A010763 -id:A010763 - cp's OEIS frontend

A014473 Pascal's triangle - 1: Triangle read by rows: T(n, k) = A007318(n, k) - 1.

0, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 3, 5, 3, 0, 0, 4, 9, 9, 4, 0, 0, 5, 14, 19, 14, 5, 0, 0, 6, 20, 34, 34, 20, 6, 0, 0, 7, 27, 55, 69, 55, 27, 7, 0, 0, 8, 35, 83, 125, 125, 83, 35, 8, 0, 0, 9, 44, 119, 209, 251, 209, 119, 44, 9, 0, 0, 10, 54, 164, 329, 461, 461, 329, 164, 54, 10, 0

Offset: 0

N. J. A. Sloane

Indexed as a square array A(n,k): If X is an (n+k)-set and Y a fixed k-subset of X then A(n,k) is equal to the number of n-subsets of X intersecting Y. - Peter Luschny, Apr 20 2012

			Triangle begins:
   0;
   0, 0;
   0, 1,  0;
   0, 2,  2,  0;
   0, 3,  5,  3,  0;
   0, 4,  9,  9,  4,  0;
   0, 5, 14, 19, 14,  5, 0;
   0, 6, 20, 34, 34, 20, 6, 0;
   ...
Seen as a square array read by antidiagonals:
  [0] 0, 0,  0,  0,   0,   0,   0,    0,    0,    0,    0,     0, ... A000004
  [1] 0, 1,  2,  3,   4,   5,   6,    7,    8,    9,   10,    11, ... A001477
  [2] 0, 2,  5,  9,  14,  20,  27,   35,   44,   54,   65,    77, ... A000096
  [3] 0, 3,  9, 19,  34,  55,  83,  119,  164,  219,  285,   363, ... A062748
  [4] 0, 4, 14, 34,  69, 125, 209,  329,  494,  714, 1000,  1364, ... A063258
  [5] 0, 5, 20, 55, 125, 251, 461,  791, 1286, 2001, 3002,  4367, ... A062988
  [6] 0, 6, 27, 83, 209, 461, 923, 1715, 3002, 5004, 8007, 12375, ... A124089

Reinhard Zumkeller, Rows n=0..100 of triangle, flattened
Milan Janjic, Two Enumerative Functions

Triangle without zeros: A014430.
Related: A323211 (A007318(n, k) + 1).
A000295 (row sums), A059841 (alternating row sums), A030662(n-1) (central terms).
Columns include A000096, A062748, A062988, A063258.
Diagonals of A(n, n+d): A030662 (d=0), A010763 (d=1), A322938 (d=2).
Cf. A000004, A001477, A007318, A030662, A059841, A109128, A124089, A129696.

a014473 n k = a014473_tabl !! n !! k
a014473_row n = a014473_tabl !! n
a014473_tabl = map (map (subtract 1)) a007318_tabl
-- Reinhard Zumkeller, Apr 10 2012

[Binomial(n,k)-1: k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 08 2024

with(combstruct): for n from 0 to 11 do seq(-1+count(Combination(n), size=m), m = 0 .. n) od; # Zerinvary Lajos, Apr 09 2008
# The rows of the square array:
Arow := proc(n, len) local gf, ser;
gf := (x - 1)^(-n - 1) + (-1)^(n + 1)/(x*(x - 1));
ser := series(gf, x, len+2): seq((-1)^(n+1)*coeff(ser, x, j), j=0..len) end:
for n from 0 to 9 do lprint([n], Arow(n, 12)) od; # Peter Luschny, Feb 13 2019

Table[Binomial[n,k] -1, {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Apr 08 2024 *)

flatten([[binomial(n,k)-1 for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Apr 08 2024

G.f.: x^2*y/((1 - x)*(1 - x*y)*(1 - x*(1 + y))). - Ralf Stephan, Jan 24 2005
T(n,k) = A109128(n,k) - A007318(n,k), 0 <= k <= n. - Reinhard Zumkeller, Apr 10 2012
T(n, k) = T(n-1, k-1) + T(n-1, k) + 1, 0 < k < n with T(n, 0) = T(n, n) = 0. - Reinhard Zumkeller, Jul 18 2015
If seen as a square array read by antidiagonals the generating function of row n is: G(n) = (x - 1)^(-n - 1) + (-1)^(n + 1)/(x*(x - 1)). - Peter Luschny, Feb 13 2019
From G. C. Greubel, Apr 08 2024: (Start)
T(n, n-k) = T(n, k).
Sum_{k=0..floor(n/2)} T(n-k, k) = A129696(n-2).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = b(n-1), where b(n) is the repeating pattern {0, 0, -1, -2, -1, 1, 1, -1, -2, -1, 0, 0}_{n=0..11}, with b(n) = b(n-12). (End)

More terms from Erich Friedman

A256117 Number T(n,k) of length 2n words such that all letters of the k-ary alphabet occur at least once and are introduced in ascending order and which can be built by repeatedly inserting doublets into the initially empty word; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 9, 5, 0, 1, 34, 56, 14, 0, 1, 125, 465, 300, 42, 0, 1, 461, 3509, 4400, 1485, 132, 0, 1, 1715, 25571, 55692, 34034, 7007, 429, 0, 1, 6434, 184232, 657370, 647920, 231868, 32032, 1430, 0, 1, 24309, 1325609, 7488228, 11187462, 6191808, 1447992, 143208, 4862

Offset: 0

Alois P. Heinz, Mar 15 2015

In general, column k>2 is asymptotic to (4*(k-1))^n / ((k-2)^2 * (k-2)! * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Jun 01 2015

			T(0,0) = 1: (the empty word).
T(1,1) = 1: aa.
T(2,1) = 1: aaaa.
T(2,2) = 2: aabb, abba.
T(3,1) = 1: aaaaaa.
T(3,2) = 9: aaaabb, aaabba, aabaab, aabbaa, aabbbb, abaaba, abbaaa, abbabb, abbbba.
T(3,3) = 5: aabbcc, aabccb, abbacc, abbcca, abccba.
Triangle T(n,k) begins:
  1;
  0, 1;
  0, 1,    2;
  0, 1,    9,      5;
  0, 1,   34,     56,     14;
  0, 1,  125,    465,    300,     42;
  0, 1,  461,   3509,   4400,   1485,    132;
  0, 1, 1715,  25571,  55692,  34034,   7007,   429;
  0, 1, 6434, 184232, 657370, 647920, 231868, 32032, 1430;
  ...

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

Columns k=0-10 give: A000007, A057427, A010763(n-1) (for n>1), A258490, A258491, A258492, A258493, A258494, A258495, A258496, A258497.
Main diagonal gives A000108.
T(n+2,n+1) gives A002055(n+5).
Row sums give A258498.
T(2n,n) gives A258499.
Cf. A183135, A256116, A256311.

A:= proc(n, k) option remember; `if`(n=0, 1, k/n*
      add(binomial(2*n, j)*(n-j)*(k-1)^j, j=0..n-1))
    end:
T:= (n, k)-> add((-1)^i*A(n, k-i)/(i!*(k-i)!), i=0..k):
seq(seq(T(n, k), k=0..n), n=0..10);

A[n_, k_] := A[n, k] = If[n == 0, 1, k/n*Sum[Binomial[2*n, j]*(n - j)*If[j == 0, 1, (k - 1)^j], {j, 0, n - 1}]];
T[n_, k_] := Sum[(-1)^i*A[n, k - i]/(i!*(k - i)!), {i, 0, k}];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* Jean-François Alcover, Feb 22 2016, after Alois P. Heinz, updated Jan 01 2021 *)

T(n,k) = Sum_{i=0..k} (-1)^i * A183135(n,k-i) / (i!*(k-i)!).

T(n,k) = A256116(n,k) / (k-1)! for k > 0.

A061691 Triangle of generalized Stirling numbers.

Original entry on oeis.org

1, 1, 2, 1, 9, 6, 1, 34, 72, 24, 1, 125, 650, 600, 120, 1, 461, 5400, 10500, 5400, 720, 1, 1715, 43757, 161700, 161700, 52920, 5040, 1, 6434, 353192, 2361016, 4116000, 2493120, 564480, 40320, 1, 24309, 2862330, 33731208, 96960024, 97161120, 39372480, 6531840, 362880

Offset: 1

N. J. A. Sloane, Jun 18 2001

The Eulerian-type number triangle associated with this triangle of generalized Stirling numbers is A192721. The table entry T(n,k) gives the number of uniform block permutations of the set {1,2,...,n} partitioned into k blocks. An example is given below. T(n,k) also gives the number of games of simple patience with n cards resulting in k piles (adapt Algorithm 1.1.22 of Lankham). [Peter Bala, Jul 14 2011]

			Triangle begins:
  1;
  1,2;
  1,9,6;
  1,34,72,24;
  1,125,650,600,120;
  ...
T(4,2) = 34:
There are 7 partitions of the set {1,2,3,4} into 2 blocks. The four partitions {1,2,3}{4}, {1,2,4}{3}, {1,3,4}{2} and {2,3,4}{1} give rise to 4*4 = 16 uniform block permutations while the remaining 3 partitions {1,2}{3,4}, {1,3}{2,4} and {1,4}{2,3} give 2!*3*3 = 18 uniform block permutations : thus in total there are 16+18 = 34 block permutations between the set partitions of {1,2,3,4} into 2 blocks.

Alois P. Heinz, Rows n = 1..141, flattened
M. Aguiar and R. C. Orellana, The Hopf algebra of uniform block permutations, 17th International Conference on Formal Power Series and Algebraic Combinatorics, Taormina, July 2005.
D. Aldous and P. Diaconis, Longest increasing subsequences: from patience sorting to the Baik-Deift-Johansson theorem, Bull. Amer. Math. Soc. 36 (1999), 413-432.
D. G. Fitzgerald, A presentation for the monoid of uniform block permutations, Bull. Austral. Math. Soc. 68 (2003), 317-324.
A. T. Irish, F. Quitin, U. Madhow, and M. Rodwell, Achieving multiple degrees of freedom in long-range mm-wave MIMO channels using randomly distributed relays, 2014.
I. P. Lankham, Patience Sorting and Its Generalizations, arXiv:0705.4524 [math.CO], 2007.
J.-M. Sixdeniers, K. A. Penson and A. I. Solomon, Extended Bell and Stirling Numbers From Hypergeometric Exponentiation, J. Integer Seqs. Vol. 4 (2001), #01.1.4.

Diagonals give A010763, A061690, A000142, A001809, A061689. Cf. A061692. A023998 (row sums), A192721, A192722.

#A061691
#J = sum {n>=0} z^n/n!^2
J := BesselJ(0, 2*i*sqrt(z)):
G := exp(x*(J(z)-1)):
Gser := simplify(series(G, z = 0, 12)):
for n from 1 to 10 do
P[n] := n!^2*sort(coeff(Gser, z, n)) od:
for n from 1 to 10 do seq(coeff(P[n],x,k), k = 1..n) od;
# yields sequence in triangular form
# second Maple program:
b:= proc(n) option remember; expand(`if`(n=0, 1,
      add(x*b(n-i)*binomial(n, i)/i!, i=1..n)))
    end:
T:= n-> (p-> seq(coeff(p, x, i)/i!, i=1..n))(b(n)*n!):
seq(T(n), n=1..12);  # Alois P. Heinz, Sep 10 2019

max = 9; g := Exp[x*(BesselI[0, 2*Sqrt[z]] - 1)]; gser = Series[g, {z, 0, max}, {x, 0, max}]; t[n_, k_] := n!^2*SeriesCoefficient[ gser // Normal, {z, 0, n}, {x, 0, k}]; Flatten[ Table[ t[n, k], {n, 1, max}, {k, 1, n}]] (* Jean-François Alcover, Apr 04 2012, after Maple *)

T(n, k) = 1/k!*Sum multinomial(n, n_1, n_2, ..n_k)^2, where the sum extends over all compositions (n_1, n_2, .., n_k) of n into exactly k nonnegative parts. - Vladeta Jovovic, Apr 23 2003
From Peter Bala, Jul 14 2011: (Start)
The table entry T(n,k) may also be expressed as a sum over (unordered) partitions of n into k parts:
T(n,k) = sum {partitions m_1*1+...+m_n*n = n, m_1+...+m_n = k} 1/(m_1!*...*m_n!)*{n!/(1!^(m_1)*...*n!^(m_n))}^2.
Generating function:
Let J(z) = sum {n>=0} z^n/n!^2. Then
exp(x*(J(z)-1)) = 1 + x*z + (x + 2*x^2)*z^2/2!^2 + (x + 9*x^2 + 6*x^3)*z^3/3!^2 + ....
Relations with other sequences:
T(n,k) = 1/k!*A192722(n,k).
Row sums [1,3,16,131,...] = A023998. (End)
The row polynomials R(n,x) satisfy the recurrence equation R(n,x) = x*( sum {k = 0..n-1} binomial(n,k)*binomial(n-1,k)*R(k,x) ) with R(0,x) = 1. Also R(n,x + y) = sum {k = 0..n} binomial(n,k)^2*R(k,x)*R(n-k,y). - Peter Bala, Sep 17 2013

More terms from Vladeta Jovovic, Apr 23 2003

A256116 Number T(n,k) of length 2n k-ary words, either empty or beginning with the first letter of the alphabet and using each letter at least once, that can be built by repeatedly inserting doublets into the initially empty word; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 9, 10, 0, 1, 34, 112, 84, 0, 1, 125, 930, 1800, 1008, 0, 1, 461, 7018, 26400, 35640, 15840, 0, 1, 1715, 51142, 334152, 816816, 840840, 308880, 0, 1, 6434, 368464, 3944220, 15550080, 27824160, 23063040, 7207200

Offset: 0

Alois P. Heinz, Mar 15 2015

			T(3,2) = 9: aaaabb, aaabba, aabaab, aabbaa, aabbbb, abaaba, abbaaa, abbabb, abbbba.
T(3,3) = 10: aabbcc, aabccb, aacbbc, aaccbb, abbacc, abbcca, abccba, acbbca, accabb, accbba.
T(4,2) = 34: aaaaaabb, aaaaabba, aaaabaab, aaaabbaa, aaaabbbb, aaabaaba, aaabbaaa, aaabbabb, aaabbbba, aabaaaab, aabaabaa, aabaabbb, aababbab, aabbaaaa, aabbaabb, aabbabba, aabbbaab, aabbbbaa, aabbbbbb, abaaaaba, abaabaaa, abaababb, abaabbba, ababbaba, abbaaaaa, abbaaabb, abbaabba, abbabaab, abbabbaa, abbabbbb, abbbaaba, abbbbaaa, abbbbabb, abbbbbba.
T(4,4) = 84: aabbccdd, aabbcddc, aabbdccd, aabbddcc, aabccbdd, aabccddb, aabcddcb, aabdccdb, aabddbcc, aabddccb, aacbbcdd, aacbbddc, aacbddbc, aaccbbdd, aaccbddb, aaccdbbd, aaccddbb, aacdbbdc, aacddbbc, aacddcbb, aadbbccd, aadbbdcc, aadbccbd, aadcbbcd, aadccbbd, aadccdbb, aaddbbcc, aaddbccb, aaddcbbc, aaddccbb, abbaccdd, abbacddc, abbadccd, abbaddcc, abbccadd, abbccdda, abbcddca, abbdccda, abbddacc, abbddcca, abccbadd, abccbdda, abccddba, abcddcba, abdccdba, abddbacc, abddbcca, abddccba, acbbcadd, acbbcdda, acbbddca, acbddbca, accabbdd, accabddb, accadbbd, accaddbb, accbbadd, accbbdda, accbddba, accdbbda, accddabb, accddbba, acdbbdca, acddbbca, acddcabb, acddcbba, adbbccda, adbbdacc, adbbdcca, adbccbda, adcbbcda, adccbbda, adccdabb, adccdbba, addabbcc, addabccb, addacbbc, addaccbb, addbbacc, addbbcca, addbccba, addcbbca, addccabb, addccbba.
Triangle T(n,k) begins:
  1;
  0, 1;
  0, 1,    2;
  0, 1,    9,    10;
  0, 1,   34,   112,     84;
  0, 1,  125,   930,   1800,   1008;
  0, 1,  461,  7018,  26400,  35640,  15840;
  0, 1, 1715, 51142, 334152, 816816, 840840, 308880;

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

Columns k=0-2 give: A000007, A057427, A010763(n-1) for n>0.
Main diagonal gives A065866(n-1) (for n>0).
Row sums give A294603.
Cf. A183135, A256117.

A:= proc(n, k) option remember; `if`(n=0, 1, k/n*
      add(binomial(2*n, j) *(n-j) *(k-1)^j, j=0..n-1))
    end:
T:= (n, k)-> add(A(n, k-i)*(-1)^i*binomial(k, i), i=0..k)/
    `if`(k=0, 1, k):
seq(seq(T(n, k), k=0..n), n=0..12);

Unprotect[Power]; 0^0 = 1; A[n_, k_] := A[n, k] = If[n==0, 1, k/n*Sum[ Binomial[2*n, j]*(n-j)*(k-1)^j, {j, 0, n-1}]];
T[n_, k_] := Sum[A[n, k-i]*(-1)^i*Binomial[k, i], {i, 0, k}]/If[k==0, 1, k]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 22 2017, translated from Maple *)

T(n,k) = (Sum_{i=0..k} (-1)^i * C(k,i) * A183135(n,k-i)) / A028310(k).

T(n,k) = (k-1)! * A256117(n,k) for k > 0.

A370366 Number A(n,k) of partitions of [k*n] into n sets of size k having no set of consecutive numbers whose maximum (if k>0) is a multiple of k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 2, 0, 0, 1, 0, 9, 8, 0, 0, 1, 0, 34, 252, 60, 0, 0, 1, 0, 125, 5672, 14337, 544, 0, 0, 1, 0, 461, 125750, 2604732, 1327104, 6040, 0, 0, 1, 0, 1715, 2857472, 488360625, 2533087904, 182407545, 79008, 0, 0

Offset: 0

Alois P. Heinz, Feb 16 2024

			A(2,3) = 9: 124|356, 125|346, 126|345, 134|256, 135|246, 136|245, 145|236, 146|235, 156|234.
Square array A(n,k) begins:
  1, 1,   1,       1,          1,             1, ...
  0, 0,   0,       0,          0,             0, ...
  0, 0,   2,       9,         34,           125, ...
  0, 0,   8,     252,       5672,        125750, ...
  0, 0,  60,   14337,    2604732,     488360625, ...
  0, 0, 544, 1327104, 2533087904, 5192229797500, ...

Alois P. Heinz, Antidiagonals n = 0..54, flattened
Wikipedia, Partition of a set

Columns k=0+1,2-3 give: A000007, A053871, A370357.
Rows n=0-2 give: A000012, A000004, A010763(n-1) for k>0.
Main diagonal gives A370367.
Antidiagonal sums give A370368.
Cf. A060540, A370363.

A:= proc(n, k) `if`(k=0,`if`(n=0, 1, 0), add(
      (-1)^(n-j)*binomial(n, j)*(k*j)!/(j!*k!^j), j=0..n))
    end:
seq(seq(A(n, d-n), n=0..d), d=0..10);

A(n,k) = A060540(n,k) - A370363(n,k) for n,k >= 1.

A260875 Square array read by ascending antidiagonals: number of m-shape complementary Bell numbers.

Original entry on oeis.org

1, 1, -1, 1, -1, 0, 1, -1, 0, -1, 1, -1, 2, 1, 1, 1, -1, 9, -1, 1, -1, 1, -1, 34, -197, -43, -2, 1, 1, -1, 125, -5281, 6841, 254, -9, -1, 1, -1, 461, -123124, 2185429, -254801, 4157, -9, 2, 1, -1, 1715, -2840293, 465693001, -1854147586, -3000807, -70981, 50, -2

Offset: 1

Peter Luschny, Aug 09 2015

A set partition of m-shape is a partition of a set with cardinality m*n for some n >= 0 such that the sizes of the blocks are m times the parts of the integer partitions of n.
M-complementary Bell numbers count the m-shape set partitions which have even length minus the number of such partitions which have odd length.
If m=0 all possible sizes are zero. Thus in this case the complementary Bell numbers count the integer partitions of n into an even number of parts minus the number of integer partitions of n into an odd number of parts (A081362).
If m=1 the set is {1,2,...,n} and the complementary Bell numbers count the set partitions which have even length minus the set partitions which have odd length (A000587).
If m=2 the set is {1,2,...,2n} and the complementary Bell numbers count the set partitions with even blocks which have even length minus the number of partitions with even blocks which have odd length (A260884).

			[ n ] [ 0   1   2      3        4            5              6]
[ m ] --------------------------------------------------------
[ 0 ] [ 1, -1,  0,    -1,       1,          -1,             1] A081362
[ 1 ] [ 1, -1,  0,     1,       1,          -2,            -9] A000587
[ 2 ] [ 1, -1,  2,    -1,     -43,         254,          4157] A260884
[ 3 ] [ 1, -1,  9,  -197,    6841,     -254801,      -3000807]
[ 4 ] [ 1, -1, 34, -5281, 2185429, -1854147586, 2755045819549]
      A010763,
For example the number of set partitions of {1,2,...,9} with sizes in [9], [6,3] and [3,3,3] are 1, 84, 280 respectively. Thus A(3,3) = -1 + 84 - 280 = -197.
Formatted as a triangle:
[1]
[1, -1]
[1, -1,   0]
[1, -1,   0,    -1]
[1, -1,   2,     1,    1]
[1, -1,   9,    -1,    1,  -1]
[1, -1,  34,  -197,  -43,  -2,  1]
[1, -1, 125, -5281, 6841, 254, -9, -1]

Cf. A000587, A010763, A081362, A260877, A260833, A260884, A260876.

def A260875(m, n):
    shapes = ([x*m for x in p] for p in Partitions(n))
    return sum((-1)^len(s)*SetPartitions(sum(s),s).cardinality() for s in shapes)
for m in (0..4): print([A260875(m,n) for n in (0..6)])

A318253 Coefficient of x of the OmegaPolynomials (A318146), T(n, k) = [x] P(n, k) with n>=1 and k>=0, square array read by ascending antidiagonals.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, 1, -2, 0, 0, 1, -9, 16, 0, 0, 1, -34, 477, -272, 0, 0, 1, -125, 11056, -74601, 7936, 0, 0, 1, -461, 249250, -14873104, 25740261, -353792, 0, 0, 1, -1715, 5699149, -2886735625, 56814228736, -16591655817, 22368256, 0, 0, 1, -6434, 132908041, -574688719793, 122209131374375, -495812444583424, 17929265150637, -1903757312, 0

Offset: 1

Peter Luschny, Aug 22 2018

Because in the case n=2 these numbers are the classical signed tangent numbers (A000182) we call T(n, k) also 'generalized tangent numbers' when studied in connection with generalized Bernoulli numbers.

			[n\k][0  1     2        3              4                   5  ...]
------------------------------------------------------------------
[1]   0, 1,    0,       0,             0,                  0, ...  [A063524]
[2]   0, 1,   -2,      16,          -272,               7936, ...  [A000182]
[3]   0, 1,   -9,     477,        -74601,           25740261, ...  [A293951]
[4]   0, 1,  -34,   11056,     -14873104,        56814228736, ...  [A273352]
[5]   0, 1, -125,  249250,   -2886735625,    122209131374375, ...  [A318258]
[6]   0, 1, -461, 5699149, -574688719793, 272692888959243481, ...
        [A010763]

Cf. A318146, A181937, A063524, A000182, A293951, A273352, A318258.

# Prints square array row-wise. The function OmegaPolynomial is defined in A318146.
for n from 1 to 6 do seq(coeff(OmegaPolynomial(n, k), x, 1), k=0..6) od;
# In the sequence format:
0, seq(seq(coeff(OmegaPolynomial(n-k+1, k), x, 1), k=0..n), n=1..9);
# Alternatively, based on the recurrence of the André numbers:
ANum := proc(m, n) option remember; if n = 0 then return 1 fi;
`if`(modp(n, m) = 0, -1, 1);  [seq(m*k, k=0..(n-1)/m)];
%%*add(binomial(n, k)*ANum(m, k), k in %) end:
TNum := proc(n,k) if k=1 then 1 elif k=0 or n=1 then 0 else ANum(n, n*k-1) fi end:
for n from 1 to 6 do seq(TNum(n, k), k = 0..6) od;

OmegaPolynomial[m_, n_] := Module[{S}, S = Series[MittagLefflerE[m, z]^x, {z, 0, 10}]; Expand[(m*n)! Coefficient[S, z, n]]];
T[n_, k_] := D[OmegaPolynomial[n, k], x] /. x -> 0;
Table[T[n - k, k], {n, 1, 10}, {k, 0, n - 1}] // Flatten (* Jean-François Alcover, Nov 27 2023 *)

# Prints the array row-wise. The function OmegaPolynomial is in A318146.
for m in (1..6):
    print([0] + [list(OmegaPolynomial(m, n))[1] for n in (1..6)])
# Alternatively, based on the recurrence of the André numbers:
@cached_function
def ANum(m, n):
    if n == 0: return 1
    t = [m*k for k in (0..(n-1)//m)]
    s = sum(binomial(n, k)*ANum(m, k) for k in t)
    return -s if m.divides(n) else s
def TNum(m, n):
    if n == 1: return 1
    if n == 0 or m == 1: return 0
    return ANum(m, m*n-1)
for m in (1..6): print([TNum(m, n) for n in (0..6)])

T(n, k) is the derivative of OmegaPolynomial(n, k) evaluated at x = 0.

Apart from the border cases n=1 and k=0 the generalized tangent numbers are a subset of the André numbers A181937; more precisely: T(n, k) is 1 if k = 1 else if k = 0 or n = 1 then T(n, k) = 0 else T(n,k) = (-1)^(n+1)*A181937(n, n*k-1).

A322938 a(n) = binomial(2*n + 2, n + 2) - 1.

Original entry on oeis.org

0, 3, 14, 55, 209, 791, 3002, 11439, 43757, 167959, 646645, 2496143, 9657699, 37442159, 145422674, 565722719, 2203961429, 8597496599, 33578000609, 131282408399, 513791607419, 2012616400079, 7890371113949, 30957699535775, 121548660036299, 477551179875951

Offset: 0

Peter Luschny, Feb 13 2019

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A001791, A014473, A030662 (d=0), A010763 (d=1), this sequence (d=2).

[Binomial(2*n+2,n+2) -1: n in [0..30]]; // G. C. Greubel, Apr 22 2024

aList := proc(len) local gf, ser; assume(Im(x) > 0);
gf := (2*x^2 - x + 1)/(2*(x - 1)*x^2) - (I*(2*x - 1))/(2*x^2*sqrt(4*x - 1));
ser := series(gf, x, len+4):
seq(coeff(ser, x, n), n=0..len) end: lprint(aList(25));

Table[Binomial[2 n + 2, n + 2] - 1, {n, 0, 25}]

[binomial(2*n+2,n+2) - 1 for n in range(31)] # G. C. Greubel, Apr 22 2024

Let G(x) = (2*x^2-x+1)/(2*(x-1)*x^2)-(I*(2*x-1))/(2*x^2*sqrt(4*x-1)) with Im(x) > 0, then a(n) = [x^n] G(x). The generating function G(x) satisfies the differential equation 9*x - 16*x^2 + 4*x^3 = (8*x^5 - 22*x^4 + 21*x^3 - 8*x^2 + x)*diff(G(x), x) + (12*x^4 - 36*x^3 + 38*x^2 - 16*x + 2)*G(x).
From Peter Bala, Feb 25 2022: (Start)
a(n) = Sum_{k = 0..n+1} binomial(n+k,k+1).
a(n) = Sum_{k = 0..n-1} binomial(n+k+2,k+1).
More generally, Sum_{k = 0..n+m} binomial(n+k,k+1) = Sum_{k = 0..n-1} binomial(n+k+m+1,k+1) = binomial(2*n+m+1,n) - 1. (End)
a(n) = A001791(n+1) - 1. - Hugo Pfoertner, Feb 26 2022
a(n) = n/(n+2) * binomial(2*n+2, n+1) * Sum_{k = 0..n+1} 1/binomial(n+k+1, k). - Peter Bala, Aug 05 2025

A352027 a(n) = binomial(2n-1,n) - n(n-1) - 1.

Original entry on oeis.org

0, 0, 3, 22, 105, 431, 1673, 6378, 24237, 92287, 352605, 1351945, 5200143, 20058117, 77558549, 300539954, 1166802837, 4537567343, 17672631557, 68923264029, 269128936799, 1052049481397, 4116715363293, 16123801840997, 63205303218275, 247959266473401, 973469712823353

Offset: 1

Enrique Navarrete, Feb 28 2022

nonn

a(n) is the number of ways to place n indistinguishable balls into n distinguishable boxes with at least 2 boxes remaining empty.

a(n) is also the number of weak compositions of n into n parts in which at least two parts are zero.

			a(4)=22 since 4 can be written as 3+1+0+0, 0+3+0+1, etc. (12 such compositions); 2+2+0+0 (6 such compositions); 4+0+0+0 (4 such compositions).

Cf. A010763, A350653.

Table[Binomial[2n-1,n]-n(n-1)-1,{n,40}] (* Harvey P. Dale, Dec 03 2022 *)

a(n) = binomial(2*n-1,n) - n*(n-1) - 1; \\ Michel Marcus, Apr 12 2022

G.f.: 2*x/(4*x - 1 + sqrt(1 - 4*x)) - (1 - 2*x + 3*x^2)/(1 - x)^3. - Stefano Spezia, Mar 01 2022

D-finite with recurrence n*a(n) +2*(-3*n+2)*a(n-1) +(9*n-14)*a(n-2) +2*(-2*n+5)*a(n-3) +2*(-9*n+22)=0. - R. J. Mathar, Jan 25 2023

A382257 a(n) is the numerator of tanh(Sum_{k=1..n-1} artanh(k/n)), where artanh is the inverse hyperbolic tangent function.

Original entry on oeis.org

0, 1, 9, 17, 125, 461, 1715, 3217, 24309, 92377, 352715, 1352077, 5200299, 20058299, 77558759, 150270097, 1166803109, 4537567649, 17672631899, 68923264409, 269128937219, 1052049481859, 4116715363799, 16123801841549, 63205303218875, 247959266474051, 973469712824055, 3824345300380219, 15033633249770519

Offset: 1

M. F. Hasler, Apr 15 2025

The value of tanh(...) is always a rational number thanks to the relation tanh(x+y) = (tanh x + tanh y)/(1 + (tanh x)*tanh y).
So actually the set of fractions {1/n, ..., (n-1)/n} is "summed up" using the operator x (+) y := (x + y)/(1 + x*y).
By Wolstenholme's theorem; if p > 3 is prime, then p^3 divides a(p). - Thomas Ordowski, Apr 27 2025

			For n=2, tanh(artanh(1/2)) = 1/2, so a(2) = numerator(1/2) = 1.
For n=3, tanh(artanh(1/3) + artanh(2/3)) = (1/3 + 2/3) / (1 + 1/3 * 2/3) = 9/11, so a(3) = 9.
Numerators of 0, 1/2, 9/11, 17/18, 125/127, 461/463, 1715/1717, 3217/3218, ...

Wikipedia contributors, Area function (inverse hyperbolic function), in: Wikipedia, the free encyclopedia. As of April 7, 2025.

Cf. A001700, A010763, A034602, A383431 (denominators).

apply( {A382257(n)=my(s=0); for(i=1, n-1, s=(s*n+i)/(n+s*i));numerator(s)}, [1..30])

from sympy import S,expand_trig as ET
tanh,artanh = S("tanh, artanh")
def A382257_test(n): # for illustration only -- slow for n >= 19
    n=S(n); return ET(tanh(sum(artanh(k/n) for k in range(1,n)))).numerator
def A382257(n):
    s=0; i=S.One/n
    for k in range(1,n): s = (s + i*k)/(1 + s*k*i)
    return s.numerator

from functools import reduce
from fractions import Fraction
def A382257(n): return reduce(lambda x,y: Fraction(x+y,1+x*y),(Fraction(i,n) for i in range(1,n)),0).numerator # Chai Wah Wu, Apr 23 2025

a(n) = (binomial(2n-1, n-1) - 1)/2 if n = 2^m or a(n) = binomial(2n-1, n-1) - 1 = A010763(n-1) otherwise, since tanh(Sum_{k=1..n-1} artanh(k/n)) = (binomial(2n-1, n-1) - 1)/(binomial(2n-1, n-1) + 1) reduced. - Thomas Ordowski, Apr 27 2025

cp's OEIS Frontend

A014473 Pascal's triangle - 1: Triangle read by rows: T(n, k) = A007318(n, k) - 1.

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A256117 Number T(n,k) of length 2n words such that all letters of the k-ary alphabet occur at least once and are introduced in ascending order and which can be built by repeatedly inserting doublets into the initially empty word; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

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A061691 Triangle of generalized Stirling numbers.

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A256116 Number T(n,k) of length 2n k-ary words, either empty or beginning with the first letter of the alphabet and using each letter at least once, that can be built by repeatedly inserting doublets into the initially empty word; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

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A370366 Number A(n,k) of partitions of [k*n] into n sets of size k having no set of consecutive numbers whose maximum (if k>0) is a multiple of k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A260875 Square array read by ascending antidiagonals: number of m-shape complementary Bell numbers.

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A318253 Coefficient of x of the OmegaPolynomials (A318146), T(n, k) = [x] P(n, k) with n>=1 and k>=0, square array read by ascending antidiagonals.

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A352027 a(n) = binomial(2n-1,n) - n(n-1) - 1.