A063490 -id:A063490 - cp's OEIS frontend

A063491 a(n) = (2n - 1)(3n^2 - 3n + 2)/2.

1, 12, 50, 133, 279, 506, 832, 1275, 1853, 2584, 3486, 4577, 5875, 7398, 9164, 11191, 13497, 16100, 19018, 22269, 25871, 29842, 34200, 38963, 44149, 49776, 55862, 62425, 69483, 77054, 85156, 93807, 103025, 112828, 123234, 134261, 145927, 158250, 171248, 184939

Offset: 1

N. J. A. Sloane, Aug 01 2001

A triangle has sides of lengths 6*n-3, 6*n^2-6*n+4, and 6*n^2-6*n+7; for n>2 its area is 6*sqrt(a(n)^2 - 1). - J. M. Bergot, Aug 30 2013

[The source of this is using (n,n+1), (n+1,n+2), and (n+2,n+3) as (a,b) in the creation of three Pythagorean triangles with sides b^2-a^2, 2*a*b, and a^2+b^2. Combine the three respective sides to create a new larger triangle, then find its area. It is not simply working backwards from the sequence. As well, the sequence has this as its first comment to show that the numbers are actually doing something to find a solution.]

T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.

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.

Cf. A005448, A004188.

[(2*n-1)*(3*n^2 -3*n +2)/2: n in [1..30]]; // G. C. Greubel, Dec 01 2017

LinearRecurrence[{4,-6,4,-1},{1,12,50,133},40] (* Harvey P. Dale, Jun 05 2016 *)
Table[(2*n-1)*(3*n^2 -3*n +2)/2, {n,1,30}] (* G. C. Greubel, Dec 01 2017 *)

a(n) = { (2*n - 1)*(3*n^2 - 3*n + 2)/2 } \\ Harry J. Smith, Aug 23 2009

my(x='x+O('x^30)); Vec(serlaplace((-2 + 4*x + 9*x^2 + 6*x^3)*exp(x)/2 + 1)) \\ G. C. Greubel, Dec 01 2017

a <- c(0, 1, 9, 38, 110)
for(n in (length(a)+1):40)
  a[n] <- +4*a[n-1]-6*a[n-2]+4*a[n-3]-a[n-4]
a [Yosu Yurramendi, Sep 04 2013]

G.f.: x*(1+x)*(1+7*x+x^2)/(1-x)^4. - Colin Barker, Apr 20 2012
a(n) = +4*a(n-1) -6*a(n-2) +4*a(n-3) -1*a(n-4) n > 3, a(1)=1, a(2)=12, a(3)=50, a(4)=133. - Yosu Yurramendi, Sep 04 2013
E.g.f.: (-2 + 4*x + 9*x^2 + 6*x^3)*exp(x)/2 + 1. - G. C. Greubel, Dec 01 2017
From Bruce J. Nicholson, Jun 17 2020: (Start)
a(n) = A005448(n) * A005408(n-1).
a(n) = A004188(n) + A004188(n-1). (End)

A063492 a(n) = (2n - 1)(11n^2 - 11n + 6)/6.

Original entry on oeis.org

1, 14, 60, 161, 339, 616, 1014, 1555, 2261, 3154, 4256, 5589, 7175, 9036, 11194, 13671, 16489, 19670, 23236, 27209, 31611, 36464, 41790, 47611, 53949, 60826, 68264, 76285, 84911, 94164, 104066, 114639, 125905, 137886, 150604, 164081, 178339, 193400, 209286, 226019

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. Reports, 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)*(11*n^2-11*n+6)/6: n in [1..40]]; // Vincenzo Librandi, Dec 16 2015

A063492:=n->(2*n - 1)*(11*n^2 - 11*n + 6)/6: seq(A063492(n), n=1..50); # Wesley Ivan Hurt, Dec 16 2015

Table[(2*n-1)*(11*n^2-11*n+6)/6, {n, 5!}] (* Vladimir Joseph Stephan Orlovsky, Sep 18 2008 *)
LinearRecurrence[{4, -6, 4, -1}, {1, 14, 60, 161}, 40] (* Vincenzo Librandi, Dec 16 2015 *)

a(n) = { (2*n - 1)*(11*n^2 - 11*n + 6)/6 } \\ Harry J. Smith, Aug 23 2009

Vec(x*(1+x)*(1+9*x+x^2)/(1-x)^4 + O(x^100)) \\ Altug Alkan, Dec 16 2015

A063492_list, m = [], [22, -11, 2, 1]
for _ in range(10**2):
    A063492_list.append(m[-1])
    for i in range(3):
        m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

G.f.: x*(1+x)*(1 + 9*x + x^2)/(1-x)^4. - Colin Barker, Apr 24 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>4. - Wesley Ivan Hurt, Dec 16 2015
E.g.f.: (-6 + 12*x + 33*x^2 + 22*x^3)*exp(x)/6 + 1. - G. C. Greubel, Dec 01 2017

A063493 a(n) = (2n-1)(13n^2-13n+6)/6.

Original entry on oeis.org

1, 16, 70, 189, 399, 726, 1196, 1835, 2669, 3724, 5026, 6601, 8475, 10674, 13224, 16151, 19481, 23240, 27454, 32149, 37351, 43086, 49380, 56259, 63749, 71876, 80666, 90145, 100339, 111274, 122976, 135471, 148785, 162944, 177974, 193901, 210751, 228550, 247324, 267099

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. Reports, 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)*(13*n^2-13*n+6)/6: n in [1..40]]; // Vincenzo Librandi, Dec 16 2015

Table[(2 n - 1) (13 n^2 - 13 n + 6)/6, {n, 1, 40}] (* Bruno Berselli, Dec 16 2015 *)
LinearRecurrence[{4,-6,4,-1}, {1,16,70,189}, 30] (* G. C. Greubel, Dec 01 2017 *)

a(n) = { (2*n - 1)*(13*n^2 - 13*n + 6)/6 } \\ Harry J. Smith, Aug 23 2009

my(x='x+O('x^30)); Vec(serlaplace((-6+12*x+39*x^2+26*x^3)*exp(x)/6 + 1)) \\ G. C. Greubel, Dec 01 2017

A063493_list, m = [], [26, -13, 2, 1]
for _ in range(10**2):
    A063493_list.append(m[-1])
    for i in range(3):
        m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

G.f.: x*(1+x)*(1+11*x+x^2)/(1-x)^4. - Colin Barker, Apr 20 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. - Vincenzo Librandi, Dec 16 2015
E.g.f.: (-6 + 12*x + 39*x^2 + 26*x^3)*exp(x)/6 + 1. - G. C. Greubel, Dec 01 2017

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

Original entry on oeis.org

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

A185505 a(n) = (7n^4 + 5n^2)/12.

Original entry on oeis.org

1, 11, 51, 156, 375, 771, 1421, 2416, 3861, 5875, 8591, 12156, 16731, 22491, 29625, 38336, 48841, 61371, 76171, 93500, 113631, 136851, 163461, 193776, 228125, 266851, 310311, 358876, 412931, 472875, 539121, 612096, 692241, 780011, 875875, 980316, 1093831, 1216931, 1350141, 1494000, 1649061, 1815891, 1995071

Offset: 1

Clark Kimberling, Jan 29 2011

a(n) is the sum of terms in the square [1,n]x[1,n] of the natural number array A000027; e.g., the [1,3]x[1,3] square is
1..2..4
3..5..8
6..9..13,
so that a(1) = 1, a(2) = 1+2+3+5 = 11, a(3) = 1+2+3+4+5+6+8+9+13 = 51.
Partial sums of A063490. - Omar E. Pol, Oct 23 2019

			a(1)=(7+5)/12, a(2)=(7*16+5*4)/12.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000027, A063490.

Table[(7*n^4+5*n^2)/12, {n,1,60}]
LinearRecurrence[{5,-10,10,-5,1},{1,11,51,156,375},50] (* Harvey P. Dale, Jan 26 2024 *)

a(n)=(7*n^4+5*n^2)/12 \\ Charles R Greathouse IV, Sep 05 2016

a(n) = (7*n^4 + 5*n^2)/12.
From Chai Wah Wu, Sep 05 2016: (Start)
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 5.
G.f.: x*(1 + x)*(1 + 5*x + x^2)/(1 - x)^5. (End)
E.g.f.: (1/12)*x*(12 + 54*x + 42*x^2 + 7*x^3)*exp(x). - G. C. Greubel, Jul 07 2017

A145904 Square array read by antidiagonals: Hilbert transform of the Narayana numbers A001263.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 5, 1, 1, 10, 16, 7, 1, 1, 15, 40, 31, 9, 1, 1, 21, 85, 105, 51, 11, 1, 1, 28, 161, 295, 219, 76, 13, 1, 1, 36, 280, 721, 771, 396, 106, 15, 1, 1, 45, 456, 1582, 2331, 1681, 650, 141, 17, 1

Offset: 0

Peter Bala, Oct 31 2008

Refer to A145905 for the definition of the Hilbert transform of a lower triangular array. For the Hilbert transform of A008459, the array of type B Narayana numbers, see A108625.

This seems to be a duplicate of A273350. - Alois P. Heinz, Jun 04 2016. This could probably be proved by showing that the g.f.s are the same. - N. J. A. Sloane, Jul 02 2016

			The array begins
n\k|..0.....1.....2.....3.....4.....5
=====================================
0..|..1.....1.....1.....1.....1.....1
1..|..1.....3.....5.....7.....9....11
2..|..1.....6....16....31....51....76
3..|..1....10....40...105...219...396
4..|..1....15....85...295...771..1681
5..|..1....21...161...721..2331..6083
...
Row 2: (1 + 3x + x^2)/(1 - x)^3 = 1 + 6x + 16x^2 + 31x^3 + ... .
Row 3: (1 + 6x + 6x^2 + x^3)/(1 - x)^4 = 1 + 10x + 40x^2 + 105x^3 + ... .

Cf. A001263, A005891 (row 2), A063490 (row 3), A108625 (Hilbert transform of h-vectors of type B associahedra).

Cf. also A273350.

Table[1/(# + 1)*Sum[Binomial[# + 1, i - 1] Binomial[# + 1, i] Binomial[# + k - i + 1, k + 1 - i], {i, 0, k + 1}] &[m - k], {m, 0, 9}, {k, 0, m}] // Flatten (* Michael De Vlieger, Jan 15 2018 *)

taylor(((y-1)*sqrt(((x^2+2*x+1)*y-x^2+2*x-1)/(y-1))+(-x-1)*y-x+1)/(2*x^2*y),x,0,10,y,0,10);
T(n,m,k):=1/(n+1)*sum(binomial(n+1,i-1)*binomial(n+1,i)*binomial(n+m-i+1,m+1-i),i,0,m+1); /* Vladimir Kruchinin, Jan 15 2018 */

Row n generating function: 1/(n+1) * 1/(1-x) * Jacobi_P(n,1,1,(1+x)/(1-x)) = N_n(x)/(1-x)^n where N_n(x) denotes the shifted Narayana polynomial N_n(x) = sum{k = 1..n} A001263(k)*x^(k-1) of degree n-1.
Conjectural column n generating function: N_n(x^2)/(1-x)^(2n+1).
The entries in row n are given by the values of a polynomial function p_n(x) at x = 0,1,2,... . The first few are p_1(x) = 2x + 1, p_2(x) = (5x^2 + 5x + 2)/2, p_3(x) = (2x + 1)*(7x^2 + 7x + 6)/6 and p_4(x) = (7x^4 + 14x^3 + 21x^2 + 14x + 4)/4. These polynomials appear to have their zeros on the line Re x = -1/2; that is, the polynomials p_n(-x) appear to satisfy a Riemann hypothesis. The corresponding result for A108625 is true (see A142995 for details).
Contribution from Paul Barry, Jan 06 2009: (Start)
The g.f. for the corresponding number triangle is:
1/(1-x-xy-x^2y/(1-x-x^2y/(1-x-xy-x^2y/(1-x-x^2y/(1-x-xy-x^2y.... (a continued fraction). (End)
This g.f. satisfies x^2*y*g^2 - (1-x-x*y)*g + 1 = 0. - R. J. Mathar, Jun 16 2016
G.f.: ((y-1)*sqrt(((x^2+2*x+1)*y-x^2+2*x-1)/(y-1))+(-x-1)*y-x+1)/(2*x^2*y). - Vladimir Kruchinin, Jan 15 2018
T(n,m) = 1/(n+1)*Sum_{i=0..m+1} C(n+1,i-1)*C(n+1,i)*C(n+m-i+1,m+1-i). - Vladimir Kruchinin, Jan 15 2018

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

A355010 Array read by ascending antidiagonals: T(n, k) is the number of n-core partitions with k corners.

Original entry on oeis.org

1, 3, 1, 6, 5, 1, 10, 16, 7, 1, 15, 40, 31, 9, 1, 21, 85, 105, 51, 11, 1, 28, 161, 295, 219, 76, 13, 1, 36, 280, 721, 771, 396, 106, 15, 1, 45, 456, 1582, 2331, 1681, 650, 141, 17, 1, 55, 705, 3186, 6244, 6083, 3235, 995, 181, 19, 1, 66, 1045, 5985, 15156, 19348, 13663, 5685, 1445, 226, 21, 1

Offset: 2

Stefano Spezia, Jun 15 2022

T(n, k) is also equal to the number of cornerless Motzkin paths of length 2*k + n - 1 with n - 1 flat steps (see Theorem 3.3 and Proposition 3.4 at pp. 13 - 14 in Cho et al.).

In proposition 3.4 in Cho et al., the Narayana number is defined as N(k, i) = binomial(k, i)*binomial(k, i-1)/k, unlike A001263.

			The array begins:
    1,  1,   1,   1,    1,    1,    1,    1, ...
    3,  5,   7,   9,   11,   13,   15,   17, ...
    6, 16,  31,  51,   76,  106,  141,  181, ...
   10, 40, 105, 219,  396,  650,  995, 1445, ...
   15, 85, 295, 771, 1681, 3235, 5685, 9325, ...
   ...

Hyunsoo Cho, JiSun Huh, Hayan Nam, and Jaebum Sohn, Combinatorics on bounded free Motzkin paths and its applications, arXiv:2205.15554 [math.CO], 2022.

Cf. A000012 (n = 2), A001263, A005408 (n = 3), A005891 (n = 4), A006007, A063490 (n = 5), A160747 (n = 6), A161680 (k = 1), A355011.

T[n_,k_]:=Sum[Binomial[k,i]Binomial[k,i-1]Binomial[n+2(k-i),2k]/k,{i,Min[k,Floor[n/2]]}]; Flatten[Table[T[n-k+1,k],{n,2,12},{k,1,n-1}]]

T(n, k) = Sum_{i=1..min(k,floor(n/2))} N(k, i)*binomial(n+2*(k-i), 2*k), where N(k, i) = binomial(k, i)*binomial(k, i-1)/k. (See proposition 3.4 in Cho et al.).

T(n, 2) = A006007(n-1).

cp's OEIS Frontend

A063491 a(n) = (2*n - 1)*(3*n^2 - 3*n + 2)/2.

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A063492 a(n) = (2*n - 1)*(11*n^2 - 11*n + 6)/6.

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A063493 a(n) = (2*n-1)*(13*n^2-13*n+6)/6.

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

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

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A132366 Partial sum of centered tetrahedral numbers A005894.

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A185505 a(n) = (7*n^4 + 5*n^2)/12.

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A063491 a(n) = (2n - 1)(3n^2 - 3n + 2)/2.

A063492 a(n) = (2n - 1)(11n^2 - 11n + 6)/6.

A063493 a(n) = (2n-1)(13n^2-13n+6)/6.

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

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

A185505 a(n) = (7n^4 + 5n^2)/12.