cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 11-19 of 19 results.

A319577 a(n) = (4/45)*n*(n - 2)*(n - 1)*(n^3 - 12*n^2 + 47*n - 15).

Original entry on oeis.org

0, 0, 0, 24, 96, 240, 544, 1288, 3136, 7392, 16320, 33528, 64416, 116688, 200928, 331240, 525952, 808384, 1207680, 1759704, 2508000, 3504816, 4812192, 6503112, 8662720, 11389600, 14797120, 19014840, 24189984, 30488976, 38099040, 47229864, 58115328, 71015296
Offset: 0

Views

Author

Peter Luschny, Oct 01 2018

Keywords

Crossrefs

Cf. A000012 (m=0), A005843 (m=1), A046092 (m=2), A130809 (m=3), A319575 (m=4), A319576 (m=5), this sequence (m=6).
Column n=6 of A122141.
Cf. A319574.

Programs

  • Maple
    a := n -> (4/45)*n*(n - 2)*(n - 1)*(n^3 - 12*n^2 + 47*n - 15):
    seq(a(n), n=0..41);
  • Mathematica
    A319577[n_]:=4/45*n*(n-2)*(n-1)*(n^3-12*n^2+47*n-15); Array[A319577, 50, 0] (*or*)
    LinearRecurrence[{7, -21, 35, -35, 21, -7, 1}, {0, 0, 0, 24, 96, 240, 544}, 50] (* Paolo Xausa, Feb 20 2024 *)
  • PARI
    concat([0,0,0], Vec(8*x^3*(3 - 9*x + 9*x^2 + 5*x^3) / (1 - x)^7 + O(x^40))) \\ Colin Barker, Oct 02 2018

Formula

a(n) = [x^6] JacobiTheta3(x)^n.
a(n) = A319574(n,6).
From Colin Barker, Oct 02 2018: (Start)
G.f.: 8*x^3*(3 - 9*x + 9*x^2 + 5*x^3) / (1 - x)^7.
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) for n>6.
(End)

A130811 If X_1,...,X_n is a partition of a 2n-set X into 2-blocks then a(n) is equal to the number of 5-subsets of X containing none of X_i, (i=1,...n).

Original entry on oeis.org

32, 192, 672, 1792, 4032, 8064, 14784, 25344, 41184, 64064, 96096, 139776, 198016, 274176, 372096, 496128, 651168, 842688, 1076768, 1360128, 1700160, 2104960, 2583360, 3144960, 3800160, 4560192, 5437152, 6444032, 7594752, 8904192
Offset: 5

Views

Author

Milan Janjic, Jul 16 2007

Keywords

Comments

Number of n permutations (n>=5) of 3 objects u,v,z, with repetition allowed, containing n-5 u's. Example: if n=5 then n-5 =(0) zero u, a(1)=32. - Zerinvary Lajos, Aug 05 2008
a(n) is the number of 4-dimensional elements in an n-cross polytope where n>=5. - Patrick J. McNab, Jul 06 2015

Crossrefs

Programs

  • Magma
    [Binomial(n,n-5)*2^5: n in [5..40]]; // Vincenzo Librandi, Jul 09 2015
  • Maple
    a:=n->binomial(2*n,5)+(2*n-4)*binomial(n,2)-n*binomial(2*n-2,3)
    seq(binomial(n,n-5)*2^5,n=5..34); # Zerinvary Lajos, Dec 07 2007
    seq(binomial(n+4, 5)*2^5, n=1..22); # Zerinvary Lajos, Aug 05 2008
  • Mathematica
    Table[Binomial[2 n, 5] + (2 n - 4) Binomial[n, 2] - n Binomial[2 n - 2, 3], {n, 5, 40}] (* Vincenzo Librandi, Jul 09 2015 *)

Formula

a(n) = binomial(2*n,5) + (2*n-4)*binomial(n,2) - n*binomial(2*n-2,3).
a(n) = C(n,n-5)*2^5, for n>=5. - Zerinvary Lajos, Dec 07 2007
G.f.: 32*x^5/(1-x)^6. - Colin Barker, Apr 14 2012

A359202 Number of (bidimensional) faces of regular m-polytopes for m >= 3.

Original entry on oeis.org

4, 6, 8, 10, 12, 20, 24, 32, 35, 56, 80, 84, 96, 120, 160, 165, 220, 240, 280, 286, 364, 448, 455, 560, 672, 680, 720, 816, 960, 969, 1140, 1200, 1320, 1330, 1540, 1760, 1771, 1792, 2024, 2288, 2300, 2600, 2912, 2925, 3276, 3640, 3654, 4060, 4480, 4495, 4608
Offset: 1

Views

Author

Marco Ripà, Dec 20 2022

Keywords

Comments

In 3 dimensions there are five (convex) regular polytopes and they have 4, 6, 8, 12, or 20 (bidimensional) faces (A053016).
In 4 dimensions there are six regular 4-polytopes and they have 10, 24, 32, 96, 720, or 1200 faces (A063925).
In m >= 5 dimensions, there are only 3 regular polytopes (i.e., the m-simplex, the m-cube, and the m-crosspolytope) so that we can sort their number of bidimensional faces in ascending order and define the present sequence.

Examples

			6 is a term since a cube has 6 faces.
		

Crossrefs

Cf. A359201 (edges), A359662 (cells).

Formula

{a(n), n >= 1} = {{12, 96, 720, 1200} U {A000292} U {A001788} U {A130809}} \ {0, 1}.

A360604 Triangle read by rows. T(n, k) = 2^binomial(n - k, 2) * binomial(n - 1, k - 1).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 8, 6, 3, 1, 0, 64, 32, 12, 4, 1, 0, 1024, 320, 80, 20, 5, 1, 0, 32768, 6144, 960, 160, 30, 6, 1, 0, 2097152, 229376, 21504, 2240, 280, 42, 7, 1, 0, 268435456, 16777216, 917504, 57344, 4480, 448, 56, 8, 1
Offset: 0

Views

Author

Peter Luschny, Feb 23 2023

Keywords

Examples

			Triangle T(n, k) starts:
[0] 1;
[1] 0,       1;
[2] 0,       1,      1;
[3] 0,       2,      2,     1;
[4] 0,       8,      6,     3,    1;
[5] 0,      64,     32,    12,    4,   1;
[6] 0,    1024,    320,    80,   20,   5,  1;
[7] 0,   32768,   6144,   960,  160,  30,  6, 1;
[8] 0, 2097152, 229376, 21504, 2240, 280, 42, 7, 1;
		

Crossrefs

Cf. A006125 (column 1), A002378 (T(n+2,n)), A130809 (T(n+3,n)), A006896 (row sums).

Programs

  • Maple
    T := (n, k) -> 2^binomial(n - k, 2) * binomial(n-1, k-1):
    for n from 0 to 9 do seq(T(n, k), k = 0..n) od;

A371532 Centered cuboctahedral numbers: the number of integer triples (x,y,z) such that max(|x|,|y|,|z|) <= n and |x|+|y|+|z| <= 2n.

Original entry on oeis.org

1, 19, 93, 263, 569, 1051, 1749, 2703, 3953, 5539, 7501, 9879, 12713, 16043, 19909, 24351, 29409, 35123, 41533, 48679, 56601, 65339, 74933, 85423, 96849, 109251, 122669, 137143, 152713, 169419, 187301, 206399, 226753, 248403, 271389, 295751, 321529, 348763
Offset: 0

Views

Author

Peter Kagey, Mar 26 2024

Keywords

Examples

			The a(1) = 19 lattice points are all permutations of the points (0,0,0), (0,0,1), and (0,1,1), where any number of the coordinates can also be made negative (e.g., (1,-1,0)).
		

Crossrefs

Programs

  • Mathematica
    Array[(20*#^3 + 24*#^2 + 10*# + 3)/3 &, 50, 0] (* or *)
    LinearRecurrence[{4, -6, 4, -1}, {1, 19, 93, 263}, 50] (* Paolo Xausa, Apr 02 2024 *)
  • Python
    def A371532(n): return n*(n*(5*n+6<<2)+10)//3+1 # Chai Wah Wu, Apr 02 2024

Formula

a(n) = (20*n^3 + 24*n^2 + 10*n + 3)/3.
a(n) = A016755(n) - A130809(n-2).
G.f.: (x^3 + 23*x^2 + 15*x + 1)/(x-1)^4. - Paolo Xausa, Apr 02 2024
From Elmo R. Oliveira, Aug 22 2025: (Start)
E.g.f.: exp(x)*(3 + 54*x + 84*x^2 + 20*x^3)/3.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). (End)

A130813 If X_1,...,X_n is a partition of a 2n-set X into 2-blocks then a(n) is equal to the number of 7-subsets of X containing none of X_i, (i=1,...n).

Original entry on oeis.org

128, 1024, 4608, 15360, 42240, 101376, 219648, 439296, 823680, 1464320, 2489344, 4073472, 6449664, 9922560, 14883840, 21829632, 31380096, 44301312, 61529600, 84198400, 113667840, 151557120, 199779840, 260582400, 336585600, 430829568
Offset: 7

Views

Author

Milan Janjic, Jul 16 2007

Keywords

Comments

Number of n permutations (n>=7) of 3 objects u,v,z, with repetition allowed, containing n-7 u's. Example: if n=7 then n-7 =(0) zero u, a(1)=128. - Zerinvary Lajos, Aug 05 2008
a(n) is the number of 6-dimensional elements in an n-cross polytope where n>=7. - Patrick J. McNab, Jul 06 2015

Crossrefs

Programs

  • Magma
    [Binomial(n,n-7)*2^7: n in [7..40]]; // Vincenzo Librandi, Jul 09 2015
  • Maple
    a:=n->binomial(2*n,7)+binomial(n,2)*binomial(2*n-4,3)-n*binomial(2*n-2,5)-(2*n-6)*binomial(n,3);
    seq(binomial(n,n-7)*2^7,n=7..32); # Zerinvary Lajos, Dec 07 2007
    seq(binomial(n+6, 7)*2^7, n=1..22); # Zerinvary Lajos, Aug 05 2008
  • Mathematica
    Table[Binomial[n, n - 7] 2^7, {n, 7, 40}] (* Vincenzo Librandi, Jul 09 2015 *)

Formula

a(n) = binomial(2*n,7) + binomial(n,2)*binomial(2*n-4,3) - n*binomial(2*n-2,5) - (2*n-6)*binomial(n,3).
a(n) = C(n,n-7)*2^7, n>=7. - Zerinvary Lajos, Dec 07 2007
G.f.: 128*x^7/(1-x)^8. - Colin Barker, Mar 18 2012
a(n) = 128*A000580(n). a(n+1) = 2*(n+1)*a(n)/(n-6) for n >= 7. - Robert Israel, Jul 08 2015

A185342 Triangle of successive recurrences in columns of A117317(n).

Original entry on oeis.org

2, 4, -4, 6, -12, 8, 8, -24, 32, -16, 10, -40, 80, -80, 32, 12, -60, 160, -240, 192, -64, 14, -84, 280, -560, 672, -448, 128, 16, -112, 448, -1120, 1792, -1792, 1024, -256, 18, -144, 672, -2016, 4032, -5376, 4608, -2304, 512, 20, -180, 960, -3360, 8064
Offset: 0

Views

Author

Paul Curtz, Jan 26 2012

Keywords

Comments

A117317 (A):
1
2 1
4 5 1
8 16 9 1
16 44 41 14 1
32 112 146 85 20 1
64 272 456 377 155 27 1
have for their columns successive signatures
(2) (4,-4) (6,-12,8) (8,-24, 32, -16) (10,-40,80,-80,32) i.e. a(n).
Take based on abs(A133156) (B):
1
2 0
4 1 0
8 4 0 0
16 12 1 0 0
32 32 6 0 0 0
64 80 24 1 0 0 0.
The recurrences of successive columns are also a(n). a(n) columns: A005843(n+1), A046092(n+1), A130809, A130810, A130811, A130812, A130813.

Examples

			Triangle T(n,k),for 1<=k<=n, begins :
2                                         (1)
4    -4                                   (2)
6   -12   8                               (3)
8   -24  32   -16                         (4)
10  -40  80   -80   32                    (5)
12  -60 160  -240  192   -64              (6)
14  -84 280  -560  672  -448  128         (7)
16 -112 448 -1120 1792 -1792 1024 -256    (8)
Successive rows can be divided by A171977.
		

Crossrefs

Cf. For (A): A053220, A056243. For (B): A000079, A001787, A001788, A001789. For A193862: A115068 (a Coxeter group). For (2): A014480 (also (3),(4),(5),..); also A053220 and A001787.
Cf. A007318.

Programs

  • Mathematica
    Table[(-1)*Binomial[n, k]*(-2)^k, {n, 1, 20}, {k, 1, n}] // Flatten (* G. C. Greubel, Jun 27 2017 *)
  • PARI
    for(n=1,20, for(k=1,n, print1((-2)^(k+1)*binomial(n,k)/2, ", "))) \\ G. C. Greubel, Jun 27 2017

Formula

Take A133156(n) without 1's or -1's double triangle (C)=
2
4
8 -4
16 -12
32 -32 6
64 -80 24
128 -192 80 -8
256 -448 240 -40
512 -1024 672 -160 10;
a(n) is increasing odd diagonals and increasing (sign changed) even diagonals. Rows sum of (C) = A201629 (?) Another link between Chebyshev polynomials and cos( ).
Absolute values: A013609(n) without 1's. Also 2*A193862 = (2*A002260)*A135278.
T(n,k) = T(n-1,k) - 2*T(n-1,k-1) for k>1, T(n,1) = 2*n = 2*T(n-1,1) - T(n-2,1). - Philippe Deléham, Feb 11 2012
T(n,k) = (-1)* Binomial(n,k)*(-2)^k, 1<=k<=n. - Philippe Deléham, Feb 11 2012

A206022 Riordan array (1, x*exp(arcsinh(-2*x))).

Original entry on oeis.org

1, 0, 1, 0, -2, 1, 0, 2, -4, 1, 0, 0, 8, -6, 1, 0, -2, -8, 18, -8, 1, 0, 0, 0, -32, 32, -10, 1, 0, 4, 8, 30, -80, 50, -12, 1, 0, 0, 0, 0, 128, -160, 72, -14, 1, 0, -10, -16, -28, -112, 350, -280, 98, -16, 1, 0, 0, 0
Offset: 0

Views

Author

Philippe Deléham, Feb 02 2012

Keywords

Comments

Riordan array (1, x*(sqrt(1+4x^2)-2x)); inverse of Riordan array (1, x/sqrt(1-4x)), see A205813.
The g.f. for row sums (1,1,-1,-1,3,1,-9,1,27,13,-81,67,243,...) is (1+2*x^2+x*sqrt(1+4*x^2))/(1+3*x^2).
Triangle T(n,k), read by rows, given by (0, -2, 1, -1, 1, -1, 1, -1, 1, -1, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Examples

			Triangle begins:
  1
  0,   1
  0,  -2,   1
  0,   2,  -4,   1
  0,   0,   8,  -6,    1,
  0,  -2,  -8,  18,   -8,    1
  0,   0,   0, -32,   32,  -10,     1
  0,   4,   8,  30,  -80,   50,   -12,    1
  0,   0,   0,   0,  128, -160,    72,  -14,    1
  0, -10, -16, -28, -112,  350,  -280,   98,  -16,   1
  0,   0,   0,   0,    0, -512,   768, -448,  128, -18,   1
  0,  28,  40,  54,   96,  420, -1512, 1470, -672, 162, -20, 1
		

Crossrefs

Cf. A104624 (column k=1).

Formula

T(n,n) = 1, T(n+1,n) = -2n = -A005843(n), T(n+2,n) = 2*n^2 = A001105(n), T(n+3,n) = -A130809(n+1), T(2n,n) = A009117(n), T(2n+3,1) = (-1)^n*2*A000108(n).
T(n,k) = T(n-2,k-2) - 4*T(n-2,k-1), for k >= 2.

A276985 Triangle read by rows: T(n,k) = number of k-dimensional elements in an n-dimensional cross-polytope, n>=1, 0<=k

Original entry on oeis.org

2, 4, 4, 6, 12, 8, 8, 24, 32, 16, 10, 40, 80, 80, 32, 12, 60, 160, 240, 192, 64, 14, 84, 280, 560, 672, 448, 128, 16, 112, 448, 1120, 1792, 1792, 1024, 256, 18, 144, 672, 2016, 4032, 5376, 4608, 2304, 512, 20, 180, 960, 3360, 8064, 13440, 15360, 11520, 5120
Offset: 1

Views

Author

Felix Fröhlich, Sep 24 2016

Keywords

Comments

It appears that this is 2*A193862 (but with a different offset) and that the sum of terms of the n-th row is A024023(n) = 3^n - 1. - Michel Marcus, Sep 29 2016

Examples

			T(4, 1..4) = 8, 24, 32, 16, because the 16-cell has 8 0-faces (vertices), 24 1-faces (edges), 32 2-faces (faces) and 16 3-faces (cells).
Triangle starts
2
4, 4
6, 12, 8
8, 24, 32, 16
10, 40, 80, 80, 32
12, 60, 160, 240, 192, 64
14, 84, 280, 560, 672, 448, 128
16, 112, 448, 1120, 1792, 1792, 1024, 256
18, 144, 672, 2016, 4032, 5376, 4608, 2304, 512
20, 180, 960, 3360, 8064, 13440, 15360, 11520, 5120, 1024
		

References

  • H. S. M. Coxeter, Regular Polytopes, Third Edition, Dover Publications, 1973, ISBN 9780486141589.

Crossrefs

Cf. A038207 (hypercube), A135278 (simplex).
Rows: A005843(n), A046092(n), A130809(n+2), A130810(n+3).
Columns: A000079(n), A001787(n), A001788(n), A001789(n+3).

Programs

  • Mathematica
    Table[2^(k + 1) Binomial[n, k + 1], {n, 10}, {k, 0, n - 1}] // Flatten (* Michael De Vlieger, Sep 25 2016 *)
  • PARI
    T(n, k) = 2^(k+1)*binomial(n, k+1)
    trianglerows(n) = for(x=1, n, for(y=0, x-1, print1(T(x, y), ", ")); print(""))
    trianglerows(10) \\ print initial 10 rows of triangle

Formula

T(n,k) = 2^(k+1) * binomial(n, k+1) (cf. Coxeter, 1973, formula 7.22).
T(n,k) = A182059(n,k) = A013609(n,k) . - R. J. Mathar, May 03 2017
G.f.: 2*x/((1 - x)*(1 - x - 2*x*y)). - Stefano Spezia, Jul 17 2025
Previous Showing 11-19 of 19 results.