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 1-4 of 4 results.

A106273 Discriminant of the polynomial x^n - x^(n-1) - ... - x - 1.

Original entry on oeis.org

1, 5, -44, -563, 9584, 205937, -5390272, -167398247, 6042477824, 249317139869, -11597205023744, -601139006326619, 34383289858207744, 2151954708695291177, -146323302326154543104, -10742330662077208945103, 846940331265064719417344, 71373256668946058057974997
Offset: 1

Views

Author

T. D. Noe, May 02 2005

Keywords

Comments

This polynomial is the characteristic polynomial of the Fibonacci and Lucas n-step sequences. These discriminants are prime for n=2, 4, 6, 26, 158 (A106274). It appears that the term a(2n+1) always has a factor of 2^(2n). With that factor removed, the discriminants are prime for odd n=3, 5, 7, 21, 99, 405. See A106275 for the combined list.
a(n) is the determinant of an r X r Hankel matrix whose entries are w(i+j) where w(n) = x1^n + x2^n + ... + xr^n where x1,x2,...xr are the roots of the titular characteristic polynomial. E.g., A000032 for n=2, A001644 for n=3, A073817 for n=4, A074048 for n=5, A074584 for n=6, A104621 for n=7, ... - Kai Wang, Jan 17 2021
Luca proves that a(n) is a term of the corresponding k-nacci sequence only for n=2 and 3. - Michel Marcus, Apr 12 2025

Crossrefs

Cf. A086797 (discriminant of the polynomial x^n-x-1), A000045, A000073, A000078, A001591, A001592 (Fibonacci n-step sequences), A000032, A001644, A073817, A074048, A074584, A104621, A105754, A105755 (Lucas n-step sequences), A086937, A106276, A106277, A106278 (number of distinct zeros of these polynomials for n=2, 3, 4, 5).

Programs

  • Mathematica
    Discriminant[p_?PolynomialQ, x_] := With[{n=Exponent[p, x]}, Cancel[((-1)^(n(n-1)/2) Resultant[p, D[p, x], x])/Coefficient[p, x, n]^(2n-1)]]; Table[Discriminant[x^n-Sum[x^i, {i, 0, n-1}], x], {n, 20}]
  • PARI
    {a(n)=(-1)^(n*(n+1)/2)*((n+1)^(n+1)-2*(2*n)^n)/(n-1)^2}  \\ Max Alekseyev, May 05 2005
    
  • PARI
    a(n)=poldisc('x^n-sum(k=0,n-1,'x^k)); \\ Joerg Arndt, May 04 2013

Formula

a(n) = (-1)^(n*(n+1)/2) * ((n+1)^(n+1)-2*(2*n)^n)/(n-1)^2. - Max Alekseyev, May 05 2005

A056788 a(n) = n^n + (n-1)^(n-1).

Original entry on oeis.org

2, 5, 31, 283, 3381, 49781, 870199, 17600759, 404197705, 10387420489, 295311670611, 9201412118867, 311791207040509, 11414881932150269, 449005897206417391, 18884637964090410991, 845687005960046315793, 40173648337182874339601, 2017766063735610126699403
Offset: 1

Views

Author

Walter Nissen, Aug 20 2000

Keywords

Comments

For even n > 1, the absolute value of the discriminant of the polynomial x^n+x-1. [Corrected by Artur Jasinski, May 07 2010]
The largest known prime in this sequence is a(4) = 283.

Examples

			a(3) = 2^2 + 3^3 = 4 + 27 = 31.
		

References

  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see equation (6.7).

Crossrefs

Cf. A000312 (n^n), A086797 (discriminant of the polynomial x^n-x-1).
Cf. A056187, A056790, A192397 (smallest & largest prime factor of a(n), records of the latter), A217435 = bigomega(a(n)).

Programs

  • Mathematica
    Join[{2}, Table[n^n+(n-1)^(n-1), {n, 2, 20}]] (* T. D. Noe, Aug 13 2004 *)
    Join[{2},Total/@Partition[Table[n^n,{n,20}],2,1]] (* Harvey P. Dale, Jun 26 2017 *)
  • PARI
    A056788(n)=n^n+(n-1)^(n-1)  \\ M. F. Hasler, Oct 02 2012

Extensions

Minor corrections by M. F. Hasler, Oct 02 2012

A238194 Conjectured numbers n for which n^n + (-1)^n (n-1)^(n-1) is not squarefree.

Original entry on oeis.org

130, 257, 487, 528, 815, 897, 1176, 1225, 1320, 1373, 1430, 2029, 2050, 2084, 2198, 2247, 2526, 2608, 2895, 2936, 2958, 3166, 3679, 3849, 3909, 3950, 4237, 4319, 4598, 4647, 4723, 4795, 5472, 5487, 5620, 5669, 5948, 6030, 6317, 6358, 6588, 6677, 6936, 7101
Offset: 1

Views

Author

T. D. Noe, Feb 24 2014

Keywords

Comments

The first case (130) yields a number divisible by 83^2. The next 5 terms yield numbers divisible by 59^2. Boyd et al. are not completely certain about the other 994 numbers up to 1000. They conjecture that 0.9934466... of numbers n^n + (-1)^n (n-1)^(n-1) are squarefree.
Boyd et al. tested the values n <= 1000 for divisibility by the squares of the first 10^4 primes. To extend the sequence, I tested the divisibility of n <= 200000 by the squares of the first 10^5 primes. - Giovanni Resta, Feb 24 2014
The heuristic chance that Resta's list is incomplete is just over 1%. This drops to 0.07% with testing to the millionth prime. - Charles R Greathouse IV, Feb 25 2014

Crossrefs

Cf. A086797 (n^n + (-1)^n (n-1)^(n-1) with signs).

Programs

Extensions

a(7)-a(44) from Giovanni Resta, Feb 24 2014

A114654 Discriminant of the polynomial x^n + x + 1.

Original entry on oeis.org

1, -3, -31, 229, 3381, -43531, -870199, 15953673, 404197705, -9612579511, -295311670611, 8630788777645, 311791207040509, -10809131718965763, -449005897206417391, 18008850183328692241, 845687005960046315793, -38519167813410200811247
Offset: 1

Views

Author

T. D. Noe, Dec 21 2005

Keywords

Comments

Except for the sign, the sequence alternates between the sum and difference of consecutive terms of A000312. x^2+x+1 divides x^n+x+1 for n=2 (mod 3).

References

  • Mohammad K. Azarian, On the Hyperfactorial Function, Hypertriangular Function, and the Discriminants of Certain Polynomials, International Journal of Pure and Applied Mathematics, Vol. 36, No. 2, 2007, pp. 251-257. Mathematical Reviews, MR2312537. Zentralblatt MATH, Zbl 1133.11012.

Crossrefs

Cf. A000312 (n^n), A007781 (n^n - (n-1)^(n-1)), A056788 (n^n + (n-1)^(n-1)), A086797 (discriminant of the polynomial x^n-x-1).

Programs

  • Mathematica
    Table[Discriminant[x^n + x + 1, x], {n, 0, 100}] (* Artur Jasinski, Oct 12 2007 *)
  • PARI
    a(n) = poldisc(x^n+x+1); \\ Michel Marcus, Aug 28 2020

Formula

for n>1, a(n) = (n^n + (-1)^(n-1) * (n-1)^(n-1)) * (-1)^floor(n/2).
a(n) = (Cos[Pi n/2]+Sin[Pi n/2])(n^n)+(Cos[Pi(n+1)/2]+Sin[Pi(n+1)/2])(n+1)^(n+1). - Artur Jasinski, Oct 12 2007
Showing 1-4 of 4 results.