A063488 -id:A063488 - cp's OEIS frontend

A063494 a(n) = (2n - 1)(7n^2 - 7n + 3)/3.

1, 17, 75, 203, 429, 781, 1287, 1975, 2873, 4009, 5411, 7107, 9125, 11493, 14239, 17391, 20977, 25025, 29563, 34619, 40221, 46397, 53175, 60583, 68649, 77401, 86867, 97075, 108053, 119829, 132431, 145887, 160225, 175473, 191659, 208811, 226957, 246125, 266343, 287639

Offset: 1

N. J. A. Sloane, Aug 01 2001

Interpret A176271 as an infinite square array read by antidiagonals, with rows 1,5,11,19,...; 3,9,17,27,... and so on. The sum of the terms in the n X n upper submatrix are s(n) = 1, 18, 93, 296, ... = n^2*(7*n^2-1)/6, and a(n) = s(n) - s(n-1) are the first differences. - J. M. Bergot, Jun 27 2013

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)*(7*n^2 - 7*n + 3)/3: n in [1..30]]; // G. C. Greubel, Dec 01 2017

Table[(2*n - 1)*(7*n^2 - 7*n + 3)/3, {n,1,30}] (* or *) LinearRecurrence[{4,-6,4,-1}, {1,17,75,203}, 30] (* G. C. Greubel, Dec 01 2017 *)

a(n) = { (2*n - 1)*(7*n^2 - 7*n + 3)/3 } \\ Harry J. Smith, Aug 23 2009

my(x='x+O('x^30)); Vec(serlaplace((-3+6*x+21*x^2+14*x^3)*exp(x)/3 + 1)) \\ G. C. Greubel, Dec 01 2017

G.f.: x*(1+x)*(1+12*x+x^2)/(1-x)^4. - Colin Barker, Mar 02 2012
E.g.f.: (-3 + 6*x + 21*x^2 + 14*x^3)*exp(x)/3 + 1. - G. C. Greubel, Dec 01 2017
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Wesley Ivan Hurt, May 11 2023

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

Original entry on oeis.org

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

A005918 Number of points on surface of square pyramid: 3*n^2 + 2 (n>0).

Original entry on oeis.org

1, 5, 14, 29, 50, 77, 110, 149, 194, 245, 302, 365, 434, 509, 590, 677, 770, 869, 974, 1085, 1202, 1325, 1454, 1589, 1730, 1877, 2030, 2189, 2354, 2525, 2702, 2885, 3074, 3269, 3470, 3677, 3890, 4109, 4334, 4565, 4802, 5045, 5294, 5549, 5810, 6077, 6350, 6629

Offset: 0

N. J. A. Sloane

Also coordination sequence of the 5-connected (or bnn) net = hexagonal net X integers.
Also (except for initial term) numbers of the form 3n^2+2 that are not squares. All numbers 3n^2+2 are == 2 (mod 3), and hence not squares. - Cino Hilliard, Mar 01 2003, modified by Franklin T. Adams-Watters, Jun 27 2014
If a 2-set Y and a 3-set Z are disjoint subsets of an n-set X then a(n-4) is the number of 4-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 08 2007
Sums of three consecutive squares: (n - 2)^2 + (n - 1)^2 + n^2 for n > 1. - Keith Tyler, Aug 10 2010

			G.f. = 1 + 5*x + 14*x^2 + 29*x^3 + 50*x^4 + 77*x^5 + 110*x^6 + 149*x^7 + ...

H. S. M. Coxeter, Polyhedral numbers, in R. S. Cohen et al., editors, For Dirk Struik. Reidel, Dordrecht, 1974, pp. 25-35.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
A. F. Wells, Three-Dimensional Nets and Polyhedra, Fig. 15.1 (e).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Branko Grünbaum, Uniform tilings of 3-space, Geombinatorics, 4 (1994), 49-56. See tiling #26.
Milan Janjic, Two Enumerative Functions.
Milan Janjic and Boris Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550, 2013. - From _N. J. A. Sloane_, Feb 13 2013
Milan Janjic and Boris Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014), Article 14.3.5.
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.
Reticular Chemistry Structure Resource (RCSR), The bnn tiling (or net).
B. K. Teo and N. J. A. Sloane, Magic numbers in polygonal and polyhedral clusters, Inorgan. Chem. 24 (1985),4545-4558.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A120328, A206399.

Partial sums give A063488.

A005918:=-(z+1)*(z**2+z+1)/(z-1)**3; # Simon Plouffe in his 1992 dissertation.

Join[{1}, Table[Plus@@(Range[n, n + 2]^2), {n, 0, 49}]] (* Alonso del Arte, Oct 27 2012 *)
CoefficientList[Series[(1 - x^2) (1 - x^3)/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Aug 07 2014 *)
LinearRecurrence[{3,-3,1},{1,5,14,29},50] (* Harvey P. Dale, Dec 12 2015 *)

sq3nsqp2(n) = { for(x=1,n, y = 3*x*x+2; print1(y, ", ") ) }

{a(n) = 3*n^2 + 2 - (n==0)}; /* Michael Somos, Aug 07 2014 */

G.f.: (1 - x^2)*(1 - x^3)/(1 - x)^5 = (1+x)*(1+x+x^2)/(1-x)^3.
Euler transform of length 3 sequence [ 5, -1, -1]. - Michael Somos, Aug 07 2014
a(-n) = a(n) for all n in Z. - Michael Somos, Aug 07 2014
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) for n>3. - Colin Barker, Aug 07 2014
a(0) = 1; for n > 0, a(n) = A120328(n-1). - Doug Bell, Aug 18 2015
E.g.f.: (2+3*x+3*x^2)*exp(x)-1. - Robert Israel, Aug 18 2015
a(n) = A005448(n) + A005448(n+1), sum of 2 consecutive centered triangular numbers. - R. J. Mathar, Apr 28 2020
a(n) = (n - 1)^2 + n^2 + (n + 1)^2. - Charlie Marion, Aug 31 2021
From Amiram Eldar, Sep 14 2022: (Start)
Sum_{n>=0} 1/a(n) = coth(sqrt(2/3)*Pi)*Pi/(2*sqrt(6)) + 3/4.
Sum_{n>=0} (-1)^n/a(n) = cosech(sqrt(2/3)*Pi)*Pi/(2*sqrt(6)) + 3/4. (End)

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

A117384 Positive integers, each occurring twice in the sequence, such that a(n) = a(k) when n+k = 4*a(n), starting with a(1)=1 and filling the next vacant position with the smallest unused number.

Original entry on oeis.org

1, 2, 1, 3, 4, 2, 5, 3, 6, 7, 4, 8, 5, 9, 6, 10, 11, 7, 12, 8, 13, 9, 14, 10, 15, 16, 11, 17, 12, 18, 13, 19, 14, 20, 15, 21, 22, 16, 23, 17, 24, 18, 25, 19, 26, 20, 27, 21, 28, 29, 22, 30, 23, 31, 24, 32, 25, 33, 26, 34, 27, 35, 28, 36, 37, 29, 38, 30, 39, 31, 40, 32, 41, 33, 42

Offset: 1

Paul D. Hanna, Mar 11 2006

Positions where n occurs are A001614(n) and 4*n-A001614(n), where A001614 is the Connell sequence: 1 odd, 2 even, 3 odd, ...
From Paolo Xausa, Aug 27 2021: (Start)
Terms can be arranged in an irregular triangle T(r,c) read by rows in which row r is a permutation P of the integers in the interval [s, s+rlen-1], where s = 1+(r-1)*(r-2)/2, rlen = 2*r-1 and r >= 1 (see example).
P is the alternating (first term > second term < third term > fourth term ...) permutation m -> 1, 1 -> 2, m+1 -> 3, 2 -> 4, m+2 -> 5, 3 -> 6, ..., rlen -> rlen, where m = ceiling(rlen/2).
The triangle has the following properties.
Row lengths are the positive odd numbers.
First column is A000124.
Terms in column c (where c >= 1) are of the form k*(k+1)/2+ceiling(c/2), for integers k >= floor((c-1)/2), each even column being equal to the column preceding it.
Row records (the positive terms of A000217) are in the right border.
Indices of row records are the positive terms of A000290.
Each row r contains r terms that are duplicated in the next row.
In each row, the sum of terms which are not already listed in the sequence gives the positive terms of A006003.
Row sums give A063488.
For rows r >= 2, row product is A057003(r)*A057003(r-1). (End)

			9 first appears at position: A001614(9) = 14;
9 next appears at position: 4*9 - A001614(9) = 22.
From _Paolo Xausa_, Aug 27 2021: (Start)
Written as an irregular triangle T(r,c) the sequence begins:
  r\c  1   2   3   4   5   6   7   8   9  10  11  12  13
  1:   1;
  2:   2,  1,  3;
  3:   4,  2,  5,  3,  6;
  4:   7,  4,  8,  5,  9,  6, 10;
  5:  11,  7, 12,  8, 13,  9, 14, 10, 15;
  6:  16, 11, 17, 12, 18, 13, 19, 14, 20, 15, 21;
  7:  22, 16, 23, 17, 24, 18, 25, 19, 26, 20, 27, 21, 28;
  ...
The triangle can be arranged as shown below so that, in every row, each odd position term is equal to the term immediately below it.
               1
            2  1  3
         4  2  5  3  6
      7  4  8  5  9  6 10
  11  7 12  8 13  9 14 10 15
             ...
(End)

Cf. A117385 (a(5*a(n)-n)=a(n)), A117386 (a(6*a(n)-n)=a(n)).
Cf. A001614 (Connell sequence).
Cf. A000124, A000217, A006003, A057003, A063488, A344482.

nterms=64;a=ConstantArray[0,nterms];For[n=1;t=1,n<=nterms,n++,If[a[[n]]==0,a[[n]]=t;If[(d=4t-n)<=nterms,a[[d]]=a[[n]]];t++]]; a (* Paolo Xausa, Aug 27 2021 *)
(* Second program, triangle rows *)
nrows = 8;Table[rlen=2r-1;Permute[Range[s=1+(r-1)(r-2)/2,s+rlen-1],Join[Range[2,rlen,2],Range[1,rlen,2]]],{r,nrows}] (* Paolo Xausa, Aug 27 2021 *)

{a(n)=local(A=vector(n),m=1); for(k=1,n,if(A[k]==0,A[k]=m;if(4*m-k<=#A,A[4*m-k]=m);m+=1));A[n]}

T(r,c) = my(k = r-1-((c+1) % 2)); k*(k+1)/2+ceil(c/2);
tabf(nn) = {for (r=1, nn, for(c = 1, 2*r-1, print1(T(r,c), ", ");); print;);} \\ Michel Marcus, Sep 09 2021

a(4*a(n)-n) = a(n).
Lim_{n->infinity} a(n)/n = 1/2.
Lim_{n->infinity} (a(n+1)-a(n))/sqrt(n) = 1.
a( A001614(n) ) = n; a( 4n - A001614(n) ) = n.
T(r,c) = k*(k+1)/2+ceiling(c/2), where k = r-1-((c+1) mod 2), r >= 1 and c >= 1. - Paolo Xausa, Sep 09 2021

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

cp's OEIS Frontend

A063494 a(n) = (2*n - 1)*(7*n^2 - 7*n + 3)/3.

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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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A005918 Number of points on surface of square pyramid: 3*n^2 + 2 (n>0).

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A063494 a(n) = (2n - 1)(7n^2 - 7n + 3)/3.

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.