A063496 -id:A063496 - cp's OEIS frontend

A063495 a(n) = (2n-1)(5n^2-5n+2)/2.

1, 18, 80, 217, 459, 836, 1378, 2115, 3077, 4294, 5796, 7613, 9775, 12312, 15254, 18631, 22473, 26810, 31672, 37089, 43091, 49708, 56970, 64907, 73549, 82926, 93068, 104005, 115767, 128384, 141886, 156303, 171665, 188002, 205344, 223721, 243163, 263700, 285362

Offset: 1

N. J. A. Sloane, Aug 01 2001

Harry J. Smith, Table of n, a(n) for n = 1..1000
T. P. Martin, Shells of atoms, Phys. Rep., 273 (1996), 199-241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

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.

[(2*n-1)*(5*n^2-5*n+2)/2: n in [1..30]]; // G. C. Greubel, Dec 01 2017

Table[(2n-1)(5n^2-5n+2)/2,{n,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{1,18,80,217},40] (* Harvey P. Dale, Dec 18 2011 *)

a(n) = (2*n - 1)*(5*n^2 - 5*n + 2)/2 \\ Harry J. Smith, Aug 23 2009

my(x='x+O('x^30)); Vec(serlaplace((-2+4*x+15*x^2+10*x^3)*exp(x)/2 + 1)) \\ G. C. Greubel, Dec 01 2017

From Harvey P. Dale, Dec 18 2011: (Start)
a(1)=1, a(2)=18, a(3)=80, a(4)=217, a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) - a(n-4).
G.f.: (x^3+14*x^2+14*x+1)/(1-x)^4. (End)
E.g.f.: (-2 + 4*x + 15*x^2 + 10*x^3)*exp(x)/2 + 1. - G. C. Greubel, Dec 01 2017

A049480 a(n) = (2n-1)(n^2 -n +6)/6.

Original entry on oeis.org

1, 4, 10, 21, 39, 66, 104, 155, 221, 304, 406, 529, 675, 846, 1044, 1271, 1529, 1820, 2146, 2509, 2911, 3354, 3840, 4371, 4949, 5576, 6254, 6985, 7771, 8614, 9516, 10479, 11505, 12596, 13754, 14981, 16279, 17650, 19096, 20619, 22221

Offset: 1

N. J. A. Sloane, Aug 01 2001

Harvey P. Dale, Table of n, a(n) for n = 1..1000
T. P. Martin, Shells of atoms, Phys. Rep., 273 (1996), 199-241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

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.

[(2*n-1)*(n^2-n+6)/6: n in [1..30]]; // G. C. Greubel, Dec 01 2017

Table[(2n-1)(n^2-n+6)/6,{n,50}] (* or *) LinearRecurrence[{4,-6,4,-1},{1,4,10,21},50] (* Harvey P. Dale, Jan 01 2012 *)

a(n)=(2*n-1)*(n^2-n+6)/6 \\ Charles R Greathouse IV, Sep 24 2015

x='x+O('x^30); Vec(serlaplace((-6 + 12*x + 3*x^2 + 2*x^3)*exp(x)/6 + 1)) \\ G. C. Greubel, Dec 01 2017

From Harvey P. Dale, Jan 01 2012: (Start)
G.f.: x*(x^3 + 1)/(x-1)^4.
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4); a(1)=1, a(2)=4, a(3)=10, a(4)=21. (End)
E.g.f.: (-6 + 12*x + 3*x^2 + 2*x^3)*exp(x)/6 + 1. - G. C. Greubel, Dec 01 2017

A132366 Partial sum of centered tetrahedral numbers A005894.

Original entry on oeis.org

1, 6, 21, 56, 125, 246, 441, 736, 1161, 1750, 2541, 3576, 4901, 6566, 8625, 11136, 14161, 17766, 22021, 27000, 32781, 39446, 47081, 55776, 65625, 76726, 89181, 103096, 118581, 135750, 154721, 175616, 198561, 223686, 251125, 281016, 313501, 348726, 386841

Offset: 0

Jonathan Vos Post, Nov 09 2007

From Robert A. Russell, Oct 09 2020: (Start)
a(n-1) is the number of achiral colorings of the 5 tetrahedral facets (or vertices) of a regular 4-dimensional simplex using n or fewer colors. An achiral arrangement is identical to its reflection. The 4-dimensional simplex is also called a 5-cell or pentachoron. Its Schläfli symbol is {3,3,3}.
There are 60 elements in the automorphism group of the 4-dimensional simplex that are not in its rotation group. Each is an odd permutation of the vertices and can be associated with a partition of 5 based on the conjugacy class of the permutation. The first formula for a(n-1) is obtained by averaging their cycle indices after replacing x_i^j with n^j according to the Pólya enumeration theorem.
  Partition  Count  Odd Cycle Indices
  41         30     x_1x_4^1
  32         20     x_2^1x_3^1
  2111       10     x_1^3x_2^1 (End)

Nathaniel Johnston, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000292, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.
Cf. A337895 (oriented), A000389(n+4) (unoriented), A000389 (chiral), A331353 (5-cell edges, faces), A337955 (8-cell vertices, 16-cell facets), A337958 (16-cell vertices, 8-cell facets), A338951 (24-cell), A338967 (120-cell, 600-cell).
a(n-1) = A325001(4,n).

Do[Print[n, " ", (n^4 + 4 n^3 + 11 n^2 + 14 n + 6)/6 ], {n, 0, 10000}]
Accumulate[Table[(2n+1)(n^2+n+3)/3,{n,0,40}]] (* or *) LinearRecurrence[ {5,-10,10,-5,1},{1,6,21,56,125},40] (* Harvey P. Dale, Feb 26 2020 *)

a(n) = (n^4 + 4*n^3 + 11*n^2 + 14*n + 6)/6 = (n^2+2*n+6)*(n+1)^2/6.
G.f.: -(x+1)*(x^2+1) / (x-1)^5. - Colin Barker, May 04 2013
From Robert A. Russell, Oct 09 2020: (Start)
a(n-1) = n^2 * (5 + n^2) / 6.
a(n-1) = binomial(n+4,5) - binomial(n,5) = A000389(n+4) - A000389(n).
a(n-1) = 1*C(n,1) + 4*C(n,2) + 6*C(n,3) + 4*C(n,4), where the coefficient of C(n,k) is the number of achiral colorings using exactly k colors.
a(n-1) = 2*A000389(n+4) - A337895(n) = A337895(n) - 2*A000389(n) .
G.f. for a(n-1): x * (x+1) * (x^2+1) / (1-x)^5. (End)
From Amiram Eldar, Feb 14 2023: (Start)
Sum_{n>=0} 1/a(n) = Pi^2/5 + 3/25 - 3*Pi*coth(sqrt(5)*Pi)/(5*sqrt(5)).
Sum_{n>=0} (-1)^n/a(n) = Pi^2/10 - 3/25 + 3*Pi*cosech(sqrt(5)*Pi)/(5*sqrt(5)). (End)
a(n) = A006007(n) + A006007(n+1) = A002415(n) + A002415(n+2). - R. J. Mathar, Jun 05 2025

Corrected offset, Mathematica program by Tomas J. Bulka (tbulka(AT)rodincoil.com), Sep 02 2009

A253011 T(n,k)=Number of nXk nonnegative integer arrays with upper left 0 and lower right its king-move distance away minus 3 and every value increasing by 0 or 1 with every step right, diagonally se or down.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 4, 1, 0, 1, 4, 10, 19, 1, 1, 19, 10, 20, 85, 54, 1, 54, 85, 20, 35, 231, 632, 124, 124, 632, 231, 35, 56, 489, 2902, 3423, 250, 3423, 2902, 489, 56, 84, 891, 8416, 33533, 14795, 14795, 33533, 8416, 891, 84, 120, 1469, 18770, 158877, 309990

Offset: 1

R. H. Hardin, Dec 25 2014

Table starts
..0....0.....0.......1........4.........10..........20............35
..0....0.....0.......1.......19.........85.........231...........489
..0....0.....0.......1.......54........632........2902..........8416
..1....1.....1.......1......124.......3423.......33533........158877
..4...19....54.....124......250......14795......309990.......2853292
.10...85...632....3423....14795......54219.....2327062......43197859
.20..231..2902...33533...309990....2327062....14697256.....541365583
.35..489..8416..158877..2853292...43197859...541365583....5713126349
.56..891.18770..490403.14125312..395069496..9617848524..196919486736
.84.1469.35564.1156178.46481352.2063349297.90094114144.3496191411901

			Some solutions for n=6 k=4
..0..0..1..1....0..0..0..1....0..0..1..2....0..0..0..1....0..0..0..0
..0..0..1..2....1..1..1..1....1..1..1..2....0..0..0..1....0..0..0..0
..1..1..1..2....1..1..2..2....1..1..1..2....0..0..0..1....0..0..0..1
..1..1..1..2....1..1..2..2....1..1..1..2....0..0..0..1....0..1..1..1
..1..1..1..2....1..2..2..2....1..2..2..2....0..0..1..1....1..1..2..2
..2..2..2..2....2..2..2..2....1..2..2..2....1..1..1..2....1..1..2..2

R. H. Hardin, Table of n, a(n) for n = 1..480

Column 1 is A000292(n-3)

Column 2 is A063496(n-3)

Empirical for column k:
k=1: a(n) = (1/6)*n^3 - 1*n^2 + (11/6)*n - 1
k=2: a(n) = (16/3)*n^3 - 56*n^2 + (590/3)*n - 231 for n>3
k=3: a(n) = (800/3)*n^3 - 3980*n^2 + (60442/3)*n - 34576 for n>6
k=4: a(n) = (65536/3)*n^3 - 422912*n^2 + (8343128/3)*n - 6208382 for n>9
k=5: a(n) = 2973696*n^3 - 70716096*n^2 + 570494664*n - 1560617160 for n>12
k=6: [polynomial of degree 3] for n>15
k=7: [polynomial of degree 3] for n>18

A083669 Number of ordered quintuples (a,b,c,d,e), -n <= a,b,c,d,e <= n, such that a+b+c+d+e = 0.

Original entry on oeis.org

1, 51, 381, 1451, 3951, 8801, 17151, 30381, 50101, 78151, 116601, 167751, 234131, 318501, 423851, 553401, 710601, 899131, 1122901, 1386051, 1692951, 2048201, 2456631, 2923301, 3453501, 4052751, 4726801, 5481631, 6323451, 7258701

Offset: 0

Benoit Cloitre, Jun 14 2003

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Row 5 of A201552.

Cf. A003215, A063496.

[1+5*n*(n+1)*(23*n^2+23*n+14)/12: n in [0..30]]; // Vincenzo Librandi, Dec 15 2018

LinearRecurrence[{5, -10, 10, -5, 1}, {1, 51, 381, 1451, 3951}, 30] (* Vincenzo Librandi, Dec 15 2018 *)

a(n)=115/12*n^4+115/6*n^3+185/12*n^2+35/6*n+1

{a(n) = polcoeff((sum(k=0, 2*n, x^k))^5, 5*n, x)} \\ Seiichi Manyama, Dec 14 2018

a(n) = 1 + 5*n*(n+1)*(23*n^2 + 23*n + 14)/12.
a(n) = (1/Pi)*Integral_{x=0..Pi} (sin((n+1/2)*x)/sin(x/2))^5. - Yalcin Aktar, Dec 03 2011
G.f.: ( -1 - 46*x - 136*x^2 - 46*x^3 - x^4 ) / (x-1)^5. - R. J. Mathar, Dec 17 2011
a(n) = [x^(5*n)] (Sum_{k=0..2*n} x^k)^5. - Seiichi Manyama, Dec 14 2018

A142993 Crystal ball sequence for the lattice C_4.

Original entry on oeis.org

1, 33, 225, 833, 2241, 4961, 9633, 17025, 28033, 43681, 65121, 93633, 130625, 177633, 236321, 308481, 396033, 501025, 625633, 772161, 943041, 1140833, 1368225, 1628033, 1923201

Offset: 0

Peter Bala, Jul 18 2008

The lattice C_4 consists of all integer lattice points v = (a,b,c,d) in Z^4 such that a + b + c + d is even, equipped with the taxicab type norm ||v|| = (1/2) * (|a| + |b| + |c| + |d|). The crystal ball sequence of C_4 gives the number of lattice points v in C_4 with ||v|| <= n for n = 0,1,2,3,... [Bacher et al.].

			a(1) = 33. The origin has norm 0. The 32 lattice points in Z^4 of norm 1 (as defined above) are +-2*e_i, 1 <= i <= 4 and (+- e_i +- e_j), 1 <= i < j <= 4, where e_1, e_2, e_3 and e_4 denotes the standard basis of Z^4. These 32 vectors form a root system of type C_4. Hence sequence begins 1, 1 + 32 = 33, ... .

Seiichi Manyama, Table of n, a(n) for n = 0..10000
R. Bacher, P. de la Harpe and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, Annales de l'Institut Fourier, Tome 49 (1999) no. 3, pp. 727-762.
R. Bacher, P. de la Harpe and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 1137-1142.
Peter Bala, The C_4 lattice and a continued fraction for log(2)
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A019560, A063496, A142992, A142994.

a := n -> (2*n+1)^2*(4*n^2+4*n+3)/3: seq(a(n), n = 0..24)

Partial sums of A019560. a(n) = (2*n+1)^2*(4*n^2+4*n+3)/3 = Sum_{k = 0..4} C(8,2k)*C(n+k,4) = Sum_{k = 0..4} C(8,2k+1)*C(n+k+1/2,4). O.g.f.: (1+28*x+70*x^2+28*x^3+x^4)/(1-x)^5 = (1/(1-x)) * T(4,(1+x)/(1-x)), where T(n,x) denotes the Chebyshev polynomial of the first kind. 2*log(2) = 17/12 - Sum_{n >= 1} 1/(n*a(n-1)*a(n)).
From Peter Bala, Mar 11 2024: (Start)
Sum_{k = 1..n+1} 1/(k*a(k)*a(k-1)) = 1/(33 - 3/(41 - 60/(57 - 315/(81 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*4^2))))).
E.g.f.: exp(x)*(1 + 32*x + 160*x^2/2! + 256*x^3/3! + 128*x^4/4!).
Note that T(8, i*sqrt(x)) = 1 + 32*x + 160*x^2 + 256*x^3 + 128*x^4. See A008310. (End)

A142994 Crystal ball sequence for the lattice C_5.

Original entry on oeis.org

1, 51, 501, 2471, 8361, 22363, 50973, 103503, 192593, 334723, 550725, 866295, 1312505, 1926315, 2751085, 3837087, 5242017, 7031507, 9279637, 12069447, 15493449, 19654139, 24664509, 30648559, 37741809, 46091811, 55858661, 67215511, 80349081

Offset: 0

Peter Bala, Jul 18 2008

The lattice C_5 consists of all integer lattice points v = (x_1,...,x_5) in Z^5 such that (x_1 + ... + x_5) is even, equipped with the taxicab type norm ||v|| = (1/2) * (|x_1| + ... + |x_5|). The crystal ball sequence of C_5 gives the number of lattice points v in C_5 with ||v|| <= n for n = 0,1,2,3,... [Bacher et al.].

Partial sums of A019561.

			a(1) = 51. The origin has norm 0. The 50 lattice points in Z^5 of norm 1 (as defined above) are +-2*e_i, 1 <= i <= 5 and (+- e_i +- e_j), 1 <= i < j <= 5, where e_1, ... , e_5 denotes the standard basis of Z^5. These 50 vectors form a root system of type C_5. Hence the sequence begins 1, 1 + 50 = 51, ... .

Seiichi Manyama, Table of n, a(n) for n = 0..10000
R. Bacher, P. de la Harpe, and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, Annales de l'Institut Fourier, Tome 49 (1999) no. 3, pp. 727-762.
R. Bacher, P. de la Harpe and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 1137-1142.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Row 5 of A142992. Cf. A019561, A063496, A142993.

[(2*n+1)*(32*n^4+64*n^3+88*n^2+56*n+15)/15: n in [0..30]]; // Vincenzo Librandi, Dec 16 2015

a := n -> (2*n+1)*(32*n^4+64*n^3+88*n^2+56*n+15)/15: seq(a(n), n = 0..20)

CoefficientList[Series[(1 + 45 x + 210 x^2 + 210 x^3 + 45 x^4 + x^5)/(1 - x)^6, {x, 0, 33}], x] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1},{1, 51, 501, 2471, 8361, 22363}, 25] (* Vincenzo Librandi, Dec 16 2015 *)

A142994_list, m = [], [512, -768, 352, -48, 2, 1]
for _ in range(10**2):
    A142994_list.append(m[-1])
    for i in range(5):
        m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

a(n) = (2*n + 1)*(32*n^4 + 64*n^3 + 88*n^2 + 56*n + 15)/15.
a(n) = Sum_{k = 0..5} binomial(10, 2*k)*binomial(n+k, 5).
a(n) = Sum_{k = 0..5} binomial(10, 2*k+1)*binomial(n+k+1/2, 5).
O.g.f.: (1 + 45*x + 210*x^2 + 210*x^3 + 45*x^4 + x^5)/(1 - x)^6 = 1/(1 - x) * T(5, (1 + x)/(1 - x)), where T(n, x) denotes the Chebyshev polynomial of the first kind.
Sum_{n >= 1} 1/(n*a(n-1)*a(n)) = 2*log(2) - 41/30.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6), for n > 5. - Vincenzo Librandi, Dec 16 2015
From Peter Bala, Mar 11 2024: (Start)
Sum_{k = 1..n+1} 1/(k*a(k)*a(k-1)) = 1/(51 - 3/(59 - 60/(75 - 315/(99 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*5^2))))).
E.g.f.: exp(x)*(1 + 50*x + 400*x^2/2! + 1120*x^3/3! + 1280*x^4/4! + 512*x^5/5!).
Note that -T(10, i*sqrt(x)) = 1 + 50*x + 400*x^2 + 1120*x^3 + 1280*x^4 + 512*x^5. See A008310. (End)

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

A381555 Triangle read by rows T(n,k) is the number of diamond coverings for a specific number of diamonds covering an even length row of triangles.

Original entry on oeis.org

1, 4, 1, 8, 4, 1, 13, 16, 4, 1, 19, 41, 24, 4, 1, 26, 85, 85, 32, 4, 1, 34, 155, 231, 145, 40, 4, 1, 43, 259, 532, 489, 221, 48, 4, 1, 53, 406, 1092, 1365, 891, 313, 56, 4, 1, 64, 606, 2058, 3333, 2926, 1469, 421, 64, 4, 1, 76, 870, 3630, 7359, 8294, 5551, 2255, 545, 72, 4

Offset: 0

Craig Knecht, Feb 27 2025

The total number of ways the diamond can cover a single row of length(n) triangles is the Fibonacci series. This total can be subdivided into categories based on the number of covering diamonds. The number of categories increase with the length of the row providing the structure of the triangle (see illustrations in the link below).
The above process provides a way to subdivide the individual Fibonacci numbers.
Comparing the diamond covering of a row of triangles shown here to the diamond corona of a hexagon A380346 or a diamond A380416 may be instructive.
A381552 provides additional graphics that help explain the diamond covering.

			Triangle begins:
  1, 4;
  1, 8, 4;
  1, 13, 16, 4;
  1, 19, 41, 24, 4;
  1, 26, 85, 85, 32, 4;
  1, 34, 155, 231, 145, 40, 4;

Craig Knecht, Diamond covering of a even length row of triangles.
Craig Knecht, Geometric bisection of the Fibonacci sequence.
Craig Knecht, Subdividing the individual Fibonacci numbers.
Walter Trump, Diamond covering producing the Fibonacci sequence.

Cf. A063496, A081219, A102083, A323847, A380346, A380416, A381552.

A125530 Number of n X n arrays with entries in 1..3 in which adjacent entries differ by 1 or less (adjacent means in x or y directions).

Original entry on oeis.org

1, 3, 35, 2021, 536453, 661407975, 3780952160847, 100240477650720585, 12324646884990466825513, 7027467861520236770627093435, 18582989792598188051914777397933755

Offset: 0

R. H. Hardin, Dec 31 2006

nonn

[Empirical] a(base+1,n,diff)=a(base,n,diff)+F(n,diff) for base>=2.diff.(n-1) and F(n,diff)= (n=2, A063496(diff+1)) (n=3, A068744(diff+1)) (n=4, A068745(diff+1)) (n=5, A068746(diff+1)) (n=6, A068747(diff+1))

cp's OEIS Frontend

A063495 a(n) = (2*n-1)*(5*n^2-5*n+2)/2.

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Magma

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PARI

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A049480 a(n) = (2*n-1)*(n^2 -n +6)/6.

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Magma

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PARI

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A132366 Partial sum of centered tetrahedral numbers A005894.

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Extensions

A253011 T(n,k)=Number of nXk nonnegative integer arrays with upper left 0 and lower right its king-move distance away minus 3 and every value increasing by 0 or 1 with every step right, diagonally se or down.

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A083669 Number of ordered quintuples (a,b,c,d,e), -n <= a,b,c,d,e <= n, such that a+b+c+d+e = 0.

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Magma

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A142993 Crystal ball sequence for the lattice C_4.

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Maple

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A142994 Crystal ball sequence for the lattice C_5.

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Magma

Maple

Mathematica

A063495 a(n) = (2n-1)(5n^2-5n+2)/2.

A049480 a(n) = (2n-1)(n^2 -n +6)/6.