A049480 -id:A049480 - 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

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.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Formula

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

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

R

Formula

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

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Python

Formula

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

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Python

Formula

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

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Formula

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.