A034856 -id:A034856 - cp's OEIS frontend

A286159 Lower triangular region of array A286156, transposed.

1, 1, 3, 1, 4, 6, 1, 4, 3, 10, 1, 4, 8, 7, 15, 1, 4, 8, 3, 6, 21, 1, 4, 8, 13, 7, 11, 28, 1, 4, 8, 13, 3, 12, 10, 36, 1, 4, 8, 13, 19, 7, 6, 16, 45, 1, 4, 8, 13, 19, 3, 12, 11, 15, 55, 1, 4, 8, 13, 19, 26, 7, 18, 17, 22, 66, 1, 4, 8, 13, 19, 26, 3, 12, 6, 10, 21, 78, 1, 4, 8, 13, 19, 26, 34, 7, 18, 11, 16, 29, 91, 1, 4, 8, 13, 19, 26, 34, 3, 12, 25, 17, 23, 28, 105

Offset: 1

Antti Karttunen, May 04 2017

			The first ten rows of this triangular array:
  1,
  1, 3,
  1, 4, 6,
  1, 4, 3, 10,
  1, 4, 8,  7, 15,
  1, 4, 8,  3,  6, 21,
  1, 4, 8, 13,  7, 11, 28,
  1, 4, 8, 13,  3, 12, 10, 36,
  1, 4, 8, 13, 19,  7,  6, 16, 45,
  1, 4, 8, 13, 19,  3, 12, 11, 15, 55

Antti Karttunen, Table of n, a(n) for n = 1..10585; the first 145 rows of the triangle

Transpose: A268158.
Cf. A000012, A000217, A268156, A268157.
Rows converge towards A034856.

Map[((#1 + #2)^2 + 3 #1 + #2)/2 & @@ # & /@ Reverse@ # &, Table[Reverse@ QuotientRemainder[n, k], {n, 14}, {k, n}]] // Flatten (* Michael De Vlieger, May 20 2017 *)

def T(a, b): return ((a + b)**2 + 3*a + b)//2
def a(n, k): return T(n%k, n//k)
for n in range(1, 21): print([a(n , k) for k in range(1, n + 1)][::-1])  # Indranil Ghosh, May 20 2017

(define (A286159 n) (A286156bi (A002024 n) (A004736 n))) ;; For A286156bi see A286156.

A302537 a(n) = (n^2 + 13*n + 2)/2.

Original entry on oeis.org

1, 8, 16, 25, 35, 46, 58, 71, 85, 100, 116, 133, 151, 170, 190, 211, 233, 256, 280, 305, 331, 358, 386, 415, 445, 476, 508, 541, 575, 610, 646, 683, 721, 760, 800, 841, 883, 926, 970, 1015, 1061, 1108, 1156, 1205, 1255, 1306, 1358, 1411, 1465, 1520, 1576

Offset: 0

Franck Maminirina Ramaharo, Apr 09 2018

Binomial transform of [1, 7, 1, 0, 0, 0, ...].

Numbers m > 0 such that 8*m + 161 is a square.

			Illustration of initial terms (by the formula a(n) = A052905(n) + 3*n):
.                                                                    o
.                                                                  o o
.                                                    o           o o o
.                                                  o o         o o o o
.                                      o         o o o       o o o o o
.                                    o o       o o o o     o o o o o o
.                          o       o o o     o o o o o   o . . . . . o
.                        o o     o o o o   o . . . . o   o . . . . . o
.                o     o o o   o . . . o   o . . . . o   o . . . . . o
.              o o   o . . o   o . . . o   o . . . . o   o . . . . . o
.        o   o . o   o . . o   o . . . o   o . . . . o   o . . . . . o
.      o o   o . o   o . . o   o . . . o   o . . . . o   o . . . . . o
.  o   o o   o o o   o o o o   o o o o o   o o o o o o   o o o o o o o
.        o     o o     o o o     o o o o     o o o o o     o o o o o o
.        o     o o     o o o     o o o o     o o o o o     o o o o o o
.        o     o o     o o o     o o o o     o o o o o     o o o o o o
----------------------------------------------------------------------
.  1     8      16        25          35            46              58

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics: A Foundation for Computer Science, Addison-Wesley, 1994.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Sequences whose n-th terms are of the form binomial(n, 2) + n*k + 1:
A152947 (k = 0); A000124 (k = 1); A000217 (k = 2); A034856 (k = 3);
A052905 (k = 4); A051936 (k = 5); A246172 (k = 6).
Cf. A000578, A002939, A004120, A008589, A056119, A060544, A162261.

A302537:= func< n | ((n+1)^2 +12*n +1)/2 >;
[A302537(n): n in [0..50]]; // G. C. Greubel, Jan 21 2025

a := n -> (n^2 + 13*n + 2)/2;
seq(a(n), n = 0 .. 100);

Table[(n^2 + 13 n + 2)/2, {n, 0, 100}]
CoefficientList[ Series[(5x^2 - 5x - 1)/(x - 1)^3, {x, 0, 50}], x] (* or *)
LinearRecurrence[{3, -3, 1}, {1, 8, 16}, 51] (* Robert G. Wilson v, May 19 2018 *)

makelist((n^2 + 13*n + 2)/2, n, 0, 100);

a(n) = (n^2 + 13*n + 2)/2; \\ Altug Alkan, Apr 12 2018

def A302537(n): return (n**2 + 13*n + 2)//2
print([A302537(n) for n in range(51)]) # G. C. Greubel, Jan 21 2025

a(n) = binomial(n + 1, 2) + 6*n + 1 = binomial(n, 2) + 7*n + 1.
a(n) = a(n-1) + n + 6.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3, where a(0) = 1, a(1) = 8 and a(2) = 16.
a(n) = 2*a(n-1) - a(n-2) + 1.
a(n) = A004120(n+1) for n > 1.
a(n) = A056119(n) + 1.
a(n) = A152947(n+1) + A008589(n).
a(n) = A060544(n+1) - A002939(n).
a(n) = A000578(n+1) - A162261(n) for n > 0.
G.f.: (1 + 5*x - 5*x^2)/(1 - x)^3.
E.g.f.: (1/2)*(2 + 14*x + x^2)*exp(x).
Sum_{n>=0} 1/a(n) = 24097/45220 + 2*Pi*tan(sqrt(161)*Pi/2) / sqrt(161) = 1.4630922534498496... - Vaclav Kotesovec, Apr 11 2018

A323221 a(n) = n(n + 5)(n + 7)/6 + 1.

Original entry on oeis.org

1, 9, 22, 41, 67, 101, 144, 197, 261, 337, 426, 529, 647, 781, 932, 1101, 1289, 1497, 1726, 1977, 2251, 2549, 2872, 3221, 3597, 4001, 4434, 4897, 5391, 5917, 6476, 7069, 7697, 8361, 9062, 9801, 10579, 11397, 12256, 13157, 14101, 15089, 16122, 17201, 18327, 19501

Offset: 0

Peter Luschny, Jan 25 2019

a(n) is related to the total angular defect of certain polytopes. See Hilton and Pedersen, Cor. 1; compare A275874.

			For n = 2 the sum formula gives:
I(2) = {{0,0}, {0,1}, {1,0}, {0,2}, {1,1}, {2,0}, {0,3}, {1,2}, {2,1}, {3,0}};
a(2) = 1 + 1 + 1 + 2 + 1 + 2 + 5 + 2 + 2 + 5 = 22.

P. Hilton and J. Pedersen, Descartes, Euler, Poincaré, Pólya and Polyhedra, L'Enseign. Math., 27 (1981), 327-343.

Çf. A323224 (column 4), A323233 (row 4), A034856 (first difference), A275874.

a := n -> n*(35 + 12*n + n^2)/6 + 1:
seq(a(n), n = 0..45);

a[n_] := n (35 + 12 n + n^2)/6 + 1;
Table[a[n], {n, 0, 45}]

Let I(n) denote the set of all tuples of length n with elements from {0, 1, 2, 3} with sum <= 3 and C(m) denote the m-th Catalan number. Then for n > 0
a(n) = Sum_{(j1,...,jn) in I(n)} C(j1)*C(j2)*...*C(jn).
a(n) = [x^n] (3*x^3 - 8*x^2 + 5*x + 1)/(x - 1)^4.
a(n) = n! [x^n] exp(x)*(x^3 + 15*x^2 + 48*x + 6)/6.
a(n) = a(n - 1)*(n*(n + 5)*(n + 7) + 6)/(n*(n + 2)*(n + 7) - 18) for n > 0.
a(n) = A323224(n, 4).
a(n) = A275874(n+4) + 1.

A323233 Coefficients of polynomials p(n, x) generating the columns of A323224, triangle read by rows, T(n, k) for n >= 1 and k >= 0.

Original entry on oeis.org

1, 2, 2, 6, 15, 3, 24, 140, 48, 4, 120, 1750, 775, 110, 5, 720, 28644, 14550, 2670, 210, 6, 5040, 588588, 323008, 68775, 7105, 357, 7, 40320, 14592864, 8388800, 1962632, 239120, 16016, 560, 8, 362880, 423227376, 250742700, 62531532, 8502921, 680904, 32130, 828, 9

Offset: 1

Peter Luschny, Jan 27 2019

			The triangle starts:
[ 1]       1;
[ 2]       2,         2;
[ 3]       6,        15,         3;
[ 4]      24,       140,        48,        4;
[ 5]     120,      1750,       775,      110,       5;
[ 6]     720,     28644,     14550,     2670,     210,      6;
[ 7]    5040,    588588,    323008,    68775,    7105,    357,     7;
[ 8]   40320,  14592864,   8388800,  1962632,  239120,  16016,   560,   8;
[ 9]  362880, 423227376, 250742700, 62531532, 8502921, 680904, 32130, 828, 9;
The first few polynomials are:
p[1](x) = 1;
p[2](x) = 2*x + 2!;
p[3](x) = 3*x*(x + 5) + 3!;
p[4](x) = 4*x*(x + 5)*(x + 7) + 4!;
p[5](x) = 5*x*(x + 5)*(x + 7)*(x + 10) + 5!;
p[6](x) = 6*x*(x + 7)*(x + 11)*(x^2 + 17*x + 62) + 6!;
p[7](x) = 7*x*(x + 6)*(x + 7)*(x + 11)*(x + 13)*(x + 14) + 7!;

Cf. A323224, A034856, A323221, A323220, A014137, A014138.

ogf[n_] := (2/(1 + Sqrt[1 - 4 x] ))^n  x/(1 - x);
ser[n_, len_] := CoefficientList[Series[ogf[n], {x, 0, (n + 1) len + 1}], x];
tab[k_, len_] := Table[{n, ser[n, k + 1][[k + 1]]}, {n, 0, len - 1}];
pol[n_] := n! InterpolatingPolynomial[tab[n, n + 1], x] // Expand;
row[n_] := CoefficientList[pol[n], x]; Table[row[n], {n, 1, 9}]

A323224(n, k) = p(k, n)/k!.
T(n, k) = [x^k] p(n, x).
p(n, 1)/n! and p(n, -1)/n! are versions of the partial sums of the Catalan numbers.

A361952 Array read by antidiagonals: T(n,k) is the number of unlabeled posets with n elements together with a function rk mapping each element to a rank between 1 and k such that whenever v covers w in the poset then rk(v) = rk(w) + 1.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 4, 1, 0, 1, 4, 8, 8, 1, 0, 1, 5, 13, 21, 17, 1, 0, 1, 6, 19, 40, 58, 38, 1, 0, 1, 7, 26, 66, 126, 172, 94, 1, 0, 1, 8, 34, 100, 228, 420, 569, 258, 1, 0, 1, 9, 43, 143, 373, 816, 1537, 2148, 815, 1, 0, 1, 10, 53, 196, 571, 1412, 3140, 6342, 9538, 3038, 1, 0

Offset: 0

Andrew Howroyd, Mar 31 2023

A poset is counted once for each admissible ranking function. This is an intermediate step in the computation of A361953 where each graded poset is counted exactly once.

			Array begins:
============================================
n/k| 0 1   2    3    4     5     6     7 ...
---+----------------------------------------
0  | 1 1   1    1    1     1     1     1 ...
1  | 0 1   2    3    4     5     6     7 ...
2  | 0 1   4    8   13    19    26    34 ...
3  | 0 1   8   21   40    66   100   143 ...
4  | 0 1  17   58  126   228   373   571 ...
5  | 0 1  38  172  420   816  1412  2272 ...
6  | 0 1  94  569 1537  3140  5631  9351 ...
7  | 0 1 258 2148 6342 13383 24410 41097 ...
  ...

Andrew Howroyd, Table of n, a(n) for n = 0..860 (first 41 antidiagonals).

Columns k=0..2 are A000007, A000012, A049312.
Rows n=0..4 are A000012, A000027, A034856, A137742.
The labeled version is A361950.
Cf. A361953.

\\ See Links in A361953 for program.
{ my(A=A361952tabl(7)); for(i=1, #A, print(A[i,])) }

A054254 a(n) is n plus the minimum of the a(i)*a(n-i) of the previous i = 1..n-1.

Original entry on oeis.org

0, 1, 2, 5, 8, 13, 19, 26, 34, 43, 53, 64, 76, 89, 103, 118, 134, 151, 169, 188, 208, 229, 251, 274, 298, 323, 349, 376, 404, 433, 463, 494, 526, 559, 593, 628, 664, 701, 739, 778, 818, 859, 901, 944, 988, 1033, 1079, 1126, 1174, 1223, 1273

Offset: 0

N. J. A. Sloane, May 04 2000

If in the Maple code "if n<=2 then n" were replaced by "if n<=1 then n", then the sequence would become the triangular numbers A000217. In general, if the Maple code were "if n<=k then n" for some given k > 0 then a(n) would be n if n <= k, n + k*(n-k) if k <= n <= 2k and n*(n+1)/2 - k*(k-1) if 2k <= n. - Henry Bottomley, Mar 30 2001

			a(3) = 3 + 1*2 = 5,
a(4) = 4 + 2*2 = 8 since 2*2 < 1*5,
a(5) = 5 + 1*8 = 13 since 1*8 < 2*5.

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

A054254 := proc(n) local i,j; option remember; if n<=2 then n else j := 10^100; for i from 1 to n-1 do if procname(i)*procname(n-i) < j then j := procname(i)*procname(n-i); end if; end do; n+j; fi; end proc;

Join[{0, 1, 2, 5}, LinearRecurrence[{3, -3, 1}, {8, 13, 19}, 50]] (* Jean-François Alcover, Apr 29 2023 *)

For n > 3: a(n) = (n^2 + n - 4)/2 = A034856(n-1) = A000217(n) - 2 = A000297(n-3) - A000297(n-2).
For n > 4: a(n) = a(n-1) + n.
G.f.: x*(x^2-x+1)*(x^3-x^2-1)/(x-1)^3. - R. J. Mathar, Dec 09 2009

More specific name from R. J. Mathar, Dec 09 2009

A077593 Table by antidiagonals where T(n,k) = Sum_{i=1..n} T(floor(n/i),k-1) starting with T(n,0)=1 if n>0 and T(0,0)=0.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 5, 4, 1, 0, 1, 5, 7, 8, 5, 1, 0, 1, 6, 9, 13, 10, 6, 1, 0, 1, 7, 11, 19, 16, 14, 7, 1, 0, 1, 8, 13, 26, 23, 25, 16, 8, 1, 0, 1, 9, 15, 34, 31, 39, 28, 20, 9, 1, 0, 1, 10, 17, 43, 40, 56, 43, 38, 23, 10, 1, 0, 1, 11, 19, 53, 50, 76, 61, 63

Offset: 0

Henry Bottomley, Nov 08 2002

			Rows start:
 0,0,0,0,0,0...;
 1,1,1,1,1,1...;
 1,2,3,4,5,6...;
 1,3,5,7,9,11...;
 1,4,8,13,19,26,...;
 ...

Columns include A057427, A001477, A006218, A061201, A061202, A061203, A061204.
Rows include (with offsets) A000004, A000012, A000027, A005408, A034856, A052905.
Cf. A077593.

T(n, k) = T(n-1, k) + A077592(n, k). Writing m as Sum_{i} p_i^e_i, T(n, k) = Sum_{m=1..n} Product_{i} C(k+e_i-1, e_i).

A104473 a(n) = binomial(n+2,2)*binomial(n+6,2).

Original entry on oeis.org

15, 63, 168, 360, 675, 1155, 1848, 2808, 4095, 5775, 7920, 10608, 13923, 17955, 22800, 28560, 35343, 43263, 52440, 63000, 75075, 88803, 104328, 121800, 141375, 163215, 187488, 214368, 244035, 276675, 312480, 351648, 394383, 440895, 491400, 546120, 605283, 669123

Offset: 0

Zerinvary Lajos, Apr 18 2005

			a(0) = C(0+2,2)*C(0+6,2) = C(2,2)*C(6,2) = 1*15 = 155.
a(6) = 1*3*5 + 2*4*6 + 3*5*7 + 4*6*8 + 5*7*9 + 6*8*10 + 7*9*11 = 1848.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217, A000579, A033275, A034856, A062264.

Subsequence of A085780.

[Binomial(n+2, 2)*Binomial(n+6, 2): n in [0..50]]; // Vincenzo Librandi, Apr 28 2014

f[n_] := Binomial[n + 2, 2] Binomial[n + 6, 2]; Table[f[n], {n,0,40}] (* Robert G. Wilson v, Apr 20 2005 *)
CoefficientList[Series[3 (5-4*x+x^2)/(1-x)^5, {x,0,40}], x] (* Vincenzo Librandi, Apr 28 2014 *)

a(n)=binomial(n+2,2)*binomial(n+6,2) \\ Charles R Greathouse IV, Jun 07 2013

def A104473(n): return binomial(n+2,2)*binomial(n+6,2)
print([A104473(n) for n in range(51)]) # G. C. Greubel, Mar 05 2025

a(n) = (1/4)*(n+1)*(n+2)*(n+5)*(n+6).
a(n) = A034856(n+2)^2 - 1. - J. M. Bergot, Dec 14 2010
G.f.: 3*(5-4*x+x^2)/(1-x)^5. - Colin Barker, Sep 21 2012
a(n) = Sum_{i=1..n+1} i*(i+2)*(i+4). - Bruno Berselli, Apr 28 2014
a(n) = A000217(n)*A000217(n+4) = 3*A033275(n+4). - R. J. Mathar, Nov 29 2015
From Amiram Eldar, Aug 30 2022: (Start)
Sum_{n>=0} 1/a(n) = 43/450.
Sum_{n>=0} (-1)^n/a(n) = 16*log(2)/15 - 154/225. (End)
From G. C. Greubel, Mar 05 2025: (Start)
a(n) = 90*A000579(n+6)/A000279(n+3).
E.g.f.: (1/4)*(60 + 192*x + 114*x^2 + 20*x^3 + x^4)*exp(x). (End)

A111694 a(1) = 1+2-3 = 0, a(2) = 4+5+6-7 = 8, a(3) = 8+9+10+11-12 = 26, a(4) = 13+14+15+16+17-18 = 57, ...

Original entry on oeis.org

0, 8, 26, 57, 104, 170, 258, 371, 512, 684, 890, 1133, 1416, 1742, 2114, 2535, 3008, 3536, 4122, 4769, 5480, 6258, 7106, 8027, 9024, 10100, 11258, 12501, 13832, 15254, 16770, 18383, 20096, 21912, 23834, 25865, 28008, 30266, 32642, 35139, 37760

Offset: 1

Amarnath Murthy, Aug 17 2005

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A034856.

[(n^3 + 4*n^2 - 3*n - 2)/2: n in [1..60]]; // Vincenzo Librandi, May 21 2011

seq(sum(j,j=binomial(n+1,2)+n-1..binomial(n+2,2)+n-2) - binomial(n+2,2)-n+1,n=1..50); # Emeric Deutsch, Aug 27 2005

Table[(n^3 + 4n^2 - 3n - 2)/2, {n, 41}] (* Robert G. Wilson v, Aug 27 2005 *)

a(n) = (n-1)*(n^2 + 5*n + 2)/2.
G.f.: x^2*(x-2)*(x-4)/(1-x)^4. - Colin Barker, Mar 18 2012
a(n) = (n-1)*A034856(n+1). - R. J. Mathar, Aug 22 2016

Edited and extended by Emeric Deutsch and Robert G. Wilson v, Aug 27 2005

A118976 Triangle read by rows: T(n,k) = binomial(n-1,k-1)binomial(n,k-1)/k + binomial(n-1,k)binomial(n,k)/(k+1) (1 <= k <= n). In other words, to each entry of the Narayana triangle (A001263) add the entry on its right.

Original entry on oeis.org

1, 2, 1, 4, 4, 1, 7, 12, 7, 1, 11, 30, 30, 11, 1, 16, 65, 100, 65, 16, 1, 22, 126, 280, 280, 126, 22, 1, 29, 224, 686, 980, 686, 224, 29, 1, 37, 372, 1512, 2940, 2940, 1512, 372, 37, 1, 46, 585, 3060, 7812, 10584, 7812, 3060, 585, 46, 1, 56, 880, 5775, 18810, 33264, 33264, 18810, 5775, 880, 56, 1

Offset: 1

Gary W. Adamson, May 07 2006

Sum of entries in row n = 2*Cat(n)-1, where Cat(n) are the Catalan numbers (A000108).

Row sums = A131428 starting (1, 3, 9, 27, 83, ...). - Gary W. Adamson, Aug 31 2007

			First few rows of the triangle:
   1;
   2,  1;
   4,  4,   1;
   7, 12,   7,  1;
  11, 30,  30, 11,  1;
  16, 65, 100, 65, 16, 1;
...
Row 4 of the triangle = (7, 12, 7, 1), derived from row 4 of the Narayana triangle, (1, 6, 6, 1): = ((1+6), (6+6), (6+1), (1)).

G. C. Greubel, Rows n = 1..100 of triangle, flattened

Cf. A001263, A034856, A000108, A131428.

B:=Binomial; Flat(List([1..12], n-> List([1..n], k-> B(n-1,k-1)*B(n,k-1)/k + B(n-1,k)*B(n,k)/(k+1) ))); # G. C. Greubel, Aug 12 2019

B:=Binomial; [B(n-1,k-1)*B(n,k-1)/k + B(n-1,k)*B(n,k)/(k+1): k in [1..n], n in [1..12]]; // G. C. Greubel, Aug 12 2019

T:=(n,k)->binomial(n-1,k-1)*binomial(n,k-1)/k+binomial(n-1,k) *binomial(n,k)/ (k+1): for n from 1 to 12 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form
# Alternatively:
gf := 1 - ((1/2)*(x + 1)*(sqrt((x*y + y - 1)^2 - 4*y^2*x) + x*y + y - 1))/(y*x):
sery := series(gf, y, 10): coeffy := n -> expand(coeff(sery, y, n)):
seq(print(seq(coeff(coeffy(n), x, k), k=1..n)), n=1..8); # Peter Luschny, Oct 21 2020

With[{B=Binomial}, Table[B[n-1,k-1]*B[n,k-1]/k + B[n-1,k]*B[n,k]/(k+1), {n,12}, {k,n}]//Flatten] (* G. C. Greubel, Aug 12 2019 *)

T(n,k) = b=binomial; b(n-1,k-1)*b(n,k-1)/k + b(n-1,k)*b(n,k)/(k+1);
for(n=1,12, for(k=1,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Aug 12 2019

def T(n, k):
    b=binomial
    return b(n-1,k-1)*b(n,k-1)/k + b(n-1,k)*b(n,k)/(k+1)
[[T(n, k) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Aug 12 2019

G.f.: A001263(x, y)*(x + x*y) + x*y. - Vladimir Kruchinin, Oct 21 2020

Edited by N. J. A. Sloane, Nov 29 2006

cp's OEIS Frontend

A286159 Lower triangular region of array A286156, transposed.

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A302537 a(n) = (n^2 + 13*n + 2)/2.

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A323221 a(n) = n*(n + 5)*(n + 7)/6 + 1.

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A323233 Coefficients of polynomials p(n, x) generating the columns of A323224, triangle read by rows, T(n, k) for n >= 1 and k >= 0.

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A361952 Array read by antidiagonals: T(n,k) is the number of unlabeled posets with n elements together with a function rk mapping each element to a rank between 1 and k such that whenever v covers w in the poset then rk(v) = rk(w) + 1.

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A054254 a(n) is n plus the minimum of the a(i)*a(n-i) of the previous i = 1..n-1.

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A077593 Table by antidiagonals where T(n,k) = Sum_{i=1..n} T(floor(n/i),k-1) starting with T(n,0)=1 if n>0 and T(0,0)=0.

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A104473 a(n) = binomial(n+2,2)*binomial(n+6,2).

A323221 a(n) = n(n + 5)(n + 7)/6 + 1.

A118976 Triangle read by rows: T(n,k) = binomial(n-1,k-1)binomial(n,k-1)/k + binomial(n-1,k)binomial(n,k)/(k+1) (1 <= k <= n). In other words, to each entry of the Narayana triangle (A001263) add the entry on its right.