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 21-29 of 29 results.

A299807 Rectangular array read by antidiagonals: T(n,k) is the number of distinct sums of k complex n-th roots of 1, n >= 1, k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 9, 10, 5, 1, 1, 6, 15, 16, 15, 6, 1, 1, 7, 19, 35, 25, 21, 7, 1, 1, 8, 28, 37, 70, 36, 28, 8, 1, 1, 9, 33, 84, 61, 126, 49, 36, 9, 1, 1, 10, 45, 96, 210, 91, 210, 64, 45, 10, 1, 1, 11, 51, 163, 225, 462, 127, 330, 81, 55, 11, 1, 1, 12, 66, 180, 477, 456, 924, 169, 495, 100, 66
Offset: 1

Views

Author

Max Alekseyev, Feb 24 2018

Keywords

Examples

			Array starts:
  n=1:  1,  1,  1,   1,   1,    1,    1,    1,     1,     1,     1, ...
  n=2:  1,  2,  3,   4,   5,    6,    7,    8,     9,    10,    11, ...
  n=3:  1,  3,  6,  10,  15,   21,   28,   36,    45,    55,    66, ...
  n=4:  1,  4,  9,  16,  25,   36,   49,   64,    81,   100,   121, ...
  n=5:  1,  5, 15,  35,  70,  126,  210,  330,   495,   715,  1001, ...
  n=6:  1,  6, 19,  37,  61,   91,  127,  169,   217,   271,   331, ...
  n=7:  1,  7, 28,  84, 210,  462,  924, 1716,  3003,  5005,  8008, ...
  n=8:  1,  8, 33,  96, 225,  456,  833, 1408,  2241,  3400,  4961, ...
  n=9:  1,  9, 45, 163, 477, 1197, 2674, 5454, 10341, 18469, 31383, ...
  n=10: 1, 10, 51, 180, 501, 1131, 2221, 3951,  6531, 10201, 15231, ...
  ...
		

Crossrefs

Rows: A000012 (n=1), A000027 (n=2), A000217 (n=3), A000290 (n=4), A000332 (n=5), A354343 (n=6), A000579 (n=7), A014820 (n=8).
Columns: A000012 (k=0), A000027 (k=1), A031940 (k=2).
Diagonal: A299754 (n=k).

Formula

From Chai Wah Wu, May 28 2018: (Start)
The following are all conjectures.
For m >= 0, the 2^(m+1)-th row are the figurate numbers based on the 2^m-dimensional regular convex polytope with g.f.: (1+x)^(2^m-1)/(1-x)^(2^m+1).
For p prime, the n=p row corresponds to binomial(k+p-1,p-1) for k = 0,1,2,3,..., which is the maximum possible (i.e., the number of combinations with repetitions of k choices from p categories) with g.f.: 1/(1-x)^p.
(End)

Extensions

Row 6 corrected by Max Alekseyev, Aug 14 2022

A061926 Square table by antidiagonals where odd rows are partial sums of previous row, even rows are sums of pairs of values in previous row and initial row is 0 and 1 alternating.

Original entry on oeis.org

0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 2, 2, 1, 0, 1, 2, 3, 3, 1, 0, 0, 3, 4, 6, 4, 1, 0, 1, 3, 5, 10, 9, 5, 1, 0, 0, 4, 6, 15, 16, 14, 6, 1, 0, 1, 4, 7, 21, 25, 30, 19, 7, 1, 0, 0, 5, 8, 28, 36, 55, 44, 26, 8, 1, 0, 1, 5, 9, 36, 49, 91, 85, 70, 33, 9, 1, 0, 0, 6, 10, 45, 64, 140, 146, 155, 96, 42, 10
Offset: 0

Views

Author

Henry Bottomley, May 17 2001

Keywords

Examples

			From _Sean A. Irvine_, Mar 14 2023: (Start)
Table begins:
  0 1 0  1  0   1   0   1
  0 1 1  2  2   3   3   4
  0 1 2  3  4   5   6   7
  0 1 3  6 10  15  21  28
  0 1 4  9 16  25  36  49
  0 1 5 14 30  55  91 140
  0 1 6 19 44  85 146 231
  0 1 7 26 70 155 301 532
(End)
		

Crossrefs

Formula

T(0, 2*k) = 0, T(0, 2*k+1) = 1, T(n, 0) = 0, T(2*n, k) = T(2*n-1, k-1) + T(2*n-1, k), T(2*n+1, k) = T(2*n+1, k-1) + T(2*n, k). - Sean A. Irvine, Mar 14 2023

A070212 Number of 5 X 5 pandiagonal magic squares with sum n.

Original entry on oeis.org

1, 10, 55, 220, 715, 2001, 4995, 11385, 24090, 47905, 90376, 162955, 282490, 473110, 768570, 1215126, 1875015, 2830620, 4189405, 6089710, 8707501, 12264175, 17035525, 23361975, 31660200, 42436251, 56300310, 73983205, 96354820, 124444540, 159463876, 202831420, 256200285, 321488190
Offset: 0

Views

Author

Sharon Sela (sharonsela(AT)hotmail.com), May 07 2002

Keywords

Comments

In contrast to other definitions, a magic square may contain here any nonnegative integers, not necessarily distinct. For example, the 10 solutions for n = 1 are the 10 permutation matrices of size 5 X 5 which are pandiagonal in the sense that any of the 10 (principal or broken) diagonals has exactly one 1 and four 0's. - M. F. Hasler, Oct 23 2018

Crossrefs

Programs

  • GAP
    a:=[1, 10, 55, 220, 715, 2001, 4995, 11385, 24090];;  for n in [10..36] do a[n]:=9*a[n-1]-36*a[n-2]+84*a[n-3]-126*a[n-4]+126*a[n-5]-84*a[n-6]+36*a[n-7]-9*a[n-8]+a[n-9]; od; a; # Muniru A Asiru, Oct 23 2018
  • Maple
    seq(coeff(series(-(x^4+x^3+x^2+x+1)/(x-1)^9,x,n+1), x, n), n = 0 .. 35); # Muniru A Asiru, Oct 23 2018
  • Mathematica
    LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{1,10,55,220,715,2001,4995,11385,24090},40] (* Harvey P. Dale, Mar 13 2018 *)
  • PARI
    apply( A070212(n)=1/8064*(n+4)*(n+3)*(n+2)*(n+1)*(n^2+5*n+8)*(n^2+5*n+42), [0..20]) \\ Edited by M. F. Hasler, Oct 23 2018
    

Formula

a(n) = (1/8064) * (n+4)*(n+3)*(n+2)*(n+1)*(n^2+5n+8)*(n^2+5n+42).
G.f.: -(x^4+x^3+x^2+x+1) / (x-1)^9. [Colin Barker, Dec 10 2012]

Extensions

More terms from Benoit Cloitre, May 12 2002
More terms from M. F. Hasler, Oct 23 2018

A269237 a(n) = (n + 1)^2*(5*n^2 + 10*n + 2)/2.

Original entry on oeis.org

1, 34, 189, 616, 1525, 3186, 5929, 10144, 16281, 24850, 36421, 51624, 71149, 95746, 126225, 163456, 208369, 261954, 325261, 399400, 485541, 584914, 698809, 828576, 975625, 1141426, 1327509, 1535464, 1766941, 2023650, 2307361, 2619904, 2963169, 3339106, 3749725, 4197096
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 09 2016

Keywords

Comments

Partial sums of centered dodecahedral numbers (A005904).

Crossrefs

Programs

  • Magma
    [(n + 1)^2*(5*n^2 + 10*n + 2)/2 : n in [0..50]]; // Wesley Ivan Hurt, Oct 15 2017
  • Maple
    A269237:=n->(n + 1)^2*(5*n^2 + 10*n + 2)/2: seq(A269237(n), n=0..50); # Wesley Ivan Hurt, Oct 15 2017
  • Mathematica
    Table[(n + 1)^2 ((5 n^2 + 10 n + 2)/2), {n, 0, 35}]
    LinearRecurrence[{5, -10, 10, -5, 1}, {1, 34, 189, 616, 1525}, 36]
  • PARI
    x='x+O('x^99); Vec((1+29*x+29*x^2+x^3)/(1-x)^5) \\ Altug Alkan, Apr 10 2016
    

Formula

G.f.: (1 + 29*x + 29*x^2 + x^3)/(1 - x)^5.
E.g.f.: exp(x)*(2 + 66*x + 122*x^2 + 50*x^3 + 5*x^4)/2.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
Sum_{n>=0} 1/a(n) = (5 - Pi^2 - sqrt(15)*Pi*cot(sqrt(3/5)*Pi))/9 = 1.0377796966... . - Vaclav Kotesovec, Apr 10 2016

A300624 Figurate numbers based on the 11-dimensional regular convex polytope called the 11-dimensional cross-polytope, or 11-dimensional hyperoctahedron.

Original entry on oeis.org

0, 1, 22, 243, 1804, 10165, 46530, 180775, 614680, 1871145, 5188590, 13286043, 31760676, 71513949, 152784282, 311603535, 609802800, 1150082385, 2098144710, 3714481475, 6399123260, 10753517061, 17664712562, 28418229623, 44847366984, 69528316025, 106032285086
Offset: 0

Views

Author

Keywords

Comments

The 11-dimensional cross-polytope is represented by the Schlaefli symbol {3, 3, 3, 3, 3, 3, 3, 3, 3, 4}. It is the dual of the 11-dimensional hypercube.

Crossrefs

Similar sequences: A005900 (m=3), A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099193 (m=7), A099195 (m=8), A099196 (m=9), A099197 (m=10).

Programs

  • Magma
    [(n*(14175 + 83754*n^2 + 50270*n^4 + 7392*n^6 + 330*n^8 + 4*n^10)) / 155925 : n in [0..40]]; // Wesley Ivan Hurt, Jul 17 2020
  • PARI
    concat(0, Vec(x*(1 + x)^10 / (1 - x)^12 + O(x^40))) \\ Colin Barker, Aug 15 2018
    
  • PARI
    a(n) = (n*(14175 + 83754*n^2 + 50270*n^4 + 7392*n^6 + 330*n^8 + 4*n^10)) / 155925 \\ Colin Barker, Aug 15 2018
    

Formula

a(n) = 11-crosspolytope(n).
From Colin Barker, Aug 15 2018: (Start)
G.f.: x*(1 + x)^10 / (1 - x)^12.
a(n) = (n*(14175 + 83754*n^2 + 50270*n^4 + 7392*n^6 + 330*n^8 + 4*n^10)) / 155925.
(End)

A354343 Number of distinct sums of n complex 6th power roots of unity.

Original entry on oeis.org

1, 6, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, 469, 547, 631, 721, 817, 919, 1027, 1141, 1261, 1387, 1519, 1657, 1801, 1951, 2107, 2269, 2437, 2611, 2791, 2977, 3169, 3367, 3571, 3781, 3997, 4219, 4447, 4681, 4921, 5167, 5419, 5677, 5941, 6211, 6487, 6769, 7057, 7351, 7651, 7957
Offset: 0

Views

Author

Max Alekseyev, Aug 15 2022

Keywords

Crossrefs

Programs

Formula

For n >= 2, a(n) = 3*n^2 + 3*n + 1 = A003215(n).
For n >= 5, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f. (1 + 3*x + 4*x^2 - 3*x^3 + x^4) / (1 - x)^3.

A128089 Denominators in inverse of triangle A128078 by rows, n * each term in n-th row of A126615.

Original entry on oeis.org

1, 4, 4, 6, 18, 9, 8, 24, 48, 16, 10, 30, 60, 100, 25, 12, 36, 72, 120, 180, 36, 14, 42, 84, 140, 210, 294, 49, 16, 48, 96, 160, 240, 336, 448, 64, 18, 54, 108, 180, 270, 378, 504, 648, 81, 20, 60, 120, 200, 300, 420, 560, 720, 900, 100
Offset: 1

Views

Author

Gary W. Adamson, Feb 14 2007

Keywords

Comments

Row sums = A014820: (1, 8, 33, 96, 225, 456, ...).
Denominators of the inverse of A128078: (1/1; 1/4, 1/4; 1/6, 1/18, 1/9; 1/8, 1/24, 1/48, 1/16; ...).
Row sums of this triangle: 1/1, 1/2, 1/3, ...; e.g., (1/8 + 1/24 + 1/48 + 1/16) = 1/4.

Examples

			First few rows of the triangle:
   1;
   4,  4;
   6, 18,  9;
   8, 24, 48,  16;
  10, 30, 60, 100,  25;
  12, 36, 72, 120, 180,  36;
  14, 42, 84, 140, 210, 294,  49;
  16, 48, 96, 160, 240, 336, 448,  64;
  ...
Row 4 = (8, 24, 48, 16) = 4 * (2, 6, 12, 4); where (2, 6, 12, 4) = row 4 of A126615.
		

Crossrefs

Formula

Denominators in inverse triangular matrix of A128078, where A128078 = A002260 * A128064, = (1; -1, 4; -1, -2, 9; -1, -2, -3, 16; ...).

A267218 a(n) is the a(n-1)-st a(n-2)-dimensional orthoplex number, starting with the terms 1, 2.

Original entry on oeis.org

1, 2, 2, 4, 16, 22016
Offset: 1

Views

Author

Robin Powell, Jan 18 2016

Keywords

Examples

			16 is the 4th 2-orthoplex number = A000290(4).
22016 is the 16th 4-orthoplex number = A014820(16).
The next term will be the 22016th 16-orthoplex number.
		

Crossrefs

A364429 a(0) = 1, a(n) = (2*n^5 + 20*n^3 + 23*n) * 2/15 for n>=1.

Original entry on oeis.org

1, 6, 36, 146, 456, 1182, 2668, 5418, 10128, 17718, 29364, 46530, 71000, 104910, 150780, 211546, 290592, 391782, 519492, 678642, 874728, 1113854, 1402764, 1748874, 2160304, 2645910, 3215316, 3878946, 4648056, 5534766, 6552092, 7713978, 9035328, 10532038, 12221028
Offset: 0

Views

Author

Steven Lu, Jul 24 2023

Keywords

Comments

a(n) is the 6th n-orthoplex (n-dimensional cross-polytope) number.

Examples

			a(3) = 146 since the 6th octahedral number is 146; A005900(6) = 146.
a(4) = 456 since the 6th 16-cell number is 456; A014820(5) = 456.
		

Crossrefs

Cf. A142978 (column 6 with an initial 1).

Programs

  • Mathematica
    Prepend[Table[2/15 (2 x^5 + 20 x^3 + 23 x), {x, 100}], 1]
  • Python
    print([1]+[(2*i**5+20*i**3+23*i)*2//15 for i in range(1,101)])

Formula

a(0) = 1, a(n) = (2*n^5 + 20*n^3 + 23*n) * 2/15 for n>=1.
G.f.: (1 + 15*x^2 + 15*x^4 + x^6)/(1 - x)^6. - Stefano Spezia, Jul 24 2023
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