Search: id:A143166 Results 1-1 of 1 results found. %I A143166 %S A143166 0,3,22,73,172,335,578,917,1368,1947,2670,3553,4612,5863,7322,9005,10928, %T A143166 13107,15558,18297,21340,24703,28402,32453,36872,41675,46878,52497,58548, %U A143166 65047,72010,79453,87392,95843,104822,114345,124428,135087,146338,158197 %N A143166 n*(8*n^2+1)/3. %C A143166 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). %C A143166 Because the summation over p*q becomes zero due to symmetry, this is also the sum over, e.g., (p+q)^2. %C A143166 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. %C A143166 The author was led to consider such sums by a (much more difficult) question asked by R. Thomale. %H A143166 W. Lang, The squares for n=0..4 %F A143166 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. %F A143166 a(n)= n*(8*n^2+1)/3. %e A143166 The total sums S(n) are: [0, 12, 100, 392, 1080, 2420, 4732, 8400, 13872, 21660, 32340,...] %e A143166 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. %e A143166 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. %Y A143166 Sequence in context: A139272 A006532 A005288 this_sequence A055550 A075204 A106150 %Y A143166 Adjacent sequences: A143163 A143164 A143165 this_sequence A143167 A143168 A143169 %K A143166 nonn,easy %O A143166 0,2 %A A143166 Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de) Sep 15 2008 Search completed in 0.001 seconds