OFFSET
1,2
COMMENTS
A triangle has sides of lengths 6*n-3, 6*n^2-6*n+4, and 6*n^2-6*n+7; for n>2 its area is 6*sqrt(a(n)^2 - 1). - J. M. Bergot, Aug 30 2013
[The source of this is using (n,n+1), (n+1,n+2), and (n+2,n+3) as (a,b) in the creation of three Pythagorean triangles with sides b^2-a^2, 2*a*b, and a^2+b^2. Combine the three respective sides to create a new larger triangle, then find its area. It is not simply working backwards from the sequence. As well, the sequence has this as its first comment to show that the numbers are actually doing something to find a solution.]
REFERENCES
T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.
LINKS
Harry J. Smith, Table of n, a(n) for n = 1..1000
T. P. Martin, Shells of atoms, Phys. Rep., 273 (1996), 199-241, eq. (10).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
FORMULA
G.f.: x*(1+x)*(1+7*x+x^2)/(1-x)^4. - Colin Barker, Apr 20 2012
a(n) = +4*a(n-1) -6*a(n-2) +4*a(n-3) -1*a(n-4) n > 3, a(1)=1, a(2)=12, a(3)=50, a(4)=133. - Yosu Yurramendi, Sep 04 2013
E.g.f.: (-2 + 4*x + 9*x^2 + 6*x^3)*exp(x)/2 + 1. - G. C. Greubel, Dec 01 2017
From Bruce J. Nicholson, Jun 17 2020: (Start)
MATHEMATICA
LinearRecurrence[{4, -6, 4, -1}, {1, 12, 50, 133}, 40] (* Harvey P. Dale, Jun 05 2016 *)
Table[(2*n-1)*(3*n^2 -3*n +2)/2, {n, 1, 30}] (* G. C. Greubel, Dec 01 2017 *)
PROG
(PARI) { for (n=1, 1000, write("b063491.txt", n, " ", (2*n - 1)*(3*n^2 - 3*n + 2)/2) ) } \\ Harry J. Smith, Aug 23 2009
(PARI) x='x+O('x^30); Vec(serlaplace((-2 + 4*x + 9*x^2 + 6*x^3)*exp(x)/2 + 1)) \\ G. C. Greubel, Dec 01 2017
(R)
a <- c(0, 1, 9, 38, 110)
for(n in (length(a)+1):40)
a[n] <- +4*a[n-1]-6*a[n-2]+4*a[n-3]-a[n-4]
a [Yosu Yurramendi, Sep 04 2013]
(Magma) [(2*n-1)*(3*n^2 -3*n +2)/2: n in [1..30]]; // G. C. Greubel, Dec 01 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Aug 01 2001
STATUS
approved