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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A053469 a(n) = n*6^(n-1).

Original entry on oeis.org

1, 12, 108, 864, 6480, 46656, 326592, 2239488, 15116544, 100776960, 665127936, 4353564672, 28298170368, 182849716224, 1175462461440, 7522959753216, 47958868426752, 304679870005248, 1929639176699904, 12187194800209920, 76779327241322496, 482612914088312832
Offset: 1

Views

Author

Barry E. Williams, Jan 13 2000

Keywords

Comments

Binomial transform of A053464. - R. J. Mathar, Oct 26 2011

Examples

			G.f. = x + 12*x^2 + 108*x^3 + 864*x^4 + 6480*x^5 + 46656*x^6 + ... - _Michael Somos_, Dec 16 2019
		

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Crossrefs

Programs

Formula

a(n) = 12*a(n-1) - 36*a(n-2), n>=3.
G.f.: x/(6x-1)^2. - Zerinvary Lajos, Apr 28 2009
E.g.f.: x*exp(6*x). - Michael Somos, Dec 16 2019
From Amiram Eldar, Oct 28 2020: (Start)
Sum_{n>=1} 1/a(n) = 6*log(6/5).
Sum_{n>=1} (-1)^(n+1)/a(n) = 6*log(7/6). (End)

Extensions

More terms from James Sellers, Feb 02 2000
More terms from Zerinvary Lajos, Oct 02 2007

A054335 A convolution triangle of numbers based on A000984 (central binomial coefficients of even order).

Original entry on oeis.org

1, 2, 1, 6, 4, 1, 20, 16, 6, 1, 70, 64, 30, 8, 1, 252, 256, 140, 48, 10, 1, 924, 1024, 630, 256, 70, 12, 1, 3432, 4096, 2772, 1280, 420, 96, 14, 1, 12870, 16384, 12012, 6144, 2310, 640, 126, 16, 1, 48620, 65536, 51480, 28672, 12012, 3840, 924, 160, 18, 1
Offset: 0

Views

Author

Wolfdieter Lang, Mar 13 2000

Keywords

Comments

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group. The g.f. for the row polynomials p(n,x) (increasing powers of x) is 1/(sqrt(1-4*z)-x*z).
Riordan array (1/sqrt(1-4*x),x/sqrt(1-4*x)). - Paul Barry, May 06 2009
The matrix inverse is apparently given by deleting the leftmost column from A206022. - R. J. Mathar, Mar 12 2013

Examples

			Triangle begins:
    1;
    2,    1;
    6,    4,   1;
   20,   16,   6,   1;
   70,   64,  30,   8,  1;
  252,  256, 140,  48, 10,  1;
  924, 1024, 630, 256, 70, 12, 1; ...
Fourth row polynomial (n=3): p(3,x) = 20 + 16*x + 6*x^2 + x^3.
From _Paul Barry_, May 06 2009: (Start)
Production matrix begins
    2,   1;
    2,   2,  1;
    0,   2,  2,  1;
   -2,   0,  2,  2,  1;
    0,  -2,  0,  2,  2,  1;
    4,   0, -2,  0,  2,  2, 1;
    0,   4,  0, -2,  0,  2, 2, 1;
  -10,   0,  4,  0, -2,  0, 2, 2, 1;
    0, -10,  0,  4,  0, -2, 0, 2, 2, 1; (End)
		

Crossrefs

Row sums: A026671.

Programs

  • GAP
    T:= function(n, k)
        if k mod 2=0 then return Binomial(2*n-k, n-Int(k/2))*Binomial(n-Int(k/2),Int(k/2))/Binomial(k,Int(k/2));
        else return 4^(n-k)*Binomial(n-Int((k-1)/2)-1, Int((k-1)/2));
        fi;
      end;
    Flat(List([0..10], n-> List([0..n], k-> T(n, k) ))); # G. C. Greubel, Jul 20 2019
  • Magma
    T:= func< n, k | (k mod 2) eq 0 select Binomial(2*n-k, n-Floor(k/2))* Binomial(n-Floor(k/2),Floor(k/2))/Binomial(k,Floor(k/2)) else 4^(n-k)*Binomial(n-Floor((k-1)/2)-1, Floor((k-1)/2)) >;
    [[T(n,k): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Jul 20 2019
    
  • Maple
    A054335 := proc(n,k)
        if k <0 or k > n then
            0 ;
        elif type(k,odd) then
            kprime := floor(k/2) ;
            binomial(n-kprime-1,kprime)*4^(n-k) ;
        else
            kprime := k/2 ;
            binomial(2*n-k,n-kprime)*binomial(n-kprime,kprime)/binomial(k,kprime) ;
        end if;
    end proc: # R. J. Mathar, Mar 12 2013
    # Uses function PMatrix from A357368. Adds column 1,0,0,0,... to the left.
    PMatrix(10, n -> binomial(2*(n-1), n-1)); # Peter Luschny, Oct 19 2022
  • Mathematica
    Flatten[ CoefficientList[#1, x] & /@ CoefficientList[ Series[1/(Sqrt[1 - 4*z] - x*z), {z, 0, 9}], z]] (* or *)
    a[n_, k_?OddQ] := 4^(n-k)*Binomial[(2*n-k-1)/2, (k-1)/2]; a[n_, k_?EvenQ] := (Binomial[n-k/2, k/2]*Binomial[2*n-k, n-k/2])/Binomial[k, k/2]; Table[a[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 08 2011, updated Jan 16 2014 *)
  • PARI
    T(n, k) = if(k%2==0, binomial(2*n-k, n-k/2)*binomial(n-k/2,k/2)/binomial(k,k/2), 4^(n-k)*binomial(n-(k-1)/2-1, (k-1)/2));
    for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Jul 20 2019
    
  • Sage
    def T(n, k):
        if (mod(k,2)==0): return binomial(2*n-k, n-k/2)*binomial(n-k/2,k/2)/binomial(k,k/2)
        else: return 4^(n-k)*binomial(n-(k-1)/2-1, (k-1)/2)
    [[T(n,k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Jul 20 2019
    

Formula

a(n, 2*k+1) = binomial(n-k-1, k)*4^(n-2*k-1), a(n, 2*k) = binomial(2*(n-k), n-k)*binomial(n-k, k)/binomial(2*k, k), k >= 0, n >= m >= 0; a(n, m) := 0 if n
Column recursion: a(n, m)=2*(2*n-m-1)*a(n-1, m)/(n-m), n>m >= 0, a(m, m) := 1.
G.f. for column m: cbie(x)*(x*cbie(x))^m, with cbie(x) := 1/sqrt(1-4*x).
G.f.: 1/(1-x*y-2*x/(1-x/(1-x/(1-x/(1-x/(1-... (continued fraction). - Paul Barry, May 06 2009
Sum_{k>=0} T(n,2*k)*(-1)^k*A000108(k) = A000108(n+1). - Philippe Deléham, Jan 30 2012
Sum_{k=0..floor(n/2)} T(n-k,n-2*k) = A098615(n). - Philippe Deléham, Feb 01 2012
T(n,k) = 4*T(n-1,k) + T(n-2,k-2) for k>=1. - Philippe Deléham, Feb 02 2012
Vertical recurrence: T(n,k) = 1*T(n-1,k-1) + 2*T(n-2,k-1) + 6*T(n-3,k-1) + 20*T(n-4,k-1) + ... for k >= 1 (the coefficients 1, 2, 6, 20, ... are the central binomial coefficients A000984). - Peter Bala, Oct 17 2015

A020920 Expansion of 1/(1-4*x)^(9/2).

Original entry on oeis.org

1, 18, 198, 1716, 12870, 87516, 554268, 3325608, 19122246, 106234700, 573667380, 3024791640, 15628090140, 79342611480, 396713057400, 1957117749840, 9540949030470, 46021048264620, 219878341708740, 1041528987041400, 4895186239094580, 22844202449108040
Offset: 0

Keywords

Comments

Also convolution of A000984 with A038846, also convolution of A000302 with A020918, also convolution of A002457 with A038845, also convolution of A002697 with A002802. - Rui Duarte, Oct 08 2011

Crossrefs

Programs

  • GAP
    List([0..30], n-> Binomial(n+4, 4)*Binomial(2*(n+4), n+4)/70) # G. C. Greubel, Jul 20 2019
  • Magma
    [(2*n+7)*(2*n+5)*(2*n+3)*(2*n+1)*Binomial(2*n, n)/105: n in [0..30]]; // Vincenzo Librandi, Jul 05 2013
    
  • Maple
    seq(binomial(2*n+8, n+4)*binomial(n+4, n)/70, n=0..30); # Zerinvary Lajos, May 05 2007
  • Mathematica
    CoefficientList[Series[1/(1-4x)^(9/2), {x,0,30}], x] (* Vincenzo Librandi, Jul 05 2013 *)
  • PARI
    vector(30, n, n--; m=n+4; binomial(m, 4)*binomial(2*m, m)/70) \\ G. C. Greubel, Jul 20 2019
    
  • Sage
    [binomial(n+4, 4)*binomial(2*(n+4), n+4)/70 for n in (0..30)] # G. C. Greubel, Jul 20 2019
    

Formula

a(n) = binomial(n+4, 4)*A000984(n+4)/A000984(4), where A000984 are the central binomial coefficients. - Wolfdieter Lang
a(n) = Sum_{ a+b+c+d+e+f+g+h+i=n} f(a)*f(b)*f(c)*f(d)*f(e)*f(f)*f(g) *f(h)*f(i) with f(n)=A000984(n). - Philippe Deléham, Jan 22 2004
a(n) = A000332(n+4)*A000984(n+4)/70. - Zerinvary Lajos, May 05 2007
From Rui Duarte, Oct 08 2011: (Start)
a(n) = ((2n+7)(2n+5)(2n+3)(2n+1)/(7*5*3*1)) * binomial(2n, n).
a(n) = binomial(2n+8, 8) * binomial(2n, n) / binomial(n+4, 4).
a(n) = binomial(n+4, 4) * binomial(2n+8, n+4) / binomial(8, 4). (End)
Boas-Buck recurrence: a(n) = (18/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+4, 4). See a comment there.
From Amiram Eldar, Mar 25 2022: (Start)
Sum_{n>=0} 1/a(n) = 1148/5 - 42*sqrt(3)*Pi.
Sum_{n>=0} (-1)^n/a(n) = 700*sqrt(5)*log(phi) - 11284/15, where phi is the golden ratio (A001622). (End)

A212698 Main transitions in systems of n particles with spin 3/2.

Original entry on oeis.org

3, 24, 144, 768, 3840, 18432, 86016, 393216, 1769472, 7864320, 34603008, 150994944, 654311424, 2818572288, 12079595520, 51539607552, 219043332096, 927712935936, 3917010173952, 16492674416640, 69269232549888, 290271069732864, 1213860837064704, 5066549580791808
Offset: 1

Author

Stanislav Sykora, May 25 2012

Keywords

Comments

Please refer to the general explanation in A212697. This particular sequence is obtained for base b=4, corresponding to spin S = (b-1)/2 = 3/2.
Let P(A) be the power set of an n-element set A and let B be the Cartesian product of P(A) with itself. Then a(n) = the sum of the size of the union of x and y for every (x,y) in B. [See Relation (28): U(n) in document of Ross La Haye in reference.] - Bernard Schott, Jan 04 2013
A002697 is the analogous sequence if "union" is replaced by "intersection" and A002699 is the analogous sequence if "union" is replaced by "symmetric difference". Here, X union Y and Y union X are considered as two distinct Cartesian products, if we want to consider that X Union Y = Y Union X are the same Cartesian product, see A133224. - Bernard Schott Jan 11 2013

Crossrefs

Cf. A001787, A212697, A212699, A212700, A212701, A212702, A212703, A212704 (for b = 2, 3, 5, 6, 7, 8, 9, 10).

Programs

  • Magma
    [3*n*4^(n-1): n in [1..30]]; // Vincenzo Librandi, Nov 29 2015
  • Mathematica
    Table[Sum[Binomial[n,i] i 3^i,{i,0,n}],{n,1,21}] (* Geoffrey Critzer, Aug 08 2013 *)
  • PARI
    mtrans(n, b) = n*(b-1)*b^(n-1);
    for (n=1, 100, write("b212698.txt", n, " ", mtrans(n, 4)))
    

Formula

a(n) = n*(b-1)*b^(n-1). For this sequence, set b=4.
a(n) = 3*n*4^(n-1).
a(n) = 3*A002697(n).
From Geoffrey Critzer, Aug 08 2013: (Start)
a(n) = Sum_{i>=0} binomial(n,i)*i*3^i.
E.g.f.: 3*x*exp(4*x). (End)
G.f.: 3*x/(4*x-1)^2. - Colin Barker, Nov 03 2014
From Elmo R. Oliveira, May 24 2025: (Start)
a(n) = 8*a(n-1) - 16*a(n-2) for n > 2.
a(n) = A008585(n)*A000302(n-1). (End)

A002699 a(n) = n*2^(2*n-1).

Original entry on oeis.org

0, 2, 16, 96, 512, 2560, 12288, 57344, 262144, 1179648, 5242880, 23068672, 100663296, 436207616, 1879048192, 8053063680, 34359738368, 146028888064, 618475290624, 2611340115968, 10995116277760, 46179488366592, 193514046488576
Offset: 0

Keywords

Comments

Right side of binomial sum Sum(i * binomial(2*n, i), i=1..n) - Yong Kong (ykong(AT)curagen.com), Dec 26 2000
Coefficients of shifted Chebyshev polynomials.
Starting with offset 1 = 4th binomial transform of [2, 8, 0, 0, 0, ...]. - Gary W. Adamson, Jul 21 2009
Let P(A) be the power set of an n-element set A and B be the Cartesian product of P(A) with itself. Then a(n) = the sum of the size of the symmetric difference of x and y for every (x,y) of B. - Ross La Haye, Jan 04 2013
It's the relation [27] with T(n) in the document of Ross. Following the last comment of Ross, A002697 is the similar sequence when replacing "symmetric difference" by "intersection" and A212698 is the similar sequence when replacing "symmetric difference" by union. - Bernard Schott, Jan 04 2013
If Delta = Symmetric difference, here, X Delta Y and Y Delta X are considered as two distinct Cartesian products, if we want to consider that X Delta Y = X Delta Y is the same Cartesian product, see A002697. - Bernard Schott, Jan 15 2013

References

  • C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 518.
  • 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

Programs

Formula

a(n) = 2 * A002697(n). - Bernard Schott, Jan 04 2013
a(n) = A212698(n) - A002697(n)
a(n) = 8*a(n-1)-16*a(n-2) with n>1, a(0)=0, a(1)=2. - Vincenzo Librandi, Mar 20 2013
G.f.: (2*x)/(1 - 4*x)^2. - Harvey P. Dale, Jul 28 2021
E.g.f.: (exp(4*x) - 1)/2. - Stefano Spezia, Aug 04 2022

A082134 Expansion of e.g.f. x*exp(3*x)*cosh(x).

Original entry on oeis.org

0, 1, 6, 30, 144, 680, 3168, 14560, 66048, 296064, 1313280, 5772800, 25178112, 109078528, 469819392, 2013388800, 8590196736, 36507779072, 154620002304, 652837519360, 2748784312320, 11544883101696, 48378534690816
Offset: 0

Author

Paul Barry, Apr 06 2003

Keywords

Comments

Binomial transform of A082133. 3rd binomial transform of (0,1,0,3,0,5,0,7,...)
Let P(A) be the power set of an n-element set A and B be the Cartesian product of P(A) with itself. Then remove (y,x) from B when (x,y) is in B and x <> y and call this R35. Then a(n) = the sum of the size of the intersection of x and y for every (x,y) of R35. - Ross La Haye, Dec 30 2007; edited Jan 05 2013
A133224 is the analogous sequence if "Intersection" is replaced by "Union" and A002697 is the analogous sequence if "Intersection" is replaced by "Symmetric difference". Here, X Intersection Y = Y Intersection X is considered as the same set [Relation (37): T_Q(n) in document of Ross La Haye in reference]. If we want to consider that X Intersection Y and Y Intersection X are two distinct formula for describing the same set, see A002697. - Bernard Schott, Jan 19 2013

Crossrefs

Cf. A057711 (x*exp(x)*cosh(x)), A082133 (x*exp(2*x)*cosh(x)).
Cf. A082135 (x*exp(4*x)*cosh(x)), A082136 (x*exp(5*x)*cosh(x)).

Programs

  • Magma
    [n*2^(n-2)*(1+2^(n-1)): n in [0..25]]; // G. C. Greubel, Feb 05 2018
    
  • Maple
    a:= n -> n*binomial(2^(n-1) +1, 2); seq(a(n), n=0..25); # G. C. Greubel, Apr 16 2020
  • Mathematica
    Table[n(2^(n-1) +4^(n-1))/2, {n, 0, 22}] (* Michael De Vlieger, Nov 29 2015 *)
    With[{nmax = 25}, CoefficientList[Series[x*Exp[3*x]*Cosh[x], {x, 0, nmax}], x]*Range[0, nmax]!] (* G. C. Greubel, Feb 05 2018 *)
  • PARI
    a(n)=n*(2^n--+4^n)/2 \\ Charles R Greathouse IV, Jan 14 2013
    
  • Sage
    [n*binomial(2^(n-1)+1, 2) for n in (0..25)] # G. C. Greubel, Apr 16 2020

Formula

a(n) = n*(2^(n-1) + 4^(n-1))/2.
E.g.f.: x*exp(3*x)*cosh(x).
Conjecture: (n+28)*a(n) + (n-282)*a(n-1) + 2*(-17*n+423)*a(n-2) + 8*(7*n-94)*a(n-3) = 0. - R. J. Mathar, Nov 29 2012
G.f.: x*(10*x^2-6*x+1) / ((2*x-1)^2*(4*x-1)^2). - Colin Barker, Dec 10 2012

A260006 a(n) = f(1,n,n), where f is the Sudan function defined in A260002.

Original entry on oeis.org

0, 3, 12, 35, 90, 217, 504, 1143, 2550, 5621, 12276, 26611, 57330, 122865, 262128, 557039, 1179630, 2490349, 5242860, 11010027, 23068650, 48234473, 100663272, 209715175, 436207590, 905969637, 1879048164, 3892314083, 8053063650, 16642998241, 34359738336
Offset: 0

Author

Natan Arie Consigli, Jul 23 2015

Keywords

Comments

f(1,n,n) = 2^n*(n+2) - (n+2) = (2^n - 1)*(n+2).
To evaluate the Sudan function see A260002 and A260003.
The numbers are alternately even and odd because for even n (2^n-1)*(n+2) is even and (2^(n+1)-1)*(n+1+2) is odd.
From Enrique Navarrete, Oct 02 2021: (Start)
a(n-2) is the number of ways we can write [n] as the union of 2 sets of sizes i, j which intersect in exactly 1 element (1 < i, j < n; i = j allowed), n>=2.
For n = 4, a(n-2) = 12 since [4] can be written as the unions:
{1,2} U {1,3,4}; {2,3} U {1,2,4}; {1,2} U {2,3,4}; {2,3} U {1,3,4};
{1,3} U {1,2,4}; {2,4} U {1,2,3}; {1,3} U {2,3,4}; {2,4} U {1,3,4};
{1,4} U {1,2,3}; {3,4} U {1,2,3}; {1,4} U {2,3,4}; {3,4} U {1,2,4}. (End)

Examples

			a(4) = (2^4 - 1)*(4 + 2) = 90.
		

Crossrefs

Cf. A000295 (f(1,0,n)), A000325 (f(1,2,n)), A005408 (f(1,n,1) = 2n+1), A001787 (n*2^(n-1)), A079583 (f(1,1,n)), A123720 (f(1,4,n)), A133124 (f(1,3,n)).

Programs

  • Magma
    [(2^n-1)*(n+2): n in [0..30]]; // Vincenzo Librandi, Aug 22 2015
    
  • Mathematica
    Table[(2^n -1)(n+2), {n, 0, 30}] (* Michael De Vlieger, Aug 22 2015 *)
    CoefficientList[Series[x(3 -6x +2x^2)/((1-x)^2 (1-2x)^2), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 22 2015 *)
    LinearRecurrence[{6,-13,12,-4},{0,3,12,35},40] (* Harvey P. Dale, Mar 04 2023 *)
  • PARI
    vector(40, n, n--; (2^n-1)*(n+2)) \\ Michel Marcus, Jul 29 2015
    
  • PARI
    concat(0, Vec(x*(3-6*x+2*x^2)/((1-x)^2*(1-2*x)^2) + O(x^40))) \\ Colin Barker, Jul 29 2015
    
  • Sage
    [(n+2)*(2^n -1) for n in (0..30)] # G. C. Greubel, Dec 30 2021

Formula

a(n) = (2^n -1)*(n+2).
a(n) = 6*a(n-1) - 13*a(n-2) + 12*a(n-3) - 4*a(n-4) for n>3. - Colin Barker, Jul 29 2015
G.f.: x*(3 - 6*x + 2*x^2) / ((1-x)^2*(1-2*x)^2). - Colin Barker, Jul 29 2015
E.g.f.: 2*(x+1)*exp(2*x) - (x+2)*exp(x). - Robert Israel, Aug 23 2015
From Enrique Navarrete, Oct 02 2021: (Start)
a(n-2) = Sum_{j=2..n/2} binomial(n,j)*j, n even > 2.
a(n-2) = (Sum_{j=2..floor(n/2)} binomial(n,j)*j) + (1/2)*binomial(n, ceiling(n/2))*ceiling(n/2), n odd > 1. (End)
From Wolfdieter Lang, Nov 12 2021: (Start)
The previous bisection becomes for a(n):
a(2*k) = 2*(A002697(k+1) - (k+1)), and a(2*k+1) = A303602(k+1) - (2*k+3)*(2 - A000984(k+1))/2 = (2*k+3)*(4^(k+1) - 2)/2, for k >= 0. (End)

A020922 Expansion of 1/(1-4*x)^(11/2).

Original entry on oeis.org

1, 22, 286, 2860, 24310, 184756, 1293292, 8498776, 53117350, 318704100, 1848483780, 10418726760, 57302997180, 308554600200, 1630931458200, 8480843582640, 43464323361030, 219878341708740, 1099391708543700, 5439095821216200, 26651569523959380, 129450480544945560
Offset: 0

Keywords

Comments

Also convolution of A000984 with A040075, also convolution of A000302 with A020920, also convolution of A002457 with A038846, also convolution of A002697 with A020918, also convolution of A002802 with A038845. - Rui Duarte, Oct 08 2011

Programs

  • GAP
    List([0..30], n-> Binomial(n+5, 5)*Binomial(2*n+10, n+5)/252); # G. C. Greubel, Jul 20 2019
  • Magma
    [(2*n+9)*(2*n+7)*(2*n+5)*(2*n+3)*(2*n+1)*Binomial(2*n, n)/945: n in [0..30]] // Vincenzo Librandi, Jul 05 2013
    
  • Mathematica
    CoefficientList[Series[1/(1-4x)^(11/2), {x,0,30}], x] (* Vincenzo Librandi, Jul 05 2013 *)
  • PARI
    vector(30, n, n--; m=n+5; binomial(m, 5)*binomial(2*m, m)/252) \\ G. C. Greubel, Jul 20 2019
    
  • Sage
    [binomial(n+5, 5)*binomial(2*n+10, n+5)/252 for n in (0..30)] # G. C. Greubel, Jul 20 2019
    

Formula

a(n) = binomial(n+5, 5)*A000984(n+5)/A000984(5), where A000984 are central binomial coefficients. - Wolfdieter Lang
From Rui Duarte, Oct 08 2011: (Start)
a(n) = ((2n+9)(2n+7)(2n+5)(2n+3)(2n+1)/(9*7*5*3*1)) * binomial(2n, n).
a(n) = binomial(2n+10, 10) * binomial(2n, n) / binomial(n+5, 5).
a(n) = binomial(n+5, 5) * binomial(2n+10, n+5) / binomial(10, 5).
a(n) = Sum_{ i_1+i_2+i_3+i_4+i_5+i_6+i_7+i_8+i_9+i_10+i_11 = n } f(i_1)* f(i_2)*f(i_3)*f(i_4)*f(i_5)*f(i_6)*f(i_7)*f(i_8)*f(i_9)*f(i_10)*f(i_11) with f(k)=A000984(k). (End)
Boas-Buck recurrence: a(n) = (22/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+5, 5). See a comment there. - Wolfdieter Lang, Aug 10 2017
From Amiram Eldar, Mar 25 2022: (Start)
Sum_{n>=0} 1/a(n) = 162*sqrt(3)*Pi - 30816/35.
Sum_{n>=0} (-1)^n/a(n) = 4500*sqrt(5)*log(phi) - 33888/7, where phi is the golden ratio (A001622). (End)

A045543 6-fold convolution of A000302 (powers of 4); expansion of 1/(1-4*x)^6.

Original entry on oeis.org

1, 24, 336, 3584, 32256, 258048, 1892352, 12976128, 84344832, 524812288, 3148873728, 18320719872, 103817412608, 574988746752, 3121367482368, 16647293239296, 87398289506304, 452414675091456, 2312341672689664, 11683410556747776, 58417052783738880, 289303499500421120
Offset: 0

Keywords

Comments

Also convolution of A020922 with A000984 (central binomial coefficients); also convolution of A040075 with A000302 (powers of 4).
With a different offset, number of n-permutations of 5 objects: u,v,z,x, y with repetition allowed, containing exactly five (5) u's. Example: a(1)=24 because we have uuuuuv uuuuvu uuuvuu uuvuuu uvuuuu vuuuuu uuuuuz uuuuzu uuuzuu uuzuuu uzuuuu zuuuuu uuuuux uuuuxu uuuxuu uuxuuu uxuuuu xuuuuu uuuuuy uuuuyu uuuyuu uuyuuu uyuuuu yuuuuu. - Zerinvary Lajos, Jun 16 2008
Also convolution of A002457 with A020920, also convolution of A002697 with A038846, also convolution of A002802 with A020918, also convolution of A038845 with A038845. - Rui Duarte, Oct 08 2011

Crossrefs

Cf. A038231.

Programs

  • GAP
    List([0..30], n-> 4^n*Binomial(n+5,5)); # G. C. Greubel, Jul 20 2019
  • Magma
    [4^n*Binomial(n+5, 5): n in [0..30]]; // Vincenzo Librandi, Oct 15 2011
    
  • Maple
    seq(seq(binomial(i+5, j)*4^i, j =i), i=0..30); # Zerinvary Lajos, Dec 03 2007
    seq(binomial(n+5,5)*4^n,n=0..30); # Zerinvary Lajos, Jun 16 2008
  • Mathematica
    CoefficientList[Series[1/(1-4x)^6,{x,0,30}],x] (* or *) LinearRecurrence[ {24,-240,1280,-3840,6144,-4096}, {1,24,336,3584,32256, 258048}, 30] (* Harvey P. Dale, Mar 24 2018 *)
  • PARI
    Vec(1/(1-4*x)^6 + O(x^30)) \\ Michel Marcus, Aug 21 2015
    
  • Sage
    [lucas_number2(n, 4, 0)*binomial(n,5)/2^10 for n in range(5, 35)] # Zerinvary Lajos, Mar 11 2009
    

Formula

a(n) = binomial(n+5, 5)*4^n.
G.f.: 1/(1-4*x)^6.
a(n) = Sum_{ i_1+i_2+i_3+i_4+i_5+i_6+i_7+i_8+i_9+i_10+i_11+i_12 = n} f(i_1)* f(i_2)*f(i_3)*f(i_4)*f(i_5)*f(i_6)*f(i_7)*f(i_8)*f(i_9)*f(i_10) *f(i_11)*f(i_12), with f(k)=A000984(k). - Rui Duarte, Oct 08 2011
E.g.f.: (15 + 120*x + 240*x^2 + 160*x^3 + 32*x^4)*exp(4*x)/3. - G. C. Greubel, Jul 20 2019
From Amiram Eldar, Mar 25 2022: (Start)
Sum_{n>=0} 1/a(n) = 1620*log(4/3) - 465.
Sum_{n>=0} (-1)^n/a(n) = 12500*log(5/4) - 8365/3. (End)

A079028 a(0) = 1, a(n) = (n + 4)*4^(n-1) for n >= 1.

Original entry on oeis.org

1, 5, 24, 112, 512, 2304, 10240, 45056, 196608, 851968, 3670016, 15728640, 67108864, 285212672, 1207959552, 5100273664, 21474836480, 90194313216, 377957122048, 1580547964928, 6597069766656, 27487790694400, 114349209288704, 474989023199232, 1970324836974592, 8162774324609024
Offset: 0

Author

Benoit Cloitre, Feb 01 2003

Keywords

Comments

a(n) = det(M(n)) where M(n) is the n X n matrix defined by m(i,i) = 5, m(i,j) = i/j.
Main diagonal of array defined by m(1,j) = j; m(i,1) = i and m(i,j) = m(i-1,j) + 3*m(i-1,j-1).
4th binomial transform of (1,1,0,0,0,0,...). - Paul Barry, Mar 07 2003
Number of independent vertex subsets of the graph obtained by attaching two pendant edges to each vertex of the complete graph K_n (see A235113). Example: a(1)=5; indeed, K_1 is the one vertex graph and after attaching two pendant vertices we obtain the path graph ABC; the independent vertex subsets are: empty, {A}, {B}, {C}, and {A,C}. - Emeric Deutsch, Jan 13 2014
Row sums of A235113.

Programs

Formula

a(n) = 8*a(n-1)-16*a(n-2), a(0) = 1, a(1) = 5. - Paul Barry, Mar 07 2003
G.f.: (1 - 3*x)/(1 - 4*x)^2. - Philippe Deléham, Dec 11 2008
From Amiram Eldar, Jan 14 2021: (Start)
Sum_{n>=0} 1/a(n) = 1024*log(4/3) - 880/3.
Sum_{n>=0} (-1)^n/a(n) = 688/3 - 1024*log(5/4). (End)
E.g.f.: exp(4*x)*(1 + x). - Stefano Spezia, Mar 05 2023

Extensions

More terms from Stefano Spezia, Mar 05 2023
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