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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A374439 Triangle read by rows: the coefficients of the Lucas-Fibonacci polynomials. T(n, k) = T(n - 1, k) + T(n - 2, k - 2) with initial values T(n, k) = k + 1 for k < 2.

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 1, 2, 2, 2, 1, 2, 3, 4, 1, 1, 2, 4, 6, 3, 2, 1, 2, 5, 8, 6, 6, 1, 1, 2, 6, 10, 10, 12, 4, 2, 1, 2, 7, 12, 15, 20, 10, 8, 1, 1, 2, 8, 14, 21, 30, 20, 20, 5, 2, 1, 2, 9, 16, 28, 42, 35, 40, 15, 10, 1, 1, 2, 10, 18, 36, 56, 56, 70, 35, 30, 6, 2
Offset: 0

Views

Author

Peter Luschny, Jul 22 2024

Keywords

Comments

There are several versions of Lucas and Fibonacci polynomials in this database. Our naming follows the convention of calling polynomials after the values of the polynomials at x = 1. This assumes a regular sequence of polynomials, that is, a sequence of polynomials where degree(p(n)) = n. This view makes the coefficients of the polynomials (the terms of a row) a refinement of the values at the unity.
A remarkable property of the polynomials under consideration is that they are dual in this respect. This means they give the Lucas numbers at x = 1 and the Fibonacci numbers at x = -1 (except for the sign). See the example section.
The Pell numbers and the dual Pell numbers are also values of the polynomials, at the points x = -1/2 and x = 1/2 (up to the normalization factor 2^n). This suggests a harmonized terminology: To call 2^n*P(n, -1/2) = 1, 0, 1, 2, 5, ... the Pell numbers (A000129) and 2^n*P(n, 1/2) = 1, 4, 9, 22, ... the dual Pell numbers (A048654).
Based on our naming convention one could call A162515 (without the prepended 0) the Fibonacci polynomials. In the definition above only the initial values would change to: T(n, k) = k + 1 for k < 1. To extend this line of thought we introduce A374438 as the third triangle of this family.
The triangle is closely related to the qStirling2 numbers at q = -1. For the definition of these numbers see A333143. This relates the triangle to A065941 and A103631.

Examples

			Triangle starts:
  [ 0] [1]
  [ 1] [1, 2]
  [ 2] [1, 2, 1]
  [ 3] [1, 2, 2,  2]
  [ 4] [1, 2, 3,  4,  1]
  [ 5] [1, 2, 4,  6,  3,  2]
  [ 6] [1, 2, 5,  8,  6,  6,  1]
  [ 7] [1, 2, 6, 10, 10, 12,  4,  2]
  [ 8] [1, 2, 7, 12, 15, 20, 10,  8,  1]
  [ 9] [1, 2, 8, 14, 21, 30, 20, 20,  5,  2]
  [10] [1, 2, 9, 16, 28, 42, 35, 40, 15, 10, 1]
.
Table of interpolated sequences:
  |  n | A039834 & A000045 | A000032 |   A000129   |   A048654  |
  |  n |     -P(n,-1)      | P(n,1)  |2^n*P(n,-1/2)|2^n*P(n,1/2)|
  |    |     Fibonacci     |  Lucas  |     Pell    |    Pell*   |
  |  0 |        -1         |     1   |       1     |       1    |
  |  1 |         1         |     3   |       0     |       4    |
  |  2 |         0         |     4   |       1     |       9    |
  |  3 |         1         |     7   |       2     |      22    |
  |  4 |         1         |    11   |       5     |      53    |
  |  5 |         2         |    18   |      12     |     128    |
  |  6 |         3         |    29   |      29     |     309    |
  |  7 |         5         |    47   |      70     |     746    |
  |  8 |         8         |    76   |     169     |    1801    |
  |  9 |        13         |   123   |     408     |    4348    |
		

Crossrefs

Triangles related to Lucas polynomials: A034807, A114525, A122075, A061896, A352362.
Triangles related to Fibonacci polynomials: A162515, A053119, A168561, A049310, A374441.
Sums include: A000204 (Lucas numbers, row), A000045 & A212804 (even sums, Fibonacci numbers), A006355 (odd sums), A039834 (alternating sign row).
Type m^n*P(n, 1/m): A000129 & A048654 (Pell, m=2), A108300 & A003688 (m=3), A001077 & A048875 (m=4).
Adding and subtracting the values in a row of the table (plus halving the values obtained in this way): A022087, A055389, A118658, A052542, A163271, A371596, A324969, A212804, A077985, A069306, A215928.
Columns include: A040000 (k=1), A000027 (k=2), A005843 (k=3), A000217 (k=4), A002378 (k=5).
Diagonals include: A000034 (k=n), A029578 (k=n-1), abs(A131259) (k=n-2).
Cf. A029578 (subdiagonal), A124038 (row reversed triangle, signed).

Programs

  • Magma
    function T(n,k) // T = A374439
      if k lt 0 or k gt n then return 0;
      elif k le 1 then return k+1;
      else return T(n-1,k) + T(n-2,k-2);
      end if;
    end function;
    [T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 23 2025
    
  • Maple
    A374439 := (n, k) -> ifelse(k::odd, 2, 1)*binomial(n - irem(k, 2) - iquo(k, 2), iquo(k, 2)):
    # Alternative, using the function qStirling2 from A333143:
    T := (n, k) -> 2^irem(k, 2)*qStirling2(n, k, -1):
    seq(seq(T(n, k), k = 0..n), n = 0..10);
  • Mathematica
    A374439[n_, k_] := (# + 1)*Binomial[n - (k + #)/2, (k - #)/2] & [Mod[k, 2]];
    Table[A374439[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* Paolo Xausa, Jul 24 2024 *)
  • Python
    from functools import cache
    @cache
    def T(n: int, k: int) -> int:
        if k > n: return 0
        if k < 2: return k + 1
        return T(n - 1, k) + T(n - 2, k - 2)
    
  • Python
    from math import comb as binomial
    def T(n: int, k: int) -> int:
        o = k & 1
        return binomial(n - o - (k - o) // 2, (k - o) // 2) << o
    
  • Python
    def P(n, x):
        if n < 0: return P(n, x)
        return sum(T(n, k)*x**k for k in range(n + 1))
    def sgn(x: int) -> int: return (x > 0) - (x < 0)
    # Table of interpolated sequences
    print("|  n | A039834 & A000045 | A000032 |   A000129   |   A048654  |")
    print("|  n |     -P(n,-1)      | P(n,1)  |2^n*P(n,-1/2)|2^n*P(n,1/2)|")
    print("|    |     Fibonacci     |  Lucas  |     Pell    |    Pell*   |")
    f = "| {0:2d} | {1:9d}         |  {2:4d}   |   {3:5d}     |    {4:4d}    |"
    for n in range(10): print(f.format(n, -P(n, -1), P(n, 1), int(2**n*P(n, -1/2)), int(2**n*P(n, 1/2))))
    
  • SageMath
    from sage.combinat.q_analogues import q_stirling_number2
    def A374439(n,k): return (-1)^((k+1)//2)*2^(k%2)*q_stirling_number2(n+1, k+1, -1)
    print(flatten([[A374439(n, k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Jan 23 2025

Formula

T(n, k) = 2^k' * binomial(n - k' - (k - k') / 2, (k - k') / 2) where k' = 1 if k is odd and otherwise 0.
T(n, k) = (1 + (k mod 2))*qStirling2(n, k, -1), see A333143.
2^n*P(n, -1/2) = A000129(n - 1), Pell numbers, P(-1) = 1.
2^n*P(n, 1/2) = A048654(n), dual Pell numbers.
T(2*n, n) = (1/2)*(-1)^n*( (1+(-1)^n)*A005809(n/2) - 2*(1-(-1)^n)*A045721((n-1)/2) ). - G. C. Greubel, Jan 23 2025

A090816 a(n) = (3*n+1)!/((2*n)! * n!).

Original entry on oeis.org

1, 12, 105, 840, 6435, 48048, 352716, 2558160, 18386775, 131231100, 931395465, 6580248480, 46312074900, 324897017760, 2272989850440, 15863901576864, 110487596768703, 768095592509700, 5330949171823275, 36945070220658600, 255702514854135195, 1767643865751234240
Offset: 0

Views

Author

Al Hakanson (hawkuu(AT)excite.com), Feb 11 2004

Keywords

Examples

			a(1) = 4!/(2!*1!) = 24/2 = 12.
		

Crossrefs

Halfdiagonal of triangle A003506.

Programs

  • Magma
    [Factorial(3*n+1)/(Factorial(n)*Factorial(2*n)): n in [0..20]]; // G. C. Greubel, Feb 03 2019
    
  • Maple
    a:=n-> binomial(3*n+1,2*n)*(n+1): seq(a(n), n=0..20); # Zerinvary Lajos, Jul 31 2006
  • Mathematica
    f[n_] := 1/Integrate[(x^2 - x^3)^n, {x, 0, 1}]; Table[ f[n], {n, 0, 19}] (* Robert G. Wilson v, Feb 18 2004 *)
    Table[1/Beta[2*n+1,n+1], {n,0,20}] (* G. C. Greubel, Feb 03 2019 *)
  • PARI
    a(n)=if(n<0,0,(3*n+1)!/(2*n)!/n!) /* Michael Somos, Feb 14 2004 */
    
  • PARI
    a(n)=if(n<0,0,1/subst(intformal((x^2-x^3)^n),x,1)) /* Michael Somos, Feb 14 2004 */
    
  • Sage
    [1/beta(2*n+1,n+1) for n in range(20)] # G. C. Greubel, Feb 03 2019

Formula

a(n) = A005809(n) * A016777(n).
a(n) = 1/(Integral_{x=0..1} (x^2 - x^3)^n dx).
G.f.: (((8 + 27*z)*(1/(4*sqrt(4 - 27*z) + 12*i*sqrt(3)*sqrt(z))^(1/3) + 1/(4*sqrt(4 - 27*z) - 12*i*sqrt(3)*sqrt(z))^(1/3)) - 3*i*sqrt(3)*sqrt(4 - 27*z)*sqrt(z)*(1/(4*sqrt(4 - 27*z) + 12*i*sqrt(3)*sqrt(z))^(1/3) - 1/(4*sqrt(4 - 27*z) - 12*i*sqrt(3)*sqrt(z))^(1/3)))*8^(1/3))/(2*(4 - 27*z)^(3/2)), where i is the imaginary unit. - Karol A. Penson, Feb 06 2024
a(n) = Sum_{k = 0..n} (-1)^(n+k) * (2*n + 2*k + 1)*binomial(2*n+k, k). This is the particular case m = 1 of the identity Sum_{k = 0..m*n} (-1)^k * (2*n + 2*k + 1) * binomial(2*n+k, k) = (-1)^(m*n) * (m*n + 1) * binomial((m+2)*n+1, 2*n). Cf. A002457 and A306290. - Peter Bala, Nov 02 2024
From Amiram Eldar, Dec 09 2024: (Start)
Sum_{n>=0} 1/a(n) = f(c) = 1.09422712102982285131..., where f(x) = (x*(x-1)/(3*x-1)) * ((3/2)*log(abs(x/(x-1))) + ((3*x-2)/sqrt(3*x^2-4*x)) * (arctan(x/sqrt(3*x^2-4*x)) + arctan((2-x)/sqrt(3*x^2-4*x)))), and c = 2/3 + (1/3)*((25+3*sqrt(69))/2)^(-1/3) + (1/3)*((25+3*sqrt(69))/2)^(1/3).
Sum_{n>=0} (-1)^n/a(n) = f(d) = 0.92513707957813718109..., where f(x) is defined above, and d = 2/3 - (1/3)*((29+3*sqrt(93))/2)^(-1/3) - (1/3)*((29+3*sqrt(93))/2)^(1/3).
Both formulas are from Batir (2013). (End)

Extensions

New definition from Vladeta Jovovic, Feb 12 2004

A119301 Triangle read by rows: T(n,k) = binomial(3*n-k,n-k).

Original entry on oeis.org

1, 3, 1, 15, 5, 1, 84, 28, 7, 1, 495, 165, 45, 9, 1, 3003, 1001, 286, 66, 11, 1, 18564, 6188, 1820, 455, 91, 13, 1, 116280, 38760, 11628, 3060, 680, 120, 15, 1, 735471, 245157, 74613, 20349, 4845, 969, 153, 17, 1, 4686825, 1562275, 480700, 134596, 33649, 7315
Offset: 0

Views

Author

Paul Barry, May 13 2006

Keywords

Comments

First column is A005809. Second column is A025174.
Row sums are A045721. Inverse is Riordan array (1-3x,x(1-x)^2), A119302.

Examples

			Triangle begins
  1,
  3, 1,
  15, 5, 1,
  84, 28, 7, 1,
  495, 165, 45, 9, 1,
  3003, 1001, 286, 66, 11, 1,
  18564, 6188, 1820, 455, 91, 13, 1,
  116280, 38760, 11628, 3060, 680, 120, 15, 1
  ...
Horizontal recurrence: T(4,1) = 1*84 + 2*28 + 3*7 + 4*1 = 165. - _Peter Bala_, Dec 29 2014
		

Crossrefs

Programs

  • Maple
    T := proc(n,k) option remember;
    `if`(n = 0, 1, add(i*T(n-1,k-2+i),i=1..n+1-k)) end:
    for n from 0 to 9 do print(seq(T(n,k),k=0..n)) od; # Peter Luschny, Dec 30 2014
  • Mathematica
    Flatten[Table[Binomial[3n-k,n-k],{n,0,10},{k,0,n}]] (* Harvey P. Dale, Jul 28 2012 *)

Formula

G.f. g(x) = 2*sin(arcsin(3*sqrt(3*x)/2)/3)/sqrt(3*x) satisfies g(x) = 1/(1-x*g(x)^2).
Riordan array (1/(1-3*x*g(x)^2),x*g(x)^2) where g(x)=1+x*g(x)^3.
'Horizontal' recurrence equation: T(n,0) = binomial(3*n,n) and for k >= 1, T(n,k) = Sum_{i = 1..n+1-k} i*T(n-1,k-2+i). - Peter Bala, Dec 28 2014
T(n, k) = Sum_{j = 0..n} binomial(n+j-1, j)*binomial(2*n-k-j, n). - Peter Bala, Jun 04 2024

A065942 Central column of triangle A065941.

Original entry on oeis.org

1, 1, 3, 4, 15, 21, 84, 120, 495, 715, 3003, 4368, 18564, 27132, 116280, 170544, 735471, 1081575, 4686825, 6906900, 30045015, 44352165, 193536720, 286097760, 1251677700, 1852482996, 8122425444, 12033222880, 52860229080, 78378960360
Offset: 0

Views

Author

Len Smiley, Nov 29 2001

Keywords

Comments

When viewed as (1,1), (3,4), (15,21), ... this represents a shallow staircase on Pascal's triangle, arranged as a square array. - Paul Barry, Mar 11 2003
Also central column of triangle A011973 (taking rows with odd number of terms only). - John Molokach, Jul 08 2013
Interleaving of A005809 and A045721. - Bruce J. Nicholson, Apr 24 2018

Examples

			G.f. = 1 + x + 3*x^2 + 4*x^3 + 15*x^4 + 21*x^5 + 84*x^6 + 120*x^7 + ... - _Michael Somos_, Jun 23 2018
		

References

  • Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001 (Chapter 14)

Crossrefs

Cf. A065941 (complete triangle), A047749.

Programs

  • GAP
    List([0..40],n->Binomial(n+Int(n/2),n)); # Muniru A Asiru, Apr 28 2018
  • Mathematica
    Array[Binomial[# + Floor[#/2], #] &, 30, 0] (* Michael De Vlieger, Apr 27 2018 *)
  • PARI
    a(n) = binomial(n+n\2, n); \\ Altug Alkan, Apr 24 2018
    

Formula

a(n) = binomial(2n-floor((n+1)/2), floor(n/2)).
a(n+1) = Sum_{k=0..ceiling(n/2)} binomial(n+k, k). - Benoit Cloitre, Mar 06 2004
a(n) = binomial(n+floor(n/2), n). - Paul Barry, May 18 2004
a(n) = Sum_{k=0..floor(n/2)} binomial(n-1+k, k). - Paul Barry, Jul 06 2004
a(2n-1) = binomial(3n-3,n-1); a(2n) = binomial(3n-2,n-1). - John Molokach, Jul 08 2013
G.f.: A(x) = x*(d/dx)[log(S(x)-1)] = x*[(d/dx) S(x)]/[S(x)-1], where S(x) is the g.f. of A047749. - Vladimir Kruchinin, Jun 12 2014.
Conjecture: 8*n*(n-1)*a(n) -36*(n-1)*(n-3)*a(n-1) +6*(-9*n^2+18*n-14)*a(n-2) +27*(3*n-7)*(3*n-8)*a(n-3)=0. - R. J. Mathar, Jun 13 2014
0 = a(n)*(+281138850*a(n+2) +729089100*a(n+3) -77071527*a(n+4) -134472793*a(n+5)) +a(n+1)*(+15618825*a(n+2) -1650969*a(n+3) -9342280*a(n+4) -1729448*a(n+5)) +a(n+2)*(-19089675*a(n+2) -61394833*a(n+3) +6470716*a(n+4) +14929796*a(n+5)) +a(n+3)*(-1291668*a(n+3) +553572*a(n+4) +246032*a(n+5)) for all n in Z. - Michael Somos, Jun 23 2018

A052227 a(n) = (4*n+1)*binomial(3*n,n)/(2*n+1).

Original entry on oeis.org

1, 5, 27, 156, 935, 5733, 35700, 224808, 1427679, 9126975, 58659315, 378658800, 2453288292, 15944020316, 103897691640, 678610095504, 4441369072335, 29120107628115, 191233066114545, 1257635016353100
Offset: 0

Views

Author

Barry E. Williams, Jan 29 2000

Keywords

Comments

T(2n,n) for A111125. - Paul Barry, Apr 19 2007
a(n) = A182584(2*n+1). - Reinhard Zumkeller, May 06 2012

Crossrefs

Programs

  • Haskell
    a052227 n = (a016813 n) * (a005809 n) `div` (a005408 n)
    -- Reinhard Zumkeller, May 06 2012
    
  • Magma
    [(4*n+1)*Binomial(3*n,n)/(2*n+1) : n in [0..30]]; // Vincenzo Librandi, Nov 13 2011
    
  • Mathematica
    Table[(4n + 1)Binomial[3n, n]/(2n + 1), {n, 0, 30}] (* Harvey P. Dale, Jan 31 2011 *)
  • Maxima
    makelist(binomial(3*n,n)*(4*n+1)/(2*n+1),n,0,100); /* Emanuele Munarini, Jun 06 2011 */
    
  • PARI
    {a(n)=binomial(3*n+1, n)+binomial(3*n, n-1)}  /* Paul D. Hanna, Jul 22 2013 */

Formula

G.f.: 4*x*F(4/3,5/3;5/2;27*x/4) + 2*sin(1/3*arcsin((3*sqrt(3*x))/2))/sqrt(3*x), where F(a;b;z) is a hypergeometric series. - Emanuele Munarini, Jun 06 2011
G.f.: (g+1)/((3*g-1)*(g-1)) where g*(1-g)^2 = x. - Mark van Hoeij, Nov 10 2011
Conjecture: 8*n*(2*n+1)*a(n) +6*(-8*n^2-25*n+13)*a(n-1) -45*(3*n-4)*(3*n-5)*a(n-2)=0. - R. J. Mathar, Nov 24 2012
a(n) = binomial(3*n+1, n) + binomial(3*n, n-1) for n>=0. - Paul D. Hanna, Jul 22 2013
G.f.: G(x)*(2*G(x) - 1) / (3 - 2*G(x)), where G(x) = 1 + x*G(x)^3 is the g.f. of A001764. - Paul D. Hanna, Jul 22 2013
a(n) is the coefficient of [x^n] in (1+x)/(1-x)^(2n+2) and forms the main diagonal in the following table of coefficients:
(1+x)/(1-x)^2: [1, 3, 5, 7, 9, 11, 13, 15, 17, ...];
(1+x)/(1-x)^4: [1, 5, 14, 30, 55, 91, 140, 204, 285, ...];
(1+x)/(1-x)^6: [1, 7, 27, 77, 182, 378, 714, 1254, ...];
(1+x)/(1-x)^8: [1, 9, 44, 156, 450, 1122, 2508, 5148, ...];
(1+x)/(1-x)^10:[1, 11, 65, 275, 935, 2717, 7007, 16445, ...];
(1+x)/(1-x)^12:[1, 13, 90, 442, 1729, 5733, 16744, 44200, ...];
(1+x)/(1-x)^14:[1, 15, 119, 665, 2940, 10948, 35700, 104652, ...];
(1+x)/(1-x)^16:[1, 17, 152, 952, 4692, 19380, 69768, 224808, ...]; ... - Paul D. Hanna, Jul 22 2013

Extensions

More terms from Harvey P. Dale, Jan 31 2011

A099578 a(n) = binomial(floor((3n+2)/2), floor(n/2)).

Original entry on oeis.org

1, 1, 4, 5, 21, 28, 120, 165, 715, 1001, 4368, 6188, 27132, 38760, 170544, 245157, 1081575, 1562275, 6906900, 10015005, 44352165, 64512240, 286097760, 417225900, 1852482996, 2707475148, 12033222880, 17620076360, 78378960360, 114955808528
Offset: 0

Views

Author

Paul Barry, Oct 23 2004

Keywords

Comments

Main diagonal of triangle A099575.
With offset 2, this is the number of compositions of n-1 into floor(n/2) parts. - T. D. Noe, Jan 05 2007
From Petros Hadjicostas, Jul 19 2018: (Start)
We clarify the above comment by T. D. Noe. The number of compositions of N into K positive parts is C(N-1, K-1). This was proved by MacMahon in 1893 (and probably by others before him). The number of compositions of N into K nonnegative parts is C(N+K-1, K-1) because for every composition b_1 + ... + b_K = N with b_i >= 0 for all i, we may create another composition c_1 + ... + c_K = N+K with c_i = b_i + 1 >= 1.
The statement of T. D. Noe above means that, for n>=2, a(n-2) is the number of compositions of N = n-1 into K = floor(n/2) nonnegative parts. Thus, a(n-2) = C(N+K-1, K-1) = C(n-1+floor(n/2)-1, floor(n/2)-1) = C(floor((3(n-2)+2)/2), floor((n-2)/2)).
This interpretation is important for T. D. Noe's comments for sequence A030077, whose unknown general formula remains an unsolved problem (as of July 2018).
It should be noted, however, that for most authors "composition" means "composition into positive parts". The phrase "weak composition" is sometimes used for a "composition into nonnegative parts".
(End)

Examples

			From _Petros Hadjicostas_, Jul 19 2018: (Start)
With n=2 there are a(2-2) = a(0) = 1 compositions of 2-1 = 1 into floor(2/2) = 1 nonnegative parts, namely 1 (only).
With n=3 there are a(3-2) = a(1) = 1 compositions of 3-1 = 2 into floor(3/2) = 1 nonnegative parts, namely 2 (only).
With n=4 there are a(4-2) = a(2) = 4 compositions of 4-1 = 3 into floor(4/2) = 2 nonnegative parts, namely 0+3, 3+0, 1+2, and 2+1.
With n=5 there are a(5-2) = a(3) = 5 compositions of 5-1 = 4 into floor(5/2) = 2 nonnegative parts, namely 0+4, 4+0, 1+3, 3+1, and 2+2.
With n=6 there are a(6-2) = a(4) = 21 compositions of 6-1 = 5 into floor(6/2) = 3 nonnegative parts, namely the 3 permutations of 1+1+3, the 3 permutations of 1+2+2, the 3 permutations of 0+0+5, the 6 permutations of 0+1+4, and the 6 permutations of 0+2+3.
(End)
		

Crossrefs

Cf. A025174 (bisection), A030077, A045721 (bisection), A099575, A127040.

Programs

  • Magma
    [(&+[Binomial(n+j, j): j in [0..Floor(n/2)]]): n in [0..40]]; // G. C. Greubel, Jul 24 2022
    
  • Maple
    A099578:=n->binomial(floor((3*n+2)/2), floor(n/2)); seq(A099578(k), k=0..50); # Wesley Ivan Hurt, Nov 01 2013
  • Mathematica
    Table[Binomial[Floor[(3n+2)/2], Floor[n/2]], {n, 0, 50}] (* Wesley Ivan Hurt, Nov 01 2013 *)
    CoefficientList[Series[-((Sqrt[4 -27 x^2] -2(Cos[1/3 ArcSin[(3 Sqrt[3] x)/2]] + Sqrt[3] Sin[2/3 ArcSin[(3 Sqrt[3] x)/2]]))/(3 x Sqrt[4 -27 x^2])), {x, 0, 20}], x] (* Benedict W. J. Irwin, Aug 15 2016 *)
    a[n_] := Binomial[2*n+1, n]*Hypergeometric2F1[-n, n+1, -2*n-1, -1]; Flatten[Table[a[n], {n, 0, 29}]] (* Detlef Meya, Dec 25 2023 *)
  • PARI
    a(n) = binomial((3*n+2)\2, n\2); \\ Michel Marcus, Nov 02 2013
    
  • SageMath
    [binomial((3*n+2)//2, n//2) for n in (0..40)] # G. C. Greubel, Jul 24 2022

Formula

a(n) = Sum_{k=0..floor(n/2)} binomial(n+k, k).
40*n*(n+1)*a(n) +36*n*(n-2)*a(n-1) -6*(45*n^2-23)*a(n-2) -27*(3*n-4)*(3*n-5)*a(n-3) = 0. - R. J. Mathar, Oct 30 2014
From Benedict W. J. Irwin, Aug 15 2016: (Start)
G.f.: -( (sqrt(4-27*x^2) - 2*(cos(arcsin(3*sqrt(3)*x/2)/3) + sqrt(3)*sin(2*arcsin(3*sqrt(3)*x/2)/3)) )/( 3*x*sqrt(4-27*x^2)) ).
E.g.f.: Hypergeometric2F3(2/3,4/3;1/2,1,3/2;27*x^2/16) + x*Hypergeometric2F3(4/3,5/3;3/2,3/2,2;27*x^2/16).
(End)
Recurrence: 4*n*(n+1)*(6*n-1)*a(n) = 18*n*a(n-1) + 3*(3*n-2)*(3*n-1)*(6*n+5)*a(n-2). - Vaclav Kotesovec, Aug 15 2016
a(n) = binomial(2*n+1, n)*hypergeom([-n, n+1], [-2*n-1], -1). - Detlef Meya, Dec 25 2023

A159841 Triangle T(n,k) = binomial(3*n+1, 2*n+k+1), read by rows.

Original entry on oeis.org

1, 4, 1, 21, 7, 1, 120, 45, 10, 1, 715, 286, 78, 13, 1, 4368, 1820, 560, 120, 16, 1, 27132, 11628, 3876, 969, 171, 19, 1, 170544, 74613, 26334, 7315, 1540, 231, 22, 1, 1081575, 480700, 177100, 53130, 12650, 2300, 300, 25, 1, 6906900, 3108105, 1184040, 376740
Offset: 0

Views

Author

Philippe Deléham, Apr 23 2009

Keywords

Comments

T(n,0) = A045721(n), T(2n,n) = A079590(n).

Examples

			Triangle begins:
     1;
     4,    1;
    21,    7,    1;
   120,   45,   10,    1;
   715,  286,   78,   13,    1;
  4368, 1820,  560,  120,   16,    1;
  ...
		

Crossrefs

Programs

  • Magma
    /* As triangle */ [[Binomial(3*n+1, 2*n+k+1): k in [0..n]]: n in [0..10]]; // G. C. Greubel, May 19 2018
  • Mathematica
    f[n_,k_]:=Binomial[3n+1,2n+k+1]; Table[ f[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Robert G. Wilson v, May 31 2009 *)
  • PARI
    for(n=0,10, for(k=0,n, print1(binomial(3*n+1, 2*n+k+1), ", "))) \\ G. C. Greubel, May 19 2018
    

Formula

T(n,0) = 4*T(n-1,0) + 5*T(n-1,1) + T(n-1,2), T(n+1,k+1) = T(n,k) + 3*T(n,k+1) + 3*T(n,k+2) + T(n,k+3) for k >= 0.

Extensions

More terms from Robert G. Wilson v, May 31 2009

A263134 a(n) = Sum_{k=0..n} binomial(3*k+1,k).

Original entry on oeis.org

1, 5, 26, 146, 861, 5229, 32361, 202905, 1284480, 8191380, 52543545, 338641305, 2191124301, 14224347181, 92603307541, 604342068085, 3952451061076, 25898039418496, 169977746765071, 1117287239602471, 7353933943361866, 48461930821297546
Offset: 0

Views

Author

Bruno Berselli, Oct 10 2015

Keywords

Comments

Primes in sequence: 5, 92603307541, 52176309488123582020412161, ...
a(n) is divisible by n for n = 1, 2, 8, 55, 82, 171, 210, 1060, 1141, ...

Crossrefs

Partial sums of A045721.
Cf. A079309: Sum_{k=0..n} binomial(2*k+1,k).
Cf. A188675: Sum_{k=0..n} binomial(3*k,k).
Cf. A087413: Sum_{k=0..n} binomial(3*k+2,k).

Programs

  • Magma
    [&+[Binomial(3*k+1,k): k in [0..n]]: n in [0..25]];
    
  • Mathematica
    Table[Sum[Binomial[3 k + 1, k], {k, 0, n}], {n, 0, 25}]
  • Maxima
    makelist(sum(binomial(3*k+1,k),k,0,n),n,0,25);
    
  • PARI
    a(n) = sum(k=0, n, binomial(3*k+1,k)) \\ Colin Barker, Oct 16 2015
  • Sage
    [sum(binomial(3*k+1,k) for k in (0..n)) for n in (0..25)]
    

Formula

Recurrence: 2*n*(2*n + 1)*a(n) = (31*n^2 + 2*n - 3)*a(n-1) - 3*(3*n - 1)*(3*n + 1)*a(n-2). - Vaclav Kotesovec, Oct 11 2015
a(n) ~ 27^(n + 3/2)/(23*sqrt(Pi*n)*4^(n + 1)). - Vaclav Kotesovec, Oct 11 2015

A144485 a(n) = (3n + 2)*binomial(3n + 1,n).

Original entry on oeis.org

2, 20, 168, 1320, 10010, 74256, 542640, 3922512, 28120950, 200300100, 1419269280, 10013421600, 70394353848, 493362138080, 3448674255840, 24051721745568, 167405449649550, 1163116182943260, 8068463611408200, 55891260077406600
Offset: 0

Views

Author

Roger L. Bagula, Oct 12 2008

Keywords

Crossrefs

Programs

  • Magma
    [(3*n+2)*Binomial(3*n+1, n): n in [0..20]]; // Vincenzo Librandi, Feb 14 2014
  • Maple
    a:= proc(n) option remember; `if`(n=0, 2,
          3*(3*n+1)*(3*n+2)*a(n-1)/(2*n*(2*n+1)))
        end:
    seq(a(n), n=0..30);  # Alois P. Heinz, Feb 01 2014
  • Mathematica
    a[k_] = (3k + 2)Binomial[3k + 1, k]; Table[a[k], {k, 0, 30}]

Formula

a(n) = (3n+2)*A045721(n). - R. J. Mathar, Feb 01 2014
a(n) = 2*A090763(n). - Alois P. Heinz, Feb 01 2014
From Amiram Eldar, Dec 07 2024: (Start)
a(n) = 2 * (n+1) * A005809(n+1) / 3.
Sum_{n>=0} 1/a(n) = (3/2) * A210453. (End)

A188911 Binomial convolution of the binomial coefficients bin(3n,n) (A005809).

Original entry on oeis.org

1, 6, 48, 438, 4356, 46056, 509106, 5814738, 68050116, 811240872, 9810384048, 119990105208, 1481115683754, 18421300391760, 230574816629310, 2901721280735838, 36688485233689668, 465774244616805624, 5934465567864915024
Offset: 0

Views

Author

Emanuele Munarini, Apr 13 2011

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n,k]Binomial[3k,k]Binomial[3n-3k,n-k], {k,0,n}], {n,0,22}]
  • Maxima
    makelist(sum(binomial(n,k)*binomial(3*k,k)*binomial(3*n-3*k,n-k),k,0,n),n,0,12);
    
  • PARI
    a(n)=sum(k=0,n,binomial(n,k)*binomial(3*k,k)*binomial(3*n-3*k,n-k));
    vector(66,n,a(n-1)) /* show terms */ /* Joerg Arndt, Apr 13 2011 */

Formula

a(n) = Sum_{k=0..n} binomial(n,k)*binomial(3*k,k)*binomial(3*n-3*k,n-k).
E.g.f.: F(1/3,2/3;1/2,1;27*x/4)^2, where F(a1,a2;b1,b2;z) is a hypergeometric series.
Recurrence: 8*n^2 * (2*n-1)^2 * (9*n^3 - 54*n^2 + 102*n - 61)*a(n) = 24*(3*n-1)*(108*n^6 - 855*n^5 + 2628*n^4 - 4059*n^3 + 3380*n^2 - 1470*n + 264)*a(n-1) - 18*(3645*n^7 - 34992*n^6 + 138348*n^5 - 291843*n^4 + 352980*n^3 - 241794*n^2 + 84684*n - 11104)*a(n-2) + 2187*(n-2)^2 * (3*n-7)*(3*n-5)*(9*n^3 - 27*n^2 + 21*n - 4)*a(n-3). - Vaclav Kotesovec, Feb 25 2014
a(n) ~ 3^(3*n+1) / (Pi * n * 2^(n+1)). - Vaclav Kotesovec, Feb 25 2014
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