A000389 -id:A000389 - cp's OEIS frontend

A005995 Alkane (or paraffin) numbers l(8,n).

1, 3, 12, 28, 66, 126, 236, 396, 651, 1001, 1512, 2184, 3108, 4284, 5832, 7752, 10197, 13167, 16852, 21252, 26598, 32890, 40404, 49140, 59423, 71253, 85008, 100688, 118728, 139128, 162384, 188496, 218025, 250971, 287964, 329004, 374794, 425334, 481404, 543004

Offset: 0

N. J. A. Sloane, Winston C. Yang (yang(AT)math.wisc.edu)

From M. F. Hasler, May 01 2009: (Start)
Also, number of 5-element subsets of {1,...,n+5} whose elements sum to an odd integer, i.e. column 5 of A159916.
A linear recurrent sequence with constant coefficients and characteristic polynomial x^9 - 3*x^8 + 8*x^6 - 6*x^5 - 6*x^4 + 8*x^3 - 3*x + 1. (End)
Equals (1/2)*((1, 6, 21, 56, 126, 252, ...) + (1, 0, 3, 0, 6, 0, 10, ...)), see A000389 and A000217.
Equals row sums of triangle A160770.
F(1,5,n) is the number of bracelets with 1 blue, 5 identical red and n identical black beads. If F(1,5,1) = 3 and F(1,5,2) = 12 taken as a base, F(1,5,n) = n(n+1)(n+2)(n+3)/24 + F(1,3,n) + F(1,5,n-2). [F(1,3,n) is the number of bracelets with 1 blue, 3 identical red and n identical black beads. If F(1,3,1) = 2 and F(1,3,2) = 6 taken as a base F(1,3,n) = n(n+1)/2 + [|n/2|] + 1 + F(1,3,n-2)], where [|x|]: if a is an integer and a<=x

S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Winston C. Yang (paper in preparation).

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Johann Cigler, Some remarks on Rogers-Szegö polynomials and Losanitsch's triangle, arXiv:1711.03340 [math.CO], 2017.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926. (Annotated scanned copy)
N. J. A. Sloane, Classic Sequences
L. Smith, Polynomial invariants of finite groups. A survey of recent developments. Bull. Amer. Math. Soc. (N.S.) 34 (1997), no. 3, 211-250.
Ata A. Uslu and Hamdi G. Ozmenekse, F(1,3,n)
Ata A. Uslu and Hamdi G. Ozmenekse, F(1,5,n)
Index entries for linear recurrences with constant coefficients, signature (3, 0, -8, 6, 6, -8, 0, 3, -1).

Cf. A160770, A053132 (bisection), A271870 (bisection), A018210 (partial sums).

a:= n-> (Matrix([[1, 0$6, -3, -9]]). Matrix(9, (i,j)-> if (i=j-1) then 1 elif j=1 then [3, 0, -8, 6, 6, -8, 0, 3, -1][i] else 0 fi)^n)[1, 1]: seq(a(n), n=0..40); # Alois P. Heinz, Jul 31 2008

a[n_?OddQ] := 1/240*(n+1)*(n+2)*(n+3)*(n+4)*(n+5); a[n_?EvenQ] := 1/240*(n+2)*(n+4)*(n+6)*(n^2+3*n+5); Table[a[n], {n, 0, 32}] (* Jean-François Alcover, Mar 17 2014, after M. F. Hasler *)
LinearRecurrence[{3, 0, -8, 6, 6, -8, 0, 3, -1},{1, 3, 12, 28, 66, 126, 236, 396, 651},40] (* Ray Chandler, Sep 23 2015 *)

a(k)= if(k%2,(k+1)*(k+3)*(k+5),(k+6)*(k^2+3*k+5))*(k+2)*(k+4)/240 \\ M. F. Hasler, May 01 2009

G.f.: (1+3*x^2)/((1-x)^3*(1-x^2)^3).
a(2n-1) = n(n+1)(n+2)(2n+1)(2n+3)/30, a(2n) = (n+1)(n+2)(n+3)(4n^2+6n+5)/ 30. [M. F. Hasler, May 01 2009]
a(n) = (1/240)*(n+1)*(n+2)*(n+3)*(n+4)*(n+5) + (1/16)*(n+2)*(n+4)*(1/2)*(1+(-1)^n). [Yosu Yurramendi, Jun 20 2013]
a(n) = A038163(n)+3*A038163(n-2). - R. J. Mathar, Mar 29 2018

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

Original entry on oeis.org

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

A128767 Number of inequivalent n-colorings of the 4D hypercube under the full orthogonal group of the cube (of order 2^4*4! = 384).

Original entry on oeis.org

1, 402, 132102, 11756666, 405385550, 7416923886, 86986719477, 735192450952, 4834517667381, 26073250910950, 119759687845446, 481750080584202, 1733588303252702, 5673534527793146, 17109303241791825, 48047227408513056

Offset: 1

Ricardo Perez-Aguila (ricardo.perez.aguila(AT)gmail.com), Apr 04 2007

I assume this refers to colorings of the vertices of the cube. - N. J. A. Sloane, Apr 06 2007

Number of unoriented colorings of the 16 tetrahedral facets of a hyperoctahedron or of the 16 vertices of a tesseract (4-D cube) using up to n colors. Each chiral pair is counted as one when enumerating unoriented arrangements. The Schläfli symbols for the tesseract and the hyperoctahedron are {4,3,3} and {3,3,4} respectively. Both figures are regular 4-D polyhedra and they are mutually dual. - Robert A. Russell, Oct 03 2020

			a(2)=402 because there are 402 inequivalent 2-colorings of the 4D hypercube.

Banks, D. C.; Linton, S. A. & Stockmeyer, P. K. Counting Cases in Substitope Algorithms. IEEE Transactions on Visualization and Computer Graphics, Vol. 10, No. 4, pp. 371-384. 2004.
Perez-Aguila, Ricardo. Enumerating the Configurations in the n-Dimensional Orthogonal Polytopes Through Polya's Counting and A Concise Representation. Proceedings of the 3rd International Conference on Electrical and Electronics Engineering and XII Conference on Electrical Engineering ICEEE and CIE 2006, pp. 63-66.
Polya, G. & Read R. C. Combinatorial Enumeration of Groups, Graphs and Chemical Compounds. Springer-Verlag, 1987.

Banks, D. C.; Linton, S. A. & Stockmeyer, P. K., Counting Cases in Substitope Algorithms, IEEE Transactions on Visualization and Computer Graphics, Vol. 10, No. 4, pp. 371-384. 2004.
Perez-Aguila, Ricardo, Orthogonal Polytopes: Study and Application, PhD Thesis. Universidad de las Americas, Puebla. November, 2006.
Perez-Aguila, Ricardo, Enumerating the Configurations in the n-Dimensional Orthogonal Polytopes Through Polya's Counting and A Concise Representation, Proceedings of the 3rd International Conference on Electrical and Electronics Engineering and XII Conference on Electrical Engineering ICEEE and CIE 2006, pp. 63-66.
Index entries for linear recurrences with constant coefficients, signature (17,-136,680,-2380,6188,-12376,19448,-24310,24310,-19448,12376,-6188,2380,-680,136,-17,1).

Cf. A337952 (oriented), A337954 (chiral), A337955 (achiral).
Other elements: A331359 (tesseract edges, hyperoctahedron faces), A331355 (tesseract faces, hyperoctahedron edges), A337957 (tesseract facets, hyperoctahedron vertices).
Other polychora: A000389(n+4) (4-simplex facets/vertices), A338949 (24-cell), A338965 (120-cell, 600-cell).
Row 4 of A325013 (orthoplex facets, orthotope vertices).

Table[(1/384)*( 48*n^2 + 180*n^4 + 48*n^6 + 83*n^8 + 12*n^10 + 12*n^12 + n^16),{n,30}]

a(n) = (1/384)*(48*n^2 + 180*n^4 + 48*n^6 + 83*n^8 + 12*n^10 + 12*n^12 + n^16)
G.f.: -x*(x +1)*(x^14 +384*x^13 +125020*x^12 +9439904*x^11 +213777216*x^10 +1821620108*x^9 +6527222787*x^8 +10098845160*x^7 +6527222787*x^6 +1821620108*x^5 +213777216*x^4 +9439904*x^3 +125020*x^2 +384*x +1) / (x -1)^17. [Colin Barker, Dec 04 2012]
From Robert A. Russell, Oct 03 2020: (Start)
a(n) = 1*C(n,1) + 400*C(n,2) + 130899*C(n,3) + 11230666*C(n,4) + 347919225*C(n,5) + 5158324560*C(n,6) + 43174480650*C(n,7) + 225086553300*C(n,8) + 775894225050*C(n,9) + 1831178115900*C(n,10) + 3008073915000*C(n,11) + 3439243962000*C(n,12) + 2685727044000*C(n,13) + 1366701336000*C(n,14) + 408648240000*C(n,15) + 54486432000*C(n,16), where the coefficient of C(n,k) is the number of unoriented colorings using exactly k colors.
a(n) = A337952(n) - A337954(n) = (A337952(n) + A337955(n)) / 2 = A337954(n) + A337955(n).
(End)

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 *)

A231303 Recurrence a(n) = a(n-2) + n^M for M=4, starting with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 16, 82, 272, 707, 1568, 3108, 5664, 9669, 15664, 24310, 36400, 52871, 74816, 103496, 140352, 187017, 245328, 317338, 405328, 511819, 639584, 791660, 971360, 1182285, 1428336, 1713726, 2042992, 2421007, 2852992, 3344528, 3901568, 4530449, 5237904

Offset: 0

Stanislav Sykora, Nov 07 2013

In physics, a(n)/2^(M-1) is the trace of the spin operator |S_z|^M for a particle with spin S=n/2. For example, when S=3/2, the S_z eigenvalues are -3/2, -1/2, +1/2, +3/2 and therefore the sum of their 4th powers is 2*82/16 = a(3)/8 (analogously for other values of M).

Partial sums of A062392. - Bruce J. Nicholson, Jun 29 2019

			a(4) = 4^4 + 2^4 = 272; a(5) = 5^4 + 3^4 + 1^4 = 707.

Stanislav Sykora, Table of n, a(n) for n = 0..9999
Stanislav Sýkora, Magnetic Resonance on OEIS, Stan's NMR Blog (Dec 31, 2014), Retrieved Nov 12, 2019.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A001477 (M=1), A000292 (M=2), A105636 (M=3), A231304 (M=5), A231305 (M=6), A231306 (M=7), A231307 (M=8), A231308 (M=9), A231309 (M=10).

Cf. A000389, A000579, A000332, A000583.

List([0..40], n-> n*(3*n^4+15*n^3+20*n^2-8)/30); # G. C. Greubel, Jul 01 2019

[1/30*n*(3*n^4+15*n^3+20*n^2-8): n in [0..40]]; // Vincenzo Librandi, Dec 23 2015

Table[SeriesCoefficient[x*(1+10*x+x^2)/(1-x)^6, {x, 0, n}], {n, 0, 40}] (* Michael De Vlieger, Dec 22 2015 *)
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 16, 82, 272, 707}, 40] (* Vincenzo Librandi, Dec 23 2015 *)

nmax=40;a = vector(nmax);a[2]=1;for(i=3,#a,a[i]=a[i-2]+(i-1)^4); print(a);

concat(0, Vec(x*(1+10*x+x^2)/(1-x)^6 + O(x^40))) \\ Colin Barker, Dec 22 2015

[n*(3*n^4+15*n^3+20*n^2-8)/30 for n in (0..40)] # G. C. Greubel, Jul 01 2019

a(n) = Sum_{k=0..floor(n/2)} (n - 2*k)^4.
From Colin Barker, Dec 22 2015: (Start)
a(n) = (1/30)*n*(3*n^4 + 15*n^3 + 20*n^2 - 8).
G.f.: x*(1 + 10*x + x^2) / (1-x)^6.
(End)
E.g.f.: x*(30 + 210*x + 185*x^2 + 45*x^3 + 3*x^4)*exp(x)/30. - G. C. Greubel, Apr 24 2016
From Bruce J. Nicholson, Jun 29 2019: (Start)
a(n) = 12*A000389(n+3) + A000292(n);
a(n) = (12*A000579(n+4)+A000332(n+3)) - (12*A000579(n+3)+A000332(n+2));
a(n) - a(n-2) = A000583(n). (End)

A005583 Coefficients of Chebyshev polynomials.

Original entry on oeis.org

2, 11, 36, 91, 196, 378, 672, 1122, 1782, 2717, 4004, 5733, 8008, 10948, 14688, 19380, 25194, 32319, 40964, 51359, 63756, 78430, 95680, 115830, 139230, 166257, 197316, 232841, 273296, 319176, 371008, 429352, 494802, 567987, 649572, 740259, 840788

Offset: 1

N. J. A. Sloane

If X is an n-set and Y a fixed 2-subset of X then a(n-5) is equal to the number of (n-5)-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007

a(n-1) = risefac(n,5)/5! - risefac(n,3)/3! is for n >= 1 also the number of independent components of a symmetric traceless tensor of rank 5 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), Table 22.7, 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..172
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. [broken link]
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [Cached copy, May 15 2013]
Index entries for sequences related to Chebyshev polynomials.

Cf. A000096, A000217, A000389, A051747, A127672.

Column 3 of A207606.

A005583:=-(-2+z)/(z-1)**6; # Simon Plouffe in his 1992 dissertation (this g.f. assumes offset 0)

conv(u,v)=local(w); w=vector(length(u),i,sum(j=1,i,u[j]*v[i+1-j])); w;
t(n)=n*(n+1)/2;
u=vector(10,i,t(i));
v=vector(10,i,t(i)-1);
conv(u,v)

a(n) = (1/120)*n*(n+9)*(n+3)*(n+2)*(n+1); \\ Joerg Arndt, Mar 05 2018

G.f.: x*(2-x)/(1-x)^6.
a(n) = binomial(n+4, n-1) + binomial(n+3, n-1) = (1/120)*n*(n+9)*(n+3)*(n+2)*(n+1).
a(n+1) = -A127672(10+n, n), n >= 0, with the coefficients of the Chebyshev C-polynomials A127672(n, k). - Wolfdieter Lang, Dec 10 2015
a(n) = Sum_{i=1..n} A000217(i)*A000096(n+1-i). - Bruno Berselli, Mar 05 2018
a(n) = binomial(n+3,5) + 2*binomial(n+3,4). - Yuchun Ji, May 23 2019
From Amiram Eldar, Feb 17 2023: (Start)
Sum_{n>=1} 1/a(n) = 40751/63504.
Sum_{n>=1} (-1)^(n+1)/a(n) = 1360*log(2)/63 - 922961/63504. (End)

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

More terms from Zerinvary Lajos, Jul 21 2006

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A338950 Number of chiral pairs of colorings of the 24 octahedral facets (or 24 vertices) of the 4-D 24-cell using subsets of a set of n colors.

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A005995 Alkane (or paraffin) numbers l(8,n).

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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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A128767 Number of inequivalent n-colorings of the 4D hypercube under the full orthogonal group of the cube (of order 2^4*4! = 384).

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