A005582 -id:A005582 - cp's OEIS frontend

A005581 a(n) = (n-1)n(n+4)/6.

0, 0, 2, 7, 16, 30, 50, 77, 112, 156, 210, 275, 352, 442, 546, 665, 800, 952, 1122, 1311, 1520, 1750, 2002, 2277, 2576, 2900, 3250, 3627, 4032, 4466, 4930, 5425, 5952, 6512, 7106, 7735, 8400, 9102, 9842, 10621, 11440, 12300, 13202, 14147, 15136, 16170

Offset: 0

N. J. A. Sloane

A class of Boolean functions of n variables and rank 2.
Also, number of inscribable triangles within a (n+4)-gon sharing with them its vertices but not its sides. - Lekraj Beedassy, Nov 14 2003
a(n) = A111808(n,3) for n > 2. - Reinhard Zumkeller, Aug 17 2005
If X is an n-set and Y a fixed 2-subset of X then a(n-2) is equal to the number of (n-3)-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007
The sequence starting with offset 2 = binomial transform of [2, 5, 4, 1, 0, 0, 0, ...]. - Gary W. Adamson, Mar 20 2009
Let I=I_n be the n X n identity matrix and P=P_n be the incidence matrix of the cycle (1,2,3,...,n). Then, for n >= 4, a(n-4) is the number of (0,1) n X n matrices A <= P^(-1) + I + P having exactly two 1's in every row and column with perA=8. - Vladimir Shevelev, Apr 12 2010
Also arises as the number of triples of edges which can be chosen as the cut-points in the "three-opt" heuristic for a traveling salesman problem on (n+4) nodes. - James McDermott, Jul 10 2015
a(n) = risefac(n, 3)/3! - n is for n >= 1 also the number of independent components of a symmetric traceless tensor of rank 3 and dimension n. Here risefac is the rising factorial. - Wolfdieter Lang, Dec 10 2015
For n >= 2, a(n) is the number of characters in a word Q formed by concatenating all 'directed' ( left to right or vice versa), unrearranged subwords, from length 1 to (n-1), of a length (n-1) word q- allowing for the appearance of repeated subwords- and simply inserting an extra character for all subwords thus concatenated. - Christopher Hohl, May 30 2019

			In hexagon ABCDEF, the "interior" triangles are ACE and BDF, and a(6-4)=a(2)=2. - _Toby Gottfried_, Nov 12 2011
G.f. = 2*x^2 + 7*x^3 + 16*x^4 + 30*x^5 + 50*x^6 + 77*x^7 + 112*x^8 + ...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), Table 22.7, p. 797.
Joseph D. Konhauser, Dan Velleman and Stan Wagon,, Which Way Did the Bicycle Go?, MAA, 1996, p. 177.
V. S. Shevelyov (Shevelev), Extension of the Moser class of four-line Latin rectangles, DAN Ukrainy, Vol. 3 (1992), pp. 15-19. - Vladimir Shevelev, Apr 12 2010
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
A. M. Yaglom and I. M. Yaglom, Challenging Mathematical Problems with Elementary Solutions. Vol. I. Combinatorial Analysis and Probability Theory. New York: Dover Publications, Inc., 1987, p. 13, #51 (the case k=3) (First published: San Francisco: Holden-Day, Inc., 1964).

T. D. Noe, Table of n, a(n) for n = 0..1000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972. [Alternative scanned copy]
Armen G. Bagdasaryan and Ovidiu Bagdasar, On some results concerning generalized arithmetic triangles, Electronic Notes in Discrete Mathematics, Vol. 67 (2018), pp. 71-77.
Beáta Bényi, Miguel Méndez, José L. Ramírez and Tanay Wakhare, Restricted r-Stirling Numbers and their Combinatorial Applications, arXiv:1811.12897 [math.CO], 2018.
John Elias, Illustration: Triangular Staircasing
Richard K. Guy, Letter to N. J. A. Sloane, 1987.
Richard K. Guy, Letter to N. J. A. Sloane, Feb 1988.
F. T. Howard and Curtis Cooper, Some identities for r-Fibonacci numbers, Fibonacci Quart., Vol. 49, No. 3 (2011), pp. 231-243.
Milan Janjic, Two Enumerative Functions.
V. Jovovic and G. Kilibarda, On the number of Boolean functions in the Post classes F^{mu}_8, Diskretnaya Matematika, Vol. 11, No. 4 (1999), pp. 127-138.
V. Jovovic and G. Kilibarda, On the number of Boolean functions in the Post classes F^{mu}_8, (English translation), Discrete Mathematics and Applications, Vol. 9, No. 6 (1999), pp. 593-605.
Kyu-Hwan Lee and Se-jin Oh, Catalan triangle numbers and binomial coefficients, arXiv:1601.06685 [math.CO], 2016.
Alice McLeod and William Moser, Counting cyclic binary strings, Math. Mag., Vol. 80, No. 1 (2007), pp. 29-37.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992, arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Dead link]
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Cached copy, May 15 2013]
Eric Weisstein's World of Mathematics, Trinomial Coefficient.
Index entries for sequences related to Boolean functions.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000217, A000292, A005582, A005586, A111808.
Cf. A176222, A000211, A052928, A128209.
Cf. A127672 (Chebyshev C).

[(n-1)*n*(n+4)/6 : n in [0..50]]; // Wesley Ivan Hurt, Jul 10 2015

A005581 := n->(n-1)*n*(n+4)/6: seq(A005581(n), n=0..50);
a:=n->sum ((j+3)*j/2,j=0..n): seq(a(n),n=-1..49); # Zerinvary Lajos, Dec 17 2006
seq((n+3)*binomial(n,3)/n, n=1..46); # Zerinvary Lajos, Feb 28 2007
A005581:=-(-2+z)/(z-1)**4; # Simon Plouffe in his 1992 dissertation
seq(sum(binomial(n,m), m=1..3)+n^2,n=-1..44); # Zerinvary Lajos, Jun 19 2008
A005581 := n -> GegenbauerC(`if`(3A005581(n)), n=0..50); # Peter Luschny, May 10 2016

Table[(n-1)*n*(n+4)/6, {n,0,50}] (* Stefan Steinerberger, Apr 10 2006 *)
LinearRecurrence[{4,-6,4,-1},{0,0,2,7},50] (* Harvey P. Dale, Sep 22 2012 *)

A005581(n):=(n-1)*n*(n+4)/6$ makelist(A005581(n),n,0,50); /* Martin Ettl, Dec 18 2012 */

{a(n) = n * (n+4) * (n-1) / 6}; /* Michael Somos, Apr 13 2007 */

concat([0, 0], Vec((x^2)*(2-x)/(1-x)^4 + O(x^50))) \\ Altug Alkan, Dec 10 2015

[(n-1)*n*(n+4)/6 for n in range(50)] # Danny Rorabaugh, Apr 20 2015

G.f.: (x^2)*(2-x)/(1-x)^4.
a(n) = binomial(n+1, n-2) + binomial(n, n-2).
a(n) = A027907(n, 3), n >= 0 (fourth column of trinomial coefficients). - N. J. A. Sloane, May 16 2003
Convolution of {1, 2, 3, ...} with {2, 3, 4, ...}. - Jon Perry, Jun 25 2003
a(n+2) = 2*te(n) - te(n-1), e.g., a(5) = 2*te(3) - te(2) = 2*20 - 10 = 30, where te(n) are the tetrahedral numbers A000292. - Jon Perry, Jul 23 2003
a(n) is the coefficient of x^3 in the expansion of (1+x+x^2)^n. For example, a(1)=0 since (1+x+x^2)^1=1+x+x^2. - Peter C. Heinig (algorithms(AT)gmx.de), Apr 09 2007
E.g.f.: (x^2 + x^3/6) * exp(x). - Michael Somos, Apr 13 2007
a(n) = - A005586(-4-n) for all n in Z. - Michael Somos, Apr 13 2007
a(n) = C(4+n,3)-(n+4)*(n+1), since C(4+n,3) = number of all triangles in (n+4)-gon, and (n+4)*(n+1)=number of triangles with at least one of the edges included. Example: n=0,in a square, all 4 possible triangles include some of the square's edges and C(4+n,3)-(n+4)*(n+1)=4-4*1=0 = number of other triangles = a(0). - Toby Gottfried, Nov 12 2011
a(n) = 2*binomial(n,2) + binomial(n,3). - Vladimir Shevelev and Peter J. C. Moses, Jun 22 2012
a(0)=0, a(1)=0, a(2)=2, a(3)=7, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Harvey P. Dale, Sep 22 2012
a(n) = A000292(n-1) + A000217(n-1) for all n in Z. - Michael Somos, Jul 29 2015
a(n+2) = -A127672(6+n, n), n >= 0, with A127672 giving the coefficients of Chebyshev's C polynomials. See the Abramowitz-Stegun reference. - Wolfdieter Lang, Dec 10 2015
a(n) = GegenbauerC(N, -n, -1/2) where N = 3 if 3Peter Luschny, May 10 2016
From Amiram Eldar, Jan 09 2022: (Start)
Sum_{n>=2} 1/a(n) = 163/200.
Sum_{n>=2} (-1)^n/a(n) = 12*log(2)/5 - 253/200. (End)

More terms from Larry Reeves (larryr(AT)acm.org), Jun 01 2000

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

N. J. A. Sloane

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

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).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Milan Janjic, Two Enumerative Functions.
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Richard K. Guy, Letter to N. J. A. Sloane, Feb 1988.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube.
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Cached copy, May 15 2013]
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A127672, A005581, A005582, A005583.

List([1..40], n-> (n+11)*Binomial(n+4,5)/6); # G. C. Greubel, Aug 27 2019

[n*(n+11)*(n+4)*(n+3)*(n+2)*(n+1)/720: n in [1..40]]; // Vincenzo Librandi, Jun 15 2011

[seq(binomial(n,6)+2*binomial(n,5), n=5..35)]; # Zerinvary Lajos, Jul 26 2006
A005584:=(-2+z)/(z-1)**7; # Simon Plouffe in his 1992 dissertation

Table[2*Binomial[n+4, 5] + Binomial[n+4, 6], {n,40}] (* Vladimir Joseph Stephan Orlovsky, Jun 14 2011, modified by G. C. Greubel, Aug 27 2019 *)
Table[(n+11)*Pochhammer[n, 5]/6!, {n,40}] (* G. C. Greubel, Aug 27 2019 *)

a(n)=n*(n+11)*(n+4)*(n+3)*(n+2)*(n+1)/720 \\ Charles R Greathouse IV, Jun 14 2011

[(n+11)*rising_factorial(n,5)/factorial(6) for n in (1..40)] # G. C. Greubel, Aug 27 2019

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)

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

A086274 Antidiagonal sums of A086272 (and of A086273).

Original entry on oeis.org

1, 4, 12, 29, 60, 111, 189, 302, 459, 670, 946, 1299, 1742, 2289, 2955, 3756, 4709, 5832, 7144, 8665, 10416, 12419, 14697, 17274, 20175, 23426, 27054, 31087, 35554, 40485, 45911, 51864, 58377, 65484, 73220, 81621, 90724, 100567, 111189

Offset: 1

Clark Kimberling, Jul 14 2003

nonn

Cf. A086272, A086273.

A005582(n-1) + n.

a(n)=n+Sum{(k+1)C(n+1-k, 2): k=1, 2, ..., n}.

n(n^3+6n^2-n+18)/24.

A213550 Rectangular array: (row n) = b**c, where b(h) = h*(h+1)/2, c(h) = n-1+h, n>=1, h>=1, and ** = convolution.

Original entry on oeis.org

1, 5, 2, 15, 9, 3, 35, 25, 13, 4, 70, 55, 35, 17, 5, 126, 105, 75, 45, 21, 6, 210, 182, 140, 95, 55, 25, 7, 330, 294, 238, 175, 115, 65, 29, 8, 495, 450, 378, 294, 210, 135, 75, 33, 9, 715, 660, 570, 462, 350, 245, 155, 85, 37, 10, 1001, 935, 825, 690, 546

Offset: 1

Clark Kimberling, Jun 16 2012

Principal diagonal:  A002418
Antidiagonal sums:  A005585
row 1,  (1,3,6,...)**(1,2,3,...):  A000332
row 2,  (1,3,6,...)**(2,3,4,...):  A005582
row 3,  (1,3,6,...)**(3,4,5,...):  A095661
row 4,  (1,3,6,...)**(4,5,6,...):  A095667
For a guide to related arrays, see A213500.

			Northwest corner (the array is read by falling antidiagonals):
1....5....15...35....70....126
2....9....25...55....105...182
3....13...35...75....140...238
4....17...45...95....175...294
5....21...55...115...210...350

Cf. A213500, A213548.

b[n_] := n (n + 1)/2; c[n_] := n
t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
r[n_] := Table[t[n, k], {k, 1, 60}]  (* A213550 *)
d = Table[t[n, n], {n, 1, 40}] (* A002418 *)
s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}]
s1 = Table[s[n], {n, 1, 50}] (* A005585 *)

T(n,k) = 5*T(n,k-1) - 10*T(n,k-2) + 10*T(n,k-3) - 5*T(n,k-4) + T(n,k-5).

G.f. for row n: f(x)/g(x), where f(x) = n-(n-1)*x and g(x) = (1 - x)^5.

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

Thomas Wieder, Aug 04 2006, Aug 06 2006

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

			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 ...

The first nine rows are: A006527, A005408, A000290, A000330, A002415, A005585, A040977, A050486, A053347.
The initial columns are: A000027, A000096, A005581, A005582, A005583, A005584.
Cf. A119800, A007318, A006527, A005408, A000290, A000330, A002415, A005585, A040977, A050486, A053347, A000027, A000096, A005581, A005582, A005583, A005584.

=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.

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.

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

A165241 Triangle T(n,k), 0 <= k <= n, read by rows, given by [1,1,0,0,0,0,0,0,0,...] DELTA [1,0,1,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 4, 9, 6, 1, 8, 24, 25, 10, 1, 16, 60, 85, 55, 15, 1, 32, 144, 258, 231, 105, 21, 1, 64, 336, 728, 833, 532, 182, 28, 1, 128, 768, 1952, 2720, 2241, 1092, 294, 36, 1, 256, 1728, 5040, 8280, 8361, 5301, 2058, 450, 45, 1

Offset: 0

Philippe Deléham, Sep 09 2009

Rows sums: A006012; Diagonal sums: A052960.

The sums of each column of A117317 with its subsequent column, treated as a lower triangular matrix with an initial null column attached, or, equivalently, the products of the row polynomials p(n,y) of A117317 with (1+y) with the initial first row below added to the final result. The reversal of A117317 is A056242 with several combinatorial interpretations. - Tom Copeland, Jan 08 2017

			Triangle begins:
  1;
  1,  1;
  2,  3,  1;
  4,  9,  6,  1;
  8, 24, 25, 10,  1; ...

Cf. A000012, A000217, A005582, A011782, A084858.

Cf. A056242, A117317.

Sum_{k=0..n} T(n,k)*x^k = A009116(n), A000007(n), A011782(n), A006012(n), A083881(n), A081335(n), A090139(n), A145301(n), A145302(n), A145303(n), A143079(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, respectively. Sum_{k=0..n} T(n,k)*x^(n-k) = A123335(n), A000007(n), A000012(n), A006012(n), A084120(n), A090965(n), A165225(n), A165229(n), A165230(n), A165231(n), A165232(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, respectively.
G.f.: (1-(1+y)*x)/(1-2(1+y)*x+(y+y^2)*x^2). - Philippe Deléham, Dec 19 2011
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2) with T(0,0) = T(1,0) = T(1,1) = 1 and T(n,k) = 0 if k<0 or if nPhilippe Deléham, Dec 19 2011

O.g.f. corrected by Tom Copeland, Jan 15 2017

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

Wolfdieter Lang, Aug 08 2014

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.

			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.

Wolfdieter Lang, First rows of the triangle.

Cf. A053120, A127672, A054977, -A000027, A000096, -A005581, A005582, -A005583, A005584, A057079, A000032.

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

Wolfdieter Lang, Dec 10 2015

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

			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)

Cf. A000096, A005581, A005582, A005583, A005584, A007946, A024482, A029651, A051924, A051960, A097613, A115241, A129869, A135278, A220101, A265612, A265613.

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 *)

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

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)

A260056 Irregular triangle read by rows: coefficients T(n, k) of certain polynomials p(n, x) with exponents in increasing order, n >= 0 and 0 <= k <= 2*n.

Original entry on oeis.org

1, 2, 1, 1, 3, 3, 4, 2, 1, 4, 6, 10, 9, 7, 3, 1, 5, 10, 20, 25, 26, 19, 11, 4, 1, 6, 15, 35, 55, 71, 70, 56, 34, 16, 5, 1, 7, 21, 56, 105, 161, 196, 197, 160, 106, 55, 22, 6, 1, 8, 28, 84, 182, 322, 462, 554, 553, 463, 321, 183, 83, 29, 7, 1, 9, 36, 120, 294, 588, 966, 1338, 1569, 1570, 1337, 967, 587, 295, 119, 37, 8, 1, 10, 45, 165, 450, 1002, 1848, 2892, 3873, 4477, 4476, 3874

Offset: 0

Werner Schulte, Nov 08 2015

The triangle is related to the triangle of trinomial coefficients.

			The irregular triangle T(n,k) begins:
n\k:  0   1   2    3    4    5    6    7    8    9   10  11  12  13  14  ...
0     1;
1     2   1   1;
2     3   3   4    2    1;
3     4   6  10    9    7    3    1;
4     5  10  20   25   26   19   11    4    1;
5     6  15  35   55   71   70   56   34   16    5    1;
6     7  21  56  105  161  196  197  160  106   55   22   6   1;
7     8  28  84  182  322  462  554  553  463  321  183  83  29   7   1;
etc.
The polynomial corresponding to row 2 is p(2,x) = 3+3*x+4*x^2+2*x^3+x^4.

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened

Cf. A000027 (col 0), A000217 (col 1), A000292 (col 2), A001590, A002426, A004524, A005582 (col 3), A008937, A027907, A095662 (col 5), A113682, A246437.

A027907[n_, k_] := Sum[Binomial[n, j]*Binomial[j, k - j], {j, 0, n}]; Table[ Sum[A027907[j, k], {j, 0, n}], {n,0,10}, {k, 0, 2*n} ] // Flatten (* G. C. Greubel, Mar 07 2017 *)

T(n,0) = n+1, and T(n,k) = 0 for k < 0 or k > 2*n, and T(n+1,k) = T(n,k) + T(n,k-1) + T(n,k-2) for k > 0.
T(n,k) = Sum_{j=0..n} A027907(j,k) for 0 <= k <= 2*n.
T(n,k) = Sum_{j=0..k} (-1)^(k-j)*A027907(n+1,j+1) for 0 <= k <= 2*n.
T(n,k) = T(n,2*n-1-k) + (-1)^k for 0 <= k < 2*n.
p(n,x) = Sum_{k=0..2*n} T(n,k)*x^k = Sum_{k=0..n} (1+x+x^2)^k for n >= 0.
p(n,x) = ((1+x+x^2)^(n+1)-1)/(x+x^2), p(n,0) = p(n,-1) = n+1 for n >= 0.
p(n+1,x) = (1+x+x^2)*p(n,x)+1 for n >= 0.
Sum_{n>=0} p(n,x)*t^n = 1/((1-t)*(1-t*(1+x+x^2))).
T(n,2*n) = 1, and T(n,n) = A113682(n) for n >= 0.
T(n,n-1) = A246437(n+1), and T(n,n-1)+T(n,n) = A002426(n+1) for n > 0.
If d(n) is n-th antidiagonal sum of the triangle then: d(n) = A008937(n+1), and d(n+2)-d(n) = A001590(n+5) for n >= 0.
Conjecture: If a(n) is n-th antidiagonal alternating sum of the triangle then: a(n) = A004524(n+3).
Sum_{k=0..2*n} (-1)^k*T(n,k)^2 = (3^(n+1)-1)/2 for n >= 0.
Sum_{k=0..2*n} (-1)^k*(y*k+1)*T(n,k) = Sum{k=0..n} y*k+1 = (n+1)*(y*n+2)/2 for real y and n >= 0.
Conjecture of linear recurrence for column k: Sum_{m=0..k+2} (-1)^m*T(n+m,k)* binomial(k+2,m) = 0 for k >= 0 and n >= 0.

A180174 Triangle read by rows of the numbers C(n,k) of k-subsets of a quadratically populated n-multiset M.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 3, 5, 7, 9, 10, 10, 10, 10, 10, 9, 7, 5, 3, 1, 1, 4, 9, 16, 25, 35, 45, 55, 65, 75, 84, 91, 96, 99, 100, 100, 100, 99, 96, 91, 84, 75, 65, 55, 45, 35, 25, 16, 9, 4, 1, 1, 5, 14, 30, 55, 90, 135, 190, 255, 330, 414, 505, 601, 700, 800, 900, 1000, 1099

Offset: 0

Thomas Wieder, Aug 15 2010

The multiplicity m(i) of the i-th element with 1 <= i <= n is m(i)=i^2.
Thus M=[1,2,2,2,2,...,i^2 x i,...,n^2 x n].
Row sum is equal to A028361.
Column for k=2 is equal to AA000096.
Column for k=3 is equal to AA005581.
Column for k=4 is equal to AA005582.
The number of coefficients C(n,k) for given n is equal to A056520.

			For n=4 one has M=[1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4].
For k=7 we have 55 subsets from M:
[1, 2, 2, 3, 3, 4, 4], [1, 2, 3, 3, 4, 4, 4], [1, 2, 3, 3, 3, 4, 4],
[1, 2, 2, 3, 4, 4, 4], [1, 2, 2, 3, 3, 3, 4], [1, 2, 2, 2, 3, 4, 4],
[1, 2, 2, 2, 3, 3, 4], [2, 2, 3, 3, 4, 4, 4], [2, 2, 3, 3, 3, 4, 4],
[2, 2, 2, 3, 3, 4, 4], [1, 2, 2, 2, 3, 3, 3], [1, 2, 2, 2, 4, 4, 4],
[1, 3, 3, 3, 4, 4, 4], [2, 3, 3, 3, 4, 4, 4], [2, 2, 2, 3, 4, 4, 4],
[2, 2, 2, 3, 3, 3, 4], [1, 2, 3, 4, 4, 4, 4], [1, 2, 3, 3, 3, 3, 4],
[1, 2, 2, 2, 2, 3, 4], [1, 2, 2, 3, 3, 3, 3], [1, 2, 2, 2, 2, 3, 3],
[1, 2, 2, 4, 4, 4, 4], [1, 2, 2, 2, 2, 4, 4], [1, 3, 3, 4, 4, 4, 4],
[1, 3, 3, 3, 3, 4, 4], [2, 3, 3, 4, 4, 4, 4], [2, 3, 3, 3, 3, 4, 4],
[2, 2, 3, 4, 4, 4, 4], [2, 2, 3, 3, 3, 3, 4], [2, 2, 2, 2, 3, 4, 4],
[2, 2, 2, 2, 3, 3, 4], [2, 2, 2, 3, 3, 3, 3], [2, 2, 2, 2, 3, 3, 3],
[2, 2, 2, 4, 4, 4, 4], [2, 2, 2, 2, 4, 4, 4], [3, 3, 3, 4, 4, 4, 4],
[3, 3, 3, 3, 4, 4, 4], [1, 2, 3, 3, 3, 3, 3], [1, 2, 4, 4, 4, 4, 4],
[1, 3, 4, 4, 4, 4, 4], [1, 3, 3, 3, 3, 3, 4], [2, 3, 4, 4, 4, 4, 4],
[2, 3, 3, 3, 3, 3, 4], [2, 2, 3, 3, 3, 3, 3], [2, 2, 4, 4, 4, 4, 4],
[3, 3, 4, 4, 4, 4, 4], [3, 3, 3, 3, 3, 4, 4], [1, 3, 3, 3, 3, 3, 3],
[1, 4, 4, 4, 4, 4, 4], [2, 3, 3, 3, 3, 3, 3], [2, 4, 4, 4, 4, 4, 4],
[3, 4, 4, 4, 4, 4, 4], [3, 3, 3, 3, 3, 3, 4], [3, 3, 3, 3, 3, 3, 3],
[4, 4, 4, 4, 4, 4, 4].

Cf. A007318, A008302, A028361, A056520, A000096, A005581, A005582.

with(combinat)
kend := 4;
Liste := NULL;
for k from 0 to kend do
Liste := Liste, `$`(k, k^2)
end do;
Liste := [Liste];
for k from 0 to 2^(kend+1)-1 do
Teilergebnis[k] := choose(Liste, k)
end do;
seq(nops(Teilergebnis[k]), k = 0 .. 2^(kend+1)-1)
' Excel VBA
Sub A180174()
Dim n As Long, nend As Long, k As Long, kk As Long, length_row As Long, length_sum As Long
Dim ATable(10, -1000 To 1000) As Double, Summe As Double
Dim offset_row As Integer, offset_column As Integer
Worksheets("Tabelle2").Select
Cells.Select
Selection.ClearContents
Range("A1").Select
offset_row = 1
offset_column = 1
nend = 7
ATable(0, 0) = 1
Cells(0 + offset_row, 0 + offset_column) = 1
For n = 1 To nend
length_row = n * (n + 1) * (2 * n + 1) / 6
length_sum = n ^ 2 + 1
For k = 0 To length_row / 2
Summe = 0
For kk = k - length_sum + 1 To k
Summe = Summe + ATable(n - 1, kk)
Next kk
ATable(n, k) = Summe
Cells(n + offset_row, k + offset_column) = ATable(n, k)
ATable(n, length_row - k) = Summe
Cells(n + offset_row, length_row - k + 0 + offset_column) = ATable(n, k)
Next k
Next n
End Sub

C(0,0) = 0.
C(n,k) = sum_{j=(k-LS+1)}^{k} C(n-1,j).
for n > 0 and k=1,...,LR with LS = n^2+1 and LR = n*(n+1)*(2*n+1)/6.
C(n,k) = C(n,LR-k).

cp's OEIS Frontend

A005581 a(n) = (n-1)*n*(n+4)/6.

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A005584 Coefficients of Chebyshev polynomials.

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A086274 Antidiagonal sums of A086272 (and of A086273).

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A213550 Rectangular array: (row n) = b**c, where b(h) = h*(h+1)/2, c(h) = n-1+h, n>=1, h>=1, and ** = convolution.

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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.

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A165241 Triangle T(n,k), 0 <= k <= n, read by rows, given by [1,1,0,0,0,0,0,0,0,...] DELTA [1,0,1,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

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

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A005581 a(n) = (n-1)n(n+4)/6.