A099721 -id:A099721 - cp's OEIS frontend

A229079 Number A(n,k) of ascending runs in {1,...,k}^n; square array A(n,k), n>=0, k>=0, read by antidiagonals.

0, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 3, 7, 3, 0, 0, 4, 15, 20, 4, 0, 0, 5, 26, 63, 52, 5, 0, 0, 6, 40, 144, 243, 128, 6, 0, 0, 7, 57, 275, 736, 891, 304, 7, 0, 0, 8, 77, 468, 1750, 3584, 3159, 704, 8, 0, 0, 9, 100, 735, 3564, 10625, 16896, 10935, 1600, 9, 0

Offset: 0

Alois P. Heinz, Sep 14 2013

			A(4,1) = 4: [1,1,1,1].
A(3,2) = 20 = 3+3+2+3+2+2+2+3: [1,1,1], [2,1,1], [1,2,1], [2,2,1], [1,1,2], [2,1,2], [1,2,2], [2,2,2].
A(2,3) = 15 = 2+2+2+1+2+2+1+1+2: [1,1], [2,1], [3,1], [1,2], [2,2], [3,2], [1,3], [2,3], [3,3].
A(1,4) = 4 = 1+1+1+1: [1], [2], [3], [4].
Square array A(n,k) begins:
  0, 0,   0,     0,     0,      0,       0,       0, ...
  0, 1,   2,     3,     4,      5,       6,       7, ...
  0, 2,   7,    15,    26,     40,      57,      77, ...
  0, 3,  20,    63,   144,    275,     468,     735, ...
  0, 4,  52,   243,   736,   1750,    3564,    6517, ...
  0, 5, 128,   891,  3584,  10625,   25920,   55223, ...
  0, 6, 304,  3159, 16896,  62500,  182736,  453789, ...
  0, 7, 704, 10935, 77824, 359375, 1259712, 3647119, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened

Columns k=0-10 give: A000004, A001477, A066373(n+1) for n>0, A229277, A229278, A229279, A229280, A229281, A229282, A229283, A229284.
Rows n=0-10 give: A000004, A001477, A005449, A099721, A229146, A229147, A229148, A229149, A229150, A229151, A229152.
Main diagonal gives A229078.

A:= (n, k)-> `if`(n=0, 0, k^(n-1)*((n+1)*k+n-1)/2):
seq(seq(A(n,d-n), n=0..d), d=0..12);

a[, 0] = a[0, ] = 0; a[n_, k_] := k^(n-1)*((n+1)*k+n-1)/2; Table[a[n-k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Dec 09 2013 *)

A(n,k) = k^(n-1)*((n+1)*k+n-1)/2 for n>0, A(0,k) = 0.

A024196 a(n) = 2nd elementary symmetric function of the first n+1 odd positive integers.

Original entry on oeis.org

3, 23, 86, 230, 505, 973, 1708, 2796, 4335, 6435, 9218, 12818, 17381, 23065, 30040, 38488, 48603, 60591, 74670, 91070, 110033, 131813, 156676, 184900, 216775, 252603, 292698, 337386, 387005, 441905, 502448, 569008, 641971, 721735, 808710, 903318

Offset: 1

Clark Kimberling

			a(8) = 8*80+7*79+6*78+5*77+4*76+3*75+2*74+1*73 = 2796. - _Bruno Berselli_, Mar 13 2012

Bruno Berselli, Table of n, a(n) for n = 1..1000
Wolfdieter Lang, On Generating functions of Diagonals Sequences of Sheffer and Riordan Number Triangles, arXiv:1708.01421 [math.NT], August 2017.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

From Johannes W. Meijer, Jun 08 2009: (Start)
Equals third right hand column of A028338 triangle.
Equals third left hand column of A109692 triangle.
Equals third right hand column of A161198 triangle divided by 2^m.
(End)
Cf. A016061.

List([1..36], n -> n*(n+1)*(3*n^2+5*n+1)/6); # Muniru A Asiru, Feb 13 2018

seq(n*(n+1)*(3*n^2+5*n+1)/6,n=1..25); # Muniru A Asiru, Feb 13 2018

f[k_] := 2 k - 1; t[n_] := Table[f[k], {k, 1, n}]
a[n_] := SymmetricPolynomial[2, t[n]]
Table[a[n], {n, 2, 50}]  (* A024196 *)
(* Clark Kimberling, Dec 31 2011 *)
Table[(n(n+1)(3n^2+5n+1))/6,{n,50}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{3,23,86,230,505},50] (* Harvey P. Dale, Jul 08 2019 *)

a(n) = n*(n+1)*(3*n^2+5*n+1)/6.
From Bruno Berselli, Mar 13 2012: (Start)
G.f.: x*(3 + 8*x + x^2)/(1 - x)^5.
a(n) = Sum_{i=1..n} (n+1-i)*((n+1)^2-i).
a(n) = n*A016061(n) - Sum_{i=0..n-1} A016061(i). (End)
a(n) - a(n-1) = A099721(n). Partial sums of A099721.- Philippe Deléham, May 07 2012
a(n) = Sum_{i=1..n} ((2*i-1)*Sum_{j=i..n} (2*j+1)) = 1*(3+5+...2*n+1) + 3*(5+7+...+2*n+1) + ... + (2*n-1)*(2*n+1). - J. M. Bergot, Apr 21 2017
a(n) = A028338(n+1, n-1), n >= 1, (third diagonal). See the crossref. below. Wolfdieter Lang, Jul 21 2017
a(n) = (A000583(n+1) - A000447(n+1))/2. - J. M. Bergot, Feb 13 2018

A143785 Antidiagonal sums of the triangle A120070.

Original entry on oeis.org

3, 8, 20, 36, 63, 96, 144, 200, 275, 360, 468, 588, 735, 896, 1088, 1296, 1539, 1800, 2100, 2420, 2783, 3168, 3600, 4056, 4563, 5096, 5684, 6300, 6975, 7680, 8448, 9248, 10115, 11016, 11988, 12996, 14079, 15200, 16400, 17640, 18963, 20328, 21780, 23276

Offset: 1

Paul Curtz, Sep 01 2008

Let b(n) be the sequence (0,0,0,3,8,20,36,...), with offset 0. Then b(n) is the number of triples (w,x,y) having all terms in {0,...,n} and w < range{w,x,y}. - Clark Kimberling, Jun 11 2012

Consider a(n) with two 0's prepended and offset 1. Call the new sequence b(n) and consider the partitions of n into two parts (p,q). Then b(n) represents the sum of all the products (p + q) * (q - p) where p <= q. - Wesley Ivan Hurt, Apr 12 2018

			First diagonal 3 = 3.
Second diagonal 8 = 8.
Third diagonal 5+15 = 20.
Fourth diagonal 24+12 = 36.

Iain Fox, Table of n, a(n) for n = 1..10000 (first 1000 terms from Colin Barker)
Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Cf. A035006, A099721 (bisections).

Cf. A120070, A002620.

[(n+2)*(2*n^2+4*n-(-1)^n+1)/8: n in [1..50]]; // Vincenzo Librandi, Jan 22 2018

Rest@ CoefficientList[Series[x (3 + 2 x + x^2)/((1 + x)^2*(x - 1)^4), {x, 0, 44}], x] (* Michael De Vlieger, Dec 22 2017 *)
LinearRecurrence[{2, 1, -4, 1, 2, -1}, {3, 8, 20, 36, 63, 96}, 60] (* Vincenzo Librandi, Jan 22 2018 *)

Vec(x*(3+2*x+x^2)/((1+x)^2*(x-1)^4) + O(x^50)) \\ Colin Barker, May 07 2016

a(n+1) - a(n) = A032438(n+2).
a(n) = A006918(n-2) + 2*A006918(n-1) + 3*A006918(n). - R. J. Mathar, Jul 01 2011
G.f.: x*(3+2*x+x^2) / ( (1+x)^2*(x-1)^4 ). - R. J. Mathar, Jul 01 2011
a(n) = (n+2)*(2*n^2 + 4*n - (-1)^n + 1)/8. - Ilya Gutkovskiy, May 07 2016
From Colin Barker, May 07 2016: (Start)
a(n) = (n^3 + 4*n^2 + 4*n)/4 for n even.
a(n) = (n^3 + 4*n^2 + 5*n + 2)/4 for n odd.
a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6) for n > 6. (End)
a(n) = Sum_{k=1..n+1} floor((n+1)*k/2). - Wesley Ivan Hurt, Apr 01 2017
a(n) = (n+2)*floor((n+1)^2/4) ( = (n+2)*A002620(n+1) ) for n > 0. - Heinrich Ludwig, Dec 22 2017
E.g.f.: e^(-x) * (-2 + x + e^(2*x)*(2 + 19*x + 14*x^2 + 2*x^3))/8. - Iain Fox, Dec 29 2017

A316989 Irregular triangle read by rows: row n consists of the coefficients in the expansion of the polynomial (x^2 + 4x + 3)(x + 1)^(2n) + (x^2 - 1)(x^2 + 3*x + 3).

Original entry on oeis.org

0, 1, 3, 3, 1, 0, 7, 14, 9, 2, 0, 13, 37, 43, 26, 8, 1, 0, 19, 72, 129, 141, 98, 42, 10, 1, 0, 25, 119, 291, 463, 504, 378, 192, 63, 12, 1, 0, 31, 178, 553, 1156, 1716, 1848, 1452, 825, 330, 88, 14, 1, 0, 37, 249, 939, 2432, 4576, 6435, 6864, 5577, 3432, 1573

Offset: 0

Franck Maminirina Ramaharo, Jul 18 2018

The triangle is related to the Kauffman bracket polynomial evaluated at the shadow diagram of the two-bridge knot with Conway's notation C(2n,3).

			The triangle T(n,k) begins:
n\k| 0   1    2    3     4     5     9     7     8     9    10   11   12  13 14
-------------------------------------------------------------------------------
0  | 0   1    3    3     1
1  | 0   7   14    9     2
2  | 0  13   37   43    26     8     1
3  | 0  19   72  129   141    98    42    10     1
4  | 0  25  119  291   463   504   378   192    63    12     1
5  | 0  31  178  553  1156  1716  1848  1452   825   330    88   14    1
6  | 0  37  249  939  2432  4576  6435  6864  5577  3432  1573  520  117  16  1
...

Ji-Young Ham and Joongul Lee, An explicit formula for the A-polynomial of the knot with Conway’s notation C(2n,3), Journal of Knot Theory and Its Ramifications 25 (2016), 1-9.
Ryo Hanaki, On scannable properties of the original knot from a knot shadow, Topology and its Applications 194 (2015), 296-305.
Bin Lu and Jianyuan K. Zhong, The Kauffman Polynomials of 2-bridge Knots, arXiv:math/0606114 [math.GT], 2006.

Row sums: 4*A004171.

Cf. A093560, A137396, A299989, A300184, A300192, A300453, A300454, A316659.

T := proc (n, k) if k = 1 then 6*n + 1 else binomial(2*n + 3, k + 1) + (binomial(2*n + 1, k)*(2*k - 2*n) + binomial(4, k)*(2*k - 3))/(k + 1) end if end proc:
for n from 0 to 12 do seq(T(n, k), k = 0 .. max(4, 2*(n + 1))) od;

row[n_] := CoefficientList[(x^2 + 4*x + 3)*(x + 1)^(2*n) + (x^2 - 1)*(x^2 + 3*x + 3), x];
Array[row, 12, 0] // Flatten

T(n, k) := binomial(2*n + 3, k + 1) + (binomial(2*n + 1, k)*(2*k - 2*n) + binomial(4, k)*(2*k - 3))/(k + 1) - kron_delta(1, k)$
for n:0 thru 12 do print(makelist(T(n, k), k, 0, max(4, 2*(n + 1))));

T(n,1) = A016921(n) and T(n,k) = C(2*n+3,k+1) + (C(2*n+1,k)*(2*k - 2*n) + C(4,k)*(2*k - 3))/(k + 1) for k > 1.
T(n,2) = A173247(2*n+1) = A300401(2*n,3).
T(n,3) = 2*A099721(n) + 3.
T(n,4) = A244730(n) - A002412(n) + 1.
T(n,k) = A093560(2*n,k) for n > 2 and k > 4.
G.f.: (x^2 + 4*x + 3)/(1 - y*(x + 1)^2) + (x^4 + 3*x^3 + 2*x^2 - 3*x - 3)/(1 - y).

A224454 The Wiener index of the linear phenylene with n hexagons.

Original entry on oeis.org

27, 180, 567, 1296, 2475, 4212, 6615, 9792, 13851, 18900, 25047, 32400, 41067, 51156, 62775, 76032, 91035, 107892, 126711, 147600, 170667, 196020, 223767, 254016, 286875, 322452, 360855, 402192, 446571, 494100, 544887, 599040, 656667, 717876, 782775, 851472, 924075, 1000692, 1081431, 1166400

Offset: 1

Emeric Deutsch, Apr 10 2013

a(2) and a(5) have been checked by the direct computation of the Wiener index (using Maple).

I. Gutman, The topological indices of linear phenylenes, J. Serb. Chem. Soc., 60, No. 2, 1995, 99-104.

G. C. Greubel, Table of n, a(n) for n = 1..1000
G. Cash, S. Klavzar, M. Petkovsek, Three methods for calculation of the hyper-Wiener index of a molecular graph, J. Chem. Inf. Comput. Sci. 42, 2002, 571-576.
L. Pavlovic, I. Gutman, Wiener numbers of phenylenes: an exact result, J. Chem. Inf. Comput. Sci. 37, 1997, 355-358.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1)

Cf. A224455.

[9*n^2*(2*n+1): n in [1..40]]; // Vincenzo Librandi, Dec 09 2016

a := proc (n) options operator, arrow: 9*n^2*(2*n+1) end proc: seq(a(n), n = 1 .. 40);

LinearRecurrence[{4, -6, 4, -1}, {27, 180, 567, 1296}, 100] (* or *) Table[9*n^2*(2*n+1), {n,1,100}] (* G. C. Greubel, Dec 08 2016 *)

Vec(9*x*(3 + 8*x + x^2)/(1-x)^4 + O(x^50)) \\ G. C. Greubel, Dec 08 2016

a(n) = 9*n^2*(2n+1) = 9*A099721(n).
G.f.: 9*x*(3 + 8*x + x^2)/(1-x)^4.
The Hosoya polynomial of the linear phenylene with n hexagons is nt(t^3-t^2-4t-8)/(t-1) + 2t(t+1)(t^(3n)-1)/(t-1)^2.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - G. C. Greubel, Dec 08 2016

A383368 Number of intercalates in pine Latin squares of order 2n.

Original entry on oeis.org

1, 12, 27, 80, 125, 252, 343, 576, 729, 1100, 1331, 1872, 2197, 2940, 3375, 4352, 4913, 6156, 6859, 8400, 9261, 11132, 12167, 14400, 15625

Offset: 1

Eduard I. Vatutin, Apr 24 2025

Pine Latin square is a none canonical composite Latin square of order N=2*K formed from specially arranged cyclic Latin squares of order K.
By construction, pine Latin square is determined one-to-one by the cyclic square used, so number of pine Latin squares of order N is equal to number of cyclic Latin squares of order N/2.
All pine Latin squares are horizontally symmetric column-inverse Latin squares.
All pine Latin squares for selected order N are isomorphic one to another as Latin squares, so they have same properties (number of transversals, intercalates, etc.).
Pine Latin squares have interesting properties, for example, maximum known number of intercalates for some orders N (at least N in {2, 4, 6, 10, 18}).
Pine Latin squares do not exist for odd orders due to they are horizontally symmetric.
Pine Latin squares of order N=2n exists for all even orders due to existing of corresponding cyclic Latin squares of order n. According to this, maximum number of intercalates in a Latin square A092237(N) >= (2k)^2 * (2k + 1) for N=4k and A092237(N) >= (2k+1)^3 for N=4k+2. Therefore, asimptotically maximum number of intercalates in Latin squares of even orders N greater or equal than o(k1*N^3), where k1 = 1/8.

			For order N=8 pine Latin square
  0 1 2 3 4 5 6 7
  1 2 3 0 7 4 5 6
  2 3 0 1 6 7 4 5
  3 0 1 2 5 6 7 4
  4 5 6 7 0 1 2 3
  5 6 7 4 3 0 1 2
  6 7 4 5 2 3 0 1
  7 4 5 6 1 2 3 0
have 80 intercalates.
.
For order N=10 pine Latin square
  0 1 2 3 4 5 6 7 8 9
  1 2 3 4 0 9 5 6 7 8
  2 3 4 0 1 8 9 5 6 7
  3 4 0 1 2 7 8 9 5 6
  4 0 1 2 3 6 7 8 9 5
  5 6 7 8 9 0 1 2 3 4
  6 7 8 9 5 4 0 1 2 3
  7 8 9 5 6 3 4 0 1 2
  8 9 5 6 7 2 3 4 0 1
  9 5 6 7 8 1 2 3 4 0
have 125 intercalates.
.
For order N=12 pine Latin square
  0 1 2 3 4 5 6 7 8 9 10 11
  1 2 3 4 5 0 11 6 7 8 9 10
  2 3 4 5 0 1 10 11 6 7 8 9
  3 4 5 0 1 2 9 10 11 6 7 8
  4 5 0 1 2 3 8 9 10 11 6 7
  5 0 1 2 3 4 7 8 9 10 11 6
  6 7 8 9 10 11 0 1 2 3 4 5
  7 8 9 10 11 6 5 0 1 2 3 4
  8 9 10 11 6 7 4 5 0 1 2 3
  9 10 11 6 7 8 3 4 5 0 1 2
  10 11 6 7 8 9 2 3 4 5 0 1
  11 6 7 8 9 10 1 2 3 4 5 0
have 252 intercalates.

R. Bean, Critical sets in Latin squares and associated structures, Ph.D. Thesis, The University of Queensland, 2001.
Eduard I. Vatutin, About the properties of pine Latin squares (in Russian).
Eduard I. Vatutin, Proving list (examples).
Index entries for sequences related to Latin squares and rectangles.

Cf. A002860, A016755, A089207, A092237, A099721, A338522, A383570.

Hypothesis: For all known pine Latin squares of orders N=4k+2 number of intercalates a(n) = a(N/2)= a(2k+1) = (N/2)^3 = (2k+1)^3 = A016755((n-1)/2) (verified for N<29).

Hypothesis: For all known pine Latin squares of orders N=4k number of intercalates a(n) = a(N/2) = a(2k) = (N/2)^2 + (N/2)^3 = 4*k^2 + 8*k^3 = (2k)^2 * (2k+1) = 2*A089207(n/2) = 4*A099721(n/2) (verified for N<29).

A089207 a(n) = 4n^3 + 2n^2.

Original entry on oeis.org

6, 40, 126, 288, 550, 936, 1470, 2176, 3078, 4200, 5566, 7200, 9126, 11368, 13950, 16896, 20230, 23976, 28158, 32800, 37926, 43560, 49726, 56448, 63750, 71656, 80190, 89376, 99238, 109800, 121086, 133120, 145926, 159528, 173950, 189216

Offset: 1

Jun Mizuki (suzuki32(AT)sanken.osaka-u.ac.jp), Dec 09 2003

Yet another parametric representation of the solutions of the Diophantine equation x^2 - y^2 = z^3 is (3n^3, n^3, 2n^2). By taking the sum x+y+z we get a(n) = 4n^3 + 2n^2.

If Y is a 3-subset of an 2n-set X then, for n>=5, a(n-2) is the number of 5-subsets of X having at least two elements in common with Y. - Milan Janjic, Dec 16 2007

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A085409, A087887.

Table[4n^3+2n^2,{n,40}] (* Harvey P. Dale, Jun 12 2020 *)

a(n) = 2*A099721(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). G.f.: 2*x*(3+8*x+x^2)/(x-1)^4. [R. J. Mathar, Apr 20 2009]

a(n) = 2 * n * A014105(n). - Richard R. Forberg, Jun 16 2013

More terms from Ray Chandler, Feb 15 2004

A190705 a(n) = 6n^2(2*n + 1).

Original entry on oeis.org

0, 18, 120, 378, 864, 1650, 2808, 4410, 6528, 9234, 12600, 16698, 21600, 27378, 34104, 41850, 50688, 60690, 71928, 84474, 98400, 113778, 130680, 149178, 169344, 191250, 214968, 240570, 268128, 297714, 329400

Offset: 0

Adi Dani, Jun 14 2011

Number of partitions of 12*n + 1 into 4 parts.

			a(1)=18: there are 18 partitions of 12*1+1=13 into 4 parts:
  [1,1,1,10], [1,1,2,9], [1,1,3,8], [1,1,4,7], [1,1,5,6],
  [1,2,2,8],  [1,2,3,7], [1,2,4,6], [1,2,5,5], [1,3,3,6],
  [1,3,4,5],  [1,4,4,4], [2,2,2,7], [2,2,3,6], [2,2,4,5],
  [2,3,3,5],  [2,3,4,4], [3,3,3,4].

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

[6*n^2*(2*n+1): n in [0..40]]; // Vincenzo Librandi, Jun 14 2011

Table[6n^2(2n + 1), {n, 0, 30}]
LinearRecurrence[{4,-6,4,-1},{0,18,120,378},40] (* Harvey P. Dale, Mar 20 2016 *)

a(n)=6*n^2*(2*n+1) \\ Charles R Greathouse IV, Aug 05 2013

a(n) = 6 * A099721(n).

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(0)=0, a(1)=18, a(2)=120, a(3)=378. - Harvey P. Dale, Mar 20 2016

cp's OEIS Frontend

A229079 Number A(n,k) of ascending runs in {1,...,k}^n; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A024196 a(n) = 2nd elementary symmetric function of the first n+1 odd positive integers.

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A143785 Antidiagonal sums of the triangle A120070.

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A316989 Irregular triangle read by rows: row n consists of the coefficients in the expansion of the polynomial (x^2 + 4*x + 3)*(x + 1)^(2*n) + (x^2 - 1)*(x^2 + 3*x + 3).

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A224454 The Wiener index of the linear phenylene with n hexagons.

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A383368 Number of intercalates in pine Latin squares of order 2n.

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A089207 a(n) = 4n^3 + 2n^2.

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A316989 Irregular triangle read by rows: row n consists of the coefficients in the expansion of the polynomial (x^2 + 4x + 3)(x + 1)^(2n) + (x^2 - 1)(x^2 + 3*x + 3).

A190705 a(n) = 6n^2(2*n + 1).