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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A001791 a(n) = binomial coefficient C(2n, n-1).

Original entry on oeis.org

0, 1, 4, 15, 56, 210, 792, 3003, 11440, 43758, 167960, 646646, 2496144, 9657700, 37442160, 145422675, 565722720, 2203961430, 8597496600, 33578000610, 131282408400, 513791607420, 2012616400080, 7890371113950, 30957699535776, 121548660036300, 477551179875952
Offset: 0

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Comments

Number of peaks at even level in all Dyck paths of semilength n+1. Example: a(2)=4 because UDUDUD, UDUU*DD, UU*DDUD, UU*DU*DD, UUUDDD, where U=(1,1), D=(1,-1) and the peaks at even level are shown by *. - Emeric Deutsch, Dec 05 2003
Also number of long ascents (i.e., ascents of length at least two) in all Dyck paths of semilength n+1. Example: a(2)=4 because in the five Dyck paths of semilength 3, namely UDUDUD, UD(UU)DD, (UU)DDUD, (UU)DUDD and (UUU)DDD, we have four long ascents (shown between parentheses). Here U=(1,1) and D=(1,-1). Also number of branch nodes (i.e., vertices of outdegree at least two) in all ordered trees with n+1 edges. - Emeric Deutsch, Feb 22 2004
Number of lattice paths from (0,0) to (n,n) with steps E=(1,0) and N=(0,1) which touch or cross the line x-y=1. Example: For n=2 these are the paths EENN, ENEN, ENNE and NEEN. - Herbert Kociemba, May 23 2004
Narayana transform (A001263) of [1, 3, 5, 7, 9, ...] = (1, 4, 15, 56, 210, ...). Row sums of triangles A136534 and A136536. - Gary W. Adamson, Jan 04 2008
Starting with offset 1 = the Catalan sequence starting (1, 2, 5, 14, ...) convolved with A000984: (1, 2, 6, 20, ...). - Gary W. Adamson, May 17 2009
Also number of peaks plus number of valleys in all Dyck n-paths. - David Scambler, Oct 08 2012
Apparently counts UDDUD in all Dyck paths of semilength n+2. - David Scambler, Apr 22 2013
Apparently the number of peaks strictly left of the midpoint in all Dyck paths of semilength n+1. - David Scambler, Apr 30 2013
For n>0, a(n) is the number of compositions of n into at most n parts if zeros are allowed as parts (so-called "weak" compositions). - L. Edson Jeffery, Jul 24 2014
Number of paths in the half-plane x >= 0, from (0,0) to (2n,2), and consisting of steps U=(1,1) and D=(1,-1). For example, for n=2, we have the 4 paths: UUUD, UUDU, UDUU, DUUU. - José Luis Ramírez Ramírez, Apr 19 2015
For n>1, 1/a(n) is the probability that when a stick is broken up at n points independently and uniformly chosen at random along its length any triple of pieces of the n+1 pieces can form a triangle. The corresponding probability for the existence of at least one triple is A339392(n)/A339393(n). - Amiram Eldar, Dec 04 2020
a(n) is the number of lattice paths of 2n steps taken from the step set {U=(1,1), D=(1,-1)} that start at the origin, never go below the x-axis, and end strictly above the x-axis; more succinctly, proper left factors of Dyck paths. For example, a(2)=4 counts UUUU, UUUD, UUDU, UDUU. - David Callan and Emeric Deutsch, Jan 25 2021
From Gus Wiseman, Jul 21 2021: (Start)
Also the number of integer compositions of 2n+1 with alternating sum -1, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(1) = 1 through a(3) = 15 compositions are:
(1,2) (2,3) (3,4)
(1,3,1) (1,4,2)
(1,1,1,2) (2,4,1)
(1,2,1,1) (1,1,2,3)
(1,2,2,2)
(1,3,2,1)
(2,1,1,3)
(2,2,1,2)
(2,3,1,1)
(1,1,1,3,1)
(1,2,1,2,1)
(1,3,1,1,1)
(1,1,1,1,1,2)
(1,1,1,2,1,1)
(1,2,1,1,1,1)
The following relate to these compositions.
- The unordered version is A000070.
- Allowing any negative alternating sum gives A000346.
- The opposite (positive 1) version is A000984.
- The version for reverse-alternating sum is also A001791 (this sequence).
- Taking alternating sum -2 instead of -1 gives A002054.
- The shifted version for alternating sum 0 is counted by A088218 and ranked by A344619.
- Ranked by A345910 (reverse: A345912).
Equivalently, a(n) counts binary numbers with 2n+1 digits and one more 0 than 1's. For example, the a(2) = 4 binary numbers are: 10001, 10010, 10100, 11000.
(End)
The diagonal of a square n X n matrix where cells of the first row are the nonnegative integers and cells of subsequent rows are sums of cells of the previous row up to and including n. - Torlach Rush, Apr 24 2024
For n>=1, a(n) is the independence number of the odd graph O_{n+1}. - Miquel A. Fiol, Jul 07 2024

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.
  • Cornelius Lanczos, Applied Analysis, Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 517.
  • R. C. Mullin, E. Nemeth and P. J. Schellenberg, The enumeration of almost cubic maps, pp. 281-295 in Proceedings of the Louisiana Conference on Combinatorics, Graph Theory and Computer Science. Vol. 1, edited R. C. Mullin et al., 1970.
  • 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 3 of triangle A100257.
First differences are in A076540.
A345197 counts compositions by length and alternating sum.

Programs

  • GAP
    List([0..30],n->Binomial(2*n,n-1)); # Muniru A Asiru, Aug 09 2018
  • Magma
    [Binomial(2*n, n-1): n in [0..30]]; // Vincenzo Librandi, Apr 20 2015
    
  • Mathematica
    Table[Binomial[2n,n-1],{n,0,30}] (* Harvey P. Dale, Jul 12 2012 *)
    CoefficientList[ Series[(1 - 2x - Sqrt[1 - 4x])/(2x*Sqrt[1 - 4x]), {x, 0, 26}], x] (* Robert G. Wilson v, Aug 10 2018 *)
  • Maxima
    A001791(n):=binomial(2*n,n-1)$
    makelist(A001791(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
    
  • PARI
    a(n)=if(n<1,0,(2*n)!/(n+1)!/(n-1)!)
    

Formula

a(n) = n*A000108(n).
G.f.: x*(d/dx)c(x) where c(x) = Catalan g.f. - Wolfdieter Lang
Convolution of A001700 (central binomial of odd order) and A000108 (Catalan): a(n+1) = Sum_{k=0..n} C(k)*binomial(2*(n-k)+1, n-k), C(k): Catalan. - Wolfdieter Lang
E.g.f.: exp(2x) * I_1(2x), where I_1 is Bessel function. - Michael Somos, Sep 08 2002
a(n) = Sum_{k=0..n} C(n, k)*C(n, k+1). - Paul Barry, May 15 2003
a(n) = Sum_{i=1..n} binomial(i+n-1, n).
G.f.: (1-2x-sqrt(1-4x))/(2x*sqrt(1-4x)). - Emeric Deutsch, Dec 05 2003
a(n) = A092956/(n!). - Amarnath Murthy, Jun 16 2004
a(n) = binomial(2n,n) - A000108(n). - Paul Barry, Apr 21 2005
a(n) = (1/(2*Pi))*Integral_{x=0..4} (x^n*(x-2)/sqrt(x(4-x))) is the moment sequence representation. - Paul Barry, Jan 11 2007
Row sums of triangle A132812 starting (1, 4, 15, 56, 210, ...). - Gary W. Adamson, Sep 01 2007
Starting (1, 4, 15, 56, 210, ...) gives the binomial transform of A025566 starting (1, 3, 8, 22, 61, 171, ...). - Gary W. Adamson, Sep 01 2007
For n >= 1, a(2^n) = 2^(n+1)*A001795(2^(n-1)). - Vladimir Shevelev, Sep 05 2010
D-finite with recurrence: (n-1)*(n+1)*a(n) = 2*n*(2n-1)*a(n-1). - R. J. Mathar, Dec 17 2011
From Sergei N. Gladkovskii, Jul 07 2012: (Start)
G.f.: -1/(2*x) - G(0) where G(k) = 1 - 1/(2*x - 8*x^3*(2*k+1)/(4*x^2*(2*k+1)- (k+1)/G(k+1))); (continued fraction, 3rd kind, 3-step);
E.g.f.: BesselI(1,2*x)*exp(2*x) = x*G(0) where G(k) = 1 + 2*x*(4*k+3)/((2*k+1)*(2*k+3) - x*(2*k+1)*(2*k+3)*(4*k+5)/(x*(4*k+5) + 2*(k+1)*(k+2)/G(k+1))); (continued fraction, 3rd kind, 3-step).
(End)
G.f.: c(x)^3/(2-c(x)) where c(x) is the g.f. for A000108. - Cheyne Homberger, May 05 2014
G.f.: z*C(z)^2/(1-2*z*C(z)), where C(z) is the g.f. of Catalan numbers. - José Luis Ramírez Ramírez, Apr 19 2015
G.f.: x*2F1(3/2,2;3;4x). - R. J. Mathar, Aug 09 2015
a(n) = Sum_{i=1..n} binomial(2*i-2,i-1)*binomial(2*(n-i+1),n-i+2)/(n-i+1). - Vladimir Kruchinin, Sep 07 2015
L.g.f.: 1/(1 - x/(1 - x/(1 - x/(1 - x/(1 - x/(1 - ...)))))) = Sum_{n>=1} a(n)*x^n/n. - Ilya Gutkovskiy, May 10 2017
Sum_{n>=1} 1/a(n) = 1/3 + 5*Pi/(9*sqrt(3)). - Amiram Eldar, Dec 04 2020
Sum_{n>=1} (-1)^(n+1)/a(n) = 1/5 + 14*sqrt(5)*log(phi)/25, where log(phi) = A002390. - Amiram Eldar, Feb 20 2021
a(n) = Product_{i=1..(n - 1)} (((4*i + 6)*i + 2)/((i + 2)*i)), for n>=1. - Neven Sajko, Oct 10 2021
a(n) = 2^(2*n)*gamma(n + 1/2)/(sqrt(Pi)*gamma(n)*(n+1)) for n > 0, and a(0) = lim_{n->0} a(n). - Karol A. Penson, Apr 24 2025

A001813 Quadruple factorial numbers: a(n) = (2n)!/n!.

Original entry on oeis.org

1, 2, 12, 120, 1680, 30240, 665280, 17297280, 518918400, 17643225600, 670442572800, 28158588057600, 1295295050649600, 64764752532480000, 3497296636753920000, 202843204931727360000, 12576278705767096320000, 830034394580628357120000, 58102407620643984998400000
Offset: 0

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Counts binary rooted trees (with out-degree <= 2), embedded in plane, with n labeled end nodes of degree 1. Unlabeled version gives Catalan numbers A000108.
Define a "downgrade" to be the permutation which places the items of a permutation in descending order. We are concerned with permutations that are identical to their downgrades. Only permutations of order 4n and 4n+1 can have this property; the number of permutations of length 4n having this property are equinumerous with those of length 4n+1. If a permutation p has this property then the reversal of this permutation also has it. a(n) = number of permutations of length 4n and 4n+1 that are identical to their downgrades. - Eugene McDonnell (eemcd(AT)mac.com), Oct 26 2003
Number of broadcast schemes in the complete graph on n+1 vertices, K_{n+1}. - Calin D. Morosan (cd_moros(AT)alumni.concordia.ca), Nov 28 2008
Hankel transform is A137565. - Paul Barry, Nov 25 2009
The e.g.f. of 1/a(n) = n!/(2*n)! is (exp(sqrt(x)) + exp(-sqrt(x)) )/2. - Wolfdieter Lang, Jan 09 2012
From Tom Copeland, Nov 15 2014: (Start)
Aerated with intervening zeros (1,0,2,0,12,0,120,...) = a(n) (cf. A123023 and A001147), the e.g.f. is e^(t^2), so this is the base for the Appell sequence with e.g.f. e^(t^2) e^(x*t) = exp(P(.,x),t) (reverse A059344, cf. A099174, A066325 also). P(n,x) = (a. + x)^n with (a.)^n = a_n and comprise the umbral compositional inverses for e^(-t^2)e^(x*t) = exp(UP(.,x),t), i.e., UP(n,P(.,t)) = x^n = P(n,UP(.,t)), e.g., (P(.,t))^n = P(n,t).
Equals A000407*2 with leading 1 added. (End)
a(n) is also the number of square roots of any permutation in S_{4*n} whose disjoint cycle decomposition consists of 2*n transpositions. - Luis Manuel Rivera Martínez, Mar 04 2015
Self-convolution gives A076729. - Vladimir Reshetnikov, Oct 11 2016
For n > 1, it follows from the formula dated Aug 07 2013 that a(n) is a Zumkeller number (A083207). - Ivan N. Ianakiev, Feb 28 2017
For n divisible by 4, a(n/4) is the number of ways to place n points on an n X n grid with pairwise distinct abscissae, pairwise distinct ordinates, and 90-degree rotational symmetry. For n == 1 (mod 4), the number of ways is a((n-1)/4) because the center point can be considered "fixed". For 180-degree rotational symmetry see A006882, for mirror symmetry see A000085, A135401, and A297708. - Manfred Scheucher, Dec 29 2017

Examples

			The following permutations of order 8 and their reversals have this property:
  1 7 3 5 2 4 0 6
  1 7 4 2 5 3 0 6
  2 3 7 6 1 0 4 5
  2 4 7 1 6 0 3 5
  3 2 6 7 0 1 5 4
  3 5 1 7 0 6 2 4
		

References

  • D. E. Knuth, The Art of Computer Programming, Vol. 4, Section 7.2.1.6, Eq. 32.
  • L. C. Larson, The number of essentially different nonattacking rook arrangements, J. Recreat. Math., 7 (No. 3, 1974), circa pages 180-181.
  • Eugene McDonnell, "Magic Squares and Permutations" APL Quote-Quad 7.3 (Fall, 1976)
  • R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976).
  • 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([0..20],n->Factorial(2*n)/Factorial(n)); # Muniru A Asiru, Nov 01 2018
    
  • Magma
    [Factorial(2*n)/Factorial(n): n in [0..20]]; // Vincenzo Librandi, Oct 09 2018
    
  • Maple
    A001813 := n->(2*n)!/n!;
    A001813 := n -> mul(k, k = select(k-> k mod 4 = 2,[$1 .. 4*n])):
    seq(A001813(n), n=0..16);  # Peter Luschny, Jun 23 2011
  • Mathematica
    Table[(2n)!/n!, {n,0,20}] (* Harvey P. Dale, May 02 2011 *)
  • Maxima
    makelist(binomial(n+n, n)*n!,n,0,30); /* Martin Ettl, Nov 05 2012 */
    
  • PARI
    a(n)=binomial(n+n,n)*n! \\ Charles R Greathouse IV, Jun 15 2011
    
  • PARI
    first(n) = x='x+O('x^n); Vec(serlaplace((1 - 4*x)^(-1/2))) \\ Iain Fox, Jan 01 2018 (corrected by Iain Fox, Jan 11 2018)
    
  • Python
    from math import factorial
    def A001813(n): return factorial(n<<1)//factorial(n) # Chai Wah Wu, Feb 14 2023
  • Sage
    [binomial(2*n,n)*factorial(n) for n in range(0, 17)] # Zerinvary Lajos, Dec 03 2009
    

Formula

E.g.f.: (1-4*x)^(-1/2).
a(n) = (2*n)!/n! = Product_{k=0..n-1} (4*k + 2) = A081125(2*n).
Integral representation as n-th moment of a positive function on a positive half-axis: a(n) = Integral_{x=0..oo} x^n*exp(-x/4)/(sqrt(x)*2*sqrt(Pi)) dx, n >= 0. This representation is unique. - Karol A. Penson, Sep 18 2001
Define a'(1)=1, a'(n) = Sum_{k=1..n-1} a'(n-k)*a'(k)*C(n, k); then a(n)=a'(n+1). - Benoit Cloitre, Apr 27 2003
With interpolated zeros (1, 0, 2, 0, 12, ...) this has e.g.f. exp(x^2). - Paul Barry, May 09 2003
a(n) = A000680(n)/A000142(n)*A000079(n) = Product_{i=0..n-1} (4*i + 2) = 4^n*Pochhammer(1/2, n) = 4^n*GAMMA(n+1/2)/sqrt(Pi). - Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
For asymptotics, see the Robinson paper.
a(k) = (2*k)!/k! = Sum_{i=1..k+1} |A008275(i,k+1)| * k^(i-1). - André F. Labossière, Jun 21 2007
a(n) = 12*A051618(a) n >= 2. - Zerinvary Lajos, Feb 15 2008
a(n) = A000984(n)*A000142(n). - Zerinvary Lajos, Mar 25 2008
a(n) = A016825(n-1)*a(n-1). - Roger L. Bagula, Sep 17 2008
a(n) = (-1)^n*A097388(n). - D. Morosan (cd_moros(AT)alumni.concordia.ca), Nov 28 2008
From Paul Barry, Jan 15 2009: (Start)
G.f.: 1/(1-2x/(1-4x/(1-6x/(1-8x/(1-10x/(1-... (continued fraction);
a(n) = (n+1)!*A000108(n). (End)
a(n) = Sum_{k=0..n} A132393(n,k)*2^(2n-k). - Philippe Deléham, Feb 10 2009
G.f.: 1/(1-2x-8x^2/(1-10x-48x^2/(1-18x-120x^2/(1-26x-224x^2/(1-34x-360x^2/(1-42x-528x^2/(1-... (continued fraction). - Paul Barry, Nov 25 2009
a(n) = A173333(2*n,n) for n>0; cf. A006963, A001761. - Reinhard Zumkeller, Feb 19 2010
From Gary W. Adamson, Jul 19 2011: (Start)
a(n) = upper left term of M^n, M = an infinite square production matrix as follows:
2, 2, 0, 0, 0, 0, ...
4, 4, 4, 0, 0, 0, ...
6, 6, 6, 6, 0, 0, ...
8, 8, 8, 8, 8, 0, ...
...
(End)
a(n) = (-2)^n*Sum_{k=0..n} 2^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
G.f.: 1/Q(0), where Q(k) = 1 + x*(4*k+2) - x*(4*k+4)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 18 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - x*(8*k+4)/(x*(8*k+4) - 1 + 8*x*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 30 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 2*x/(2*x + 1/(2*k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
D-finite with recurrence: a(n) = (4*n-6)*a(n-2) + (4*n-3)*a(n-1), n>=2. - Ivan N. Ianakiev, Aug 07 2013
Sum_{n>=0} 1/a(n) = (exp(1/4)*sqrt(Pi)*erf(1/2) + 2)/2 = 1 + A214869, where erf(x) is the error function. - Ilya Gutkovskiy, Nov 10 2016
Sum_{n>=0} (-1)^n/a(n) = 1 - sqrt(Pi)*erfi(1/2)/(2*exp(1/4)), where erfi(x) is the imaginary error function. - Amiram Eldar, Feb 20 2021
a(n) = 1/([x^n] hypergeom([1], [1/2], x/4)). - Peter Luschny, Sep 13 2024
a(n) = 2^n*n!*JacobiP(n, -1/2, -n, 3). - Peter Luschny, Jan 22 2025
G.f.: 2F0(1,1/2;;4x). - R. J. Mathar, Jun 07 2025

Extensions

More terms from James Sellers, May 01 2000

A039599 Triangle formed from even-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 9, 5, 1, 14, 28, 20, 7, 1, 42, 90, 75, 35, 9, 1, 132, 297, 275, 154, 54, 11, 1, 429, 1001, 1001, 637, 273, 77, 13, 1, 1430, 3432, 3640, 2548, 1260, 440, 104, 15, 1, 4862, 11934, 13260, 9996, 5508, 2244, 663, 135, 17, 1
Offset: 0

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T(n,k) is the number of lattice paths from (0,0) to (n,n) with steps E = (1,0) and N = (0,1) which touch but do not cross the line x - y = k and only situated above this line; example: T(3,2) = 5 because we have EENNNE, EENNEN, EENENN, ENEENN, NEEENN. - Philippe Deléham, May 23 2005
The matrix inverse of this triangle is the triangular matrix T(n,k) = (-1)^(n+k)* A085478(n,k). - Philippe Deléham, May 26 2005
Essentially the same as A050155 except with a leading diagonal A000108 (Catalan numbers) 1, 1, 2, 5, 14, 42, 132, 429, .... - Philippe Deléham, May 31 2005
Number of Grand Dyck paths of semilength n and having k downward returns to the x-axis. (A Grand Dyck path of semilength n is a path in the half-plane x>=0, starting at (0,0), ending at (2n,0) and consisting of steps u=(1,1) and d=(1,-1)). Example: T(3,2)=5 because we have u(d)uud(d),uud(d)u(d),u(d)u(d)du,u(d)duu(d) and duu(d)u(d) (the downward returns to the x-axis are shown between parentheses). - Emeric Deutsch, May 06 2006
Riordan array (c(x),x*c(x)^2) where c(x) is the g.f. of A000108; inverse array is (1/(1+x),x/(1+x)^2). - Philippe Deléham, Feb 12 2007
The triangle may also be generated from M^n*[1,0,0,0,0,0,0,0,...], where M is the infinite tridiagonal matrix with all 1's in the super and subdiagonals and [1,2,2,2,2,2,2,...] in the main diagonal. - Philippe Deléham, Feb 26 2007
Inverse binomial matrix applied to A124733. Binomial matrix applied to A089942. - Philippe Deléham, Feb 26 2007
Number of standard tableaux of shape (n+k,n-k). - Philippe Deléham, Mar 22 2007
From Philippe Deléham, Mar 30 2007: (Start)
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y):
(0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970
(1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877;
(1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598;
(2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954;
(3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791;
(4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. (End)
The table U(n,k) = Sum_{j=0..n} T(n,j)*k^j is given in A098474. - Philippe Deléham, Mar 29 2007
Sequence read mod 2 gives A127872. - Philippe Deléham, Apr 12 2007
Number of 2n step walks from (0,0) to (2n,2k) and consisting of step u=(1,1) and d=(1,-1) and the path stays in the nonnegative quadrant. Example: T(3,0)=5 because we have uuuddd, uududd, ududud, uduudd, uuddud; T(3,1)=9 because we have uuuudd, uuuddu, uuudud, ududuu, uuduud, uduudu, uudduu, uduuud, uududu; T(3,2)=5 because we have uuuuud, uuuudu, uuuduu, uuduuu, uduuuu; T(3,3)=1 because we have uuuuuu. - Philippe Deléham, Apr 16 2007, Apr 17 2007, Apr 18 2007
Triangular matrix, read by rows, equal to the matrix inverse of triangle A129818. - Philippe Deléham, Jun 19 2007
Let Sum_{n>=0} a(n)*x^n = (1+x)/(1-mx+x^2) = o.g.f. of A_m, then Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n. Related expansions of A_m are: A099493, A033999, A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, A097783, A077416, A126866, A028230, A161591, for m=-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, respectively. - Philippe Deléham, Nov 16 2009
The Kn11, Kn12, Fi1 and Fi2 triangle sums link the triangle given above with three sequences; see the crossrefs. For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 20 2011
4^n = (n-th row terms) dot (first n+1 odd integer terms). Example: 4^4 = 256 = (14, 28, 20, 7, 1) dot (1, 3, 5, 7, 9) = (14 + 84 + 100 + 49 + 9) = 256. - Gary W. Adamson, Jun 13 2011
The linear system of n equations with coefficients defined by the first n rows solve for diagonal lengths of regular polygons with N= 2n+1 edges; the constants c^0, c^1, c^2, ... are on the right hand side, where c = 2 + 2*cos(2*Pi/N). Example: take the first 4 rows relating to the 9-gon (nonagon), N = 2*4 + 1; with c = 2 + 2*cos(2*Pi/9) = 3.5320888.... The equations are (1,0,0,0) = 1; (1,1,0,0) = c; (2,3,1,0) = c^2; (5,9,5,1) = c^3. The solutions are 1, 2.53208..., 2.87938..., and 1.87938...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. (Cf. comment in A089942 which uses the analogous operations but with c = 1 + 2*cos(2*Pi/9).) - Gary W. Adamson, Sep 21 2011
Also called the Lobb numbers, after Andrew Lobb, are a natural generalization of the Catalan numbers, given by L(m,n)=(2m+1)*Binomial(2n,m+n)/(m+n+1), where n >= m >= 0. For m=0, we get the n-th Catalan number. See added reference. - Jayanta Basu, Apr 30 2013
From Wolfdieter Lang, Sep 20 2013: (Start)
T(n, k) = A053121(2*n, 2*k). T(n, k) appears in the formula for the (2*n)-th power of the algebraic number rho(N):= 2*cos(Pi/N) = R(N, 2) in terms of the odd-indexed diagonal/side length ratios R(N, 2*k+1) = S(2*k, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310):
rho(N)^(2*n) = Sum_{k=0..n} T(n, k)*R(N, 2*k+1), n >= 0, identical in N > = 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears.
For the odd powers of rho(n) see A039598. (End)
Unsigned coefficients of polynomial numerators of Eqn. 2.1 of the Chakravarty and Kodama paper, defining the polynomials of A067311. - Tom Copeland, May 26 2016
The triangle is the Riordan square of the Catalan numbers in the sense of A321620. - Peter Luschny, Feb 14 2023

Examples

			Triangle T(n, k) begins:
  n\k     0     1     2     3     4     5    6   7   8  9
  0:      1
  1:      1     1
  2:      2     3     1
  3:      5     9     5     1
  4:     14    28    20     7     1
  5:     42    90    75    35     9     1
  6:    132   297   275   154    54    11    1
  7:    429  1001  1001   637   273    77   13   1
  8:   1430  3432  3640  2548  1260   440  104  15   1
  9:   4862 11934 13260  9996  5508  2244  663 135  17  1
  ... Reformatted by _Wolfdieter Lang_, Dec 21 2015
From _Paul Barry_, Feb 17 2011: (Start)
Production matrix begins
  1, 1,
  1, 2, 1,
  0, 1, 2, 1,
  0, 0, 1, 2, 1,
  0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 0, 1, 2, 1 (End)
From _Wolfdieter Lang_, Sep 20 2013: (Start)
Example for rho(N) = 2*cos(Pi/N) powers:
n=2: rho(N)^4 = 2*R(N,1) + 3*R(N,3) + 1*R(N, 5) =
  2 + 3*S(2, rho(N)) + 1*S(4, rho(N)), identical in N >= 1. For N=4 (the square with only one distinct diagonal), the degree delta(4) = 2, hence R(4, 3) and R(4, 5) can be reduced, namely to R(4, 1) = 1 and R(4, 5) = -R(4,1) = -1, respectively. Therefore, rho(4)^4 =(2*cos(Pi/4))^4 = 2 + 3 -1 = 4. (End)
		

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.
  • 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.

Crossrefs

Row sums: A000984.
Triangle sums (see the comments): A000958 (Kn11), A001558 (Kn12), A088218 (Fi1, Fi2).

Programs

  • Magma
    /* As triangle */ [[Binomial(2*n, k+n)*(2*k+1)/(k+n+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 16 2015
    
  • Maple
    T:=(n,k)->(2*k+1)*binomial(2*n,n-k)/(n+k+1): for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form # Emeric Deutsch, May 06 2006
    T := proc(n, k) option remember; if k = n then 1 elif k > n then 0 elif k = 0 then T(n-1, 0) + T(n-1,1) else T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1) fi end:
    seq(seq(T(n, k), k = 0..n), n = 0..9) od; # Peter Luschny, Feb 14 2023
  • Mathematica
    Table[Abs[Differences[Table[Binomial[2 n, n + i], {i, 0, n + 1}]]], {n, 0,7}] // Flatten (* Geoffrey Critzer, Dec 18 2011 *)
    Join[{1},Flatten[Table[Binomial[2n-1,n-k]-Binomial[2n-1,n-k-2],{n,10},{k,0,n}]]] (* Harvey P. Dale, Dec 18 2011 *)
    Flatten[Table[Binomial[2*n,m+n]*(2*m+1)/(m+n+1),{n,0,9},{m,0,n}]] (* Jayanta Basu, Apr 30 2013 *)
  • PARI
    a(n, k) = (2*n+1)/(n+k+1)*binomial(2*k, n+k)
    trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(a(y, x), ", ")); print(""))
    trianglerows(10) \\ Felix Fröhlich, Jun 24 2016
  • Sage
    # Algorithm of L. Seidel (1877)
    # Prints the first n rows of the triangle
    def A039599_triangle(n) :
        D = [0]*(n+2); D[1] = 1
        b = True ; h = 1
        for i in range(2*n-1) :
            if b :
                for k in range(h,0,-1) : D[k] += D[k-1]
                h += 1
            else :
                for k in range(1,h, 1) : D[k] += D[k+1]
            if b : print([D[z] for z in (1..h-1)])
            b = not b
    A039599_triangle(10)  # Peter Luschny, May 01 2012
    

Formula

T(n,k) = C(2*n-1, n-k) - C(2*n-1, n-k-2), n >= 1, T(0,0) = 1.
From Emeric Deutsch, May 06 2006: (Start)
T(n,k) = (2*k+1)*binomial(2*n,n-k)/(n+k+1).
G.f.: G(t,z)=1/(1-(1+t)*z*C), where C=(1-sqrt(1-4*z))/(2*z) is the Catalan function. (End)
The following formulas were added by Philippe Deléham during 2003 to 2009: (Start)
Triangle T(n, k) read by rows; given by A000012 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
T(n, k) = C(2*n, n-k)*(2*k+1)/(n+k+1). Sum(k>=0; T(n, k)*T(m, k) = A000108(n+m)); A000108: numbers of Catalan.
T(n, 0) = A000108(n); T(n, k) = 0 if k>n; for k>0, T(n, k) = Sum_{j=1..n} T(n-j, k-1)*A000108(j).
T(n, k) = A009766(n+k, n-k) = A033184(n+k+1, 2k+1).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+1) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
T(0, 0) = 1, T(n, k) = 0 if n<0 or n=1, T(n, k) = T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1).
a(n) + a(n+1) = 1 + A000108(m+1) if n = m*(m+3)/2; a(n) + a(n+1) = A039598(n) otherwise.
T(n, k) = A050165(n, n-k).
Sum_{j>=0} T(n-k, j)*A039598(k, j) = A028364(n, k).
Matrix inverse of the triangle T(n, k) = (-1)^(n+k)*binomial(n+k, 2*k) = (-1)^(n+k)*A085478(n, k).
Sum_{k=0..n} T(n, k)*x^k = A000108(n), A000984(n), A007854(n), A076035(n), A076036(n) for x = 0, 1, 2, 3, 4.
Sum_{k=0..n} (2*k+1)*T(n, k) = 4^n.
T(n, k)*(-2)^(n-k) = A114193(n, k).
Sum_{k>=h} T(n,k) = binomial(2n,n-h).
Sum_{k=0..n} T(n,k)*5^k = A127628(n).
Sum_{k=0..n} T(n,k)*7^k = A115970(n).
T(n,k) = Sum_{j=0..n-k} A106566(n+k,2*k+j).
Sum_{k=0..n} T(n,k)*6^k = A126694(n).
Sum_{k=0..n} T(n,k)*A000108(k) = A007852(n+1).
Sum_{k=0..floor(n/2)} T(n-k,k) = A000958(n+1).
Sum_{k=0..n} T(n,k)*(-1)^k = A000007(n).
Sum_{k=0..n} T(n,k)*(-2)^k = (-1)^n*A064310(n).
T(2*n,n) = A126596(n).
Sum_{k=0..n} T(n,k)*(-x)^k = A000007(n), A126983(n), A126984(n), A126982(n), A126986(n), A126987(n), A127017(n), A127016(n), A126985(n), A127053(n) for x=1,2,3,4,5,6,7,8,9,10 respectively.
Sum_{j>=0} T(n,j)*binomial(j,k) = A116395(n,k).
T(n,k) = Sum_{j>=0} A106566(n,j)*binomial(j,k).
T(n,k) = Sum_{j>=0} A127543(n,j)*A038207(j,k).
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A101490(n+1).
T(n,k) = A053121(2*n,2*k).
Sum_{k=0..n} T(n,k)*sin((2*k+1)*x) = sin(x)*(2*cos(x))^(2*n).
T(n,n-k) = Sum_{j>=0} (-1)^(n-j)*A094385(n,j)*binomial(j,k).
Sum_{j>=0} A110506(n,j)*binomial(j,k) = Sum_{j>=0} A110510(n,j)*A038207(j,k) = T(n,k)*2^(n-k).
Sum_{j>=0} A110518(n,j)*A027465(j,k) = Sum_{j>=0} A110519(n,j)*A038207(j,k) = T(n,k)*3^(n-k).
Sum_{k=0..n} T(n,k)*A001045(k) = A049027(n), for n>=1.
Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n if Sum_{k>=0} a(k)*x^k = (1+x)/(x^2-m*x+1).
Sum_{k=0..n} T(n,k)*A040000(k) = A001700(n).
Sum_{k=0..n} T(n,k)*A122553(k) = A051924(n+1).
Sum_{k=0..n} T(n,k)*A123932(k) = A051944(n).
Sum_{k=0..n} T(n,k)*k^2 = A000531(n), for n>=1.
Sum_{k=0..n} T(n,k)*A000217(k) = A002457(n-1), for n>=1.
Sum{j>=0} binomial(n,j)*T(j,k)= A124733(n,k).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000984(n), A089022(n), A035610(n), A130976(n), A130977(n), A130978(n), A130979(n), A130980(n), A131521(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
Sum_{k=0..n} T(n,k)*A005043(k) = A127632(n).
Sum_{k=0..n} T(n,k)*A132262(k) = A089022(n).
T(n,k) + T(n,k+1) = A039598(n,k).
T(n,k) = A128899(n,k)+A128899(n,k+1).
Sum_{k=0..n} T(n,k)*A015518(k) = A076025(n), for n>=1. Also Sum_{k=0..n} T(n,k)*A015521(k) = A076026(n), for n>=1.
Sum_{k=0..n} T(n,k)*(-1)^k*x^(n-k) = A033999(n), A000007(n), A064062(n), A110520(n), A132863(n), A132864(n), A132865(n), A132866(n), A132867(n), A132869(n), A132897(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively.
Sum_{k=0..n} T(n,k)*(-1)^(k+1)*A000045(k) = A109262(n), A000045:= Fibonacci numbers.
Sum_{k=0..n} T(n,k)*A000035(k)*A016116(k) = A143464(n).
Sum_{k=0..n} T(n,k)*A016116(k) = A101850(n).
Sum_{k=0..n} T(n,k)*A010684(k) = A100320(n).
Sum_{k=0..n} T(n,k)*A000034(k) = A029651(n).
Sum_{k=0..n} T(n,k)*A010686(k) = A144706(n).
Sum_{k=0..n} T(n,k)*A006130(k-1) = A143646(n), with A006130(-1)=0.
T(n,2*k)+T(n,2*k+1) = A118919(n,k).
Sum_{k=0..j} T(n,k) = A050157(n,j).
Sum_{k=0..2} T(n,k) = A026012(n); Sum_{k=0..3} T(n,k)=A026029(n).
Sum_{k=0..n} T(n,k)*A000045(k+2) = A026671(n).
Sum_{k=0..n} T(n,k)*A000045(k+1) = A026726(n).
Sum_{k=0..n} T(n,k)*A057078(k) = A000012(n).
Sum_{k=0..n} T(n,k)*A108411(k) = A155084(n).
Sum_{k=0..n} T(n,k)*A057077(k) = 2^n = A000079(n).
Sum_{k=0..n} T(n,k)*A057079(k) = 3^n = A000244(n).
Sum_{k=0..n} T(n,k)*(-1)^k*A011782(k) = A000957(n+1).
(End)
T(n,k) = Sum_{j=0..k} binomial(k+j,2j)*(-1)^(k-j)*A000108(n+j). - Paul Barry, Feb 17 2011
Sum_{k=0..n} T(n,k)*A071679(k+1) = A026674(n+1). - Philippe Deléham, Feb 01 2014
Sum_{k=0..n} T(n,k)*(2*k+1)^2 = (4*n+1)*binomial(2*n,n). - Werner Schulte, Jul 22 2015
Sum_{k=0..n} T(n,k)*(2*k+1)^3 = (6*n+1)*4^n. - Werner Schulte, Jul 22 2015
Sum_{k=0..n} (-1)^k*T(n,k)*(2*k+1)^(2*m) = 0 for 0 <= m < n (see also A160562). - Werner Schulte, Dec 03 2015
T(n,k) = GegenbauerC(n-k,-n+1,-1) - GegenbauerC(n-k-1,-n+1,-1). - Peter Luschny, May 13 2016
T(n,n-2) = A014107(n). - R. J. Mathar, Jan 30 2019
T(n,n-3) = n*(2*n-1)*(2*n-5)/3. - R. J. Mathar, Jan 30 2019
T(n,n-4) = n*(n-1)*(2*n-1)*(2*n-7)/6. - R. J. Mathar, Jan 30 2019
T(n,n-5) = n*(n-1)*(2*n-1)*(2*n-3)*(2*n-9)/30. - R. J. Mathar, Jan 30 2019

Extensions

Corrected by Philippe Deléham, Nov 26 2009, Dec 14 2009

A002878 Bisection of Lucas sequence: a(n) = L(2*n+1).

Original entry on oeis.org

1, 4, 11, 29, 76, 199, 521, 1364, 3571, 9349, 24476, 64079, 167761, 439204, 1149851, 3010349, 7881196, 20633239, 54018521, 141422324, 370248451, 969323029, 2537720636, 6643838879, 17393796001, 45537549124, 119218851371, 312119004989, 817138163596, 2139295485799
Offset: 0

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Keywords

Comments

In any generalized Fibonacci sequence {f(i)}, Sum_{i=0..4n+1} f(i) = a(n)*f(2n+2). - Lekraj Beedassy, Dec 31 2002
The continued fraction expansion for F((2n+1)*(k+1))/F((2n+1)*k), k>=1 is [a(n),a(n),...,a(n)] where there are exactly k elements (F(n) denotes the n-th Fibonacci number). E.g., continued fraction for F(12)/F(9) is [4, 4,4]. - Benoit Cloitre, Apr 10 2003
See A135064 for a possible connection with Galois groups of quintics.
Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(5)). - Thomas Baruchel, Sep 15 2003
All positive integer solutions of Pell equation a(n)^2 - 5*b(n)^2 = -4 together with b(n)=A001519(n), n>=0.
a(n) = L(n,-3)*(-1)^n, where L is defined as in A108299; see also A001519 for L(n,+3).
Inverse binomial transform of A030191. - Philippe Deléham, Oct 04 2005
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) = x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
Let r = (2n+1), then a(n), n>0 = Product_{k=1..floor((r-1)/2)} (1 + sin^2 k*Pi/r); e.g., a(3) = 29 = (3.4450418679...)*(4.801937735...)*(1.753020396...). - Gary W. Adamson, Nov 26 2008
a(n+1) is the Hankel transform of A001700(n)+A001700(n+1). - Paul Barry, Apr 21 2009
a(n) is equal to the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(5)'s along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. - John M. Campbell, Jun 09 2011
Conjecture: for n > 0, a(n) = sqrt(Fibonacci(4*n+3) + Sum_{k=2..2*n} Fibonacci(2*k)). - Alex Ratushnyak, May 06 2012
Pisano period lengths: 1, 3, 4, 3, 2, 12, 8, 6, 12, 6, 5, 12, 14, 24, 4, 12, 18, 12, 9, 6, ... . - R. J. Mathar, Aug 10 2012
The continued fraction [a(n); a(n), a(n), ...] = phi^(2n+1), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 05 2013
Solutions (x, y) = (a(n), a(n+1)) satisfying x^2 + y^2 = 3xy + 5. - Michel Lagneau, Feb 01 2014
Conjecture: except for the number 3, a(n) are the numbers such that a(n)^2+2 are Lucas numbers. - Michel Lagneau, Jul 22 2014
Comment on the preceding conjecture: It is clear that all a(n) satisfy a(n)^2 + 2 = L(2*(2*n+1)) due to the identity (17 c) of Vajda, p. 177: L(2*n) + 2*(-1)^n = L(n)^2 (take n -> 2*n+1). - Wolfdieter Lang, Oct 10 2014
Limit_{n->oo} a(n+1)/a(n) = phi^2 = phi + 1 = (3+sqrt(5))/2. - Derek Orr, Jun 18 2015
If d[k] denotes the sequence of k-th differences of this sequence, then d[0](0), d[1](1), d[2](2), d[3](3), ... = A048876, cf. message to SeqFan list by P. Curtz on March 2, 2016. - M. F. Hasler, Mar 03 2016
a(n-1) and a(n) are the least phi-antipalindromic numbers (A178482) with 2*n and 2*n+1 digits in base phi, respectively. - Amiram Eldar, Jul 07 2021
Triangulate (hyperbolic) 2-space such that around every vertex exactly 7 triangles touch. Call any 7 triangles having a common vertex the first layer and let the (n+1)-st layer be all triangles that do not appear in any of the first n layers and have a common vertex with the n-th layer. Then the n-th layer contains 7*a(n-1) triangles. E.g., the first layer (by definition) contains 7 triangles, the second layer (the "ring" of triangles around the first layer) consists of 28 triangles, the third layer (the next "ring") consists of 77 triangles, and so on. - Nicolas Nagel, Aug 13 2022

Examples

			G.f. = 1 + 4*x + 11*x^2 + 29*x^3 + 76*x^4 + 199*x^5 + 521*x^6 + ... - _Michael Somos_, Jan 13 2019
		

References

  • J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
  • 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).
  • Steven Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

Crossrefs

Cf. A000204. a(n) = A060923(n, 0), a(n)^2 = A081071(n).
Cf. A005248 [L(2n) = bisection (even n) of Lucas sequence].
Cf. A001906 [F(2n) = bisection (even n) of Fibonacci sequence], A000045, A002315, A004146, A029907, A113224, A153387, A153416, A178482, A192425, A285992 (prime subsequence).
Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n listed in A264080.

Programs

  • GAP
    List([0..40], n-> Lucas(1,-1,2*n+1)[2] ); # G. C. Greubel, Jul 15 2019
    
  • Haskell
    a002878 n = a002878_list !! n
    a002878_list = zipWith (+) (tail a001906_list) a001906_list
    -- Reinhard Zumkeller, Jan 11 2012
    
  • Magma
    [Lucas(2*n+1): n in [0..40]]; // Vincenzo Librandi, Apr 16 2011
    
  • Maple
    A002878 := proc(n)
        option remember;
        if n <= 1 then
            op(n+1,[1,4]);
        else
            3*procname(n-1)-procname(n-2) ;
        end if;
    end proc: # R. J. Mathar, Apr 30 2017
  • Mathematica
    a[n_]:= FullSimplify[GoldenRatio^n - GoldenRatio^-n]; Table[a[n], {n, 1, 40, 2}]
    a[1]=1; a[2]=4; a[n_]:=a[n]= 3a[n-1] -a[n-2]; Array[a, 40]
    LinearRecurrence[{3, -1}, {1, 4}, 41] (* Jean-François Alcover, Sep 23 2017 *)
    Table[Sum[(-1)^Floor[k/2] Binomial[n -Floor[(k+1)/2], Floor[k/2]] 3^(n - k), {k, 0, n}], {n, 0, 40}] (* L. Edson Jeffery, Feb 26 2018 *)
    a[ n_] := Fibonacci[2n] + Fibonacci[2n+2]; (* Michael Somos, Jul 31 2018 *)
    a[ n_]:= LucasL[2n+1]; (* Michael Somos, Jan 13 2019 *)
  • PARI
    a(n)=fibonacci(2*n)+fibonacci(2*n+2) \\ Charles R Greathouse IV, Jun 16 2011
    
  • PARI
    for(n=1,40,q=((1+sqrt(5))/2)^(2*n-1);print1(contfrac(q)[1],", ")) \\ Derek Orr, Jun 18 2015
    
  • PARI
    Vec((1+x)/(1-3*x+x^2) + O(x^40)) \\ Altug Alkan, Oct 26 2015
    
  • Python
    a002878 = [1, 4]
    for n in range(30): a002878.append(3*a002878[-1] - a002878[-2])
    print(a002878) # Gennady Eremin, Feb 05 2022
  • Sage
    [lucas_number2(2*n+1,1,-1) for n in (0..40)] # G. C. Greubel, Jul 15 2019
    

Formula

a(n+1) = 3*a(n) - a(n-1).
G.f.: (1+x)/(1-3*x+x^2). - Simon Plouffe in his 1992 dissertation
a(n) = S(2*n, sqrt(5)) = S(n, 3) + S(n-1, 3); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 3) = A001906(n+1) (even-indexed Fibonacci numbers).
a(n) ~ phi^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then (-1)^n*q(n, -1) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = A005248(n+1) - A005248(n) = -1 + Sum_{k=0..n} A005248(k). - Lekraj Beedassy, Dec 31 2002
a(n) = 2^(-n)*A082762(n) = 4^(-n)*Sum_{k>=0} binomial(2*n+1, 2*k)*5^k; see A091042. - Philippe Deléham, Mar 01 2004
a(n) = (-1)^n*Sum_{k=0..n} (-5)^k*binomial(n+k, n-k). - Benoit Cloitre, May 09 2004
From Paul Barry, May 27 2004: (Start)
Both bisection and binomial transform of A000204.
a(n) = Fibonacci(2n) + Fibonacci(2n+2). (End)
Sequence lists the numerators of sinh((2*n-1)*psi) where the denominators are 2; psi=log((1+sqrt(5))/2). Offset 1. a(3)=11. - Al Hakanson (hawkuu(AT)gmail.com), Mar 25 2009
a(n) = A001906(n) + A001906(n+1). - Reinhard Zumkeller, Jan 11 2012
a(n) = floor(phi^(2n+1)), where phi is the golden ratio, A001622. - Thomas Ordowski, Jun 10 2012
a(n) = A014217(2*n+1) = A014217(2*n+2) - A014217(2*n). - Paul Curtz, Jun 11 2013
Sum_{n >= 0} 1/(a(n) + 5/a(n)) = 1/2. Compare with A005248, A001906, A075796. - Peter Bala, Nov 29 2013
a(n) = lim_{m->infinity} Fibonacci(m)^(4n+1)*Fibonacci(m+2*n+1)/ Sum_{k=0..m} Fibonacci(k)^(4n+2). - Yalcin Aktar, Sep 02 2014
From Peter Bala, Mar 22 2015: (Start)
The aerated sequence (b(n))n>=1 = [1, 0, 4, 0, 11, 0, 29, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -1, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
b(n) = (1/2)*((-1)^n - 1)*F(n) + (1 + (-1)^(n-1))*F(n+1), where F(n) is a Fibonacci number. The o.g.f. is x*(1 + x^2)/(1 - 3*x^2 + x^4).
Exp( Sum_{n >= 1} 2*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*x^n.
Exp( Sum_{n >= 1} (-2)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 2*F(n)*(-x)^n.
Exp( Sum_{n >= 1} 4*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*x^n.
Exp( Sum_{n >= 1} (-4)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 4*A029907(n)*(-x)^n. Cf. A002315, A004146, A113224 and A192425. (End)
a(n) = sqrt(5*F(2*n+1)^2-4), where F(n) = A000045(n). - Derek Orr, Jun 18 2015
For n > 1, a(n) = 5*F(2*n-1) + L(2*n-3) with F(n) = A000045(n). - J. M. Bergot, Oct 25 2015
For n > 0, a(n) = L(n-1)*L(n+2) + 4*(-1)^n. - J. M. Bergot, Oct 25 2015
For n > 2, a(n) = a(n-2) + F(n+2)^2 + F(n-3)^2 = L(2*n-3) + F(n+2)^2 + F(n-3)^2. - J. M. Bergot, Feb 05 2016 and Feb 07 2016
E.g.f.: ((sqrt(5) - 5)*exp((3-sqrt(5))*x/2) + (5 + sqrt(5))*exp((3+sqrt(5))*x/2))/(2*sqrt(5)). - Ilya Gutkovskiy, Apr 24 2016
a(n) = Sum_{k=0..n} (-1)^floor(k/2)*binomial(n-floor((k+1)/2), floor(k/2))*3^(n-k). - L. Edson Jeffery, Feb 26 2018
a(n)*F(m+2n-1) = F(m+4n-2)-F(m), with Fibonacci number F(m), empirical observation. - Dan Weisz, Jul 30 2018
a(n) = -a(-1-n) for all n in Z. - Michael Somos, Jul 31 2018
Sum_{n>=0} 1/a(n) = A153416. - Amiram Eldar, Nov 11 2020
a(n) = Product_{k=1..n} (1 + 4*sin(2*k*Pi/(2*n+1))^2). - Seiichi Manyama, Apr 30 2021
Sum_{n>=0} (-1)^n/a(n) = (1/sqrt(5)) * A153387 (Carlitz, 1967). - Amiram Eldar, Feb 05 2022
The continued fraction [a(n);a(n),a(n),...] = phi^(2*n+1), with phi = A001622. - A.H.M. Smeets, Feb 25 2022
a(n) = 2*sinh((2*n + 1)*arccsch(2)). - Peter Luschny, May 25 2022
This gives the sequence with 2 1's prepended: b(1)=b(2)=1 and, for k >= 3, b(k) = Sum_{j=1..k-2} (2^(k-j-1) - 1)*b(j). - Neal Gersh Tolunsky, Oct 28 2022 (formula due to Jon E. Schoenfield)
For n > 0, a(n) = 1 + 1/(Sum_{k>=1} F(k)/phi^(2*n*k + k)). - Diego Rattaggi, Nov 08 2023
From Peter Bala, Apr 16 2025: (Start)
a(3*n+1) = a(n)^3 + 3*a(n).
a(5*n+2) = a(n)^5 + 5*a(n)^3 + 5*a(n).
a(7*n+3) = a(n)^7 + 7*a(n)^5 + 14*a(n)^3 + 7*a(n).
For the coefficients see A034807.
The general result is: for k >= 0, a(k*n + (k-1)/2) = 2 * T(k, a(n)/2), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind and a(n) = ((1 + sqrt(5))/2)^(2*n+1) + ((1 - sqrt(5))/2)^(2*n+1).
Sum_{n >= 0} (-1)^n/a(n) = (1/4)* (theta_3(phi) - theta_3(phi^2)) = 0.815947983588122..., where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122) and phi = (sqrt(5) - 1)/2. See Borwein and Borwein, Exercise 3 a, p. 94 and Carlitz, 1967. (End)
From Peter Bala, May 15 2025: (Start)
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/5 (telescoping series: 5/(a(n) - 1/a(n)) = 1/A001906(n+1) + 1/A001906(n) ).
More generally, for k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - s(k)/a(k*n)) = 1/(1 + a(k)) where s(k) = a(0) + a(1) + ... + a(k-1) = Lucas(2*k) - 2.
For k >= 1, Sum_{n >= 1} (-1)^(n+1)/(a(n) + L(2*k)^2/a(n)) = (1/5) * A064170(k+2).
Sum_{n >= 1} 1/(a(n) + 9/a(n)) = 3/10 (follows from 1/(a(n) + 9/a(n)) = L(2*n)/A081076(n) - L(2*n+2)/A081076(n+1) ).
More generally, it appears that for k >= 1, Sum_{n >= 1} 1/(a(n) + L(2*k)^2/a(n)) is rational.
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(5) [telescoping product: Product_{k = 1..n} ((a(k) + 1)/(a(k) - 1))^2 = 5*(1 - 4/A240926(n+1)) ]. (End)

Extensions

Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

A036987 Fredholm-Rueppel sequence.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 0

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Author

Keywords

Comments

Binary representation of the Kempner-Mahler number Sum_{k>=0} 1/2^(2^k) = A007404.
a(n) = (product of digits of n; n in binary notation) mod 2. This sequence is a transformation of the Thue-Morse sequence (A010060), since there exists a function f such that f(sum of digits of n) = (product of digits of n). - Ctibor O. Zizka, Feb 12 2008
a(n-1), n >= 1, the characteristic sequence for powers of 2, A000079, is the unique solution of the following formal product and formal power series identity: Product_{j>=1} (1 + a(j-1)*x^j) = 1 + Sum_{k>=1} x^k = 1/(1-x). The product is therefore Product_{l>=1} (1 + x^(2^l)). Proof. Compare coefficients of x^n and use the binary representation of n. Uniqueness follows from the recurrence relation given for the general case under A147542. - Wolfdieter Lang, Mar 05 2009
a(n) is also the number of orbits of length n for the map x -> 1-cx^2 on [-1,1] at the Feigenbaum critical value c=1.401155... . - Thomas Ward, Apr 08 2009
A054525 (Mobius transform) * A001511 = A036987 = A047999^(-1) * A001511 = the inverse of Sierpiński's gasket * the ruler sequence. - Gary W. Adamson, Oct 26 2009 [Of course this is only vaguely correct depending on how the fuzzy indexing in these formulas is made concrete. - R. J. Mathar, Jun 20 2014]
Characteristic function of A000225. - Reinhard Zumkeller, Mar 06 2012
Also parity of the Catalan numbers A000108. - Omar E. Pol, Jan 17 2012
For n >= 2, also the largest exponent k >= 0 such that n^k in binary notation does not contain both 0 and 1. Unlike for the decimal version of this sequence, A062518, where the terms are only conjectural, for this sequence the values of a(n) can be proved to be the characteristic function of A000225, as follows: n^k will contain both 0 and 1 unless n^k = 2^r-1 for some r. But this is a special case of Catalan's equation x^p = y^q-1, which was proved by Preda Mihăilescu to have no nontrivial solution except 2^3 = 3^2 - 1. - Christopher J. Smyth, Aug 22 2014
Image, under the coding a,b -> 1; c -> 0, of the fixed point, starting with a, of the morphism a -> ab, b -> cb, c -> cc. - Jeffrey Shallit, May 14 2016
Number of nonisomorphic Boolean algebras of order n+1. - Jianing Song, Jan 23 2020

Examples

			G.f. = 1 + x + x^3 + x^7 + x^15 + x^31 + x^63 + x^127 + x^255 + x^511 + ...
a(7) = 1 since 7 = 2^3 - 1, while a(10) = 0 since 10 is not of the form 2^k - 1 for any integer k.
		

Crossrefs

The first row of A073346. Occurs for first time in A073202 as row 6 (and again as row 8).
Congruent to any of the sequences A000108, A007460, A007461, A007463, A007464, A061922, A068068 reduced modulo 2. Characteristic function of A000225.
If interpreted with offset=1 instead of 0 (i.e., a(1)=1, a(2)=1, a(3)=0, a(4)=1, ...) then this is the characteristic function of 2^n (A000079) and as such occurs as the first row of A073265. Also, in that case the INVERT transform will produce A023359.
This is Guy Steele's sequence GS(1, 3), also GS(3, 1) (see A135416).
Cf. A054525, A047999. - Gary W. Adamson, Oct 26 2009

Programs

  • Haskell
    a036987 n = ibp (n+1) where
       ibp 1 = 1
       ibp n = if r > 0 then 0 else ibp n' where (n',r) = divMod n 2
    a036987_list = 1 : f [0,1] where f (x:y:xs) = y : f (x:xs ++ [x,x+y])
    -- Same list generator function as for a091090_list, cf. A091090.
    -- Reinhard Zumkeller, May 19 2015, Apr 13 2013, Mar 13 2013
    
  • Maple
    A036987:= n-> `if`(2^ilog2(n+1) = n+1, 1, 0):
    seq(A036987(n), n=0..128);
  • Mathematica
    RealDigits[ N[ Sum[1/10^(2^n), {n, 0, Infinity}], 110]][[1]]
    (* Recurrence: *)
    t[n_, 1] = 1; t[1, k_] = 1;
    t[n_, k_] := t[n, k] =
      If[n < k, If[n > 1 && k > 1, -Sum[t[k - i, n], {i, 1, n - 1}], 0],
       If[n > 1 && k > 1, Sum[t[n - i, k], {i, 1, k - 1}], 0]];
    Table[t[n, k], {k, n, n}, {n, 104}]
    (* Mats Granvik, Jun 03 2011 *)
    mb2d[n_]:=1 - Module[{n2 = IntegerDigits[n, 2]}, Max[n2] - Min[n2]]; Array[mb2d, 120, 0] (* Vincenzo Librandi, Jul 19 2019 *)
    Table[PadRight[{1},2^k,0],{k,0,7}]//Flatten (* Harvey P. Dale, Apr 23 2022 *)
  • PARI
    {a(n) =( n++) == 2^valuation(n, 2)}; /* Michael Somos, Aug 25 2003 */
    
  • PARI
    a(n) = !bitand(n, n+1); \\ Ruud H.G. van Tol, Apr 05 2023
    
  • Python
    from sympy import catalan
    def a(n): return catalan(n)%2 # Indranil Ghosh, May 25 2017
    
  • Python
    def A036987(n): return int(not(n&(n+1))) # Chai Wah Wu, Jul 06 2022

Formula

1 followed by a string of 2^k - 1 0's. Also a(n)=1 iff n = 2^m - 1.
a(n) = a(floor(n/2)) * (n mod 2) for n>0 with a(0)=1. - Reinhard Zumkeller, Aug 02 2002 [Corrected by Mikhail Kurkov, Jul 16 2019]
Sum_{n>=0} 1/10^(2^n) = 0.110100010000000100000000000000010...
1 if n=0, floor(log_2(n+1)) - floor(log_2(n)) otherwise. G.f.: (1/x) * Sum_{k>=0} x^(2^k) = Sum_{k>=0} x^(2^k-1). - Ralf Stephan, Apr 28 2003
a(n) = 1 - A043545(n). - Michael Somos, Aug 25 2003
a(n) = -Sum_{d|n+1} mu(2*d). - Benoit Cloitre, Oct 24 2003
Dirichlet g.f. for right-shifted sequence: 2^(-s)/(1-2^(-s)).
a(n) = A000108(n) mod 2 = A001405(n) mod 2. - Paul Barry, Nov 22 2004
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n,k)*Sum_{j=0..k} binomial(k, 2^j-1). - Paul Barry, Jun 01 2006
A000523(n+1) = Sum_{k=1..n} a(k). - Mitch Harris, Jul 22 2011
a(n) = A209229(n+1). - Reinhard Zumkeller, Mar 07 2012
a(n) = Sum_{k=1..n} A191898(n,k)*cos(Pi*(n-1)*(k-1))/n; (conjecture). - Mats Granvik, Mar 04 2013
a(n) = A000035(A000108(n)). - Omar E. Pol, Aug 06 2013
a(n) = 1 iff n=2^k-1 for some k, 0 otherwise. - M. F. Hasler, Jun 20 2014
a(n) = ceiling(log_2(n+2)) - ceiling(log_2(n+1)). - Gionata Neri, Sep 06 2015
From John M. Campbell, Jul 21 2016: (Start)
a(n) = (A000168(n-1) mod 2).
a(n) = (A000531(n+1) mod 2).
a(n) = (A000699(n+1) mod 2).
a(n) = (A000891(n) mod 2).
a(n) = (A000913(n-1) mod 2), for n>1.
a(n) = (A000917(n-1) mod 2), for n>0.
a(n) = (A001142(n) mod 2).
a(n) = (A001246(n) mod 2).
a(n) = (A001246(n) mod 4).
a(n) = (A002057(n-2) mod 2), for n>1.
a(n) = (A002430(n+1) mod 2). (End)
a(n) = 2 - A043529(n). - Antti Karttunen, Nov 19 2017
a(n) = floor(1+log(n+1)/log(2)) - floor(log(2n+1)/log(2)). - Adriano Caroli, Sep 22 2019
This is also the decimal expansion of -Sum_{k>=1} mu(2*k)/(10^k - 1), where mu is the Möbius function (A008683). - Amiram Eldar, Jul 12 2020

Extensions

Edited by M. F. Hasler, Jun 20 2014

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

Original entry on oeis.org

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

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Author

Keywords

Comments

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

Examples

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

References

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

Crossrefs

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

Programs

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

Formula

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

Extensions

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

A005773 Number of directed animals of size n (or directed n-ominoes in standard position).

Original entry on oeis.org

1, 1, 2, 5, 13, 35, 96, 267, 750, 2123, 6046, 17303, 49721, 143365, 414584, 1201917, 3492117, 10165779, 29643870, 86574831, 253188111, 741365049, 2173243128, 6377181825, 18730782252, 55062586341, 161995031226, 476941691177, 1405155255055, 4142457992363
Offset: 0

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This sequence, with first term a(0) deleted, appears to be determined by the conditions that the diagonal and first superdiagonal of U are {1,1,1,1,...} and {2,3,4,5,...,n+1,...} respectively, where A=LU is the LU factorization of the Hankel matrix A given by [{a(1),a(2),...}, {a(2),a(3),...}, ..., {a(n),a(n+1),...}, ...]. - John W. Layman, Jul 21 2000
Also the number of base 3 n-digit numbers (not starting with 0) with digit sum n. For the analogous sequence in base 10 see A071976, see example. - John W. Layman, Jun 22 2002
Also number of paths in an n X n grid from (0,0) to the line x=n-1, using only steps U=(1,1), H=(1,0) and D=(1,-1) (i.e., left factors of length n-1 of Motzkin paths, palindromic Motzkin paths of length 2n-2 or 2n-1). Example: a(3)=5, namely, HH, UD, HU, UH and UU. Also number of ordered trees with n edges and having nonroot nodes of outdegree at most 2. - Emeric Deutsch, Aug 01 2002
Number of symmetric Dyck paths of semilength 2n-1 with no peaks at even level. Example: a(3)=5 because we have UDUDUDUDUD, UDUUUDDDUD, UUUUUDDDDD, UUUDUDUDDD and UUUDDUUDDD, where U=(1,1) and D=(1,-1). Also number of symmetric Dyck paths of semilength 2n with no peaks at even level. Example: a(3)=5 because we have UDUDUDUDUDUD, UDUUUDUDDDUD, UUUDUDUDUDDD, UUUUUDUDDDDD and UUUDDDUUUDDD. - Emeric Deutsch, Nov 21 2003
a(n) is the sum of the (n-1)-st central trinomial coefficient and its predecessor. Example: a(4) = 6 + 7 and (1 + x + x^2)^3 = ... + 6*x^2 + 7*x^3 + ... . - David Callan, Feb 07 2004
a(n) is the number of UDU-free paths of n upsteps (U) and n downsteps (D) that start U (n>=1). Example: a(2)=2 counts UUDD, UDDU. - David Callan, Aug 18 2004
a(n) is also the number of Grand-Dyck paths of semilength n starting with an up-step and avoiding the pattern DUD. - David Bevan, Nov 19 2019
Hankel transform of a(n+1) = [1,2,5,13,35,96,...] gives A000012 = [1,1,1,1,1,1,...]. - Philippe Deléham, Oct 24 2007
Equals row sums of triangle A136787 starting (1, 2, 5, 13, 35, ...). - Gary W. Adamson, Jan 21 2008
a(n) is the number of permutations on [n] that avoid the patterns 1-23-4 and 1-3-2, where the omission of a dash in a pattern means the permutation entries must be adjacent. Example: a(4) = 13 counts all 14 (Catalan number) (1-3-2)-avoiding permutations on [4] except 1234. - David Callan, Jul 22 2008
a(n) is also the number of involutions of length 2n-2 which are invariant under the reverse-complement map and have no decreasing subsequences of length 4. - Eric S. Egge, Oct 21 2008
Hankel transform is A010892. - Paul Barry, Jan 19 2009
a(n) is the number of Dyck words of semilength n with no DUUU. For example, a(4) = 14-1 = 13 because there is only one Dyck 4-word containing DUUU, namely UDUUUDDD. - Eric Rowland, Apr 21 2009
Inverse binomial transform of A024718. - Philippe Deléham, Dec 13 2009
Let w(i, j, n) denote walks in N^2 which satisfy the multivariate recurrence
w(i, j, n) = w(i - 1, j, n - 1) + w(i, j - 1, n - 1) + w(i + 1, j - 1,n - 1) with boundary conditions w(0,0,0) = 1 and w(i,j,n) = 0 if i or j or n is < 0. Let alpha(n) the number of such walks of length n, alpha(n) = Sum_{i = 0..n, j=0..n} w(i, j, n). Then a(n+1) = alpha(n). - Peter Luschny, May 21 2011
Number of length-n strings [d(0),d(1),d(2),...,d(n-1)] where 0 <= d(k) <= k and abs(d(k) - d(k-1)) <= 1 (smooth factorial numbers, see example). - Joerg Arndt, Nov 10 2012
a(n) is the number of n-multisets of {1,...,n} containing no pair of consecutive integers (e.g., 111, 113, 133, 222, 333 for n=3). - David Bevan, Jun 10 2013
a(n) is also the number of n-multisets of [n] in which no integer except n occurs exactly once (e.g., 111, 113, 222, 223, 333 for n=3). - David Bevan, Nov 19 2019
Number of minimax elements in the affine Weyl group of the Lie algebra so(2n+1) or the Lie algebra sp(2n). See Panyushev 2005. Cf. A245455. - Peter Bala, Jul 22 2014
The shifted, signed array belongs to an interpolated family of arrays associated to the Catalan A000108 (t=1), and Riordan, or Motzkin sums A005043 (t=0), with the interpolating (here t=-2) o.g.f. G(x,t) = (1-sqrt(1-4x/(1+(1-t)x)))/2 and inverse o.g.f. Ginv(x,t) = x(1-x)/(1+(t-1)x(1-x)) (A057682). See A091867 for more info on this family. - Tom Copeland, Nov 09 2014
Alternatively, this sequence corresponds to the number of positive walks with n steps {-1,0,1} starting at the origin, ending at any altitude, and staying strictly above the x-axis. - David Nguyen, Dec 01 2016
Let N be a squarefree number with n prime factors: p_1 < p_2 < ... < p_n. Let D be its set of divisors, E the subset of D X D made of the (d_1, d_2) for which, provided that we know which p_i are in d_1, which p_i are in d_2, d_1 <= d_2 is provable without needing to know the numerical values of the p_i. It appears that a(n+1) is the number of (d_1, d_2) in E such that d_1 and d_2 are coprime. - Luc Rousseau, Aug 21 2017
Number of ordered rooted trees with n non-root nodes and all non-root nodes having outdegrees 1 or 2. - Andrew Howroyd, Dec 04 2017
a(n) is the number of compositions (ordered partitions) of n where there are A001006(k-1) sorts of part k (see formula by Andrew Howroyd, Dec 04 2017). - Joerg Arndt, Jan 26 2024

Examples

			G.f. = 1 + x + 2*x^2 + 5*x^3 + 13*x^4 + 35*x^5 + 96*x^6 + 267*x^7 + ...
a(3) = 5, a(4) = 13; since the top row of M^3 = (5, 5, 2, 1, ...)
From _Eric Rowland_, Sep 25 2021: (Start)
There are a(4) = 13 directed animals of size 4:
  O
  O    O    O    OO              O         O
  O    O    OO   O    OO   O    OO   OOO   O    O    OO    O
  O    OO   O    O    OO   OOO  O    O    OO   OOO  OO   OOO  OOOO
(End)
From _Joerg Arndt_, Nov 10 2012: (Start)
There are a(4)=13 smooth factorial numbers of length 4 (dots for zeros):
[ 1]   [ . . . . ]
[ 2]   [ . . . 1 ]
[ 3]   [ . . 1 . ]
[ 4]   [ . . 1 1 ]
[ 5]   [ . . 1 2 ]
[ 6]   [ . 1 . . ]
[ 7]   [ . 1 . 1 ]
[ 8]   [ . 1 1 . ]
[ 9]   [ . 1 1 1 ]
[10]   [ . 1 1 2 ]
[11]   [ . 1 2 1 ]
[12]   [ . 1 2 2 ]
[13]   [ . 1 2 3 ]
(End)
From _Joerg Arndt_, Nov 22 2012: (Start)
There are a(4)=13 base 3 4-digit numbers (not starting with 0) with digit sum 4:
[ 1]   [ 2 2 . . ]
[ 2]   [ 2 1 1 . ]
[ 3]   [ 1 2 1 . ]
[ 4]   [ 2 . 2 . ]
[ 5]   [ 1 1 2 . ]
[ 6]   [ 2 1 . 1 ]
[ 7]   [ 1 2 . 1 ]
[ 8]   [ 2 . 1 1 ]
[ 9]   [ 1 1 1 1 ]
[10]   [ 1 . 2 1 ]
[11]   [ 2 . . 2 ]
[12]   [ 1 1 . 2 ]
[13]   [ 1 . 1 2 ]
(End)
		

References

  • J. E. Goodman and J. O'Rourke, editors, Handbook of Discrete and Computational Geometry, CRC Press, 1997, p. 237.
  • T. Mansour, Combinatorics of Set Partitions, Discrete Mathematics and Its Applications, CRC Press, 2013, p. 377.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 6.46a.
  • R. P. Stanley, Catalan Numbers, Cambridge, 2015, p. 132.

Crossrefs

See also A005775. Inverse of A001006. Also sum of numbers in row n+1 of array T in A026300. Leading column of array in A038622.
The right edge of the triangle A062105.
Column k=3 of A295679.
Interpolates between Motzkin numbers (A001006) and Catalan numbers (A000108). Cf. A054391, A054392, A054393, A055898.
Except for the first term a(0), sequence is the binomial transform of A001405.
a(n) = A002426(n-1) + A005717(n-1) if n > 0. - Emeric Deutsch, Aug 14 2002

Programs

  • Haskell
    a005773 n = a005773_list !! n
    a005773_list = 1 : f a001006_list [] where
       f (x:xs) ys = y : f xs (y : ys) where
         y = x + sum (zipWith (*) a001006_list ys)
    -- Reinhard Zumkeller, Mar 30 2012
    
  • Magma
    R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( 2*x/(3*x-1+Sqrt(1-2*x-3*x^2)) )); // G. C. Greubel, Apr 05 2019
  • Maple
    seq( sum(binomial(i-1, k)*binomial(i-k, k), k=0..floor(i/2)), i=0..30 ); # Detlef Pauly (dettodet(AT)yahoo.de), Nov 09 2001
    A005773:=proc(n::integer)
    local i, j, A, istart, iend, KartProd, Liste, Term, delta;
        A:=0;
        for i from 0 to n do
            Liste[i]:=NULL;
            istart[i]:=0;
            iend[i]:=n-i+1:
            for j from istart[i] to iend[i] do
                Liste[i]:=Liste[i], j;
            end do;
            Liste[i]:=[Liste[i]]:
        end do;
        KartProd:=cartprod([seq(Liste[i], i=1..n)]);
        while not KartProd[finished] do
            Term:=KartProd[nextvalue]();
            delta:=1;
            for i from 1 to n-1 do
                if (op(i, Term) - op(i+1, Term))^2 >= 2 then
                    delta:=0;
                    break;
                end if;
            end do;
            A:=A+delta;
        end do;
    end proc; # Thomas Wieder, Feb 22 2009:
    # n -> [a(0),a(1),..,a(n)]
    A005773_list := proc(n) local W, m, j, i;
    W := proc(i, j, n) option remember;
    if min(i, j, n) < 0 or max(i, j) > n then 0
    elif n = 0 then if i = 0 and j = 0 then 1 else 0 fi
    else W(i-1,j,n-1)+W(i,j-1,n-1)+W(i+1,j-1,n-1) fi end:
    [1,seq(add(add(W(i,j,m),i=0..m),j=0..m),m=0..n-1)] end:
    A005773_list(27); # Peter Luschny, May 21 2011
    A005773 := proc(n)
        option remember;
        if n <= 1 then
            1 ;
        else
            2*n*procname(n-1)+3*(n-2)*procname(n-2) ;
            %/n ;
        end if;
    end proc:
    seq(A005773(n),n=0..10) ; # R. J. Mathar, Jul 25 2017
  • Mathematica
    CoefficientList[Series[(2x)/(3x-1+Sqrt[1-2x-3x^2]), {x,0,40}], x] (* Harvey P. Dale, Apr 03 2011 *)
    a[0]=1; a[n_] := Sum[k/n*Sum[Binomial[n, j]*Binomial[j, 2*j-n-k], {j, 0, n}], {k, 1, n}]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Mar 31 2015, after Vladimir Kruchinin *)
    A005773[n_] := 2 (-1)^(n+1) JacobiP[n - 1, 3, -n -1/2, -7] / (n^2 + n); A005773[0] := 1; Table[A005773[n], {n, 0, 27}] (* Peter Luschny, May 25 2021 *)
  • PARI
    a(n)=if(n<2,n>=0,(2*n*a(n-1)+3*(n-2)*a(n-2))/n)
    
  • PARI
    for(n=0, 27, print1(if(n==0, 1, sum(k=0, n-1, (-1)^(n - 1 + k)*binomial(n - 1, k)*binomial(2*k + 1, k + 1))),", ")) \\ Indranil Ghosh, Mar 14 2017
    
  • PARI
    Vec(1/(1-serreverse(x*(1-x)/(1-x^3) + O(x*x^25)))) \\ Andrew Howroyd, Dec 04 2017
    
  • Sage
    def da():
        a, b, c, d, n = 0, 1, 1, -1, 1
        yield 1
        yield 1
        while True:
            yield b + (-1)^n*d
            n += 1
            a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)//((n+1)*(n-1))
            c, d = d, (3*(n-1)*c-(2*n-1)*d)//n
    A005773 = da()
    print([next(A005773) for  in range(28)]) # _Peter Luschny, May 16 2016
    
  • Sage
    (2*x/(3*x-1+sqrt(1-2*x-3*x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Apr 05 2019
    

Formula

G.f.: 2*x/(3*x-1+sqrt(1-2*x-3*x^2)). - Len Smiley
Also a(0)=1, a(n) = Sum_{k=0..n-1} M(k)*a(n-k-1), where M(n) are the Motzkin numbers (A001006).
D-finite with recurrence n*a(n) = 2*n*a(n-1) + 3*(n-2)*a(n-2), a(0)=a(1)=1. - Michael Somos, Feb 02 2002
G.f.: 1/2+(1/2)*((1+x)/(1-3*x))^(1/2). Related to Motzkin numbers A001006 by a(n+1) = 3*a(n) - A001006(n-1) [see Yaqubi Lemma 2.6].
a(n) = Sum_{q=0..n} binomial(q, floor(q/2))*binomial(n-1, q) for n > 0. - Emeric Deutsch, Aug 15 2002
From Paul Barry, Jun 22 2004: (Start)
a(n+1) = Sum_{k=0..n} (-1)^(n+k)*C(n, k)*C(2*k+1, k+1).
a(n) = 0^n + Sum_{k=0..n-1} (-1)^(n+k-1)*C(n-1, k)*C(2*k+1, k+1). (End)
a(n+1) = Sum_{k=0..n} (-1)^k*3^(n-k)*binomial(n, k)*A000108(k). - Paul Barry, Jan 27 2005
Starting (1, 2, 5, 13, ...) gives binomial transform of A001405 and inverse binomial transform of A001700. - Gary W. Adamson, Aug 31 2007
Starting (1, 2, 5, 13, 35, 96, ...) gives row sums of triangle A132814. - Gary W. Adamson, Aug 31 2007
G.f.: 1/(1-x/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-... (continued fraction). - Paul Barry, Jan 19 2009
G.f.: 1+x/(1-2*x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-.... (continued fraction). - Paul Barry, Jan 19 2009
a(n) = Sum_{l_1=0..n+1} Sum_{l_2=0..n}...Sum_{l_i=0..n-i}...Sum_{l_n=0..1} delta(l_1,l_2,...,l_i,...,l_n) where delta(l_1,l_2,...,l_i,...,l_n) = 0 if any (l_i - l_(i+1))^2 >= 2 for i=1..n-1 and delta(l_1,l_2,..., l_i,...,l_n) = 1 otherwise. - Thomas Wieder, Feb 25 2009
INVERT transform of offset Motzkin numbers (A001006): (a(n)){n>=1}=(1,1,2,4,9,21,...). - _David Callan, Aug 27 2009
A005773(n) = ((n+3)*A001006(n+1) + (n-3)*A001006(n)) * (n+2)/(18*n) for n > 0. - Mark van Hoeij, Jul 02 2010
a(n) = Sum_{k=1..n} (k/n * Sum_{j=0..n} binomial(n,j)*binomial(j,2*j-n-k)). - Vladimir Kruchinin, Sep 06 2010
a(0) = 1; a(n+1) = Sum_{t=0..n} n!/((n-t)!*ceiling(t/2)!*floor(t/2)!). - Andrew S. Hays, Feb 02 2011
a(n) = leftmost column term of M^n*V, where M = an infinite quadradiagonal matrix with all 1's in the main, super and subdiagonals, [1,0,0,0,...] in the diagonal starting at position (2,0); and rest zeros. V = vector [1,0,0,0,...]. - Gary W. Adamson, Jun 16 2011
From Gary W. Adamson, Jul 29 2011: (Start)
a(n) = upper left term of M^n, a(n+1) = sum of top row terms of M^n; M = an infinite square production matrix in which the main diagonal is (1,1,0,0,0,...) as follows:
1, 1, 0, 0, 0, 0, ...
1, 1, 1, 0, 0, 0, ...
1, 1, 0, 1, 0, 0, ...
1, 1, 1, 0, 1, 0, ...
1, 1, 1, 1, 0, 1, ...
1, 1, 1, 1, 1, 0, ... (End)
Limit_{n->oo} a(n+1)/a(n) = 3.0 = lim_{n->oo} (1 + 2*cos(Pi/n)). - Gary W. Adamson, Feb 10 2012
a(n) = A025565(n+1) / 2 for n > 0. - Reinhard Zumkeller, Mar 30 2012
With first term deleted: E.g.f.: a(n) = n! * [x^n] exp(x)*(BesselI(0, 2*x) + BesselI(1, 2*x)). - Peter Luschny, Aug 25 2012
G.f.: G(0)/2 + 1/2, where G(k) = 1 + 2*x*(4*k+1)/( (2*k+1)*(1+x) - x*(1+x)*(2*k+1)*(4*k+3)/(x*(4*k+3) + (1+x)*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 24 2013
a(n) ~ 3^(n-1/2)/sqrt(Pi*n). - Vaclav Kotesovec, Jul 30 2013
For n > 0, a(n) = (-1)^(n+1) * hypergeom([3/2, 1-n], [2], 4). - Vladimir Reshetnikov, Apr 25 2016
a(n) = GegenbauerC(n-2,-n+1,-1/2) + GegenbauerC(n-1,-n+1,-1/2) for n >= 1. - Peter Luschny, May 12 2016
0 = a(n)*(+9*a(n+1) + 18*a(n+2) - 9*a(n+3)) + a(n+1)*(-6*a(n+1) + 7*a(n+2) - 2*a(n+3)) + a(n+2)*(-2*a(n+2) + a(n+3)) for n >= 0. - Michael Somos, Dec 01 2016
G.f.: 1/(1-x*G(x)) where G(x) is g.f. of A001006. - Andrew Howroyd, Dec 04 2017
a(n) = (-1)^(n + 1)*2*JacobiP(n - 1, 3, -n - 1/2, -7)/(n^2 + n). - Peter Luschny, May 25 2021
a(n+1) = A005043(n) + 2*A005717(n) for n >= 1. - Peter Bala, Feb 11 2022
a(n) = Sum_{k=0..n-1} A064189(n-1,k) for n >= 1. - Alois P. Heinz, Aug 29 2022

A031443 Digitally balanced numbers: positive numbers that in base 2 have the same number of 0's as 1's.

Original entry on oeis.org

2, 9, 10, 12, 35, 37, 38, 41, 42, 44, 49, 50, 52, 56, 135, 139, 141, 142, 147, 149, 150, 153, 154, 156, 163, 165, 166, 169, 170, 172, 177, 178, 180, 184, 195, 197, 198, 201, 202, 204, 209, 210, 212, 216, 225, 226, 228, 232, 240, 527, 535, 539, 541, 542, 551
Offset: 1

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Keywords

Comments

Also numbers k such that the binary digital mean dm(2, k) = (Sum_{i=1..d} 2*d_i - 1) / (2*d) = 0, where d is the number of digits in the binary representation of k and d_i the individual digits. - Reikku Kulon, Sep 21 2008
From Reikku Kulon, Sep 29 2008: (Start)
Each run of values begins with 2^(2k + 1) + 2^(k + 1) - 2^k - 1. The initial values increase according to the sequence {2^(k - 1), 2^(k - 2), 2^(k - 3), ..., 2^(k - k)}.
After this, the values follow a periodic sequence of increases by successive powers of two with single odd values interspersed.
Each run ends with an odd increase followed by increases of {2^(k - k), ..., 2^(k - 2), 2^(k - 1), 2^k}, finally reaching 2^(2k + 2) - 2^(k + 1).
Similar behavior occurs in other bases. (End)
Numbers k such that A000120(k)/A070939(k) = 1/2. - Ctibor O. Zizka, Oct 15 2008
Subsequence of A053754; A179888 is a subsequence. - Reinhard Zumkeller, Jul 31 2010
A000120(a(n)) = A023416(a(n)); A037861(a(n)) = 0.
A001700 gives number of terms having length 2*n in binary representation: A001700(n-1) = #{m: A070939(a(m))=2*n}. - Reinhard Zumkeller, Jun 08 2011
The number of terms below 2^k is A079309(floor(k/2)) for k > 1. - Amiram Eldar, Nov 21 2020

Examples

			9 is a term because '1001' contains 2 '0's and 2 '1's.
		

Crossrefs

Subsequence of A053754.
Row n = 2 of A378000.
Terms of binary width n are enumerated by A001700.

Programs

  • Haskell
    -- See link, showing that Ulrich Schimkes formula provides a very efficient algorithm. Reinhard Zumkeller, Jun 15 2011
    
  • Magma
    [ n: n in [2..250] | Multiplicity({* z: z in Intseq(n,2) *}, 0) eq &+Intseq(n,2) ];  // Bruno Berselli, Jun 07 2011
    
  • Maple
    a:=proc(n) local nn, n1, n0: nn:=convert(n,base,2): n1:=add(nn[i],i=1..nops(nn)): n0:=nops(nn)-n1: if n0=n1 then n else end if end proc: seq(a(n), n = 1..240); # Emeric Deutsch, Jul 31 2008
  • Mathematica
    Select[Range[250],DigitCount[#,2,1]==DigitCount[#,2,0]&] (* Harvey P. Dale, Jul 22 2013 *)
    FromDigits[#,2]&/@DeleteCases[Flatten[Permutations/@Table[PadRight[{},2n,{1,0}],{n,5}],1],?(#[[1]]==0&)]//Sort (* _Harvey P. Dale, May 30 2016 *)
  • PARI
    for(n=1,100,b=binary(n); l=length(b); if(sum(i=1,l, component(b,i))==l/2,print1(n,",")))
    
  • PARI
    is(n)=hammingweight(n)==hammingweight(bitneg(n,#binary(n))) \\ Charles R Greathouse IV, Mar 29 2013
    
  • PARI
    is(n)=2*hammingweight(n)==exponent(n)+1 \\ Charles R Greathouse IV, Apr 18 2020
    
  • Perl
    for my $half ( 1 .. 4 ) {
      my $N = 2 * $half;  # only even widths apply
      my $vector = (1 << ($N-1)) | ((1 << ($N/2-1)) - 1);  # first key
      my $n = 1; $n *= $_ for 2 .. $N;    # N!
      my $d = 1; $d *= $_ for 2 .. $N/2;  # (N/2)!
      for (1 .. $n/($d*$d*2)) {
        print "$vector, ";
        my ($v, $d) = ($vector, 0);
        until ($v & 1 or !$v) { $d = ($d << 1)|1; $v >>= 1 }
        $vector += $d + 1 + (($v ^ ($v + 1)) >> 2);  # next key
      }
    } # Ruud H.G. van Tol, Mar 30 2014
    
  • Python
    from sympy.utilities.iterables import multiset_permutations
    A031443_list = [int('1'+''.join(p),2) for n in range(1,10) for p in multiset_permutations('0'*n+'1'*(n-1))] # Chai Wah Wu, Nov 15 2019

Formula

a(n+1) = a(n) + 2^k + 2^(m-1) - 1 + floor((2^(k+m) - 2^k)/a(n))*(2^(2*m) + 2^(m-1)) where k is the largest integer such that 2^k divides a(n) and m is the largest integer such that 2^m divides a(n)/2^k+1. - Ulrich Schimke (UlrSchimke(AT)aol.com)
A145037(a(n)) = 0. - Reikku Kulon, Oct 02 2008

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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Author

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

Views

Author

Keywords

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)
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