A002817 -id:A002817 - cp's OEIS frontend

A081720 Triangle T(n,k) read by rows, giving number of bracelets (turnover necklaces) with n beads of k colors (n >= 1, 1 <= k <= n).

1, 1, 3, 1, 4, 10, 1, 6, 21, 55, 1, 8, 39, 136, 377, 1, 13, 92, 430, 1505, 4291, 1, 18, 198, 1300, 5895, 20646, 60028, 1, 30, 498, 4435, 25395, 107331, 365260, 1058058, 1, 46, 1219, 15084, 110085, 563786, 2250311, 7472984, 21552969, 1, 78, 3210, 53764, 493131, 3037314

Offset: 1

N. J. A. Sloane, based on information supplied by Gary W. Adamson, Apr 05 2003

From Petros Hadjicostas, Nov 29 2017: (Start)
The formula given below is clear from the programs given in the Maple and Mathematica sections, while the g.f. for column k can be obtained using standard techniques.
If we differentiate the column k g.f. m times, then we can get a formula for row m. (For this sequence, we only need to use this row m formula for 1 <= k <= m, but it is valid even for k>m.) For example, to get the formula for row 8, we have T(n=8,k) = d^8/dx^8 (column k g.f.)/8! evaluated at x=0. Here, "d^8/dx^8" means "8th derivative w.r.t. x" of the column k g.f. Doing so, we get T(n=8, k) = (k^6 - k^5 + k^4 + 3*k^3 + 2*k^2 - 2*k + 4)*(k + 1)*k/16, which is the formula given for sequence A060560. (Here, we use this formula only for 1 <= k <= 8.)
(End)

			1;                                                (A000027)
1,  3;                                            (A000217)
1,  4,  10;                                       (A000292)
1,  6,  21,   55;                                 (A002817)
1,  8,  39,  136,   377;                          (A060446)
1, 13,  92,  430,  1505,   4291;                  (A027670)
1, 18, 198, 1300,  5895,  20646,  60028;          (A060532)
1, 30, 498, 4435, 25395, 107331, 365260, 1058058; (A060560)
...
For example, when n=k=3, we have the following T(3,3)=10 bracelets of 3 beads using up to 3 colors: 000, 001, 002, 011, 012, 022, 111, 112, 122, and 222. (Note that 012 = 120 = 201 = 210 = 102 = 021.) _Petros Hadjicostas_, Nov 29 2017

N. Zagaglia Salvi, Ordered partitions and colourings of cycles and necklaces, Bull. Inst. Combin. Appl., 27 (1999), 37-40.

Andrew Howroyd, Table of n, a(n) for n = 1..1275
Yi Hu, Numerical Transfer Matrix Method of Next-nearest-neighbor Ising Models, Master's Thesis, Duke Univ. (2021).
Yi Hu and Patrick Charbonneau, Numerical transfer matrix study of frustrated next-nearest-neighbor Ising models on square lattices, arXiv:2106.08442 [cond-mat.stat-mech], 2021, cites the 4th column.

Cf. A321791 (extension to n >= 0, k >= 0).

Cf. A081721 (diagonal), A081722 (row sums), column sequences k=2..6: A000029, A027671, A032275, A032276, A056341.

A081720 := proc(n, k)
    local d, t1;
    t1 := 0;
    if n mod 2 = 0 then
        for d from 1 to n do
            if n mod d = 0 then
                t1 := t1+numtheory[phi](d)*k^(n/d);
            end if;
        end do:
        (t1+(n/2)*(1+k)*k^(n/2)) /(2*n) ;
    else
        for d from 1 to n do
            if n mod d = 0 then
                t1 := t1+numtheory[phi](d)*k^(n/d);
            end if;
        end do;
        (t1+n*k^((n+1)/2)) /(2*n) ;
    end if;
end proc:
seq(seq(A081720(n,k),k=1..n),n=1..10) ;

t[n_, k_] := (For[t1 = 0; d = 1, d <= n, d++, If[Mod[n, d] == 0, t1 = t1 + EulerPhi[d]*k^(n/d)]]; If[EvenQ[n], (t1 + (n/2)*(1 + k)*k^(n/2))/(2*n), (t1 + n*k^((n + 1)/2))/(2*n)]); Table[t[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Sep 13 2012, after Maple, updated Nov 02 2017 *)
Needs["Combinatorica`"]; Table[Table[NumberOfNecklaces[n,k,Dihedral],{k,1,n}],{n,1,8}]//Grid  (* Geoffrey Critzer, Oct 07 2012, after code by T. D. Noe in A027671 *)

See Maple code.
From Petros Hadjicostas, Nov 29 2017: (Start)
T(n,k) = ((1+k)*k^{n/2}/2 + (1/n)*Sum_{d|n} phi(n/d)*k^d)/2, if n is even, and = (k^{(n+1)/2} + (1/n)*Sum_{d|n} phi(n/d)*k^d)/2, if n is odd.
G.f. for column k: (1/2)*((k*x+k*(k+1)*x^2/2)/(1-k*x^2) - Sum_{n>=1} (phi(n)/n)*log(1-k*x^n)) provided we chop off the Taylor expansion starting at x^k (and ignore all the terms x^n with n(End)
2*n*T(n,k) = A054618(n,k)+n*(1+k)^(n/2)/2 if n even, = A054618(n,k)+n*k^((n+1)/2) if n odd. - R. J. Mathar, Jan 23 2022

Name edited by Petros Hadjicostas, Nov 29 2017

A185787 Sum of first k numbers in column k of the natural number array A000027; by antidiagonals.

Original entry on oeis.org

1, 7, 25, 62, 125, 221, 357, 540, 777, 1075, 1441, 1882, 2405, 3017, 3725, 4536, 5457, 6495, 7657, 8950, 10381, 11957, 13685, 15572, 17625, 19851, 22257, 24850, 27637, 30625, 33821, 37232, 40865, 44727, 48825, 53166, 57757, 62605, 67717, 73100, 78761, 84707, 90945, 97482, 104325, 111481, 118957, 126760, 134897, 143375

Offset: 1

Clark Kimberling, Feb 03 2011

This is one of many interesting sequences and arrays that stem from the natural number array A000027, of which a northwest corner is as follows:
  1....2.....4.....7...11...16...22...29...
  3....5.....8....12...17...23...30...38...
  6....9....13....18...24...31...39...48...
  10...14...19....25...32...40...49...59...
  15...20...26....33...41...50...60...71...
  21...27...34....42...51...61...72...84...
  28...35...43....52...62...73...85...98...
Blocking out all terms below the main diagonal leaves columns whose sums comprise A185787.  Deleting the main diagonal and then summing give A185787.  Analogous treatments to the left of the main diagonal give A100182 and A101165.  Further sequences obtained directly from this array are easily obtained using the following formula for the array: T(n,k)=n+(n+k-2)(n+k-1)/2.
Examples:
row 1:  A000124
row 2:  A022856
row 3:  A016028
row 4:  A145018
row 5:  A077169
col 1:  A000217
col 2:  A000096
col 3:  A034856
col 4:  A055998
col 5:  A046691
col 6:  A052905
col 7:  A055999
diag. (1,5,...) ...... A001844
diag. (2,8,...) ...... A001105
diag. (4,12,...)...... A046092
diag. (7,17,...)...... A056220
diag. (11,23,...) .... A132209
diag. (16,30,...) .... A054000
diag. (22,38,...) .... A090288
diag. (3,9,...) ...... A058331
diag. (6,14,...) ..... A051890
diag. (10,20,...) .... A005893
diag. (15,27,...) .... A097080
diag. (21,35,...) .... A093328
antidiagonal sums:  (1,5,15,34,...)=A006003=partial sums of A002817.
Let S(n,k) denote the n-th partial sum of column k.  Then
S(n,k)=n*(n^2+3k*n+3*k^2-6*k+5)/6.
S(n,1)=n(n+1)(n+2)/6
S(n,2)=n(n+1)(n+5)/6
S(n,3)=n(n+2)(n+7)/6
S(n,4)=n(n^2+12n+29)/6
S(n,5)=n(n+5)(n+10)/6
S(n,6)=n(n+7)(n+11)/6
S(n,7)=n(n+10)(n+11)/6
Weight array of T: A144112
Accumulation array of T: A185506
Second rectangular sum array of T: A185507
Third rectangular sum array of T: A185508
Fourth rectangular sum array of T: A185509

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

Cf. A000027, A185788, A100182, A101165, A079824.

[n*(7*n^2-6*n+5)/6: n in [1..50]]; // Vincenzo Librandi, Jul 04 2012

f[n_,k_]:=n+(n+k-2)(n+k-1)/2;
s[k_]:=Sum[f[n,k],{n,1,k}];
Factor[s[k]]
Table[s[k],{k,1,70}]  (* A185787 *)
CoefficientList[Series[(3*x^2+3*x+1)/(1-x)^4,{x,0,50}],x] (* Vincenzo Librandi, Jul 04 2012 *)

a(n)=n*(7*n^2-6*n+5)/6.

G.f.: x*(3*x^2+3*x+1)/(1-x)^4. - Vincenzo Librandi, Jul 04 2012

Edited by Clark Kimberling, Feb 25 2023

A006528 a(n) = (n^4 + n^2 + 2*n)/4.

Original entry on oeis.org

0, 1, 6, 24, 70, 165, 336, 616, 1044, 1665, 2530, 3696, 5226, 7189, 9660, 12720, 16456, 20961, 26334, 32680, 40110, 48741, 58696, 70104, 83100, 97825, 114426, 133056, 153874, 177045, 202740, 231136, 262416, 296769, 334390, 375480, 420246, 468901, 521664, 578760

Offset: 0

N. J. A. Sloane

Number of ways to color vertices of a square using <= n colors, allowing only rotations.

Also product of first and last terms in n-th row of a triangle of form: row(1)= 1; row(2)= 2,3; row(3) = 4, 5, 6, ... . - Dave Durgin, Aug 17 2012

Nick Baxter, The Burnside di-lemma: combinatorics and puzzle symmetry, in Tribute to a Mathemagician, Peters, 2005, pp. 199-210.
M. Gardner, New Mathematical Diversions from Scientific American. Simon and Schuster, NY, 1966, p. 246.
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 = 0..1000
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.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Row n=2 of A343095.

Cf. A002817 (square colorings).

I:=[0, 1, 6, 24, 70]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..40]]; // Vincenzo Librandi, Apr 30 2012

A006528:=-z*(1+z+4*z**2)/(z-1)**5; # Simon Plouffe in his 1992 dissertation
a:=n->add(n+add(binomial(n,2), j=1..n),j=0..n):seq(a(n)/2, n=0..35); # Zerinvary Lajos, Aug 26 2008

Table[CycleIndex[CyclicGroup[4],t]/.Table[t[i]->n,{i,1,4}],{n,0,20}] (* Geoffrey Critzer, Mar 13 2011*)
Table[(n^4+n^2+2*n)/4,{n,0,40}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{0,1,6,24,70},40] (* Harvey P. Dale, Jan 13 2019 *)

a(n) = n*(n+1)*(n^2-n+2)/4; /* Joerg Arndt, Apr 30 2012 */

a(n) = n*(n+1)*(n^2-n+2)/4.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Vincenzo Librandi, Apr 30 2012
From Stefano Spezia, Oct 11 2023: (Start)
O.g.f.:  x*(1 + x + 4*x^2)/(1 - x)^5.
E.g.f.: exp(x)*x*(4 + 8*x + 6*x^2 + x^3)/4. (End)

A128766 Number of inequivalent n-colorings of the vertices of the 3D cube under full orthogonal group of the cube (of order 48).

Original entry on oeis.org

1, 22, 267, 1996, 10375, 41406, 135877, 384112, 966141, 2212750, 4693711, 9340332, 17610307, 31703686, 54839625, 91604416, 148382137, 233880102, 359762131, 541403500, 798782271, 1157522542, 1650105997, 2317268976, 3209603125

Offset: 1

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

The formula was obtained by computing the cycle index of the group of geometric transformations, in 3D space, generated by all possible compositions of the 3 main reflections and the 3 main rotations and their inverses, in any order, with repetition of these geometric transformations allowed.
I assume this refers to colorings of the vertices of the cube. - N. J. A. Sloane, Apr 06 2007
Also the number of ways to color the faces of a regular octahedron with n colors, counting each pair of mirror images as one.

			a(2)=22 because there are 22 inequivalent 2-colorings of the 3D cube, including two for which all of the vertices have the same color.

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.

Cf. A000616, A002817.

Cf. A000543 Number when mirror images are counted separately.

A[n_] := (1/48)*(20*n^2 + 21*n^4 + 6*n^6 + n^8)
(*or*)
Drop[Table[CycleIndex[GraphData[{"Hypercube",3},"Automorphisms"],s]/.Table[s[i]->n,{i,1,8}],{n,0,25}],1]  (* Geoffrey Critzer, Mar 31 2013 *)

a(n) = (1/48)*(20*n^2 + 21*n^4 + 6*n^6 + n^8).
G.f.: x*(1+x)*(1+12*x+93*x^2+208*x^3+93*x^4+12*x^5+x^6)/(1-x)^9. [Colin Barker, Mar 08 2012]
Cycle Index is (1/48)*(s[1]^8 + 6*s[1]^4*s[2]^2 + 13*s[2]^4 + 8*s[1]^2*s[3]^2 + 12*s[4]^2 + 8*s[2]*s[6]) - Geoffrey Critzer, Mar 31 2013
a(n)=C(n,1)+20C(n,2)+204C(n,3)+1056C(n,4)+2850C(n,5)+4080C(n,6)+2940C(n,7)+840C(n,8). Each term indicates the number of ways to use n colors to color the cube vertices (octahedron faces) with exactly 1, 2, 3, 4, 5, 6, 7, or 8 colors.

A343097 Array read by antidiagonals: T(n,k) is the number of k-colorings of an n X n grid, up to rotations and reflections.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 6, 1, 0, 1, 4, 21, 102, 1, 0, 1, 5, 55, 2862, 8548, 1, 0, 1, 6, 120, 34960, 5398083, 4211744, 1, 0, 1, 7, 231, 252375, 537157696, 105918450471, 8590557312, 1, 0, 1, 8, 406, 1284066, 19076074375, 140738033618944, 18761832172500795, 70368882591744, 1, 0

Offset: 0

Andrew Howroyd, Apr 14 2021

			Array begins:
====================================================================
n\k | 0 1       2            3               4                 5
----+---------------------------------------------------------------
  0 | 1 1       1            1               1                 1 ...
  1 | 0 1       2            3               4                 5 ...
  2 | 0 1       6           21              55               120 ...
  3 | 0 1     102         2862           34960            252375 ...
  4 | 0 1    8548      5398083       537157696       19076074375 ...
  5 | 0 1 4211744 105918450471 140738033618944 37252918396015625 ...
  ...

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

Rows 0..5 are A000012, A001477, A002817, A217331, A217338, A283033.
Columns 0..10 are A000007, A000012, A054247, A054739, A054751, A054752, A286392, A286393, A286394, A286396, A286397.
Main diagonal is A001326.
Cf. A246106, A343095, A343875.

T(n,k) = {(k^(n^2) + 2*k^((n^2 + 3*(n%2))/4) + k^((n^2 + (n%2))/2) + 2*k^(n*(n+1)/2) + 2*k^(n*(n+n%2)/2) )/8}

T(n,k) = (k^(n^2) + 2*k^((n^2 + 3*(n mod 2))/4) + k^((n^2 + (n mod 2))/2) + 2*k^(n*(n+1)/2) + 2*k^(n*(n + n mod 2)/2) )/8.

A011863 Nearest integer to (n/2)^4.

Original entry on oeis.org

0, 0, 1, 5, 16, 39, 81, 150, 256, 410, 625, 915, 1296, 1785, 2401, 3164, 4096, 5220, 6561, 8145, 10000, 12155, 14641, 17490, 20736, 24414, 28561, 33215, 38416, 44205, 50625, 57720, 65536, 74120, 83521, 93789, 104976, 117135, 130321, 144590

Offset: 0

R. K. Guy

First differences are in A019298.
The bisections are A000583 and A219086.
Number of ways to put n-1 copies of 1,2,3 into sets. [Zeilberger?]
s(n) is the number of 4-tuples (w,x,y,z) with all terms in {1,...,n} and |w-x| >= w + |y-z|; see A186707. - Clark Kimberling, May 24 2012

N. J. A. Sloane, Table of n, a(n) for n = 0..10000 (first 2000 terms from Vicenzo Librandi)
A. J. Guttmann, Indicators of solvability for lattice models, Discrete Math., 217 (2000), 167-189 (H_2 for square lattice of Section 6).
Doron Zeilberger, In How Many Ways Can You Reassemble Several Russian Dolls?, The Personal Journal of Shalosh B. Ekhad and Doron Zeilberger (2009); Local copy. [PDF file only, no active links]
Index entries for linear recurrences with constant coefficients, signature (4,-5,0,5,-4,1).

Cf. A000583, A002817, A019298, A106707, A212714, A219086.

[ (2*n^4-(1-(-1)^n))/32: n in [0..50] ];

Maple
```
seq(round((n/2)^4), n=0..40);
```

Round[(Range[40]/2)^4] (* or *) LinearRecurrence[{4,-5,0,5,-4,1},{0,1,5,16,39,81},40] (* Harvey P. Dale, Feb 07 2015 *)

a(n)=round((n/2)^4) \\ Charles R Greathouse IV, Jun 23 2011

G.f.: x^2*(1 + x + x^2)/((1 - x)^5*(1+x)).
a(n) = +4*a(n-1) -5*a(n-2) +5*a(n-4) -4*a(n-5) +a(n-6). - R. J. Mathar, Dec 07 2010
a(n)+a(n+1) = A002817(n). - R. J. Mathar, Dec 19 2008
a(n) = n^4/16 - 1/32 + (-1)^n/32 - R. J. Mathar, Dec 07 2010, adapted to added a(0) by Hugo Pfoertner, Dec 29 2019
a(n) = (2*A000583(n) + (-1)^n - 1)/32. - Bruno Berselli, Dec 07 2010, adapted to added a(0) by Hugo Pfoertner, Dec 29 2019
n*(n^2+n+2)*a(n+1) = 4*(n^2+2*n+2)*a(n)+(n+2)*(n^2+3*n+4)*a(n-1). Holonomic Ansatz with smallest order of recurrence. - Thotsaporn Thanatipanonda, Dec 12 2010
a(n) = floor(n^4/8)/2. - Gary Detlefs, Feb 19 2011, adapted to added a(0) by Hugo Pfoertner, Dec 29 2019
a(n) = A212714(n)/2, n >= 0. - Wolfdieter Lang, Oct 03 2016, adapted to added a(0) by Hugo Pfoertner, Dec 29 2019
E.g.f.: (1/32)*exp(-x)*(1 + exp(2*x)*(-1 + 2*x + 14*x^2 + 12*x^3 + 2*x^4)). - Stefano Spezia, Dec 29 2019
Sum_{n>=2} 1/a(n) = 6 + Pi^4/90 - 2*Pi*tanh(Pi/2). - Amiram Eldar, Aug 13 2022

Missing a(0) added by N. J. A. Sloane, Dec 29 2019. As a result some of the comments and formulas will need to be adjusted.

A118235 Smallest positive number starting an interval of consecutive integers with element sum n.

Original entry on oeis.org

1, 2, 1, 4, 2, 1, 3, 8, 2, 1, 5, 3, 6, 2, 1, 16, 8, 3, 9, 2, 1, 4, 11, 7, 3, 5, 2, 1, 14, 4, 15, 32, 3, 7, 2, 1, 18, 8, 4, 6, 20, 3, 21, 2, 1, 10, 23, 15, 4, 8, 6, 3, 26, 2, 1, 5, 7, 13, 29, 4, 30, 14, 3, 64, 2, 1, 33, 5, 9, 7, 35, 4, 36, 17, 3, 6, 2, 1, 39, 14, 5, 19, 41, 7, 4, 20, 12, 3, 44, 2, 1, 8

Offset: 1

Reinhard Zumkeller, Apr 18 2006

Right border of A299765. - Omar E. Pol, Jul 24 2018

In other words: a(n) is smallest part of the partitions of n into consecutive parts. - Omar E. Pol, Mar 12 2019

			a(3)=1 since 3 = 1+2; a(5)=2 since 5 = 2+3; a(6)=1 since 6 = 1+2+3; etc.

Alois P. Heinz, Table of n, a(n) for n = 1..20000 (first 1000 terms from Paul D. Hanna)

Cf. A000079, A000217, A001227, A002817, A104512, A109814, A212652, A299765.

a:= proc(n) local j, k, s; j, k, s:= 1$3;
      while s<>n do
         if sAlois P. Heinz, Aug 05 2018

a[n_] := Module[{j = 1, k = 1, s = 1}, While[True, If[s == n, Break[]]; If[s < n, k = k+1; s = s+k, s = s-j; j = j+1]]; j];
Array[a, 100] (* Jean-François Alcover, Mar 12 2019, after Alois P. Heinz *)

{a(n)=local(A=n);for(j=1,n,for(k=j,n+1,if(n==k*(k-1)/2-j*(j-1)/2,A=j;k=j=2*n+1)));A} /* Paul D. Hanna, Oct 28 2011 */

A109814(n) * (A109814(n) + 2*a(n) - 1) / 2 = n.
a(m) = n iff m = 2^k: a(A000079(n)) = A000079(n);
a(m) = 1 iff m = k*(k+1)/2: a(A000217(n)) = 1.
a(A002817(n-1)+1) = n; i.e., a(m) = n if m = k*(k-1)/2 + 1 and k = n*(n-1)/2 + 1. - Paul D. Hanna, Oct 28 2011
a(m) = 2 iff m = k*(k+3)/2: a(A000096(n)) = 2. - Bernard Schott, Mar 12 2019

A331436 Array read by antidiagonals: A(n,k) is the number of n element multisets of n element multisets of a k-set.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 6, 1, 0, 1, 4, 21, 20, 1, 0, 1, 5, 55, 220, 70, 1, 0, 1, 6, 120, 1540, 3060, 252, 1, 0, 1, 7, 231, 7770, 73815, 53130, 924, 1, 0, 1, 8, 406, 30856, 1088430, 5461512, 1107568, 3432, 1, 0, 1, 9, 666, 102340, 11009376, 286243776, 581106988, 26978328, 12870, 1, 0

Offset: 0

Andrew Howroyd, Jan 17 2020

			Array begins:
==================================================================
n\k | 0 1   2       3         4            5              6
----+-------------------------------------------------------------
  0 | 1 1   1       1         1            1              1 ...
  1 | 0 1   2       3         4            5              6 ...
  2 | 0 1   6      21        55          120            231 ...
  3 | 0 1  20     220      1540         7770          30856 ...
  4 | 0 1  70    3060     73815      1088430       11009376 ...
  5 | 0 1 252   53130   5461512    286243776     8809549056 ...
  6 | 0 1 924 1107568 581106988 127860662755 13949678575756 ...
    ...
The A(2,2) = 6 multisets are:
   {{1,1}, {1,1}},
   {{1,1}, {1,2}},
   {{1,1}, {2,2}},
   {{1,2}, {1,2}},
   {{1,2}, {2,2}},
   {{2,2}, {2,2}}.

Andrew Howroyd, Table of n, a(n) for n = 0..1325

Rows n=0..3 are A000012, A001477, A002817, A140236.
Columns k=0..10 are A000007, A000012, A000984, A099121, A099122, A099123, A099124, A099125, A099126, A099127, A099128.
Min diagonal is A331477.

T(n,k)={binomial(binomial(n + k - 1, n) + n - 1, n)}
{ for(n=0, 7, for(k=0, 7, print1(T(n,k), ", ")); print) }

A(n,k) = binomial(binomial(n + k - 1, n) + n - 1, n).

A236770 a(n) = n(n + 1)(3n^2 + 3n - 2)/8.

Original entry on oeis.org

0, 1, 12, 51, 145, 330, 651, 1162, 1926, 3015, 4510, 6501, 9087, 12376, 16485, 21540, 27676, 35037, 43776, 54055, 66045, 79926, 95887, 114126, 134850, 158275, 184626, 214137, 247051, 283620, 324105, 368776, 417912, 471801, 530740, 595035, 665001, 740962

Offset: 0

Bruno Berselli, Jan 31 2014

After 0, first trisection of A011779 and right border of A177708.

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

Partial sums of A004188.
Cf. A000217, A000332, A011779, A177708.
Cf. similar sequences on the polygonal numbers: A002817(n) = A000217(A000217(n)); A000537(n) = A000290(A000217(n)); A037270(n) = A000217(A000290(n)); A062392(n) = A000384(A000217(n)).
Cf. sequences of the form A000217(m)+k*A000332(m+2): A062392 (k=12); A264854 (k=11); A264853 (k=10); this sequence (k=9); A006324 (k=8); A006323 (k=7); A000537 (k=6); A006322 (k=5); A006325 (k=4), A002817 (k=3), A006007 (k=2), A006522 (k=1).
Cf. A232713, A260810.

[n*(n+1)*(3*n^2+3*n-2)/8: n in [0..40]];

Table[n (n + 1) (3 n^2 + 3 n - 2)/8, {n, 0, 40}]
LinearRecurrence[{5,-10,10,-5,1},{0,1,12,51,145},40] (* Harvey P. Dale, Aug 22 2016 *)

for(n=0, 40, print1(n*(n+1)*(3*n^2+3*n-2)/8", "));

G.f.: x*(1 + 7*x + x^2)/(1 - x)^5.
a(n) = a(-n-1) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A000326(A000217(n)).
a(n) = A000217(n) + 9*A000332(n+2).
Sum_{n>=1} 1/a(n) = 2 + 4*sqrt(3/11)*Pi*tan(sqrt(11/3)*Pi/2) = 1.11700627139319... . - Vaclav Kotesovec, Apr 27 2016

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cp's OEIS Frontend

A001496 Number of 4 X 4 matrices with nonnegative integer entries and row and column sums equal to n.

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A081720 Triangle T(n,k) read by rows, giving number of bracelets (turnover necklaces) with n beads of k colors (n >= 1, 1 <= k <= n).

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A185787 Sum of first k numbers in column k of the natural number array A000027; by antidiagonals.

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A006528 a(n) = (n^4 + n^2 + 2*n)/4.

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A128766 Number of inequivalent n-colorings of the vertices of the 3D cube under full orthogonal group of the cube (of order 48).

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A343097 Array read by antidiagonals: T(n,k) is the number of k-colorings of an n X n grid, up to rotations and reflections.

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A011863 Nearest integer to (n/2)^4.

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A236770 a(n) = n(n + 1)(3n^2 + 3n - 2)/8.