A005902 -id:A005902 - cp's OEIS frontend

A117066 Partial sums of cupolar numbers (1/3)(n+1)(5n^2+7n+3) (A096000).

1, 11, 48, 140, 325, 651, 1176, 1968, 3105, 4675, 6776, 9516, 13013, 17395, 22800, 29376, 37281, 46683, 57760, 70700, 85701, 102971, 122728, 145200, 170625, 199251, 231336, 267148, 306965, 351075, 399776, 453376, 512193, 576555, 646800, 723276, 806341

Offset: 1

Jonathan Vos Post, Apr 17 2006

Consider the partitions of 2n into two parts (p,q) where p <= q. Then a(n) is the total volume of the family of rectangular prisms with dimensions p, p and q. - Wesley Ivan Hurt, Apr 15 2018

			For n=6, 9*a(6) = 6^3 + 7^3 + 8^3 + 9^3 + 10^3 + 11^3 +12^3 = 9*651. - _Bruno Berselli_, Apr 01 2014

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A005902, A096000.

List([1..40], n-> n^2*(n+1)*(5*n+1)/12); # G. C. Greubel, Jul 05 2019

[n^2*(n+1)*(5*n+1)/12: n in [1..40]]; // Vincenzo Librandi, Apr 16 2018

a:=n->sum ((n+j)^3, j=0..n): seq(a(n)/9, n=1..40);# Zerinvary Lajos, Dec 17 2008

Table[Sum[n i (n + i), {i, 0, n}]/2, {n, 40}] (* Vladimir Joseph Stephan Orlovsky, Jun 03 2011 *)
Accumulate[Table[((n+1)(5n^2+7n+3))/3,{n,0,50}]] (* or *) LinearRecurrence[{5,-10,10,-5,1},{1,11,48,140,325},50] (* Harvey P. Dale, Jan 03 2024 *)

my(x='x+O('x^40)); Vec(x*(3*x^2+6*x+1)/(1-x)^5) \\ G. C. Greubel, Jul 19 2017

a(n) = n^2*(n+1)*(5*n+1)/12; \\ Altug Alkan, Apr 16 2018

[n^2*(n+1)*(5*n+1)/12 for n in (1..40)] # G. C. Greubel, Jul 05 2019

a(n) = Sum_{i=1..n} A096000(i).
a(n) = Sum_{i=1..n} (1/3)*(i+1)*(5*i^2+7*i+3).
a(n) = Sum_{i=1..n} (1/2)*(Q(i) + 3*i^2 + 3*i + 1), where Q(i) are the cuboctahedral numbers (A005902).
a(n) = Sum_{i=0..n} A073254(n,i)*i. - Peter Luschny, Oct 29 2011
G.f.: x*(1+6*x+3*x^2) / (1-x)^5. - Colin Barker, May 08 2013
9*a(n) = Sum_{i=0..n} (n+i)^3, see Maple code by Zerinvary Lajos. - Bruno Berselli, Apr 01 2014
a(n) = n^2*(n+1)*(5*n+1)/12. - Vaclav Kotesovec, Jan 03 2017
E.g.f.: (x/12)*(12 + 54*x + 36*x^2 + 5*x^3)*exp(x). - G. C. Greubel, Jul 19 2017
Another identity: ..., a(4) = (1/2)*(7*(2+4+6+8)+5*(4+6+8)+3*(6+8)+1*(8)) = 140, a(5) = (1/2)*(9*(2+4+6+8+10)+7*(4+6+8+10)+5*(6+8+10)+3*(8+10)+1*(10)) = 325, ... - J. M. Bergot, Aug 25 2022

Terms corrected by Colin Barker, May 08 2013

A008354 a(n) = (5n^2 + 1)n^2 / 6.

Original entry on oeis.org

0, 1, 14, 69, 216, 525, 1086, 2009, 3424, 5481, 8350, 12221, 17304, 23829, 32046, 42225, 54656, 69649, 87534, 108661, 133400, 162141, 195294, 233289, 276576, 325625, 380926, 442989, 512344, 589541, 675150, 769761, 873984, 988449, 1113806, 1250725, 1399896

Offset: 0

N. J. A. Sloane and J. H. Conway

Partial sums of A005902. - Jonathan Vos Post, Mar 14 2006

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

Cf. A005901, A005902.

List([0..30], n -> (5*n^2+1)*n^2/6); # Muniru A Asiru, Feb 12 2018

a:= n-> 5*n^4/6 + n^2/6: seq(a(n), n=0..45);

Table[n^2 (5 n^2 + 1)/6, {n, 0, 30}] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 14, 69, 216}, 30] (* Harvey P. Dale, Feb 12 2015 *)

From R. J. Mathar, Aug 10 2008: (Start)
O.g.f.: x*(1 + x)*(x^2 + 8*x + 1)/(1 - x)^5.
a(n) = n*A004068(n). (End)
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>4, a(0)=0, a(1)=1, a(2)=14, a(3)=69, a(4)=216. - Harvey P. Dale, Feb 12 2015

Definition corrected by R. J. Mathar, Aug 10 2008

A193218 Number of vertices in truncated tetrahedron with faces that are centered polygons.

Original entry on oeis.org

1, 21, 95, 259, 549, 1001, 1651, 2535, 3689, 5149, 6951, 9131, 11725, 14769, 18299, 22351, 26961, 32165, 37999, 44499, 51701, 59641, 68355, 77879, 88249, 99501, 111671, 124795, 138909, 154049, 170251, 187551, 205985, 225589, 246399, 268451, 291781, 316425

Offset: 1

Craig Ferguson, Jul 18 2011

The sequence starts with a central vertex and expands outward with (n-1) centered polygonal pyramids producing a truncated tetrahedron. Each iteration requires the addition of (n-2) edges and (n-1) vertices to complete the centered polygon in each face. For centered triangles see A005448 and centered hexagons A003215.
This sequence is the 18th in the series (1/12)*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496 and t = 36. While adjusting for offsets, the beginning sequence A049480 is generated by adding the square pyramidal numbers A000330 to the odd numbers A005408 and each subsequent sequence is found by adding another set of square pyramidals A000330. (T/2) * A000330(n) + A005408(n). At 30 * A000330 + A005408 = centered dodecahedral numbers, 36 * A000330 + A005408 = A193228 truncated octahedron and 90 * A000330 + A005408 = A193248 = truncated icosahedron and dodecahedron. All five of the "Centered Platonic Solids" numbers sequences are in this series of sequences. Also 4 out of five of the "truncated" platonic solid number sequences are in this series. - Bruce J. Nicholson, Jul 06 2018
It would be good to have a detailed description of how the sequence is constructed. Maybe in the Examples section? - N. J. A. Sloane, Sep 07 2018

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
OEIS, (Centered_polygons) pyramidal numbers
Wikipedia, Tetrahedral number
Wikipedia, Triangular number
Wikipedia, Centered polygonal number
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A260810 (partial sums).

[6*n^3-9*n^2+5*n-1: n in [1..40]]; // Vincenzo Librandi, Aug 30 2011

Table[6 n^3 - 9 n^2 + 5 n - 1, {n, 35}] (* Alonso del Arte, Jul 18 2011 *)
CoefficientList[Series[(1+x)*(x^2+16*x+1)/(1-x)^4, {x, 0, 50}], x] (* Stefano Spezia, Sep 04 2018 *)

a(n) = 6*n^3 - 9*n^2 + 5*n - 1.
G.f.: x*(1+x)*(x^2+16*x+1) / (1-x)^4. - R. J. Mathar, Aug 26 2011
a(n) = 18 * A000330(n-1) + A005408(n-1) = A063496(n) + A006331(n-1). - Bruce J. Nicholson, Jul 06 2018

A142997 a(0) = 0, a(1) = 1, a(n+1) = (2n^2+2n+13)a(n) - n^4a(n-1).

Original entry on oeis.org

0, 1, 17, 409, 13756, 624364, 36981072, 2777988240, 258456976128, 29199105421056, 3939691125888000, 625956978121344000, 115709065165486080000, 24625602280458786816000

Offset: 0

Peter Bala, Jul 18 2008, Oct 16 2008

This is the case m = 3 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n^2+2*n+m^2+m+1)*a(n) - n^4*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series sum {k = 1..inf} 1/k^2 for the constant zeta(2). See A142995 for remarks on the general case.

Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Cf. A001819, A005902, A142995, A142996, A142998.

p := n -> (10*n^3+15*n^2+11*n+3)/3: a := n -> n!^2*p(n)*sum (1/(k^2*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..20);

a(n) = n!^2*p(n)*sum {k = 1..n} 1/(k^2*p(k-1)*p(k)), where p(n) = (10*n^3+15*n^2+11*n+3)/3 = A005902(n) is the polynomial that gives the crystal ball sequence for the A_3 lattice. Recurrence: a(0) = 1, a(1) = 1, a(n+1) = (2*n^2+2*n+13)*a(n) - n^4*a(n-1). The sequence b(n):= n!^2*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 13. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(13-1^4/(17-2^4/(25-3^4/(37-...-(n-1)^4/(2*n^2-2*n+13))))), for n >= 2. Thus the behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = 1/(13-1^4/(17-2^4/(25-3^4/(37-...-n^4/((2*n^2+2*n+13)-...))))) = sum {k = 1..inf} 1/(k^2*p(k-1)*p(k)) = 2*(1-1/4+1/9) - zeta(2). The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 31] (replace x by 2x+1 in the corollary and apply Entry 14).

A274973 Centered cubohemioctahedral numbers: a(n) = 2n^3+9n^2+n+1.

Original entry on oeis.org

1, 13, 55, 139, 277, 481, 763, 1135, 1609, 2197, 2911, 3763, 4765, 5929, 7267, 8791, 10513, 12445, 14599, 16987, 19621, 22513, 25675, 29119, 32857, 36901, 41263, 45955, 50989, 56377, 62131, 68263, 74785, 81709, 89047, 96811, 105013, 113665, 122779, 132367

Offset: 0

Steven Beard, Jul 13 2016

A faceting of the cuboctahedron, sharing the same square faces. The cubohemioctahedron has the same edge and vertex arrangement as the cuboctahedron. Beginning with the fourth term, the eight tetrahedral faces are each now "missing" a tetrahedron of size 1,4,10,20,35...(A000292). See A274974 centered octahemioctahedron for similar cuboctahedral faceting but with the square faces "missing."

Wikipedia, Cubohemioctahedron
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A005902 (centered cuboctahedral numbers), A274974 (centered octahemioctahedral numbers).

Table[2 n^3 + 9 n^2 + n + 1, {n, 40}] (* Michael De Vlieger, Jul 13 2016 *)
LinearRecurrence[{4,-6,4,-1},{1,13,55,139},40] (* Harvey P. Dale, Feb 18 2024 *)

a(n)=2*n^3+9*n^2+n+1 \\ Charles R Greathouse IV, Jul 14 2016

a(n) = 2*n^3+9*n^2+n+1.

G.f.: (-7*x^3+9*x^2+9*x+1)/(x-1)^4.

A274974 Centered octahemioctahedral numbers: a(n) = (4n^3+24n^2+8*n+3)/3.

Original entry on oeis.org

1, 13, 49, 117, 225, 381, 593, 869, 1217, 1645, 2161, 2773, 3489, 4317, 5265, 6341, 7553, 8909, 10417, 12085, 13921, 15933, 18129, 20517, 23105, 25901, 28913, 32149, 35617, 39325, 43281, 47493, 51969, 56717, 61745, 67061, 72673, 78589, 84817, 91365, 98241

Offset: 0

Steven Beard, Jul 13 2016

Related to a faceting of the cuboctahedron, sharing the same triangular faces. The octahemioctahedron has the same edge and vertex arrangement as the cuboctahedron (as does A274973). Beginning with the third term, the six square faces are each now "missing" a square pyramid of size 1, 5, 14, 30, 55, 91...(A000330). See A274973 centered cubohemioctahedron for similar cuboctahedral faceting but without the triangular faces.

Steven Beard, Music track made with this sequence
Wikipedia, Octahemioctahedron

Cf. A005902 (centered cuboctahedral numbers), A274973 (centered cubohemioctahedral numbers).

CoefficientList[Series[(-5 x^3 + 3 x^2 + 9 x + 1)/(x - 1)^4, {x, 0, 40}], x] (* or *)
Table[(4 n^3 + 24 n^2 + 8 n+3)/3, {n, 41}] (* Michael De Vlieger, Jul 13 2016 *)

a(n)=(4*n^3+24*n^2+8*n+3)/3 \\ Charles R Greathouse IV, Nov 03 2017

a(n) = (4*n^3+24*n^2+8*n+3)/3.

G.f.: (-5*x^3+3*x^2+9*x+1)/(x-1)^4.

A290396 a(n) = 32^n + 34^n + 6^(n+1) + 1.

Original entry on oeis.org

13, 55, 277, 1513, 8593, 49825, 292417, 1729153, 10275073, 61254145, 365945857, 2189371393, 13111037953, 78565515265, 470990340097, 2824331231233, 16939544543233, 101611496669185, 609565899227137, 3656983075356673, 21940249178406913, 131634897988091905, 789782999624318977

Offset: 0

Steven Beard, Jul 29 2017

One of several types of cuboctahedra with Sierpinski recursion. This Sierpinski cuboctahedron is a truncation of the six corners of the Sierpinski octahedron (A279512). The resulting Sierpinski cuboctahedron contains six square pyramids joined along the edges thus it can also be constructed by joining Sierpinski square pyramids (A281698).
Each face of the Sierpinski octahedron is a Sierpinski sieve with all triangular spaces completely formed. The triangles opening within the sieve, on each face, become tetrahedral excavations (cf. A067771). The third term above, for example, begins with an octahedron of 489 counters from A005900. From each of the eight faces a tetrahedron of four counters is removed. Thus the corresponding Sierpinski octahedron has 457 counters at this stage of recursion. Next, to construct the Sierpinski cuboctahedron, the truncation of the six corners is 30 counters each, thus 457 minus 180 equals 277, the third term of the Sierpinski cuboctahedral sequence shown above. For the next term in the sequence, two different sizes of tetrahedra are removed (and not only seen on the outer eight faces but also on the newly opened internal faces, now excavated in like manner.
To arrive at the next term, 1513, there are eight tetrahedra of 56 counters AND forty-eight tetrahedra of 4 counters each excavated from the octahedron, and then the six corners of 188 counters each are truncated thus: (56*8) + (4*48) = 640 total counters removed from the octahedral number 3281 (in A005900) to yield the Sierpinski octahedral number 2641 (in A279512, note that for subsequent terms the number of different size excavations of the octahedron increases, thus the next term would require three different sized excavations for the associated octahedron: 8 tetrahedra of 560 counters, 48 tetrahedra of 56 counters, and 288 tetrahedra of 4 counters).
Next truncate the six corners of 188 counters each: (6*188) = 1128. Sierpinski octahedron 2641 - 1128 = 1513, the fourth term in the Sierpinski cuboctahedral sequence above. The square pyramid of 188 counters truncated from each corner is found in A281698 by, in this case, taking the fourth term of 269 and subtracting its base of 81 counters.
The base of the square pyramid that is not counted is actually the square face of the remaining cuboctahedron. It is therefore easier to construct this Sierpinski cuboctahedral geometry by taking the six square pyramids of 269 counters each, subtracting 5 for the overlapping apices and then subtracting 96 to account for the edges where the square pyramids are overlapped: 6*269 = 1614 - 5 = 1609, - 96 = 1513, the fourth term shown in the Sierpinski cuboctahedral sequence above.

Colin Barker, Table of n, a(n) for n = 0..1000
Wikipedia, , Octahedron flake and Sierpinski tetrahedron.
Index entries for linear recurrences with constant coefficients, signature (13,-56,92,-48).

Cf. A005902, A279512, A274973, A281698, A067771.

Table[3*2^n + 3*4^n + 6^(n + 1) + 1, {n, 0, 22}] (* Michael De Vlieger, Jul 29 2017 *)

Vec((13 - 114*x + 290*x^2 - 204*x^3) / ((1 - x)*(1 - 2*x)*(1 - 4*x)*(1 - 6*x)) + O(x^30)) \\ Colin Barker, Jul 29 2017

a(n) = 3*2^n + 3*4^n + 6^(n+1) + 1 \\ Charles R Greathouse IV, Nov 03 2017

a(n) = 3*2^n + 3*4^n + 6^(n+1) + 1.
From Colin Barker, Jul 29 2017: (Start)
G.f.: (13 - 114*x + 290*x^2 - 204*x^3) / ((1 - x)*(1 - 2*x)*(1 - 4*x)*(1 - 6*x)).
a(n) = 13*a(n-1) - 56*a(n-2) + 92*a(n-3) - 48*a(n-4) for n>3.
(End)

A099608 Table of crystal ball sequences for A_n lattices read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 7, 1, 1, 7, 19, 13, 1, 1, 9, 37, 55, 21, 1, 1, 11, 61, 147, 131, 31, 1, 1, 13, 91, 309, 471, 271, 43, 1, 1, 15, 127, 561, 1251, 1281, 505, 57, 1, 1, 17, 169, 923, 2751, 4251, 3067, 869, 73, 1, 1, 19, 217, 1415, 5321, 11253, 12559, 6637, 1405, 91, 1

Offset: 0

Michael Somos, Oct 25 2004

Rows (n=0..10) are A000012, A005408, A003215, A005902, A008384, A008386, A008388, A008390, A008392, A008394, A008396.

Row reverse of A108625. - Peter Bala, Jul 18 2008

T(n, n)=A005258(n), T(2n, n)=A099601(n).

Cf. A108625.

T(n,k)=sum(j=0,min(n,k),binomial(n+j,j)*binomial(n,j)*binomial(k,j))

T(n, k)=hypergeom([n+1, -n, -k], [1, 1], 1).

A160249 Triangle of "less regular" face-centered-cubic sphere pack cuboctahedron numbers read by rows.

Original entry on oeis.org

1, 4, 13, 10, 28, 55, 20, 50, 92, 147, 35, 80, 140, 216, 309, 56, 119, 200, 300, 420, 561, 84, 168, 273, 400, 550, 724, 923, 120, 228, 360, 517, 700, 910, 1148, 1415, 165, 300, 462, 652, 871, 1120, 1400, 1712, 2057

Offset: 1

Chris G. Spies-Rusk (chaosorder4(AT)gmail.com), May 05 2009, May 11 2009

These solids each have 6 same rectangles, if one includes the first column (tetrahedra), where the rectangle is 1*(row number). They each have 2 tetrahedral quartets of same equilateral triangles, if one includes again the first column, for whom one or both quartets may be a single unit.
The first column is the series of tetrahedral numbers, (essentially A000292) and represents increasing edge length of one quartet of equilateral triangles. The column number represents the other quartet. The 2 quartet edges are interchangeable, because this triangle array is one half of a square table reflectable about the diagonal.
So the final column of each row represents a point where the 2 quartets' edge lengths are equal and the result is a regular cuboctahedron (A005902).
It seems remarkable that 24 of these shapes have even-hundred quantities of 10,000 or less, including 10,000 itself! This is far more than in any other figurate number series I've encountered. Additionally, 300 occurs twice, at 2,9 and 4,6 as does 9100, at 7,24 and 13,16. It suggests to me that this class of shapes may be useful in illustrating large numbers attractively.

Main Title: Polyhedra primer / Peter Pearce and Susan Pearce. Published/Created: New York : Van Nostrand Reinhold, c1978. Description: viii, 134 p. : ill. ; 24 cm. ISBN: 0442264968
Main Title: The book of numbers / John H. Conway, Richard K. Guy. Published/Created: New York, NY : Copernicus c1996. Description: ix, 310 p. : ill. (some col.) ; 24 cm. ISBN: 038797993X

Paste the following formula into cell C3, and fill down and right to desired table size. All volumes 10,000 and under are covered by column Q and row 38. The cells above the diagonal can be cleared since they are duplicates of content below the diagonal.
=((ROW()-2)^3+(COLUMN()-2)^3+9*(ROW()-2)^2*(COLUMN()-2)+9*(ROW()-2)*(COLUMN()-2)^2-6*(ROW()-2)^2-6*(COLUMN()-2)^2-18*(ROW()-2)*(COLUMN()-2)+11*(ROW()-2)+11*(COLUMN()-2)-6)/6

v=(y^3+x^3+9*y^2*x+9*y*x^2-6*y^2-6*x^2-18*y*x+11*y+11*x-6)/6

Improvement of the definition's precision by Chris G. Spies-Rusk (chaosorder4(AT)gmail.com), May 19 2009

A242941 a(n) is the number of convex uniform tessellations in dimension n.

Original entry on oeis.org

1, 11, 28, 143

Offset: 1

Felix Fröhlich, May 27 2014

Terms for n > 4 have not been determined so far. Alfredo Andreini in 1905 gave a value of 25 for a(3), later found to be incorrect. The value 28 for a(3) was given by Norman Johnson in 1991 and later in 1994 independently by Branko Grünbaum. The value for a(4) was given by George Olshevsky in 2006.
Deza and Shtogrin (2000) agree that the value of a(3) is 28, although the authors do not provide a proof. - Felix Fröhlich, Nov 29 2014
From Felix Fröhlich, Feb 03 2019: (Start)
The 11 convex uniform tilings are all illustrated in Kepler, 1619. For an argument that exactly 11 such tilings exist, see Grünbaum, Shephard, 1977.
In dimension 2, the definition of "uniform polytope" usually seems to be equivalent to the regular polygons in order to exclude polygons that alternate two different edge-lengths. Applying this principle retroactively to dimension 1 (as done, as I assume, by Coxeter, see Coxeter, 1973, p. 129) yields a(1) = 1. (End)

H. S. M. Coxeter, Regular Polytopes, Third Edition, Dover Publications, 1973, ISBN 9780486614809.
B. Grünbaum, Uniform tilings of 3-space, Geombinatorics, Vol. 4, No. 2 (1994), 49-56.
N. W. Johnson, Uniform Polytopes, [To appear, cf. Weiss, Stehle, 2017].

A. Andreini, Sulle reti di poliedri regolari e semiregolari e sulle correspondenti reto correlative [On the regular and semiregular nets of polyhedra and on the corresponding correlative nets], Mem. Società Italiana della Scienze, Ser.3, 14 (1905), 75-129.
M. Deza and M. Shtogrin, Uniform Partitions of 3-space, their Relatives and Embedding, European Journal of Combinatorics, Vol. 21, No. 6 (2000), 807-814.
B. Grünbaum and G. C. Shephard, Tilings by Regular Polygons, Mathematics Magazine, Vol. 50, No. 5 (1977), 227-247.
J. Kepler, Harmonices Mundi [The Harmony of the World] (1619).
G. Olshevsky, Uniform Panoploid Tetracombs (2006)
A. I. Weiss and E. M. Stehle, Norman W. Johnson (12 November 1930 to 13 July 2017), The Art of Discrete and Applied Mathematics, Vol. 1, No. 1 (2018).
Wikipedia, Convex uniform honeycomb
Wikipedia, List of convex uniform tilings

Cf. A068599.
List of coordination sequences for the 11 uniform 2D tilings: A008458(the planar net 3.3.3.3.3.3), A008486 (6^3), A008574 (4.4.4.4 and 3.4.6.4), A008576 (4.8.8), A008579 (3.6.3.6), A008706(3.3.3.4.4), A072154 (4.6.12), A219529 (3.3.4.3.4), A250120(3.3.3.3.6), A250122 (3.12.12).
List of coordination sequences for the 28 uniform 3D tilings: cab: A299266, A299267; crs: A299268, A299269; fcu: A005901, A005902; fee: A299259, A299265; flu-e: A299272, A299273; fst: A299258, A299264; hal: A299274, A299275; hcp: A007899, A007202; hex: A005897, A005898; kag: A299256, A299262; lta: A008137, A299276; pcu: A005899, A001845; pcu-i: A299277, A299278; reo: A299279, A299280; reo-e:  A299281, A299282; rho: A008137, A299276; sod: A005893, A005894; sve: A299255, A299261; svh: A299283, A299284; svj: A299254, A299260; svk: A010001, A063489; tca: A299285, A299286; tcd: A299287, A299288; tfs: A005899, A001845; tsi: A299289, A299290; ttw: A299257, A299263; ubt: A299291, A299292; bnn: A007899, A007202. See the Proserpio link in A299266 for overview.

Edited by N. J. A. Sloane, Feb 15 2018

Edited by Felix Fröhlich, Feb 03-10 2019

cp's OEIS Frontend

A117066 Partial sums of cupolar numbers (1/3)*(n+1)*(5*n^2+7*n+3) (A096000).

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A008354 a(n) = (5*n^2 + 1)*n^2 / 6.

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A193218 Number of vertices in truncated tetrahedron with faces that are centered polygons.

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A142997 a(0) = 0, a(1) = 1, a(n+1) = (2*n^2+2*n+13)*a(n) - n^4*a(n-1).

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A274973 Centered cubohemioctahedral numbers: a(n) = 2*n^3+9*n^2+n+1.

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A274974 Centered octahemioctahedral numbers: a(n) = (4*n^3+24*n^2+8*n+3)/3.

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A290396 a(n) = 3*2^n + 3*4^n + 6^(n+1) + 1.

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A117066 Partial sums of cupolar numbers (1/3)(n+1)(5n^2+7n+3) (A096000).

A008354 a(n) = (5n^2 + 1)n^2 / 6.

A142997 a(0) = 0, a(1) = 1, a(n+1) = (2n^2+2n+13)a(n) - n^4a(n-1).

A274973 Centered cubohemioctahedral numbers: a(n) = 2n^3+9n^2+n+1.

A274974 Centered octahemioctahedral numbers: a(n) = (4n^3+24n^2+8*n+3)/3.

A290396 a(n) = 32^n + 34^n + 6^(n+1) + 1.