A000580 -id:A000580 - cp's OEIS frontend

A056005 Number of 3-element ordered antichains on an unlabeled n-element set; T_1-hypergraphs with 3 labeled nodes and n hyperedges.

0, 0, 0, 2, 19, 90, 302, 820, 1926, 4068, 7920, 14454, 25025, 41470, 66222, 102440, 154156, 226440, 325584, 459306, 636975, 869858, 1171390, 1557468, 2046770, 2661100, 3425760, 4369950, 5527197, 6935814, 8639390, 10687312, 13135320, 16046096, 19489888

Offset: 0

Vladeta Jovovic, Goran Kilibarda, Jul 24 2000

T_1-hypergraph is a hypergraph (not necessarily without empty hyperedges or multiple hyperedges) which for every ordered pair of distinct nodes have a hyperedge containing one but not the other node.

			There are 19 3-element ordered antichains on an unlabeled 4-element set: ({4},{3},{2}), ({4},{3},{1,2}), ({4},{2,3},{1}), ({4},{2,3},{1,3}), ({3,4},{2},{1}), ({3,4},{2},{1,4}), ({3,4},{2,4},{2,3}), ({3,4},{2,4},{1}), ({3,4},{2,4},{1,4}), ({3,4},{2,4},{1,3}), ({3,4},{2,4},{1,2}), ({3,4},{2,4},{1,2,3}), ({3,4},{1,2},{2,4}), ({3,4},{1,2,4},{2,3}), ({3,4},{1,2,4},{1,2,3}), ({2,3,4},{1,4},{1,3}), ({2,3,4},{1,4},{1,2,3}), ({2,3,4},{1,3,4},{1,2}), ({2,3,4},{1,3,4},{1,2,4}).

G. C. Greubel, Table of n, a(n) for n = 0..1000
K. S. Brown, Dedekind's problem
V. Jovovic and G. Kilibarda, On the number of Boolean functions in the Post classes F^{mu}_8, (in Russian), Diskretnaya Matematika, 11 (1999), no. 4, 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, 9, (1999), no. 6.
G. Kilibarda and V. Jovovic, Enumeration of some classes of T_0-hypergraphs, arXiv:1411.4187 [math.CO], 2014.
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A047707 for 3-element (unordered) antichains on a labeled n-element set.

Cf. A056069, A056070, A056071, A056073, A056163.

[n*(n-2)*(n-1)*(n+1)*(n^3 + 30*n^2 + 131*n - 270)/5040: n in [0..30]]; // G. C. Greubel, Nov 22 2017

Table[Binomial[n+7,7]-6Binomial[n+5,5]+6Binomial[n+4,4]+3Binomial[n+3,3]- 6Binomial[n+2,2]+ 2Binomial[n+1,1],{n,0,40}] (* or *) LinearRecurrence[ {8,-28,56,-70,56,-28,8,-1},{0,0,0,2,19,90,302,820},40] (* Harvey P. Dale, Jul 27 2011 *)

x='x+O('x^50); concat([0,0,0], Vec(x^3*(2+3*x-6*x^2+2*x^3)/(1-x)^8)) \\ G. C. Greubel, Oct 06 2017

a(n) = C(n+7, 7) - 6*C(n+5, 5) + 6*C(n+4, 4) + 3*C(n+3, 3) - 6*C(n+2, 2) + 2*C(n+1, 1).
a(n) = n*(n-2)*(n-1)*(n+1)*(n^3 + 30*n^2 + 131*n - 270)/5040.
G.f.: 1/(1-x)^8 - 6/(1-x)^6 + 6/(1-x)^5 + 3/(1-x)^4 - 6/(1-x)^3 + 2/(1-x)^2.
G.f.: x^3*(2 + 3*x - 6*x^2 + 2*x^3)/(1-x)^8.
Recurrence: a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8).
Generally, recurrence for the number of m-element ordered antichains on an unlabeled n-element set is a(m, n) = C(2^m, 1)*a(m, n - 1) - C(2^m, 2)*a(m, n - 2) + C(2^m, 3)*a(m, n - 3) + ... + ( - 1)^(k - 1)*C(2^m, k)*a(m, n - k) + ... - a(m, n - 2^m).
a(n) = A000580(n+7) - 6*A000389(n+5) + 6*A000332(n+4) + 3*A000292(n+1) - 6*A000217(n+1) + 2*A000027(n+1). - R. J. Mathar, Nov 16 2007

More terms from Harvey P. Dale, Jul 27 2011

A087108 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,4). The p-th row (p>=1) contains a(i,p) for i=1 to 4p-3, where a(i,p) satisfies Sum_{i=1..n} C(i+3,4)^p = 5 C(n+4,5) * Sum_{i=1..4p-3} a(i,p) C(n-1,i-1)/(i+4).

Original entry on oeis.org

1, 1, 4, 6, 4, 1, 1, 24, 176, 624, 1251, 1500, 1070, 420, 70, 1, 124, 3126, 33124, 191251, 681000, 1596120, 2543520, 2780820, 2058000, 987000, 277200, 34650, 1, 624, 49376, 1350624, 18308751, 146500500, 763418870, 2749648020, 7101675070, 13440210000

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 11 2018: (Start)

The table entries T(n,k) are the coefficients when expressing the polynomial C(x+4,4)^p of degree 4*p in terms of falling factorials: C(x+4,4)^p = Sum_{k = 0..4*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+4,4)^p = Sum_{k = 0..4*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,24,176,...,70, so Sum_{i=1..n} C(i+3,4)^3 = 5 * C(n+4,5) * [ a(1,3)/5 + a(2,3)*C(n-1,1)/6 + a(3,3)*C(n-1,2)/7 + ... + a(9,3)*C(n-1,8)/13 ] = 5 * C(n+4,5) * [ 1/5 + 24*C(n-1,1)/6 + 176*C(n-1,2)/7 + ... + 70*C(n-1,8)/13 ]. Cf. A086024 for more details.
From _Peter Bala_, Mar 11 2018: (Start)
Table begins
  n = 0 | 1
  n = 1 | 1   4    6     4      1
  n = 2 | 1  24  176   624   1251   1500    1070  420  70
  n = 3 | 1 124 3126 33124 191251 681000 1596120 ...
  ...
Row 2: C(i+4,4)^2 = C(i,0) + 24*C(i,1) + 176*C(i,2) + 624*C(i,3) + 1251*C(i,4) + 1500*C(i,5) + 1070*C(i,6) + 420*C(i,7) + 70*C(i,8). Hence, Sum_{i = 0..n-1} C(i+4,4)^2 =  C(n,1) + 24*C(n,2) + 176*C(n,3) + 624*C(n,4) + 1251*C(n,5) + 1500*C(n,6) + 1070*C(n,7) + 420*C(n,8) + 70*C(n,9) .(End)

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A236463.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+4, 4)^n, i = 0..k), k = 0..4*n), n = 0..6); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 5, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 4, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 4*p - 3}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 5, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 4, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 4*p-3, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+5, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+4, i-2*k)^(p-1) ]
From Peter Bala, Mar 11 2018: (Start)
The following remarks assume the row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * binomial(i+4,4)^n. Equivalently, let v_n denote the sequence (1, 5^n, 15^n, 35^n, ...) regarded as an infinite column vector, where 1, 5, 15, 35, ... is the sequence binomial(n+4,4) - see A000332. Then the n-th row of this table is determined by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318.
Recurrence: T(n+1,k) = C(k+4,4)*T(n,k) + 4*C(k+3,4)*T(n,k-1) + 6*C(k+2,4)*T(n,k-2) + 4*C(k+1,4)*T(n,k-3) + C(k,4)*T(n,k-4) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 4*n.
n-th row polynomial R(n,x) = (1 + x)^4 o (1 + x)^4 o ... o (1 + x)^4 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n,x) = Sum_{i >= 0} binomial(i+4,4)^n*x^i/(1 + x)^(i+1).
R(n+1,x) = 1/4! * (1 + x)^4 * (d/dx)^4(x^4*R(n,x)).
(1 - x)^(4*n)*R(n,x/(1 - x)) appears to equal the n-th row polynomial of A236463. (End)

Edited by Dean Hickerson, Aug 16 2003

A087109 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,5). The p-th row (p>=1) contains a(i,p) for i=1 to 5p-4, where a(i,p) satisfies Sum_{i=1..n} C(i+4,5)^p = 6 C(n+5,6) * Sum_{i=1..5p-4} a(i,p) C(n-1,i-1)/(i+5).

Original entry on oeis.org

1, 1, 5, 10, 10, 5, 1, 1, 35, 370, 1920, 5835, 11253, 14240, 11830, 6230, 1890, 252, 1, 215, 8830, 148480, 1352615, 7665757, 29224020, 78518790, 152794740, 218270220, 229279512, 175227360, 94864770, 34504470, 7567560, 756756, 1, 1295, 191890

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 11 2018: (Start)

The table entries T(n,k) are the coefficients when expressing the polynomial C(x+5,5)^p of degree 5*p in terms of falling factorials: C(x+5,5)^p = Sum_{k = 0..5*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+5,5)^p = Sum_{k = 0..5*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,35,370,...,252, so Sum_{i=1..n} C(i+4,5)^3 = 6 * C(n+5,6) * [ a(1,3)/6 + a(2,3)*C(n-1,1)/7 + a(3,3)*C(n-1,2)/8 + ... + a(11,3)*C(n-1,10)/16 ] = 6 * C(n+5,6) * [ 1/6 + 35*C(n-1,1)/7 + 370*C(n-1,2)/8 + ... + 252*C(n-1,10)/16 ]. Cf. A086026 for more details.
From _Peter Bala_, Mar 11 2018: (Start)
Table begins
1
1  5  10   10    5     1
1 35 370 1920 5835 11253 14240 11830 6230 1890 252
...
Row 2: C(i+5,5)^2 = C(i,0) + 35*C(i,1) + 370*C(i,2) + 1920*C(i,3) + 5835*C(i,4) + 11253*C(i,5) + 14240*C(i,6) + 11830*C(i,7) + 6230*C(i,8) + 1890*C(i,9) + 252*C(i,10). Hence, Sum_{i = 0..n-1} C(i+5,5)^2 = C(n,1) + 35*C(n,2) + 370*C(n,3) + 1920*C(n,4) + 5835*C(n,5) + 11253*C(n,6) + 14240*C(n,7) + 11830*C(n,8) + 6230*C(n,9) + 1890*C(n,10) + 252*C(n,11). (End)

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A237202.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+5, 5)^n, i = 0..k), k = 0..5*n), n = 0..5); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 6, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 5, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 5*p - 4}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 6, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 5, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 5*p-4, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+6, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+5, i-2*k)^(p-1) ]
From Peter Bala, Mar 11 2018: (Start)
The following remarks assume the row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * binomial(i+5,5)^n. Equivalently, let v_n denote the sequence (1, 6^n, 21^n, 56^n, ...) regarded as an infinite column vector, where 1, 6, 21, 56, ... is the sequence binomial(n+5,5) - see A000389. Then the n-th row of this table is determined by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318.
Recurrence: T(n+1,k) = Sum_{i = 0..5} C(5,i)*C(k+5-i,5)*T(n,k-i) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 5*n.
n-th row polynomial R(n,x) = (1 + x)^5 o (1 + x)^5 o ... o (1 + x)^5 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n+1,x) = 1/5!*(1 + x)^5 * (d/dx)^5(x^5*R(n,x)).
R(n,x) = Sum_{i >= 0} binomial(i+5,5)^n*x^i/(1 + x)^(i+1).
(1 - x)^(5*n)*R(n,x/(1 - x)) appears to equal the n-th row polynomial of A237202. (End)

Edited by Dean Hickerson, Aug 16 2003

A087110 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,6). The p-th row (p>=1) contains a(i,p) for i=1 to 6p-5, where a(i,p) satisfies Sum_{i=1..n} C(i+5,6)^p = 7 C(n+6,7) * Sum_{i=1..6p-5} a(i,p) C(n-1,i-1)/(i+6).

Original entry on oeis.org

1, 1, 6, 15, 20, 15, 6, 1, 1, 48, 687, 4850, 20385, 55908, 104959, 137886, 127050, 80640, 33642, 8316, 924, 1, 342, 21267, 527876, 7020525, 58015362, 324610399, 1297791264, 3839203452, 8595153000, 14760228672, 19560928464, 19987430694

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 11 2018: (Start)

The table entries T(n,k) are the coefficients when expressing the polynomial C(x+6,6)^p of degree 6*p in terms of falling factorials: C(x+6,6)^p = Sum_{k = 0..6*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+6,6)^p = Sum_{k = 0..6*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,48,687,...,924, so Sum_{i=1..n} C(i+5,6)^3 = 7 * C(n+6,7) * [ a(1,3)/7 + a(2,3)*C(n-1,1)/8 + a(3,3)*C(n-1,2)/9 + ... + a(13,3)*C(n-1,12)/19 ] = 7 * C(n+6,7) * [ 1/7 + 48*C(n-1,1)/8 + 687*C(n-1,2)/9 + ... + 924*C(n-1,12)/19 ]. Cf. A086028 for more details.

G. C. Greubel, Table of n, a(n) for the first 40 rows, flattened
Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A237252.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+6, 6)^n, i = 0..k), k = 0..6*n), n = 0..5); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 7, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 6, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 6*p - 5}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 7, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 6, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 6*p-5, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+7, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+6, i-2*k)^(p-1) ]
From Peter Bala, Mar 11 2018: (Start)
The following remarks assume the row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * binomial(i+6,6)^n. Equivalently, let v_n denote the sequence (1, 7^n, 28^n, 84^n, ...) regarded as an infinite column vector, where 1, 7, 28, 84, ... is the sequence binomial(n+6,6) - see A000579. Then the n-th row of this table is determined by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318.
Recurrence: T(n+1,k) = Sum_{i = 0..6} C(6,i)*C(k+6-i,6)*T(n,k-i) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 6*n.
n-th row polynomial R(n,x) = (1 + x)^6 o (1 + x)^6 o ... o (1 + x)^6 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n+1,x) = 1/6!*(1 + x)^6 * (d/dx)^6(x^6*R(n,x)).
R(n,x) = Sum_{i >= 0} binomial(i+6,6)^n*x^i/(1 + x)^(i+1).
(1 - x)^(6*n)*R(n,x/(1 - x)) appears to equal the n-th row polynomial of A237252. (End)

Edited by Dean Hickerson, Aug 16 2003

A221857 Number A(n,k) of shapes of balanced k-ary trees with n nodes, where a tree is balanced if the total number of nodes in subtrees corresponding to the branches of any node differ by at most one; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 3, 1, 1, 0, 1, 1, 4, 3, 4, 1, 0, 1, 1, 5, 6, 1, 4, 1, 0, 1, 1, 6, 10, 4, 9, 4, 1, 0, 1, 1, 7, 15, 10, 1, 27, 1, 1, 0, 1, 1, 8, 21, 20, 5, 16, 27, 8, 1, 0, 1, 1, 9, 28, 35, 15, 1, 96, 81, 16, 1, 0, 1, 1, 10, 36, 56, 35, 6, 25, 256, 81, 32, 1, 0

Offset: 0

Alois P. Heinz, Apr 10 2013

			: A(2,2) = 2  : A(2,3) = 3      : A(3,3) = 3          :
:   o     o   :   o    o    o   :   o      o      o   :
:  / \   / \  :  /|\  /|\  /|\  :  /|\    /|\    /|\  :
: o         o : o      o      o : o o    o   o    o o :
:.............:.................:.....................:
: A(3,4) = 6                                          :
:    o        o        o        o       o        o    :
:  /( )\    /( )\    /( )\    /( )\   /( )\    /( )\  :
: o o      o   o    o     o    o o     o   o      o o :
Square array A(n,k) begins:
  1, 1, 1,  1,   1,   1,  1,  1,  1,   1,   1, ...
  1, 1, 1,  1,   1,   1,  1,  1,  1,   1,   1, ...
  0, 1, 2,  3,   4,   5,  6,  7,  8,   9,  10, ...
  0, 1, 1,  3,   6,  10, 15, 21, 28,  36,  45, ...
  0, 1, 4,  1,   4,  10, 20, 35, 56,  84, 120, ...
  0, 1, 4,  9,   1,   5, 15, 35, 70, 126, 210, ...
  0, 1, 4, 27,  16,   1,  6, 21, 56, 126, 252, ...
  0, 1, 1, 27,  96,  25,  1,  7, 28,  84, 210, ...
  0, 1, 8, 81, 256, 250, 36,  1,  8,  36, 120, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened
Jeffrey Barnett, Counting Balanced Tree Shapes, 2007
Samuele Giraudo, Intervals of balanced binary trees in the Tamari lattice, arXiv:1107.3472 [math.CO], Apr 2012

Columns k=1-10 give: A000012, A110316, A131889, A131890, A131891, A131892, A131893, A229393, A229394, A229395.
Rows n=0+1, 2-3, give: A000012, A001477, A179865.
Diagonal and upper diagonals give: A028310, A000217, A000292, A000332, A000389, A000579, A000580, A000581, A000582, A001287, A001288.
Lower diagonals give: A000012, A000290, A092364(n) for n>1.

A:= proc(n, k) option remember; local m, r; if n<2 or k=1 then 1
      elif k=0 then 0 else r:= iquo(n-1, k, 'm');
      binomial(k, m)*A(r+1, k)^m*A(r, k)^(k-m) fi
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);

a[n_, k_] := a[n, k] = Module[{m, r}, If[n < 2 || k == 1, 1, If[k == 0, 0, {r, m} = QuotientRemainder[n-1, k]; Binomial[k, m]*a[r+1, k]^m*a[r, k]^(k-m)]]]; Table[a[n, d-n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, Apr 17 2013, translated from Maple *)

A101104 a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4.

Original entry on oeis.org

1, 12, 23, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24

Offset: 1

Cecilia Rossiter, Dec 15 2004

Original name: The first summation of row 4 of Euler's triangle - a row that will recursively accumulate to the power of 4.

D. J. Pengelley, The bridge between the continuous and the discrete via original sources in Study the Masters: The Abel-Fauvel Conference [pdf], Kristiansand, 2002, (ed. Otto Bekken et al), National Center for Mathematics Education, University of Gothenburg, Sweden, in press.
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, Link to section of MathWorld: Worpitzky's Identity of 1883
Eric Weisstein, Link to section of MathWorld: Eulerian Number
Eric Weisstein, Link to section of MathWorld: Nexus number
Eric Weisstein, Link to section of MathWorld: Finite Differences
Index entries for linear recurrences with constant coefficients, signature (1).

For other sequences based upon MagicNKZ(n,k,z):
..... |  n = 1  |  n = 2  |  n = 3  |  n = 4  |  n = 5  |  n = 6  |  n = 7
---------------------------------------------------------------------------
z = 0 | A000007 | A019590 | .......MagicNKZ(n,k,0) = A008292(n,k+1) .......
z = 1 | A000012 | A040000 | A101101 | thisSeq | A101100 | ....... | .......
z = 2 | A000027 | A005408 | A008458 | A101103 | A101095 | ....... | .......
z = 3 | A000217 | A000290 | A003215 | A005914 | A101096 | ....... | .......
z = 4 | A000292 | A000330 | A000578 | A005917 | A101098 | ....... | .......
z = 5 | A000332 | A002415 | A000537 | A000583 | A022521 | ....... | A255181
z = 6 | A000389 | A005585 | A024166 | A000538 | A000584 | A022522 | A255177
z = 7 | A000579 | A040977 | A101094 | A101089 | A000539 | A001014 | A022523
z = 8 | A000580 | A050486 | A101097 | A101090 | A101092 | A000540 | A001015
z = 9 | A000581 | A053347 | A101102 | A101091 | A101099 | A101093 | A000541
Cf. A101095 for an expanded table and more about MagicNKZ.

MagicNKZ = Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 4, 4}, {z, 1, 1}, {k, 0, 34}]
Join[{1, 12, 23},LinearRecurrence[{1},{24},56]] (* Ray Chandler, Sep 23 2015 *)

a(k) = MagicNKZ(4,k,1) where MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n+1-z,j)*(k-j+1)^n (cf. A101095). That is, a(k) = Sum_{j=0..k+1} (-1)^j*binomial(4, j)*(k-j+1)^4.
a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4. - Joerg Arndt, Nov 30 2014
G.f.: x*(1+11*x+11*x^2+x^3)/(1-x). - Colin Barker, Apr 16 2012

New name from Joerg Arndt, Nov 30 2014

Original Formula edited and Crossrefs table added by Danny Rorabaugh, Apr 22 2015

A238801 Triangle T(n,k), read by rows, given by T(n,k) = C(n+1, k+1)*(1-(k mod 2)).

Original entry on oeis.org

1, 2, 0, 3, 0, 1, 4, 0, 4, 0, 5, 0, 10, 0, 1, 6, 0, 20, 0, 6, 0, 7, 0, 35, 0, 21, 0, 1, 8, 0, 56, 0, 56, 0, 8, 0, 9, 0, 84, 0, 126, 0, 36, 0, 1, 10, 0, 120, 0, 252, 0, 120, 0, 10, 0, 11, 0, 165, 0, 462, 0, 330, 0, 55, 0, 1, 12, 0, 220, 0, 792, 0, 792, 0, 220, 0, 12, 0

Offset: 0

Philippe Deléham, Mar 05 2014

Row sums are powers of 2.

			Triangle begins:
1;
2, 0;
3, 0, 1;
4, 0, 4, 0;
5, 0, 10, 0, 1;
6, 0, 20, 0, 6, 0;
7, 0, 35, 0, 21, 0, 1;
8, 0, 56, 0, 56, 0, 8, 0;
9, 0, 84, 0, 126, 0, 36, 0, 1;
10, 0, 120, 0, 252, 0, 120, 0, 10, 0; etc.

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

Cf. Columns: A000027, A000292, A000389, A000580, A000582, A001288, A010966, A010968, A010970, A010972, A010974, A010976, A010980, A010982, A010984, A010986, A010988, A010990, A010992, A010994, A010996, A010998, A011000, A017713, A017715, A017717, A017719, A017721, A017723, A017725, A017727, A017729, A017731, A017733, A017735, A017737, A017739, A017741, A017743, A017745, A017747, A017749, A017751, A017753, A017755, A017757, A017759, A017761, A017763.

Cf. A095704, A178616.

Table[Binomial[n + 1, k + 1]*(1 - Mod[k , 2]), {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Nov 22 2017 *)

T(n,k) = binomial(n+1, k+1)*(1-(k % 2));
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n,k), ", ")); print); \\ Michel Marcus, Nov 23 2017

G.f.: 1/((1+(y-1)*x)*(1-(y+1)*x)).
T(n,k) = 2*T(n-1,k) + T(n-2,k-2) - T(n-2,k), T(0,0) = 1, T(1,0) = 2, T(1,1) = 0, T(n,k) = 0 if k<0 or if k>n.
Sum_{k=0..n} T(n,k)*x^k = A000027(n+1), A000079(n), A015518(n+1), A003683(n+1), A079773(n+1), A051958(n+1), A080920(n+1), A053455(n), A160958(n+1) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8 respectively.

A278385 T(n,k)=Number of nXk 0..1 arrays with rows and columns in lexicographic nondecreasing order but with exactly three mistakes.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 3, 3, 0, 0, 40, 74, 40, 0, 1, 267, 1220, 1220, 267, 1, 8, 1350, 12910, 23640, 12910, 1350, 8, 36, 5936, 100807, 368421, 368421, 100807, 5936, 36, 120, 23565, 652343, 4703562, 8632118, 4703562, 652343, 23565, 120, 330, 84912, 3750182

Offset: 1

R. H. Hardin, Nov 20 2016

Table starts
...0......0........0...........0..............0................1
...0......0........3..........40............267.............1350
...0......3.......74........1220..........12910...........100807
...0.....40.....1220.......23640.........368421..........4703562
...0....267....12910......368421........8632118........179716850
...1...1350...100807.....4703562......179716850.......6204309386
...8...5936...652343....50473056.....3325788157.....198563803019
..36..23565..3750182...474255829....54735436424....5851197688577
.120..84912.19784428..4047341159...813247916326..157794170262819
.330.278422.96786947.32112086692.11132424779200.3912513274701995

			Some solutions for n=4 k=4
..0..1..1..0. .0..1..1..0. .1..0..1..0. .0..0..0..1. .0..0..0..0
..0..1..0..0. .0..1..1..1. .1..0..0..0. .1..1..1..1. .0..1..0..0
..1..0..0..1. .1..1..0..0. .1..0..1..0. .1..1..0..1. .1..1..1..0
..1..0..1..1. .0..0..1..0. .1..1..1..0. .0..1..1..1. .1..1..0..1

R. H. Hardin, Table of n, a(n) for n = 1..219

Column 1 is A000580(n+1).

Empirical for column k:
k=1: [polynomial of degree 7]
k=2: [polynomial of degree 15]
k=3: [polynomial of degree 31]
k=4: [polynomial of degree 63]
k=5: [polynomial of degree 127]

A278855 T(n,k)=Number of nXk 0..1 arrays with rows in nondecreasing lexicographic order and columns in nonincreasing lexicographic order, but with exactly three mistakes.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 25, 66, 25, 0, 1, 239, 1348, 1348, 239, 1, 8, 1486, 15066, 30772, 15066, 1486, 8, 36, 7072, 118158, 449677, 449677, 118158, 7072, 36, 120, 27828, 731444, 4950399, 9317846, 4950399, 731444, 27828, 120, 330, 94720, 3813048

Offset: 1

R. H. Hardin, Nov 29 2016

Table starts
...0.....0........0..........0............0..............1................8
...0.....0........1.........25..........239...........1486.............7072
...0.....1.......66.......1348........15066.........118158...........731444
...0....25.....1348......30772.......449677........4950399.........44571591
...0...239....15066.....449677......9317846......154090507.......2167109763
...1..1486...118158....4950399....154090507.....4026240722......92891826961
...8..7072...731444...44571591...2167109763....92891826961....3637198028018
..36.27828..3813048..344302564..26799758534..1935392159582..131770344218055
.120.94720.17413532.2353207764.297417275848.36831668321915.4438698421228261

			Some solutions for n=4 k=4
..1..0..1..0. .1..1..1..0. .1..0..0..0. .1..0..0..0. .1..1..1..1
..0..0..1..0. .0..1..1..0. .1..0..1..0. .1..0..1..0. .0..1..0..0
..0..0..0..1. .0..0..0..1. .1..0..0..0. .1..0..0..1. .1..1..1..0
..1..0..1..1. .0..1..0..1. .0..1..0..0. .1..0..0..0. .0..1..1..1

R. H. Hardin, Table of n, a(n) for n = 1..219

Column 1 is A000580(n+1).

Empirical for column k:
k=1: [polynomial of degree 7]
k=2: [polynomial of degree 14]
k=3: [polynomial of degree 27]
k=4: [polynomial of degree 52]
k=5: [polynomial of degree 101]

A001779 Expansion of 1/((1+x)(1-x)^8).

Original entry on oeis.org

1, 7, 29, 91, 239, 553, 1163, 2269, 4166, 7274, 12174, 19650, 30738, 46782, 69498, 101046, 144111, 201993, 278707, 379093, 508937, 675103, 885677, 1150123, 1479452, 1886404, 2385644, 2993972

Offset: 0

N. J. A. Sloane

a(n) is the number of positive terms in the expansion of (a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 - z)^n. Also the convolution of A001769 and A000012; A001753 and A001477; A001752 and A000217; A002623 and A000292; A002620 and A000332; A004526 and A000389. - Sergio Falcon (sfalcon(AT)dma.ulpgc.es), Feb 13 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (7,-20,28,-14,-14,28,-20,7,-1).

Cf. A001769 (first differences), A169795 (binomial transf.)

[1/80640*(2*n+9) *(4*n^6 +108*n^5 +1138*n^4 +5904*n^3 +15628*n^2 +19638*n +8925)+(-1)^n/256 : n in [0..30]]; // Vincenzo Librandi, Oct 08 2011

A001779 := proc(n) 1/80640*(2*n+9) *(4*n^6 +108*n^5 +1138*n^4 +5904*n^3 +15628*n^2 +19638*n +8925)+(-1)^n/256 ; end proc:
seq(A001779(n),n=0..50) ; # R. J. Mathar, Mar 22 2011

CoefficientList[Series[1/((1 + x) (1 - x)^8), {x, 0, 50}], x] (* G. C. Greubel, Nov 24 2017 *)
LinearRecurrence[{7,-20,28,-14,-14,28,-20,7,-1},{1,7,29,91,239,553,1163,2269,4166},30] (* Harvey P. Dale, Jan 21 2023 *)

a(n)=(2*n+9)*(4*n^6+108*n^5+1138*n^4+5904*n^3+15628*n^2+19638*n + 8925)/80640 +(-1)^n/256 \\ Charles R Greathouse IV, Apr 17 2012

a(n) = (-1)^{7-n}*Sum_{i=0..n} ((-1)^(7-i)*binomial(7+i,i)). - Sergio Falcon, Feb 13 2007

a(n)+a(n+1) = A000580(n+8). - R. J. Mathar, Jan 06 2021

cp's OEIS Frontend

A056005 Number of 3-element ordered antichains on an unlabeled n-element set; T_1-hypergraphs with 3 labeled nodes and n hyperedges.

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A221857 Number A(n,k) of shapes of balanced k-ary trees with n nodes, where a tree is balanced if the total number of nodes in subtrees corresponding to the branches of any node differ by at most one; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A101104 a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4.

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A238801 Triangle T(n,k), read by rows, given by T(n,k) = C(n+1, k+1)*(1-(k mod 2)).

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