A117561 - OEIS

%I #13 Mar 05 2023 13:30:49

%S 3,15,38,73,124,194,286,403,548,724,934,1181,1468,1798,2174,2599,3076,

%T 3608,4198,4849,5564,6346,7198,8123,9124,10204,11366,12613,13948,

%U 15374,16894,18511,20228,22048,23974,26009,28156,30418,32798,35299,37924

%N a(n) = floor(n*(n^3-n-3)/(2*(n-1))).

%C a[n-1] is one approximation for the upper bound of the "antimagic constant" of an antimagic square of order n. The antimagic constant here is defined as the least integer in the set of consecutive integers to which the rows, columns and diagonals of the square sum. By analogy with the magic constant. This approximation follows from the observation that Sum[m + k, {k, 0, 2*n + 1}] <= (2*Sum[k, {k, 1, n^2}]) + (2*m) + (2*m + 1) where m is the antimagic constant for an antimagic square of order n. Stricter bounds seem likely to exist. See A117560 for the lower bounds. Note there exist no antimagic squares of order two or three, but the values are indexed here for completeness.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/AntimagicSquare.html">Antimagic Square</a>.

%F a(n) = floor(n*(n^3-n-3)/(2*(n-1))).

%F G.f.: x^2*(3+3*x-4*x^2-x^3+3*x^4-x^5)/(1-x)^4. - _Colin Barker_, Mar 29 2012

%e a[3] = 38 because the antimagic constant of an antimagic square of order 4 cannot exceed 38 (see comments)

%t Table[Floor[n(n^3-n-3)/(2*(n-1))], {n, 2, 50}]

%Y Cf. A117560, A050257, A049475.

%K easy,nonn

%O 2,1

%A Joseph Biberstine (jrbibers(AT)indiana.edu), Mar 29 2006