. - Bruno Berselli, Apr 27 2012
A generating function in terms of a (labyrinthine) solution to a depressed quartic equation is given in the Copeland link for signed A005810. - Tom Copeland, Oct 10 2012
Conjecture: a(n) == 4 (mod n^3) iff n is prime. - Gary Detlefs, Apr 03 2013
For prime p, the congruence a(p) = binomial(4*p,p) = 4 (mod p^3) is a known generalization of Wolstenholme's theorem. See Mestrovic, Section 6, equation 35. - Peter Bala, Dec 28 2014

Examples

			G.f. = 1 + 4*x + 28*x^2 + 220*x^3 + 1820*x^4 + 15504*x^5 + 134596*x^6 + ...
		

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. 828.
  • Umberto Scarpis, Sui numeri primi e sui problemi dell'analisi indeterminata in Questioni riguardanti le matematiche elementari, Nicola Zanichelli Editore (1924-1927, third Edition), page 11.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row 4 of A060539.
binomial(k*n,n): A000984 (k = 2), A005809 (k = 3), A001449 (k = 5), A004355 (k = 6), A004368 (k = 7), A004381 (k = 8), A169958 - A169961 (k = 9 thru 12).

Programs

Formula

a(n) is asymptotic to c*(256/27)^n/sqrt(n) with c = sqrt(2 / (3 Pi)) = 0.460658865961780639... - Benoit Cloitre, Jan 26 2003; corrected by Charles R Greathouse IV, Dec 14 2006
a(n) = Sum_{k=0..2n} binomial(2n,k) * binomial(2n,k-n). - Paul Barry, Mar 08 2011
G.f.: g/(4-3*g) where g = 1+x*g^4 is the g.f. of A002293. - Mark van Hoeij, Nov 11 2011
D-finite with recurrence: 3*n*(3*n-1)*(3*n-2)*a(n) - 8*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1) = 0. - R. J. Mathar, Dec 02 2012
a(n) = binomial(4*n,n-1)*(3*n+1)/n. - Gary Detlefs, Apr 03 2013
a(n) = C(4*n-1,n-1)*C(16*n^2,2)/(3*n*C(4*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
a(n) = Sum_{i,j,k = 0..n} binomial(n,i)*binomial(n,j)*binomial(n,k)* binomial(n,i+j+k). - Peter Bala, Dec 28 2014
a(n) = GegenbauerC(n, -2*n, -1). - Peter Luschny, May 07 2016
From Ilya Gutkovskiy, Nov 22 2016: (Start)
O.g.f.: 3F2(1/4,1/2,3/4; 1/3,2/3; 256*x/27).
E.g.f.: 3F3(1/4,1/2,3/4; 1/3,2/3,1; 256*x/27). (End)
a(n) = hypergeom([-3*n, -1*n], [1], 1). - Peter Luschny, Mar 19 2018
RHS of the identity Sum_{k = 0..2*n} (-1)^(n+k)*binomial(4*n, k)* binomial(4*n, 2*n-k) = binomial(4*n,n). - Peter Bala, Oct 07 2021
From Peter Bala, Feb 20 2022: (Start)
The o.g.f. A(x) satisfies the differential equation
(-256*x^3 + 27*x^2)*A(x)''' + (-1152*x^2 + 54*x)*A(x)'' + (-816*x + 6)*A(x)' - 24*A(x) = 0 with A(0) = 1, A'(0) = 4 and A''(0) = 56.
Algebraic equation: (1 - A(x))*(1 + 3*A(x))^3 + 256*x*A(x)^4 = 0.
Sum_{n >= 1} a(n)*( x*(3*x + 4)^3/(256*(1 + x)^4) )^n = x. (End)
From Amiram Eldar, Dec 07 2024: (Start)
Sum_{n>=1} 1/a(n) = A378806.
Sum_{n>=1} (-1)^n/a(n) = A378807. (End)
From Peter Bala, Jun 29 2025: (Start)
a(n) = (1/8)^n * Sum_{k = n..4*n} binomial(k, n) * binomial(4*n, k).
Sum_{n >= 0 } a(n)*(1/128)^n = (1/5)*(sqrt(2) + sqrt(7 + 5*sqrt(2))). (End)
From Seiichi Manyama, Aug 16 2025: (Start)
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(4*n+1,k).
G.f.: 1/(1 - 4*x*g^3) where g = 1+x*g^4 is the g.f. of A002293. (End)

Extensions

More terms from Henry Bottomley, Oct 06 2000
Corrected by T. D. Noe, Jan 16 2007
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