OFFSET
0,2
COMMENTS
One fourth of the sum of p^2 + q^2 over the discrete square frame of length 2*n centered around the origin (called the 2n-frame).See the W. Lang link below.
Because the summation over p*q becomes zero due to symmetry, this is also the sum over, e.g., (p+q)^2.
The total number of sites (vertices) s(n) of a square around (0,0) with length 2*n, is (2*n+1)^2. The 2n-frame borders 8*n = s(n) - s(n-1) sites, for n>=1. For n=0 only the site (0,0) is considered.
The author was led to consider such sums by a (much more difficult) question asked by R. Thomale.
Convolution of 4*j-1 with 4*j-3, j=1..n. For n=4: [1,5,9,13] convolved with [3,7,11,15] gives a(4) = 1*(15) + 5*(11) + 9*(7) + 13*(3) = 172. - J. M. Bergot, May 27 2017
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
W. Lang, The squares for n=0..4
FORMULA
a(n) = (1/4)*(S(n) - S(n-1)), with a(0)=0 and S(n):=sum(sum(p^2+q^2,p=-n..+n),q=-n..+n) = 2*sum(sum((p^2,p=-n..+n), q=-n..+n) = 2*sum(p^2,p=-n..+n)*sum(1,q=-n..n) = 2*2*(n*(n+1)*(2*n+1))/6)*(2*n+1) = (2/3)*n*(n+1)*(2*n+1)^2.
a(n) = n*(8*n^2 + 1)/3.
G.f.: x*(3 + 10*x + 3*x^2)/(1-x)^4. - Vincenzo Librandi, Feb 05 2014
a(n) = T(n, 0) + 2*(Sum_{k=1..n-1}T(n,k)) + T(n,k), with the triangle T(n, k) = A069011(n, k) = n^2 + k^2. - Wolfdieter Lang, Aug 15 2019
EXAMPLE
The total sums S(n) are [0, 12, 100, 392, 1080, 2420, 4732, 8400, 13872, 21660, 32340, ...].
The 2n-frame sums are 4*a(n) = [0, 12, 88, 292, 688, 1340, 2312, 3668, 5472, 7788, 10680, 14212, 18448, 23452, 29288, 36020, 43712, 52428, 62232, 73188, 85360]. The sum is over 8*n numbers.
For n=1 the 8 numbers of the 2-frame are 2,1,2; 1,1; 2,1,2, summing to 4*a(1)=12.
MAPLE
MATHEMATICA
Table[n (8 n^2 + 1)/3, {n, 0, 50}] (* Wesley Ivan Hurt, Feb 03 2014 *)
CoefficientList[Series[x (3 + 10 x + 3 x^2)/(1 - x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Feb 05 2014 *)
PROG
(Magma) [n*(8*n^2+1)/3: n in [0..40]]; // Vincenzo Librandi, Feb 05 2014
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Sep 15 2008
STATUS
approved