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.

Showing 1-3 of 3 results.

A059826 a(n) = (n^2 - n + 1)*(n^2 + n + 1).

Original entry on oeis.org

1, 3, 21, 91, 273, 651, 1333, 2451, 4161, 6643, 10101, 14763, 20881, 28731, 38613, 50851, 65793, 83811, 105301, 130683, 160401, 194923, 234741, 280371, 332353, 391251, 457653, 532171, 615441, 708123, 810901, 924483, 1049601, 1187011, 1337493, 1501851, 1680913
Offset: 0

Views

Author

N. J. A. Sloane, Feb 24 2001

Keywords

Comments

Main diagonal of A082039. - Paul Barry, Apr 02 2003
The base of the natural logarithms e = 2*Sum_{n>=0} 1/(a(n)*n!) and zeta(2) = Pi^2/6 = 1 + 2*Sum_{n>=1} (-1)^(n+1)/(a(n)*n^2). - Peter Bala, Jan 20 2008

Links

Crossrefs

Cf. A002061, A062159.
Main diagonal of A082039.

Programs

  • Magma
    [n^4+n^2+1 : n in [0..50]]; // Wesley Ivan Hurt, Jun 09 2014
  • Maple
    with(combinat): seq(fibonacci(3,n)+n^4, n=0..40); # Zerinvary Lajos, May 25 2008
  • Mathematica
    Table[n^4 + n^2 + 1, {n, 0, 50}] (* Wesley Ivan Hurt, Jun 09 2014 *)
  • PARI
    a(n) = { my(f=n^2 + 1); (f - n)*(f + n) } \\ Harry J. Smith, Jun 29 2009

Formula

a(n) = n^4+n^2+1. - Paul Barry, Apr 02 2003
a(n) = (n^2-n+1) * (n^2+n+1) = A002061(n) * A002061(n+1), products of two consecutive central polygonal numbers. a(n) = (n^6-1)/(n^2-1), n>1. a(n) = (n^5-n^4+n^3-n^2+n-1)/(n-1) = A062159(n)/(n-1), n>1. - Alexander Adamchuk, Apr 12 2006
O.g.f.: (-1+2*x-16*x^2-6*x^3-3*x^4) / (x-1)^5. - R. J. Mathar, Feb 26 2008
a(n) = A219069(n,1), for n > 0. - Reinhard Zumkeller, Nov 11 2012
a(n+2) = (n^2+3n+3) * (n^2+5n+7) = (t(n)+t(n+2)) * (t(n+1)+t(n+3)), where t=A000217 are triangular numbers. For n>=1, a(n+2) = t(2*t(n+2)+t(n)) -t(t(n)-1). - J. M. Bergot, Nov 29 2012
4*a(n) = (n^2+n+1)^2+(n^2-n+1)^2+(n^2+n-1)^2+(n^2-n-1)^2. - Bruno Berselli, Jul 03 2014
a(n) = A002061(n^2). - Franklin T. Adams-Watters, Aug 01 2014
Sum_{n>=0} 1/a(n) = 1/2 + sqrt(3)*Pi*tanh(sqrt(3)*Pi/2)/6. - Amiram Eldar, Feb 14 2021

A062158 a(n) = n^3 - n^2 + n - 1 = (n-1) * (n^2 + 1).

Original entry on oeis.org

-1, 0, 5, 20, 51, 104, 185, 300, 455, 656, 909, 1220, 1595, 2040, 2561, 3164, 3855, 4640, 5525, 6516, 7619, 8840, 10185, 11660, 13271, 15024, 16925, 18980, 21195, 23576, 26129, 28860, 31775, 34880, 38181, 41684, 45395, 49320, 53465, 57836, 62439, 67280, 72365, 77700, 83291, 89144, 95265, 101660
Offset: 0

Views

Author

Henry Bottomley, Jun 08 2001

Keywords

Comments

Number of walks of length 4 between any two distinct vertices of the complete graph K_{n+1} (n >= 1). Example: a(2) = 5 because in the complete graph ABC we have the following walks of length 4 between A and B: ABACB, ABCAB, ACACB, ACBAB and ACBCB. - Emeric Deutsch, Apr 01 2004
1/a(n) for n >= 2, is in base n given by 0.repeat(0,0,1,1), due to (1/n^3 + 1/n^4)*(1/(1-1/n^4)) = 1/((n-1)*(n^2+1)). - Wolfdieter Lang, Jun 20 2014
For n>3, a(n) is 1220 in base n-1. - Bruno Berselli, Jan 26 2016
For odd n, a(n) * (n+1) / 2 + 1 also represents the first integer in a sum of n^4 consecutive integers that equals n^8. - Patrick J. McNab, Dec 26 2016

Examples

			a(4) = 4^3 - 4^2 + 4 - 1 = 64 - 16 + 4 - 1 = 51.

Links

Crossrefs

Cf. A002061, A023443, A053698, A060884, A060888, A062159, A062160, A268086.

Programs

  • Magma
    [n^3 - n^2 + n - 1 : n in [0..50]]; // Wesley Ivan Hurt, Dec 26 2016
  • Maple
    [seq(n^3-n^2+n-1,n=0..49)]; # Zerinvary Lajos, Jun 29 2006
a:=n->sum(1+sum(n, k=1..n), k=2..n):seq(a(n), n=0...43); # Zerinvary Lajos, Aug 24 2008
  • Mathematica
    Table[n^3 - n^2 + n - 1, {n, 0, 49}] (* Alonso del Arte, Apr 30 2014 *)
  • PARI
    a(n) = { n*(n*(n - 1) + 1) - 1 } \\ Harry J. Smith, Aug 02 2009

Formula

a(n) = round(n^4/(n+1)) for n >= 2.
a(n) = A062160(n, 4), for n > 2.
G.f.: (4*x-1)*(1+x^2)/(1-x)^4 (for the signed sequence). - Emeric Deutsch, Apr 01 2004
a(n) = floor(n^5/(n^2+n)) for n > 0. - Gary Detlefs, May 27 2010
a(n) = -A053698(-n). - Bruno Berselli, Jan 26 2016
Sum_{n>=2} 1/a(n) = A268086. - Amiram Eldar, Nov 18 2020
E.g.f.: exp(x)*(x^3 + 2*x^2 + x - 1). - Stefano Spezia, Apr 22 2023

Extensions

More terms from Emeric Deutsch, Apr 01 2004

A062160 Square array T(n,k) = (n^k - (-1)^k)/(n+1), n >= 0, k >= 0, read by falling antidiagonals.

Original entry on oeis.org

0, 1, 0, -1, 1, 0, 1, 0, 1, 0, -1, 1, 1, 1, 0, 1, 0, 3, 2, 1, 0, -1, 1, 5, 7, 3, 1, 0, 1, 0, 11, 20, 13, 4, 1, 0, -1, 1, 21, 61, 51, 21, 5, 1, 0, 1, 0, 43, 182, 205, 104, 31, 6, 1, 0, -1, 1, 85, 547, 819, 521, 185, 43, 7, 1, 0, 1, 0, 171, 1640, 3277, 2604, 1111, 300, 57, 8, 1, 0, -1, 1, 341, 4921, 13107, 13021, 6665, 2101, 455, 73, 9, 1, 0
Offset: 0

Views

Author

Henry Bottomley, Jun 08 2001

Keywords

Comments

For n >= 1, T(n, k) equals the number of walks of length k between any two distinct vertices of the complete graph K_(n+1). - Peter Bala, May 30 2024

Examples

			From _Seiichi Manyama_, Apr 12 2019: (Start)
Square array begins:
   0, 1, -1,  1,  -1,    1,    -1,      1, ...
   0, 1,  0,  1,   0,    1,     0,      1, ...
   0, 1,  1,  3,   5,   11,    21,     43, ...
   0, 1,  2,  7,  20,   61,   182,    547, ...
   0, 1,  3, 13,  51,  205,   819,   3277, ...
   0, 1,  4, 21, 104,  521,  2604,  13021, ...
   0, 1,  5, 31, 185, 1111,  6665,  39991, ...
   0, 1,  6, 43, 300, 2101, 14706, 102943, ... (End)

Links

Crossrefs

Rows include A062157, A000035, A001045, A015518, A015521, A015531, A015540, A015552, A015565, A015577, A015585, A015592, A015609.
Columns include A000004, A000012, A023443, A002061, A062158, A060884, A062159, A060888.
Related to repunits in negative bases (cf. A055129 for positive bases).
Main diagonal gives A081216.
Cf. A109502.

Programs

  • Maple
    seq(print(seq((n^k - (-1)^k)/(n+1), k = 0..10)), n = 0..10); # Peter Bala, May 31 2024
  • Mathematica
    T[n_,k_]:=(n^k - (-1)^k)/(n+1); Join[{0},Table[Reverse[Table[T[n-k,k],{k,0,n}]],{n,12}]]//Flatten (* Stefano Spezia, Feb 20 2024 *)

Formula

T(n, k) = n^(k-1) - n^(k-2) + n^(k-3) - ... + (-1)^(k-1) = n^(k-1) - T(n, k-1) = n*T(n, k-1) - (-1)^k = (n - 1)*T(n, k-1) + n*T(n, k-2) = round[n^k/(n+1)] for n > 1.
T(n, k) = (-1)^(k+1) * resultant( n*x + 1, (x^k-1)/(x-1) ). - Max Alekseyev, Sep 28 2021
G.f. of row n: x/((1+x) * (1-n*x)). - Seiichi Manyama, Apr 12 2019
E.g.f. of row n: (exp(n*x) - exp(-x))/(n+1). - Stefano Spezia, Feb 20 2024
From Peter Bala, May 31 2024: (Start)
Binomial transform of the m-th row: Sum_{k = 0..n} binomial(n, k)*T(m, k) = (m + 1)^(n-1) for n >= 1.
Let R(m, x) denote the g.f. of the m-th row of the square array. Then R(m_1, x) o R(m_2, x) = R(m_1 + m_2 + m_1*m_2, x), where o denotes the black diamond product of power series as defined by Dukes and White. Cf. A109502.
T(m_1 + m_2 + m_1*m_2, k) = Sum_{i = 0..k} Sum_{j = i..k} binomial(k, i)* binomial(k-i, j-i)*T(m_1, j)*T(m_2, k-i). (End)
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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