OFFSET
1,1
COMMENTS
Sum_{(e(n)+j)^2,j=0..3} = a(n)*(a(n)+1)/2=t(a(n)) give the Pell equation c(n)^2 - 32*d(n)^2 = 41 with 2*a(n) + 1 = c(n) and e(n) + 1.5 = d(n). e(n) = A201633(n).
In general, for the sum of the squares of k consecutive numbers, one get an analog sequence with k in {4, 5, 6, 7, 11, 15, 17, 19, 23,...}. It gives the Pell equation c(n)^2 - 8k*d(n)^2 = 4*binomial((k+1),3) + 1 with 2*a(n) + 1 = c(n) and e(n) + (k-1)/2 = d(n).
LINKS
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).
FORMULA
G.f.: (35*x+48*x^2-70*x^3+3*x^5)/((1-x)*(1-34*x^2+x^4)).
a(n+4) = 34*a(n+2) - a(n) + 16.
a(n+5) = a(n+4) + 34*a(n+3) - 34*a(n+2) - a(n+1) + a(n).
eigenvalues ej: {1,(3+2r),-(3+2r),(3-2r),-(3-2r)}.
a(n+1) = (k1*e1 + k2*e2^n + k3*e3^n + k4*e4^n + k5*e5^n)/4 for k1=-2; k2=50+35r; k3=21+15r; k4=50-35r; k5=21-15r, where r = sqrt(2).
EXAMPLE
For n=2: a(2)=83; t(83)=83*84/2=3486.
A201633(2)=e(2)=28; 28^2+29^2+30^2+31^2=3486.
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul Weisenhorn, Jan 09 2013
EXTENSIONS
Corrected by R. J. Mathar, Jun 14 2016
STATUS
approved