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.

Showing 1-8 of 8 results.

A000389 Binomial coefficients C(n,5).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 6, 21, 56, 126, 252, 462, 792, 1287, 2002, 3003, 4368, 6188, 8568, 11628, 15504, 20349, 26334, 33649, 42504, 53130, 65780, 80730, 98280, 118755, 142506, 169911, 201376, 237336, 278256, 324632, 376992, 435897, 501942, 575757, 658008, 749398
Offset: 0

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Author

Keywords

Comments

a(n+4) is the number of inequivalent ways of coloring the vertices of a regular 4-dimensional simplex with n colors, under the full symmetric group S_5 of order 120, with cycle index (x1^5 + 10*x1^3*x2 + 20*x1^2*x3 + 15*x1*x2^2 + 30*x1*x4 + 20*x2*x3 + 24*x5)/120.
Figurate numbers based on 5-dimensional regular simplex. According to Hyun Kwang Kim, it appears that every nonnegative integer can be represented as the sum of g = 10 of these 5-simplex(n) numbers (compared with g=3 for triangular numbers, g=5 for tetrahedral numbers and g=8 for pentatope numbers). - Jonathan Vos Post, Nov 28 2004
The convolution of the nonnegative integers (A001477) with the tetrahedral numbers (A000292), which are the convolution of the nonnegative integers with themselves (making appropriate allowances for offsets of all sequences). - Graeme McRae, Jun 07 2006
a(n) is the number of terms in the expansion of (a_1 + a_2 + a_3 + a_4 + a_5 + a_6)^n. - Sergio Falcon, Feb 12 2007
Product of five consecutive numbers divided by 120. - Artur Jasinski, Dec 02 2007
Equals binomial transform of [1, 5, 10, 10, 5, 1, 0, 0, 0, ...]. - Gary W. Adamson, Feb 02 2009
Equals INVERTi transform of A099242 (1, 7, 34, 153, 686, 3088, ...). - Gary W. Adamson, Feb 02 2009
For a team with n basketball players (n>=5), this sequence is the number of possible starting lineups of 5 players, without regard to the positions (center, forward, guard) of the players. - Mohammad K. Azarian, Sep 10 2009
a(n) is the number of different patterns, regardless of order, when throwing (n-5) 6-sided dice. For example, one die can display the 6 numbers 1, 2, ..., 6; two dice can display the 21 digit-pairs 11, 12, ..., 56, 66. - Ian Duff, Nov 16 2009
Sum of the first n pentatope numbers (1, 5, 15, 35, 70, 126, 210, ...), see A000332. - Paul Muljadi, Dec 16 2009
Sum_{n>=0} a(n)/n! = e/120. Sum_{n>=4} a(n)/(n-4)! = 501*e/120. See A067764 regarding the second ratio. - Richard R. Forberg, Dec 26 2013
For a set of integers {1,2,...,n}, a(n) is the sum of the 2 smallest elements of each subset with 4 elements, which is 3*C(n+1,5) (for n>=4), hence a(n) = 3*C(n+1,5) = 3*A000389(n+1). - Serhat Bulut, Mar 11 2015
a(n) = fallfac(n,5)/5! is also the number of independent components of an antisymmetric tensor of rank 5 and dimension n >= 1. Here fallfac is the falling factorial. - Wolfdieter Lang, Dec 10 2015
Number of compositions (ordered partitions) of n+1 into exactly 6 parts. - Juergen Will, Jan 02 2016
Number of weak compositions (ordered weak partitions) of n-5 into exactly 6 parts. - Juergen Will, Jan 02 2016
a(n+3) could be the general number of all geodetic graphs of diameter n>=2 homeomorphic to the Petersen Graph. - Carlos Enrique Frasser, May 24 2018
From Robert A. Russell, Dec 24 2020: (Start)
a(n) is the number of chiral pairs of colorings of the 5 tetrahedral facets (or vertices) of the regular 4-D simplex (5-cell, pentachoron, Schläfli symbol {3,3,3}) using subsets of a set of n colors. Each member of a chiral pair is a reflection but not a rotation of the other.
a(n+4) is the number of unoriented colorings of the 5 tetrahedral facets of the regular 4-D simplex (5-cell, pentachoron) using subsets of a set of n colors. Each chiral pair is counted as one when enumerating unoriented arrangements. (End)
For integer m and positive integer r >= 4, the polynomial a(n) + a(n + m) + a(n + 2*m) + ... + a(n + r*m) in n has its zeros on the vertical line Re(n) = (4 - r*m)/2 in the complex plane. - Peter Bala, Jun 02 2024

Examples

			G.f. = x^5 + 6*x^6 + 21*x^7 + 56*x^8 + 126*x^9 + 252*x^10 + 462*x^11 + ...
For A={1,2,3,4}, the only subset with 4 elements is {1,2,3,4}; sum of 2 minimum elements of this subset: a(4) = 1+2 = 3 = 3*C(4+1,5).
For A={1,2,3,4,5}, the subsets with 4 elements are {1,2,3,4}, {1,2,3,5}, {1,2,4,5}, {1,3,4,5}, {2,3,4,5}; sum of 2 smallest elements of each subset: a(5) = (1+2)+(1+2)+(1+2)+(1+3)+(2+3) = 18 = 3*C(5+1,5). - _Serhat Bulut_, Mar 11 2015
a(6) = 6 from the six independent components of an antisymmetric tensor A of rank 5 and dimension 6: A(1,2,3,4,5), A(1,2,3,4,6), A(1,2,3,5,6), A(1,2,4,5,6), A(1,3,4,5,6), A(2,3,4,5,6). See the Dec 10 2015 comment. - _Wolfdieter Lang_, Dec 10 2015
		

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.
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 196.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 7.
  • Gupta, Hansraj; Partitions of j-partite numbers into twelve or a smaller number of parts. Collection of articles dedicated to Professor P. L. Bhatnagar on his sixtieth birthday. Math. Student 40 (1972), 401-441 (1974).
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • 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. A099242. - Gary W. Adamson, Feb 02 2009
Cf. A242023. A104712 (fourth column, k=5).
5-cell colorings: A337895 (oriented), A132366(n-1) (achiral).
Unoriented colorings: A063843 (5-cell edges, faces), A128767 (8-cell vertices, 16-cell facets), A337957 (16-cell vertices, 8-cell facets), A338949 (24-cell), A338965 (600-cell vertices, 120-cell facets).
Chiral colorings: A331352 (5-cell edges, faces), A337954 (8-cell vertices, 16-cell facets), A234249 (16-cell vertices, 8-cell facets), A338950 (24-cell), A338966 (600-cell vertices, 120-cell facets).

Programs

  • Haskell
    a000389 n = a000389_list !! n
    a000389_list = 0 : 0 : f [] a000217_list where
       f xs (t:ts) = (sum $ zipWith (*) xs a000217_list) : f (t:xs) ts
    -- Reinhard Zumkeller, Mar 03 2015, Apr 13 2012
    
  • Magma
    [Binomial(n, 5): n in [0..40]]; // Vincenzo Librandi, Mar 12 2015
  • Maple
    f:=n->(1/120)*(n^5-10*n^4+35*n^3-50*n^2+24*n): seq(f(n), n=0..60);
    ZL := [S, {S=Prod(B,B,B,B,B,B), B=Set(Z, 1 <= card)}, unlabeled]: seq(combstruct[count](ZL, size=n+1), n=0..42); # Zerinvary Lajos, Mar 13 2007
    A000389:=1/(z-1)**6; # Simon Plouffe, 1992 dissertation
  • Mathematica
    Table[Binomial[n, 5], {n, 5, 50}] (* Stefan Steinerberger, Apr 02 2006 *)
    CoefficientList[Series[x^5 / (1 - x)^6, {x, 0, 40}], x] (* Vincenzo Librandi, Mar 12 2015 *)
    LinearRecurrence[{6,-15,20,-15,6,-1},{0,0,0,0,0,1},50] (* Harvey P. Dale, Jul 17 2016 *)
  • PARI
    (conv(u,v)=local(w); w=vector(length(u),i,sum(j=1,i,u[j]*v[i+1-j])); w);
    (t(n)=n*(n+1)/2); u=vector(10,i,t(i)); conv(u,u)
    

Formula

G.f.: x^5/(1-x)^6.
a(n) = n*(n-1)*(n-2)*(n-3)*(n-4)/120.
a(n) = (n^5-10*n^4+35*n^3-50*n^2+24*n)/120. (Replace all x_i's in the cycle index with n.)
a(n+2) = Sum_{i+j+k=n} i*j*k. - Benoit Cloitre, Nov 01 2002
Convolution of triangular numbers (A000217) with themselves.
Partial sums of A000332. - Alexander Adamchuk, Dec 19 2004
a(n) = -A110555(n+1,5). - Reinhard Zumkeller, Jul 27 2005
a(n+3) = (1/2!)*(d^2/dx^2)S(n,x)|A049310.%20-%20_Wolfdieter%20Lang">{x=2}, n>=2, one half of second derivative of Chebyshev S-polynomials evaluated at x=2. See A049310. - _Wolfdieter Lang, Apr 04 2007
a(n) = A052787(n+5)/120. - Zerinvary Lajos, Apr 26 2007
Sum_{n>=5} 1/a(n) = 5/4. - R. J. Mathar, Jan 27 2009
For n>4, a(n) = 1/(Integral_{x=0..Pi/2} 10*(sin(x))^(2*n-9)*(cos(x))^9). - Francesco Daddi, Aug 02 2011
Sum_{n>=5} (-1)^(n + 1)/a(n) = 80*log(2) - 655/12 = 0.8684411114... - Richard R. Forberg, Aug 11 2014
a(n) = -a(4-n) for all n in Z. - Michael Somos, Oct 07 2014
0 = a(n)*(+a(n+1) + 4*a(n+2)) + a(n+1)*(-6*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Oct 07 2014
a(n) = 3*C(n+1, 5) = 3*A000389(n+1). - Serhat Bulut, Mar 11 2015
From Ilya Gutkovskiy, Jul 23 2016: (Start)
E.g.f.: x^5*exp(x)/120.
Inverse binomial transform of A054849. (End)
From Robert A. Russell, Dec 24 2020: (Start)
a(n) = A337895(n) - a(n+4) = (A337895(n) - A132366(n-1)) / 2 = a(n+4) - A132366(n-1).
a(n+4) = A337895(n) - a(n) = (A337895(n) + A132366(n-1)) / 2 = a(n) + A132366(n-1).
a(n+4) = 1*C(n,1) + 4*C(n,2) + 6*C(n,3) + 4*C(n,4) + 1*C(n,5), where the coefficient of C(n,k) is the number of unoriented pentachoron colorings using exactly k colors. (End)

Extensions

Corrected formulas that had been based on other offsets. - R. J. Mathar, Jun 16 2009
I changed the offset to 0. This will require some further adjustments to the formulas. - N. J. A. Sloane, Aug 01 2010

A005584 Coefficients of Chebyshev polynomials.

Original entry on oeis.org

2, 13, 49, 140, 336, 714, 1386, 2508, 4290, 7007, 11011, 16744, 24752, 35700, 50388, 69768, 94962, 127281, 168245, 219604, 283360, 361790, 457470, 573300, 712530, 878787, 1076103, 1308944, 1582240, 1901416, 2272424, 2701776, 3196578, 3764565, 4414137, 5154396
Offset: 1

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Author

Keywords

Comments

If X is an n-set and Y a fixed 2-subset of X then a(n-6) is equal to the number of (n-6)-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007
a(n-1) = risefac(n+1,6)/6! - risefac(n+1,4)/4! is for n >=1 also the number of independent components of a symmetric traceless tensor of rank 6 and dimension n. Here risefac is the rising factorial. Put a(0) = 0. - Wolfdieter Lang, Dec 10 2015

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. 797.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

Formula

G.f.: x*(2-x) / (1-x)^7.
a(n) = binomial(n+5, n-1) + binomial(n+4, n-1) = 1/720*n*(n+11)*(n+4)*(n+3)*(n+2)*(n+1).
a(n) = binomial(n,6) + 2*binomial(n,5), n >= 5. - Zerinvary Lajos, Jul 26 2006
a(n+1) = A127672(12+n, n), n >= 0, where A127672 gives the coefficients of Chebyshev's C polynomials. See the Abramowitz-Stegun reference. - Wolfdieter Lang, Dec 10 2015
From G. C. Greubel, Aug 27 2019: (Start)
a(n) = (n+11)*Pochhammer(n, 5)/6!.
E.g.f.: x*(1440 +3240*x +1920*x^2 +420*x^3 +36*x^4 +x^5)*exp(x)/6!. (End)
From Amiram Eldar, Feb 17 2023: (Start)
Sum_{n>=1} 1/a(n) = 1303391/2134440.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4160*log(2)/77 - 78994697/2134440. (End)

Extensions

More terms from Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 07 1999

A051747 a(n) = n*(n+1)*(n+2)*(n^2+7*n+32)/120.

Original entry on oeis.org

2, 10, 31, 76, 161, 308, 546, 912, 1452, 2222, 3289, 4732, 6643, 9128, 12308, 16320, 21318, 27474, 34979, 44044, 54901, 67804, 83030, 100880, 121680, 145782, 173565, 205436, 241831, 283216, 330088, 382976, 442442, 509082, 583527, 666444, 758537
Offset: 1

Views

Author

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 07 1999

Keywords

Crossrefs

Programs

  • Magma
    [n*(n+1)*(n+2)*(n^2+7*n+32)/120: n in [1..40]]; // Vincenzo Librandi, Jun 15 2011
    
  • Mathematica
    Table[(1/120)*n*(n + 1)*(n + 2)*(n^2 + 7*n + 32), {n, 60}] (* Vladimir Joseph Stephan Orlovsky, Jun 14 2011 *)
    LinearRecurrence[{6,-15,20,-15,6,-1},{2,10,31,76,161,308},60] (* Harvey P. Dale, Oct 03 2012 *)
  • PARI
    conv(u,v)=local(w); w=vector(length(u),i,sum(j=1,i,u[j]*v[i+1-j])); w; t(n)=n*(n+1)/2; u=vector(10,i,t(i)); v=vector(10,i,t(i)+1); conv(u,v)
    
  • PARI
    Vec(x*(x^2-2*x+2)/(x-1)^6 + O(x^100)) \\ Colin Barker, Mar 18 2015

Formula

a(n) = binomial(n+4, n-1)+binomial(n+2, n-1).
Convolution of triangular numbers with triangular numbers + 1, i.e. [1, 3, 6, 10, 15, 21, ...] with [2, 4, 7, 11, 16, 22, ...].
a(1)=2, a(2)=10, a(3)=31, a(4)=76, a(5)=161, a(6)=308, a(n)=6*a(n-1)- 15*a(n-2)+ 20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6). - Harvey P. Dale, Oct 03 2012
G.f.: x*(x^2-2*x+2) / (x-1)^6. - Colin Barker, Mar 18 2015

A121306 Array read by antidiagonals: a(m,n) = a(m,n-1)+a(m-1,n) but with initialization values a(0,0)=0, a(m>=1,0)=1, a(0,1)=1, a(0,n>1)=0.

Original entry on oeis.org

2, 2, 3, 2, 5, 4, 2, 7, 9, 5, 2, 9, 16, 14, 6, 2, 11, 25, 30, 20, 7, 2, 13, 36, 55, 50, 27, 8, 2, 15, 49, 91, 105, 77, 35, 9, 2, 17, 64, 140, 196, 182, 112, 44, 10, 19, 81, 204, 336, 378, 294, 156, 54, 100, 285, 540, 714, 672, 450, 210, 385, 825, 1254, 1386, 1122
Offset: 0

Views

Author

Thomas Wieder, Aug 04 2006, Aug 06 2006

Keywords

Comments

For a(1,0)=1, a(m>1,0)=0 and a(0,n>=0)=0 one gets Pascal's triangle A007318.

Examples

			Array begins
2 2 2 2 2 2 2 2 2 ...
3 5 7 9 11 13 15 17 19 ...
4 9 16 25 36 49 64 81 100 ...
5 14 30 55 91 140 204 285 385 ...
6 20 50 105 196 336 540 825 1210 ...
7 27 77 182 378 714 1254 2079 3289 ...
		

Crossrefs

Programs

  • Excel
    =Z(-1)S+ZS(-1). The very first row (not included into the table) contains the initialization values: a(0,1)=1, a(0,n>=2)=0. The very first column (not included into the table) contains the initialization values: a(m>=1,0)=1. The value a(0,0)=0 does not enter into the table.

Formula

a(m,n) = a(m,n-1)+a(m-1,n), a(0,0)=0, a(m>=1,0)=1, a(0,1)=1, a(0,n>1)=0.

Extensions

Edited by N. J. A. Sloane, Sep 15 2006

A244422 Quasi-Riordan triangle ((2-z)/(1-z), -z^2/(1-z)). Row reversed monic Chebyshev T-polynomials without vanishing columns.

Original entry on oeis.org

2, 1, 0, 1, -2, 0, 1, -3, 0, 0, 1, -4, 2, 0, 0, 1, -5, 5, 0, 0, 0, 1, -6, 9, -2, 0, 0, 0, 1, -7, 14, -7, 0, 0, 0, 0, 1, -8, 20, -16, 2, 0, 0, 0, 0, 1, -9, 27, -30, 9, 0, 0, 0, 0, 0, 1, -10, 35, -50, 25, -2, 0, 0, 0, 0, 0, 1, -11, 44, -77, 55, -11, 0, 0, 0, 0, 0, 0, 1, -12, 54, -112, 105, -36, 2, 0, 0, 0, 0, 0, 0
Offset: 0

Views

Author

Wolfdieter Lang, Aug 08 2014

Keywords

Comments

This is a signed version of the triangle A061896.
The coefficient table for the monic Chebyshev polynomials of the first kind R(n, x) = 2*T(n, x/2) is given in A127672. For the T-polynomials see A053120. The present table is obtained from the row reversed coefficient table A127672 by deleting all odd numbered columns which have only zeros, and appending in the rows numbered n >= 1 zeros in order to obtain a triangle. This becomes the quasi-Riordan triangle T = ((2-z)/(1-z), -z^2/(1-z)). This means that the o.g.f. of the row polynomials Rrev(n, x) := sqrt(x)^n*R(n, 1/sqrt(x)) = Sum_{k=0..n} T(n, k)*x^k have o.g.f. (2-z)/(1 - z + x*z^2) like for ordinary Riordan triangles. However this is not a Riordan triangle (or lower triangular infinite dimensional matrix) in the usual sense because it is not invertible. Therefore, this lower triangular matrix is not a member of the Riordan group.
The row sums give repeat(2,1,-1,-2,-1) which is A057079(n+1), n >= 0. The alternating row sums give the Lucas numbers A000032.

Examples

			The triangle T(n,k) begins:
  n\k  0   1   2     3    4     5  6   7  8  9 10 11
  0:   2
  1:   1   0
  2:   1  -2   0
  3:   1  -3   0     0
  4:   1  -4   2     0    0
  5:   1  -5   5     0    0     0
  6:   1  -6   9    -2    0     0  0
  7:   1  -7  14    -7    0     0  0   0
  8:   1  -8  20   -16    2     0  0   0  0
  9:   1  -9  27   -30    9     0  0   0  0  0
  10:  1 -10  35   -50   25    -2  0   0  0  0  0
  11:  1 -11  44   -77   55   -11  0   0  0  0  0  0
  ...
Rrev(3, x) = 1 - 3*x = sqrt(x)^3*R(3,1/sqrt(x)) = sqrt(x)^3*(-3/sqrt(x) + 1/sqrt(x)^3 ) = -3*x + 1.
Rrev(4, x) = 1 - 4*x + 2*x^2 = sqrt(x)^4*(2 - 4/sqrt(x)^2 + 1/sqrt(x)^4) = 2*x^2 - 4*x + 1.
Recurrence: T(4,1) = T(3, 1) - T(2, 0) = -3 -1 = -4.
		

Crossrefs

Formula

T(n,k) = [x^k] Rrev(n, x), k=0, 1, ..., n, with the row polynomials Rrev(n, x) = sqrt(x)^n*R(n,1/sqrt(x)), with R(n, x) given in A127672 (monic Chebyshev polynomials of the first kind).
O.g.f. row polynomials Rrev(n,x) = Sum_{k=0..n} T(n,k)*x^k: (2-z)/(1 - z + x*z^2) (quasi-Riordan).
O.g.f. for column number k entries with leading zeros: ((2-x)/(1-x))*(-x^2/(1-x))^k, k > = 0. See A054977, -A000027, A000096, -A005581, A005582, -A005583, A005584.
Recurrence: T(n,k) = T(n-1, k) - T(n-2, k-1), n >= k >= 1, T(n,k) = 0 if n < k, T(0,0) = 2, T(n,0) = 1 if n>=1, (Compare with A061896).
For n >= 1 the entries without trailing zeros are given by T(n,k) = (-1)^k*(n/(n-k))*binomial(n-k,k) where k=0..floor(n/2).

A264357 Array A(r, n) of number of independent components of a symmetric traceless tensor of rank r and dimension n, written as triangle T(n, r) = A(r, n-r+2), n >= 1, r = 2..n+1.

Original entry on oeis.org

0, 2, 0, 5, 2, 0, 9, 7, 2, 0, 14, 16, 9, 2, 0, 20, 30, 25, 11, 2, 0, 27, 50, 55, 36, 13, 2, 0, 35, 77, 105, 91, 49, 15, 2, 0, 44, 112, 182, 196, 140, 64, 17, 2, 0, 54, 156, 294, 378, 336, 204, 81, 19, 2, 0
Offset: 1

Views

Author

Wolfdieter Lang, Dec 10 2015

Keywords

Comments

A (totally) symmetric traceless tensor of rank r >= 2 and dimension n >= 1 is irreducible.
The array of the number of independent components of a rank r symmetric traceless tensor A(r, n), for r >= 2 and n >=1, is given by risefac(n,r)/r! - risefac(n,r-2)/(r-2)!, where the first term gives the number of independent components of a symmetric tensors of rank r (see a Dec 10 2015 comment under A135278) and the second term is the number of constraints from the tracelessness requirement. The tensor has to be traceless in each pair of indices.
The first rows of the array A, or the first columns (without the first r-2 zeros) of the triangle T are for r = 2..6: A000096, A005581, A005582, A005583, A005584.
Equals A115241 with the first column of positive integers removed. - Georg Fischer, Jul 26 2023

Examples

			The array A(r, n) starts:
   r\n 1 2  3   4   5    6    7     8     9    10 ...
   2:  0 2  5   9  14   20   27    35    44    54
   3:  0 2  7  16  30   50   77   112   156   210
   4:  0 2  9  25  55  105  182   294   450   660
   5:  0 2 11  36  91  196  378   672  1122  1782
   6:  0 2 13  49 140  336  714  1386  2508  4290
   7:  0 2 15  64 204  540 1254  2640  5148  9438
   8:  0 2 17  81 285  825 2079  4719  9867 19305
   9:  0 2 19 100 385 1210 3289  8008 17875 37180
  10:  0 2 21 121 506 1716 5005 13013 30888 68068
  ...
The triangle T(n, r) starts:
   n\r  2   3   4   5   6   7  8  9 10 11 ...
   1:   0
   2:   2   0
   3:   5   2   0
   4:   9   7   2   0
   5:  14  16   9   2   0
   6:  20  30  25  11   2   0
   7:  27  50  55  36  13   2  0
   8:  35  77 105  91  49  15  2  0
   9:  44 112 182 196 140  64 17  2  0
  10:  54 156 294 378 336 204 81 19  2  0
  ...
A(r, 1) = 0 , r >= 2, because a symmetric rank r tensor t of dimension one has one component t(1,1,...,1) (r 1's) and if the traces vanish then t vanishes.
A(3, 2) = 2 because a symmetric rank 3 tensor t with three indices taking values from 1 or 2 (n=2) has the four independent components t(1,1,1), t(1,1,2), t(1,2,2), t(2,2,2), and (invoking symmetry) the vanishing traces are Sum_{j=1..2} t(j,j,1) = 0 and Sum_{j=1..2} t(j,j,2) = 0. These are two constraints, which can be used to eliminate, say, t(1,1,1) and t(2,2,2), leaving 2 = A(3, 2) independent components, say, t(1,1,2) and t(1,2,2).
From _Peter Luschny_, Dec 14 2015: (Start)
The diagonals diag(n, k) start:
   k\n  0       1       2       3       4      5       6
   0:   0,      2,      9,     36,    140,   540,   2079, ... A007946
   1:   2,      7,     25,     91,    336,  1254,   4719, ... A097613
   2:   5,     16,     55,    196,    714,  2640,   9867, ... A051960
   3:   9,     30,    105,    378,   1386,  5148,  19305, ... A029651
   4:  14,     50,    182,    672,   2508,  9438,  35750, ... A051924
   5:  20,     77,    294,   1122,   4290, 16445,  63206, ... A129869
   6:  27,    112,    450,   1782,   7007, 27456, 107406, ... A220101
   7:  35,    156,    660,   2717,  11011, 44200, 176358, ... A265612
   8:  44,    210,    935,    4004, 16744, 68952, 281010, ... A265613
  A000096,A005581,A005582,A005583,A005584.
(End)
		

Crossrefs

Programs

  • Mathematica
    A[r_, n_] := Pochhammer[n, r]/r! - Pochhammer[n, r-2]/(r-2)!;
    T[n_, r_] := A[r, n-r+2];
    Table[T[n, r], {n, 1, 10}, {r, 2, n+1}] (* Jean-François Alcover, Jun 28 2019 *)
  • Sage
    A = lambda r, n: rising_factorial(n,r)/factorial(r) - rising_factorial(n,r-2)/factorial(r-2)
    for r in (2..10): [A(r,n) for n in (1..10)] # Peter Luschny, Dec 13 2015

Formula

T(n, r) = A(r, n-r+2) with the array A(r, n) = risefac(n,r)/r! - risefac(n,r-2)/(r-2)! where the rising factorial risefac(n,k) = Product_{j=0..k-1} (n+j) and risefac(n,0) = 1.
From Peter Luschny, Dec 14 2015: (Start)
A(n+2, n+1) = A007946(n-1) = CatalanNumber(n)*3*n*(n+1)/(n+2) for n>=0.
A(n+2, n+2) = A024482(n+2) = A097613(n+2) = CatalanNumber(n+1)*(3*n+4)/2 for n>=0.
A(n+2, n+3) = A051960(n+1) = CatalanNumber(n+1)*(3*n+5) for n>=0.
A(n+2, n+4) = A029651(n+2) = CatalanNumber(n+1)*(6*n+9) for n>=0.
A(n+2, n+5) = A051924(n+3) = CatalanNumber(n+2)*(3*n+7) for n>=0.
A(n+2, n+6) = A129869(n+4) = CatalanNumber(n+2)*(3*n+8)*(2*n+5)/(n+4) for n>=0.
A(n+2, n+7) = A220101(n+4) = CatalanNumber(n+3)*(3*(n+3)^2)/(n+5) for n>=0.
A(n+2, n+8) = CatalanNumber(n+4)*(n+3)*(3*n+10)/(2*n+12) for n>=0.
Let for n>=0 and k>=0 diag(n,k) = A(k+2,n+k+1) and G(n,k) = 2^(k+2*n)*Gamma((3-(-1)^k+2*k+4*n)/4)/(sqrt(Pi)*Gamma(k+n+0^k)) then
diag(n,0) = G(n,0)*(n*3)/(n+2),
diag(n,1) = G(n,1)*(3*n+4)/((n+1)*(n+2)),
diag(n,2) = G(n,2)*(3*n+5)/(n+2),
diag(n,3) = G(n,3)*3,
diag(n,4) = G(n,4)*(3*n+7),
diag(n,5) = G(n,5)*(3*n+8),
diag(n,6) = G(n,6)*3*(3+n)^2,
diag(n,7) = G(n,7)*(3+n)*(10+3*n). (End)

A191532 Triangle T(n,k) read by rows: T(n,n) = 2n+1, T(n,k)=k for k

Original entry on oeis.org

1, 0, 3, 0, 1, 5, 0, 1, 2, 7, 0, 1, 2, 3, 9, 0, 1, 2, 3, 4, 11, 0, 1, 2, 3, 4, 5, 13, 0, 1, 2, 3, 4, 5, 6, 15, 0, 1, 2, 3, 4, 5, 6, 7, 17, 0, 1, 2, 3, 4, 5, 6, 7, 8, 19, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 21, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 23, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 25, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 27
Offset: 0

Views

Author

Paul Curtz, Jun 05 2011

Keywords

Comments

We can build products of linear polynomials with these T(n,k) defining the absolute terms:
1+n = A000027(1+n) =2, 3, 4, 5, 6, 7,
n*(3+n)/2 = A000096(1+n) =2, 5, 9, 14, 20, 27,
n*(1+n)*(5+n)/6 = A005581(2+n) =2, 7, 16, 30, 50, 77,
n*(1+n)*(2+n)*(7+n)/24 = A005582(1+n) =2, 9, 25, 55, 105, 182,
n*(1+n)*(2+n)*(3+n)*(9+n)/120 = A005583(n) =2, 11, 36, 91, 196, 378,
n*(1+n)*(2+n)*(3+n)*(4+n)*(11+n)/720 = A005584(n)=2, 13, 49, 140, 336, 714,

Examples

			1;
0,3;
0,1,5;
0,1,2,7;
0,1,2,3,9;
0,1,2,3,4,11;
		

Crossrefs

Cf. A191302.

Formula

T(n,k) = A002262(n-1,k).
sum_{k=0..n} T(n,k) = A000217(1+n).

A238156 Triangle T(n,k), 0<=k<=n, read by rows, given by (0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 2, 0, 2, 3, 0, 2, 7, 4, 0, 2, 11, 16, 5, 0, 2, 15, 36, 30, 6, 0, 2, 19, 64, 91, 50, 7, 0, 2, 23, 100, 204, 196, 77, 8, 0, 2, 27, 144, 385, 540, 378, 112, 9, 0, 2, 31, 196, 650, 1210, 1254, 672, 156, 10, 0, 2, 35, 256, 1015, 2366, 3289, 2640, 1122, 210, 11
Offset: 0

Views

Author

Philippe Deléham, Feb 18 2014

Keywords

Comments

Row sums are A001519(n+1) = A122367(n).
Diagonal sums are A052969(n).

Examples

			Triangle begins:
1;
0, 2;
0, 2, 3;
0, 2, 7, 4;
0, 2, 11, 16, 5;
0, 2, 15, 36, 30, 6;
0, 2, 19, 64, 91, 50, 7;
0, 2, 23, 100, 204, 196, 77, 8;
0, 2, 27, 144, 385, 540, 278, 112, 9;
0, 2, 31, 196, 650, 1210, 1254, 672, 156, 10;
0, 2, 35, 256, 1015, 2366, 3289, 2640, 1122, 210, 11;
...
		

Crossrefs

Formula

G.f.: (1-x)/(1-x-2*x*y+x^2*y^2).
Sum_{k=0..n} T(n,k)*2^k = 4^n = A000302(n).
T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-2), T(0,0) = 1, T(1,0) = 0, T(1,1) = 2, T(n,k) = 0 if k<0 or if k>n.
Showing 1-8 of 8 results.