A058031 a(n) = n^4 - 2*n^3 + 3*n^2 - 2*n + 1, the Alexander polynomial for reef and granny knots.

1, 1, 9, 49, 169, 441, 961, 1849, 3249, 5329, 8281, 12321, 17689, 24649, 33489, 44521, 58081, 74529, 94249, 117649, 145161, 177241, 214369, 257049, 305809, 361201, 423801, 494209, 573049, 660969, 758641, 866761, 986049, 1117249, 1261129, 1418481, 1590121

Offset: 0

Author

Jason Earls, Nov 21 2000

Keywords

• nonn
• easy

Comments

"The standard knot invariant, in the pre-Jones era of knot theory, was the Alexander polynomial, invented in 1926. This assigns to each knot a polynomial in a variable t, which can be calculated by following a standard procedure." See Courant and Robbins, p. 503.

First differences are in A105374. - Wesley Ivan Hurt, Apr 18 2016

References

• Richard Courant and Herbert Robbins, What Is Mathematics?, 2nd Ed. 1996, pp. 501-505.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A000290, A002061, A002378, A002522, A062938, A099254, A105374.

Programs

Magma

[(n^2 - n + 1)^2 : n in [0..50]]; // Wesley Ivan Hurt, Jun 19 2014

Maple

A058031:=n->(n^2 - n + 1)^2; seq(A058031(n), n=0..50); # Wesley Ivan Hurt, Jun 19 2014

Mathematica

Table[(n^2 - n + 1)^2, {n, 0, 50}] (* Wesley Ivan Hurt, Jun 19 2014 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {1, 1, 9, 49, 169}, 50] (* Vincenzo Librandi, Apr 11 2017 *)

PARI

a(n)=n^4-2*n^3+3*n^2-2*n+1 \\ Charles R Greathouse IV, Jun 19 2014

PARI

lista(nn) = for(n=0, nn, print1((n^2-n+1)^2, ", ")); \\ Altug Alkan, Apr 16 2016

Python

def a(n): return n**4 - 2*n**3 + 3*n**2 - 2*n + 1 # Indranil Ghosh, Apr 06 2017

Formula

G.f.: (1-4*x+14*x^2+4*x^3+9*x^4)/(1-x)^5. - Colin Barker, Jan 17 2012

a(n) = (n^2-n+1)^2. - Carmine Suriano, Feb 16 2012

3*a(n+3) = A062938(n) + A062938(n+1) + A062938(n+2). - Bruno Berselli, Feb 16 2012

a(n) = (n-2)*(n-1)*n*(n+1) + (2*n-1)^2. - Charlie Marion, Apr 11 2013

a(n) = A002061(n)^2. - Richard R. Forberg, Sep 03 2013

a(n) = (n*(n-1))^2 + (n-1)^2 + n^2, sum of three squares. - Carmine Suriano, Jun 16 2014

a(n) = A002378(A002378(n-1))+A002378(n-1)+1, where A002378(-1)=0. [Bruno Berselli, May 28 2015]

E.g.f.: exp(x)*(1 + 4*x^2 + 4*x^3 + x^4). - Ilya Gutkovskiy, Apr 16 2016

a(n) = (n-1)^4 + 2*(n-1)^3 + 3*(n-1)^2 + 2*(n-1) + 1. - Bruce J. Nicholson, Apr 07 2017

For n>0 a(n) = A002522(n)*A002522(n-1) - 1. - Bruce J. Nicholson, Jul 02 2017

Extensions

Name corrected by Andrey Zabolotskiy, Nov 21 2017