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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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

Views

Author

N. J. A. Sloane

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).

Links

Crossrefs

Cf. A000984 (first differences), A097933, A038874, A132310.
Cf. A006135, A006136, A045912.
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.
Cf. A000975, A024718, A057552, A079309, A327481.

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

A068555 Triangle read by rows in which row n contains (2i)!*(2j)!/(i!*j!*(i+j)!) for i + j = n, i = 0..n.

Original entry on oeis.org

1, 2, 2, 6, 2, 6, 20, 4, 4, 20, 70, 10, 6, 10, 70, 252, 28, 12, 12, 28, 252, 924, 84, 28, 20, 28, 84, 924, 3432, 264, 72, 40, 40, 72, 264, 3432, 12870, 858, 198, 90, 70, 90, 198, 858, 12870, 48620, 2860, 572, 220, 140, 140, 220, 572, 2860, 48620, 184756, 9724
Offset: 0

Views

Author

N. J. A. Sloane, Mar 23 2002

Keywords

Comments

One of three infinite families of integral factorial ratio sequences of height 1 (see Bober, Theorem 1.2). The other two are A007318 and A046521. A related table is A182073. - Peter Bala, Apr 10 2012

Examples

			From _Bruno Berselli_, Apr 27 2012: (Start)
Triangle begins:
       1;
       2,    2;
       6,    2,    6;
      20,    4,    4,  20;
      70,   10,    6,  10,  70;
     252,   28,   12,  12,  28, 252;
     924,   84,   28,  20,  28,  84, 924;
    3432,  264,   72,  40,  40,  72, 264, 3432;
   12870,  858,  198,  90,  70,  90, 198,  858, 12870;
   48620, 2860,  572, 220, 140, 140, 220,  572,  2860, 48620;
  184756, 9724, 1716, 572, 308, 252, 308,  572,  1716,  9724, 184756; ...
(End)
T(4,0) = A000984(4) = 70, T(4,1) = 4*20 - 70 = 10, T(4,2) = 4*4 - 10 = 6, T(4,3) = 4*4 - 6 = 10, T(4,4) = 4*20 - 10 = 70. - _Philippe Deléham_, Mar 10 2014

References

  • R. K. Guy and Cal Long, Email to N. J. A. Sloane, Feb 22, 2002.
  • Peter J. Larcombe and David R. French, On the integrality of the Catalan-Larcombe-French sequence 1,8,80,896,10816,.... Proceedings of the Thirty-second Southeastern International Conference on Combinatorics, Graph Theory and Computing (Baton Rouge, LA, 2001). Congr. Numer. 148 (2001), 65-91. MR1887375
  • 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.

Links

Crossrefs

Apart perhaps from signs, diagonals give A000984, A002420, A078718.
Cf. A007318, A046521, A182073.
Cf. A182411, A082590 (row sums).

Programs

  • Magma
    [Factorial(2*i)*Factorial(2*(n-i))/(Factorial(i)*Factorial(n)*Factorial(n-i)): i in [0..n], n in [0..10]]; // Bruno Berselli, Apr 27 2012
  • Maple
    A068555 := proc(n,i)
    j := n-i ;
    (2*i)!*(2*j)!/(i!*j!*(i+j)!) ;
end proc: # R. J. Mathar, May 31 2016
  • Mathematica
    Flatten[ Table[ Table[ (2i)!*(2(n - i))!/(i!*(n - i)!*n!), {i, 0, n}], {n, 0, 9}]]
  • PARI
    a(n,k)=if(n<0 || k<0,0,(2*n)!*(2*k)!/n!/k!/(n+k)!);

Formula

The square array defined by f := (a, b)->add(binomial(2*a, k)*binomial(2*b, a+b-k)*(-1)^(a+b-k), k=0..2*a); and read by antidiagonals gives a signed version. See Sprugnoli, 3.38.
Let f(x) = 1/sqrt(1 - 4*x) denote the o.g.f for A000984. The o.g.f. for this table is (f(x) + f(y))*f(x)*f(y)*(1/(1 + f(x)*f(y))) = (1 + 2*x + 6*x^2 + 20*x^3 + ...) + (2 + 2*x + 4*x^2 + 10*x^3 + ...)*y + (6 + 4*x + 6*x^2 + 12*x^3 + ...)*y^2 + .... - Peter Bala, Apr 10 2012
T(n,0) = A000984(n), T(n,k) = 4*T(n-1,k-1) - T(n,k-1) for k = 1..n. - Philippe Deléham, Mar 10 2014

A135573 Array T(n,m) of super ballot numbers read along ascending antidiagonals.

Original entry on oeis.org

1, 3, 1, 10, 2, 2, 35, 5, 3, 5, 126, 14, 6, 6, 14, 462, 42, 14, 10, 14, 42, 1716, 132, 36, 20, 20, 36, 132, 6435, 429, 99, 45, 35, 45, 99, 429, 24310, 1430, 286, 110, 70, 70, 110, 286, 1430, 92378, 4862, 858, 286, 154, 126, 154, 286, 858, 4862
Offset: 0

Views

Author

R. J. Mathar, Feb 23 2008

Keywords

Comments

First row is A000108. 2nd row is A007054. 3rd row and 4th column are essentially A007272.
1st column is A001700. 2nd column is essentially A000108. 3rd column is A007054.
Main diagonal is A000984.

Examples

			Array with rows n >= 0 and columns m >= 0 starts:
[n\m]  0    1    2    3    4    5    6     7     8  ...
-------------------------------------------------------
[0]    1    1    2    5   14   42  132   429  1430  ...  [A000108]
[1]    3    2    3    6   14   36   99   286   858  ...  [A007054]
[2]   10    5    6   10   20   45  110   286   780  ...  [A007272]
[3]   35   14   14   20   35   70  154   364   910  ...  [A348893]
[4]  126   42   36   45   70  126  252   546  1260  ...  [A348898]
[5]  462  132   99  110  154  252  462   924  1980  ...  [A348899]
[6] 1716  429  286  286  364  546  924  1716  3432  ...
...
Seen as a triangle:
[0] 1;
[1] 3,    1;
[2] 10,   2,   2;
[3] 35,   5,   3,  5;
[4] 126,  14,  6,  6,  14;
[5] 462,  42,  14, 10, 14, 42;
[6] 1716, 132, 36, 20, 20, 36, 132;
[7] 6435, 429, 99, 45, 35, 45, 99,  429.
.
T(20, 100000) = 2.442634...*10^60129. Asymptotic formula: 2.442627..*10^60129.

Links

Crossrefs

Cf. A000108, A007054, A000984, A348893, A348898, A348899.
Cf. A000984 (main diagonal), A001700 (column 0), A082590 (sum of antidiagonals).

Programs

  • Maple
    T := proc(n,m) (2*n+1)!/n!*(2*m)!/m!/(m+n+1)! ; end proc:
for d from 0 to 12 do for c from 0 to d do printf("%d, ",T(d-c,c)) ; od: od:
# Alternatively, printed as rows:
A135573 := (n, m) -> (1/(2*Pi))*int(x^m*(4-x)^(n+1/2)*x^(-1/2), x=0..4):
for n from 0 to 9 do seq(A135573(n, m), m = 0..9) od; # Peter Luschny, Nov 03 2021
  • Mathematica
    T[n_, m_] := (2*n+1)!/n!*(2*m)!/m!/(m+n+1)!; Table[T[n-m, m], {n, 0, 12}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jan 06 2014, after Maple *)
T[n_, m_] := 4^(m+n) Hypergeometric2F1[1/2+n, 1/2-m, 3/2+n, 1] / ((2 n + 1) Pi);
Table[T[n - m + 1, m], {n, 0, 9}, {m, 0, n}] // Flatten (* Peter Luschny, Nov 03 2021 *)
  • Sage
    def T(n, m): return (-1)^m*4^(n + 1 + m)*binomial(n + 1/2, n + 1 + m)/2
for n in range(7): print([T(n, m) for m in range(9)]) # Peter Luschny, Nov 04 2021

Formula

T(n, m) = (2*n + 1)!*(2*m)! / (n!*m!*(m + n + 1)!).
From Peter Luschny, Nov 03 2021: (Start)
T(n, m) = (1/(2*Pi))*Integral_{x=0..4} x^m*(4 - x)^(n + 1/2)*x^(-1/2). These are integral representations of the n-th moment of a positive function on [0, 4]. The representations are unique.
T(n, m) = 4^(m + n)*hypergeom([1/2 + n, 1/2 - m], [3/2 + n], 1)/((2*n + 1)*Pi).
For fixed n and m -> oo: T(n, m) ~ (1/(2*Pi))*4^(n + m + 1)*(Gamma(3/2 + n) / m^(3/2 + n))*(1 - (2*n + 3)^2 / (8*m)) . (End)
T(n, m) = (-1)^m*4^(n + 1 + m)*binomial(n + 1/2, n + 1 + m)/2. - Peter Luschny, Nov 04 2021
From Peter Bala, Mar 12 2023: (Start)
T(n,m) = 2*(2*n + 1 )/(n + m + 1) * T(n-1,m) with T(0,m) = Catalan(m), where Catalan(m) = A000108(m).
T(n,m) = Sum_{k = 0..n} (-1)^k*4^(n-k)*binomial(n,k)*Catalan(m+k) (easily verified using Maple's sumrecursion command). Thus T(n,m) is an integer. (End)

A144635 a(n) = 5^n*Sum_{ k=0..n } binomial(2*k,k)/5^k.

Original entry on oeis.org

1, 7, 41, 225, 1195, 6227, 32059, 163727, 831505, 4206145, 21215481, 106782837, 536618341, 2693492305, 13507578125, 67693008145, 339066121115, 1697664211795, 8497396194275, 42522326235175, 212749477704695, 1064285646397915, 5323532330953295, 26625895085494075
Offset: 0

Views

Author

N. J. A. Sloane, Jan 21 2009

Keywords

Crossrefs

Cf. A000032, A006134, A082590, A132310, A002457.

Programs

  • Mathematica
    Table[5^n Sum[Binomial[2k,k]/5^k,{k,0,n}],{n,0,30}] (* Harvey P. Dale, Aug 08 2011 *)
Round@Table[5^(n + 1/2) - 2^(n + 1) (2 n + 1)!! Hypergeometric2F1[1, n + 3/2, n + 2, 4/5]/(5 (n + 1)!), {n, 0, 20}] (* Round is equivalent to FullSimplify here, but is much faster - Vladimir Reshetnikov, Oct 14 2016 *)
  • PARI
    a(n) = 5^n*sum(k=0, n, binomial(2*k,k)/5^k); \\ Michel Marcus, Oct 14 2016

Formula

From Vaclav Kotesovec, Jun 12 2013: (Start)
G.f.: 1/((1-5*x)*sqrt(1-4*x)).
Recurrence: n*a(n) = (9*n-2)*a(n-1) - 10*(2*n-1)*a(n-2).
a(n) ~ 5^(n+1/2). (End)
a(n) = 5^(n+1/2) - 2^(n+1)*(2*n+1)!!*hypergeom([1,n+3/2], [n+2], 4/5)/(5*(n+1)!). - Vladimir Reshetnikov, Oct 14 2016
a(n) = Sum_{k=0..n} C(2*n+1,n-k)*A000032(2*k+1). - Vladimir Kruchinin Jan 14 2025

A004074 a(n) = 2*A004001(n) - n, where A004001 is the Hofstadter-Conway $10000 sequence.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 1, 0, 1, 2, 3, 2, 3, 2, 1, 0, 1, 2, 3, 4, 3, 4, 5, 4, 5, 4, 3, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 4, 5, 6, 7, 6, 7, 8, 7, 8, 7, 6, 7, 8, 7, 8, 7, 6, 7, 6, 5, 4, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 5, 6, 7, 8, 9, 8, 9, 10, 11, 10, 11, 12, 11, 12, 11, 10, 11, 12, 13, 12, 13, 14, 13, 14, 13, 12
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

The sequence is 0 at 2^n for n = 1, 2, 3, ... The maximum value between 2^n and 2^(n+1) appears to be A072100(n). - T. D. Noe, Jun 04 2012
Hofstadter shows the plot of sequence A004001(n)-(n/2) at point 10:52 of the part two of DIMACS lecture. This sequence is obtained by doubling those values, thus producing only integers. Cf. also A249071. - Antti Karttunen, Oct 22 2014

Links

Crossrefs

Cf. A004001, A082590, A213057.
Cf. also A249071 (gives the even bisection halved), A233270 (also has a similar Blancmange curve appearance).

Programs

  • Mathematica
    Clear[a]; a[1] = 1; a[2] = 1; a[n_] := a[n] = a[a[n - 1]] + a[n - a[n - 1]]; Table[2*a[n] - n, {n, 100}] (* T. D. Noe, Jun 04 2012 *)
  • Scheme
    (define (A004074 n) (- (* 2 (A004001 n)) n)) ;; Other code as in A004001. - Antti Karttunen, Oct 22 2014

Formula

a(2^n)=0; for n >= 1, Sum_{i=2^(n-1)..2^n} a(i) = A082590(n-2). - Benoit Cloitre, Jun 04 2004

Extensions

More terms from Benoit Cloitre, Jun 04 2004

A126966 Expansion of sqrt(1 - 4*x)/(1 - 2*x).

Original entry on oeis.org

1, 0, -2, -8, -26, -80, -244, -752, -2362, -7584, -24892, -83376, -284324, -984672, -3455144, -12259168, -43908026, -158531392, -576352364, -2107982128, -7750490636, -28629222112, -106190978264, -395347083808, -1476813394916, -5533435084480, -20790762971864, -78316232088032
Offset: 0

Views

Author

N. J. A. Sloane, Mar 22 2007

Keywords

Comments

Hankel transform is 2^n*(-1)^binomial(n+1, 2) = A120617(n). - Paul Barry, Feb 08 2008

Links

Crossrefs

Cf. A000108, A028329, A082590, A126967.

Programs

  • GAP
    List([0..30], n-> (-1)*Sum([0..n], j-> 2^j*Binomial(2*(n-j), n-j)/(2*(n-j) -1) )); # G. C. Greubel, Jan 29 2020
  • Magma
    R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( Sqrt(1-4*x)/(1-2*x) )); // G. C. Greubel, Jan 29 2020
  • Maple
    a := n -> -add(2^j*binomial(2*n-2*j,n-j)/(2*n-2*j-1), j=0..n):
seq(a(n),n=0..30); # Emeric Deutsch, Mar 25 2007
# second Maple program:
CatalanNumber := n -> binomial(2*n, n)/(n+1):
a := n -> 2^n*I + CatalanNumber(n)*simplify(hypergeom([1, n + 1/2], [n + 2], 2)):
seq(a(n), n=0..26); # Peter Luschny, Aug 04 2020
# third program:
A126966 := n -> 2*binomial(2*n, n) - add(2^(n-k)*binomial(2*k,k), k=0..n):
seq(A126966(n), n = 0 .. 27); # Mélika Tebni, Mar 08 2024
  • Mathematica
    CoefficientList[Series[Sqrt[1-4*x]/(1-2*x), {x,0,30}], x] (* G. C. Greubel, Jan 31 2017 *)
  • PARI
    Vec(sqrt(1-4*x)/(1-2*x) + O(x^30)) \\ G. C. Greubel, Jan 31 2017
  • Sage
    def A126966_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( sqrt(1-4*x)/(1-2*x) ).list()
A126966_list(30) # G. C. Greubel, Jan 29 2020

Formula

a(n) = -Sum_{j=0..n} ( 2^j*binomial(2n-2j, n-j)/(2n-2j-1) ). - Emeric Deutsch, Mar 25 2007
D-finite with recurrence: n*a(n) + 6*(1-n)*a(n-1) + 4*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 14 2011, corrected Feb 17 2020
a(n) ~ -4^n/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Jun 29 2013
a(n) = 2^n*i + CatalanNumber(n)*hypergeom([1, n + 1/2], [n + 2], 2). - Peter Luschny, Aug 04 2020
a(n) = A028329(n) - A082590(n). - Mélika Tebni, Mar 08 2024

A167713 a(n) = 16^n * Sum_{k=0..n} binomial(2*k, k) / 16^k.

Original entry on oeis.org

1, 18, 294, 4724, 75654, 1210716, 19372380, 309961512, 4959397062, 79350401612, 1269606610548, 20313706474200, 325019306291356, 5200308911062296, 83204942617113336, 1331279082028930896, 21300465313063974726, 340807445011357201836, 5452919120190790364676
Offset: 0

Views

Author

Alexander Adamchuk, Nov 10 2009

Keywords

Comments

p^2 divides a((p-3)/2) for prime p of the form p = 6k + 1 (A002476).
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), a(n) = A167713 (B=16).

Links

Crossrefs

Cf. A000984, A002476, A066796, A006134, A082590, A132310, A002457, A144635.

Programs

  • Maple
    A167713 := proc(n) coeftayl( 1/(1-16*x)/sqrt(1-4*x),x=0,n) ; end proc: seq(A167713(n),n=0..40) ; # R. J. Mathar, Nov 13 2009
  • Mathematica
    Table[ 16^n * Sum[ (2k)!/(k!)^2 / 16^k, {k,0,n} ], {n,0,20} ]
CoefficientList[Series[1 / ((1 - 16 x) Sqrt[1 - 4 x]), {x, 0, 20}], x] (* Vincenzo Librandi, May 27 2013 *)

Formula

a(n) = 16^n * Sum_{k=0..n} ((2k)!/(k!)^2) / 16^k.
a(n) = 16^n * Sum_{k=0..n} binomial(2k,k) / 16^k.
G.f.: 1/((1-16*x)*sqrt(1-4*x)). - R. J. Mathar, Nov 13 2009
From Vaclav Kotesovec, Oct 20 2012: (Start)
Recurrence: n*a(n) = 2*(10*n-1)*a(n-1) - 32*(2*n-1)*a(n-2).
a(n) ~ 2^(4*n+1)/sqrt(3). (End)

Extensions

Extended by R. J. Mathar, Nov 13 2009

A167859 a(n) = 4^n * Sum_{k=0..n} binomial(2*k, k)^2 / 4^k.

Original entry on oeis.org

1, 8, 68, 672, 7588, 93856, 1229200, 16695424, 232418596, 3293578784, 47309094672, 686870685312, 10059942413584, 148412250014336, 2202990595617344, 32873407393419776, 492791264816231204
Offset: 0

Views

Author

Alexander Adamchuk, Nov 13 2009

Keywords

Comments

Every a(n) from a((p-1)/2) to a(p-1) is divisible by prime p for p = {7, 47, 191, 383, 439, 1151, 1399, 2351, 2879, 3119, 3511, 3559, ...} = A167860, apparently a subset of primes of the form 8n+7 (A007522).
7^3 divides a(13) and 7^2 divides a(10)-a(13).
Every a(n) from a(kp-1 - (p-1)/2) to a(kp-1) is divisible by prime p from A167860.
Every a(n) from a((p^2-1)/2) to a(p^2-1) is divisible by prime p from A167860. For p=7 every a(n) from a((p^3-1)/2) to a(p^3-1) and from a((p^4-1)/2) to a(p^4-1)is divisible by p^2.

Links

Crossrefs

Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167860, A007522.

Programs

  • Maple
    A167859 := proc(n)
    add( (binomial(2*k,k)/2^k)^2,k=0..n) ;
    4^n*% ;
end proc:
seq(A167859(n),n=0..20) ; # R. J. Mathar, Sep 21 2016
  • Mathematica
    Table[4^n*Sum[Binomial[2*k,k]^2/4^k,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 20 2012 *)
  • PARI
    a(n) = 4^n*sum(k=0,n, binomial(2*k,k)^2/4^k) \\ Charles R Greathouse IV, Sep 21 2016

Formula

Recurrence: n^2*a(n) = 4*(5*n^2 - 4*n + 1)*a(n-1) - 16*(2*n - 1)^2*a(n-2). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 2^(4*n+2)/(3*Pi*n). - Vaclav Kotesovec, Oct 20 2012
G.f.: 2*EllipticK(4*sqrt(x))/(Pi*(1-4*x)), where EllipticK is the complete elliptic integral of the first kind, using the Gradshteyn and Ryzhik convention, also used by Maple. In the convention of Abramowitz and Stegun, used by Mathematica, this would be written as 2*K(16*x)/(Pi*(1-4*x)). - Robert Israel, Sep 21 2016

Extensions

More terms from Sean A. Irvine, Apr 14 2010
Further terms from Jon E. Schoenfield, May 09 2010

A120617 Hankel transform of g.f. 1/sqrt(1+4x^2).

Original entry on oeis.org

1, -2, -4, 8, 16, -32, -64, 128, 256, -512, -1024, 2048, 4096, -8192, -16384, 32768, 65536, -131072, -262144, 524288, 1048576, -2097152, -4194304, 8388608, 16777216, -33554432, -67108864, 134217728, 268435456, -536870912, -1073741824, 2147483648, 4294967296, -8589934592
Offset: 0

Views

Author

Paul Barry, Jun 17 2006

Keywords

Comments

Hankel transform of e.g.f. Bessel_I(0,2*sqrt(-1)*x) or (1,0,-2,0,6,0,-20,...).
Hankel transform of Sum{k=0..n} (-1)^(n-k)*C(n,k)^2.
Hankel transform of A098331.
Hankel transform of A082590. - Paul Barry, Apr 26 2009

Links

Crossrefs

Cf. A082590, A098331.

Programs

  • Mathematica
    LinearRecurrence[{0,-4},{1,-2},40] (* or *) CoefficientList[ Series[ (1-2x)/(1+4x^2),{x,0,40}],x] (* Harvey P. Dale, Oct 12 2011 *)

Formula

G.f.: (1-2*x)/(1+4*x^2); a(n) = 2^n*(cos(Pi*(n+1)/2)+sin(Pi*(n+1)/2)).
a(0)=1, a(1)=-2, a(n)=-4*a(n-2). - Harvey P. Dale, Oct 12 2011
a(n) = ( 2*i^(n+1) )^n, where i=sqrt(-1). - Bruno Berselli, Oct 12 2011
E.g.f.: cos(2*x) - sin(2*x). - Arkadiusz Wesolowski, Aug 31 2012

A167867 a(n) = 2^n * Sum_{k=0..n} binomial(2*k,k)^3 / 2^k.

Original entry on oeis.org

1, 10, 236, 8472, 359944, 16722896, 822334816, 42068907200, 2215884717400, 119364801362800, 6545334930678816, 364137834051739200, 20502307365808906816, 1166063313963833813632, 66893439680369963627264
Offset: 0

Views

Author

Alexander Adamchuk, Nov 14 2009

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), A167713 (B=16).
The expression a(n) = B^n*Sum_{ k=0..n } binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64).

Links

Crossrefs

Cf. A079727, A167867, A167868, A167869, A167870, A167872.
Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859.

Programs

  • Mathematica
    Table[2^n Sum[Binomial[2k,k]^3/2^k,{k,0,n}],{n,0,30}] (* Vincenzo Librandi, Mar 26 2012 *)

Formula

a(n) = 2^n * Sum_{k=0..n} binomial(2*k,k)^3 / 2^k.
Recurrence: n^3*a(n) = 2*(33*n^3 - 48*n^2 + 24*n - 4)*a(n-1) - 16*(2*n-1)^3*a(n-2). - Vaclav Kotesovec, Aug 13 2013
a(n) ~ 2^(6*n+5)/(31*(Pi*n)^(3/2)). - Vaclav Kotesovec, Aug 13 2013

Extensions

More terms from Sean A. Irvine, Apr 27 2010
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