|
| |
| |
|
|
|
0, 3, 22, 73, 172, 335, 578, 917, 1368, 1947, 2670, 3553, 4612, 5863, 7322, 9005, 10928, 13107, 15558, 18297, 21340, 24703, 28402, 32453, 36872, 41675, 46878, 52497, 58548, 65047, 72010, 79453, 87392, 95843, 104822, 114345, 124428, 135087, 146338, 158197
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,2
|
|
|
COMMENTS
| One fourth of sum of p^2+q^2 over the square frame of length 2*n and width 1 centered around the origin (called 2n-frame).
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.
The author was led to consider such sums by a (much more difficult) question asked by R. Thomale.
|
|
|
LINKS
| 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.
|
|
|
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;1,0,1;2,1,2, summing to 4*a(1)=12.
|
|
|
CROSSREFS
| Sequence in context: A006532 A178492 A005288 * A055550 A075204 A106150
Adjacent sequences: A143163 A143164 A143165 * A143167 A143168 A143169
|
|
|
KEYWORD
| nonn,easy
|
|
|
AUTHOR
| Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de) Sep 15 2008
|
| |
|
|
