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 11-18 of 18 results.

A081441 a(n) = (2*n^3 - n^2 - n + 2)/2.

Original entry on oeis.org

1, 1, 6, 22, 55, 111, 196, 316, 477, 685, 946, 1266, 1651, 2107, 2640, 3256, 3961, 4761, 5662, 6670, 7791, 9031, 10396, 11892, 13525, 15301, 17226, 19306, 21547, 23955, 26536, 29296, 32241, 35377, 38710, 42246, 45991, 49951, 54132, 58540, 63181
Offset: 0

Views

Author

Paul Barry, Mar 21 2003

Keywords

Comments

Diagonal in array of n-gonal numbers A081422.

Links

Crossrefs

Cf. A005564, A081422, A081436, A081437, A081438.

Programs

  • GAP
    List([0..40], n-> (2*n^3-n^2-n+2)/2); # G. C. Greubel, Aug 14 2019
  • Magma
    [(2*n^3-n^2-n+2)/2: n in [0..50]]; // Vincenzo Librandi, Aug 08 2013
  • Maple
    a:= n-> (2*n^3-n^2-n+2)/2: seq(a(n), n=0..50); # Zerinvary Lajos, Sep 13 2006
  • Mathematica
    Table[(2n^3-n^2-n+2)/2,{n,0,40}] (* Harvey P. Dale, May 29 2012 *)
CoefficientList[Series[(1 - 4 x + 11 x^2 - 8 x^3) / (1 - x)^5, {x, 0, 50}],x] (* Vincenzo Librandi, Aug 08 2013 *)
  • PARI
    vector(40, n, n--; (2*n^3-n^2-n+2)/2) \\ G. C. Greubel, Aug 14 2019
  • Sage
    [(2*n^3-n^2-n+2)/2 for n in (0..40)] # G. C. Greubel, Aug 14 2019

Formula

G.f.: (1 -4*x +11*x^2 -8*x^3)/(1-x)^5.
a(n) = (n + 1)*(2*n^2 - 3*n + 2)/2 = (n-1)*A005564(n+1) - n*A005564(n), where A005564(0..2) = 0, -1, 0. - Bruno Berselli, May 19 2015
E.g.f.: (2 + 5*x^2 + 2*x^3)*exp(x)/2. - G. C. Greubel, Aug 14 2019

A153258 n^3 - (n+2)^2.

Original entry on oeis.org

-4, -8, -8, 2, 28, 76, 152, 262, 412, 608, 856, 1162, 1532, 1972, 2488, 3086, 3772, 4552, 5432, 6418, 7516, 8732, 10072, 11542, 13148, 14896, 16792, 18842, 21052, 23428, 25976, 28702, 31612, 34712, 38008, 41506, 45212, 49132, 53272, 57638
Offset: 1

Views

Author

Vladimir Joseph Stephan Orlovsky, Dec 22 2008

Keywords

Links

Crossrefs

Cf. A045991, A081437, A069778, A153257

Programs

  • Mathematica
    a[n_] := n^3-(n+2)^2; lst={}; Do[AppendTo[lst, a[n]], {n, 0, 5!}]; lst

Formula

G.f.: 2*x*(x^3+4*x-2)/(x-1)^4. [Colin Barker, Oct 08 2012]

A153259 a(n)=n^3-(3*(n+3))^2.

Original entry on oeis.org

-81, -143, -217, -297, -377, -451, -513, -557, -577, -567, -521, -433, -297, -107, 143, 459, 847, 1313, 1863, 2503, 3239, 4077, 5023, 6083, 7263, 8569, 10007, 11583, 13303, 15173, 17199, 19387, 21743, 24273, 26983, 29879, 32967, 36253, 39743
Offset: 1

Views

Author

Vladimir Joseph Stephan Orlovsky, Dec 22 2008

Keywords

Links

Crossrefs

Cf. A045991, A081437, A069778, A153257, A153258.

Programs

  • Mathematica
    a[n_]:=n^3-(3*(n+3))^2;lst={};Do[AppendTo[lst,a[n]],{n,0,5!}];lst
Table[n^3-(3(n+3))^2,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{-81,-143,-217,-297},40] (* Harvey P. Dale, Jul 10 2013 *)
  • PARI
    a(n)=n^3-(3*n+9)^2 \\ Charles R Greathouse IV, Oct 18 2022

Formula

a(1)=-81, a(2)=-143, a(3)=-217, a(4)=-297, a(n)=4*a(n-1)-6*a(n-2)+ 4*a(n-3)- a(n-4). - Harvey P. Dale, Jul 10 2013

A214731 a(n) = n^3 - 2*n^2 - 1.

Original entry on oeis.org

-2, -1, 8, 31, 74, 143, 244, 383, 566, 799, 1088, 1439, 1858, 2351, 2924, 3583, 4334, 5183, 6136, 7199, 8378, 9679, 11108, 12671, 14374, 16223, 18224, 20383, 22706, 25199, 27868, 30719, 33758, 36991, 40424, 44063, 47914, 51983, 56276, 60799, 65558, 70559
Offset: 1

Views

Author

Marco Piazzalunga, Jul 27 2012

Keywords

Links

Crossrefs

Cf. A080859, A085490, A144390 (first differences), A152619.
Similar sequences: A152015 (of the type m^3+2m^2-1), A081437 (m^3-2m^2+1).

Programs

Formula

From Bruno Berselli, Jul 27 2012: (Start)
G.f.: -x*(2-7*x-x^3)/(1-x)^4.
a(n) = A085490(n-1) + 2.
a(n) = A152619(n-2) - 1 for n>1.
a(n) - a(n-2) = A080859(n-2) - 1 for n>2. (End)
E.g.f.: 1 - (1-x)*(1+x)^2*exp(x). - G. C. Greubel, Dec 31 2023

Extensions

a(3) corrected by Charles R Greathouse IV, Jul 27 2012

A348462 Size of largest bipartite biregular Moore graph of diameter 6 and degrees n and 2.

Original entry on oeis.org

12, 35, 78, 147, 248
Offset: 2

Views

Author

N. J. A. Sloane, Oct 31 2021

Keywords

Comments

a(7) >= 387, a(8) = 570, a(9) = 803, a(10) = 1092.

Links

Crossrefs

Cf. A027620, A081437, A348461, A348463.

Formula

Empirical observation: For the terms a(2)-a(6) and a(8)-a(10) a(n) = A081437(n-1) + 2. It is unknown whether this is also valid for n = 7 and for n > 10. - Hugo Pfoertner, Oct 31 2021
a(n) <= A027620(n-2) + 3 = A081437(n-1) + 2 (the Moore bound). - Pontus von Brömssen, Oct 31 2021

A153260 a(n) = n^3 - 3*(n+3)^2.

Original entry on oeis.org

-27, -47, -67, -81, -83, -67, -27, 43, 149, 297, 493, 743, 1053, 1429, 1877, 2403, 3013, 3713, 4509, 5407, 6413, 7533, 8773, 10139, 11637, 13273, 15053, 16983, 19069, 21317, 23733, 26323, 29093, 32049, 35197, 38543, 42093, 45853, 49829, 54027
Offset: 0

Views

Author

Vladimir Joseph Stephan Orlovsky, Dec 22 2008

Keywords

Links

Crossrefs

Cf. A045991, A081437, A069778, A153257, A153258, A153259.

Programs

  • Magma
    [n^3-3*(n+3)^2: n in [0..40] ]; // Vincenzo Librandi, Aug 25 2011
  • Mathematica
    a[n_]:=n^3-3*(n+3)^2; a/@ Range[0, 50]
Table[n^3-3(n+3)^2,{n,0,50}] (* or *) LinearRecurrence[{4,-6,4,-1},{-27,-47,-67,-81},51] (* Harvey P. Dale, Aug 24 2011 *)
  • PARI
    vector(40, n, n--; n^3-3*(n+3)^2) \\ G. C. Greubel, Nov 10 2018

Formula

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(0)=-27, a(1)=-47, a(2)=-67, a(3)=-81. - Harvey P. Dale, Aug 24 2011
G.f.: (x*(x*(13*x - 41) + 61) - 27)/(x-1)^4. - Harvey P. Dale, Aug 24 2011
E.g.f.: (-27 - 20*x + x^3)*exp(x). - G. C. Greubel, Nov 10 2018

Extensions

Offset changed from 1 to 0 by Vincenzo Librandi, Aug 25 2011

A342240 Table read by upward antidiagonals: T(n,k) is the number of strings of length k over an n-letter alphabet that have a bifix; n, k >= 1.

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 0, 3, 4, 1, 0, 4, 9, 10, 1, 0, 5, 16, 33, 20, 1, 0, 6, 25, 76, 99, 44, 1, 0, 7, 36, 145, 304, 315, 88, 1, 0, 8, 49, 246, 725, 1264, 945, 182, 1, 0, 9, 64, 385, 1476, 3725, 5056, 2883, 364, 1, 0, 10, 81, 568, 2695, 9036, 18625, 20404, 8649, 740, 1
Offset: 1

Views

Author

Peter Kagey, Mar 07 2021

Keywords

Comments

A bifix is a nonempty substring that is both a prefix and a suffix.

Examples

			Table begins:
n\k | 1 2  3   4    5     6      7       8        9
----+----------------------------------------------
  1 | 0 1  1   1    1     1      1       1        1
  2 | 0 2  4  10   20    44     88     182      364
  3 | 0 3  9  33   99   315    945    2883     8649
  4 | 0 4 16  76  304  1264   5056   20404    81616
  5 | 0 5 25 145  725  3725  18625   93605   468025
  6 | 0 6 36 246 1476  9036  54216  326346  1958076
  7 | 0 7 49 385 2695 19159 134113  940807  6585649
  8 | 0 8 64 568 4544 36800 294400 2358728 18869824
For n = 2, k = 4, the A(2,4) = 10 length-4 strings over a 2-letter alphabet with a bifix are:
0000 with prefix and suffix 0,
0010 with prefix and suffix 0,
0100 with prefix and suffix 0,
0101 with prefix and suffix 01,
0110 with prefix and suffix 0,
1001 with prefix and suffix 1,
1010 with prefix and suffix 10,
1011 with prefix and suffix 1,
1101 with prefix and suffix 1, and
1111 with prefix and suffix 1.

Links

Crossrefs

Cf. A342239.
Rows: A094536 (n=2), A094538 (n=3), A094559 (n=4).
Columns: A000290 (k=3), A081437 (k=4).

Programs

  • Python
    from itertools import product
def has_bifix(s): return any(s[:i] == s[-i:] for i in range(1, len(s)//2+1))
def T(n, k): return sum(has_bifix(s) for s in product(range(n), repeat=k))
def atodiag(maxd): # maxd antidiagonals
  return [T(n, d-n+1) for d in range(1, maxd+1) for n in range(d, 0, -1)]
print(atodiag(11)) # Michael S. Branicky, Mar 07 2021

Formula

T(n,k) = n^k - A342239(n,k).

A348482 Triangle read by rows: T(n,k) = (Sum_{i=k..n} i!)/(k!) for 0 <= k <= n.

Original entry on oeis.org

1, 2, 1, 4, 3, 1, 10, 9, 4, 1, 34, 33, 16, 5, 1, 154, 153, 76, 25, 6, 1, 874, 873, 436, 145, 36, 7, 1, 5914, 5913, 2956, 985, 246, 49, 8, 1, 46234, 46233, 23116, 7705, 1926, 385, 64, 9, 1, 409114, 409113, 204556, 68185, 17046, 3409, 568, 81, 10, 1
Offset: 0

Views

Author

Werner Schulte, Oct 20 2021

Keywords

Comments

The matrix inverse M = T^(-1) has terms M(n,n) = 1 for n >= 0, M(n,n-1) = -(n+1) for n > 0, and M(n,n-2) = n for n > 1, otherwise 0.

Examples

			The triangle T(n,k) for 0 <= k <= n starts:
n\k :       0       1       2      3      4     5    6   7   8  9
=================================================================
  0 :       1
  1 :       2       1
  2 :       4       3       1
  3 :      10       9       4      1
  4 :      34      33      16      5      1
  5 :     154     153      76     25      6     1
  6 :     874     873     436    145     36     7    1
  7 :    5914    5913    2956    985    246    49    8   1
  8 :   46234   46233   23116   7705   1926   385   64   9   1
  9 :  409114  409113  204556  68185  17046  3409  568  81  10  1
  etc.

Links

Crossrefs

Cf. A109398, A094587, A002104 (row sums), A173184 (alt. row sums), A000012 (main diagonal), A000027(1st subdiagonal), A000290 (2nd subdiagonal), A081437 (3rd subdiagonal), A192398 (4th subdiagonal), A003422 (column 0), A007489 (column 1), A345889 (column 2), A143122.

Programs

  • Mathematica
    T[n_, k_] := Sum[i!, {i, k, n}]/k!; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, Oct 20 2021 *)

Formula

T(n,n) = 1 and T(2*n,n) = A109398(n) for n >= 0; T(n,n-1) = n+1 for n > 0; T(n,n-2) = n^2 for n > 1.
T(n,k) - T(n-1,k) = (n!) / (k!) = A094587(n,k) for 0 <= k < n.
T(n,k) = (k+2) * (T(n,k+1) - T(n,k+2)) for 0 <= k < n-1.
T(n,k) = (T(n,k-1) - 1) / k for 0 < k <= n.
T(n,k) * T(n-1,k-1) - T(n-1,k) * T(n,k-1) = (n!) / (k!) for 0 < k < n.
T(n,1) = T(n,0)-1 = Sum_{k=0..n-1} T(n,k)/(k+2) for n > 0 (conjectured).
Sum_{k=0..n} binomial(k+r,k) * (1-k) * T(n+r,k+r) = binomial(n+r+1,n) for n >= 0 and r >= 0.
Sum_{k=0..n} (-1)^k * (k+1) * T(n,k) = (1 + (-1)^n) / 2 for n >= 0.
Sum_{k=0..n} (-1)^k * (k!) * T(n,k) = Sum_{k=0..n} (k!) * (1+(-1)^k) / 2 for n >= 0.
The row polynomials p(n,x) = Sum_{k=0..n} T(n,k) * x^k for n >= 0 satisfy the following equations:
(a) p(n,x) - p'(n,x) = (x^(n+1)-1) / (x-1) for n >= 0, where p' is the first derivative of p;
(b) p(n,x) - (n+1) * p(n-1,x) + n * p(n-2,x) = x^n for n > 1.
(c) p(n,x) = (x+1) * p(n-1,x) + 1 + Sum_{i=1..n-1} (d/dx)^i p(n-1,x) for n > 0 (conjectured).
Row sums p(n,1) equal A002104(n+1) for n >= 0.
Alternating row sums p(n,-1) equal A173184(n) for n >= 0 (conjectured).
The three conjectures stated above are true. See links. - Sela Fried, Jul 11 2024.
From Peter Luschny, Jul 11 2024: (Start)
T(n, k) = (t(k) - t(n + 1)) / k!, where t(n) = (-1)^(n + 1) * Gamma(n + 1) * Subfactorial(-(n + 1)).
T(n, k) = A143122(n, k) / k!. (End)
Previous Showing 11-18 of 18 results.

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