OFFSET
3,1
COMMENTS
For n > 2, take the set [3*(n-1)] and form three subsets all of which: a) have cardinality of n, b) have the same sum of elements, and c) share one element with the other subset and another element with the third subset. a(n) is the sum of the elements of each subset.
a(n) is the minimum value of the magic constant in a normal magic triangle of order n (see formula 5 in Trotter). - Stefano Spezia, Feb 18 2021
REFERENCES
a(4) is mentioned in: Gary Gruber, "The World's 200 Hardest Brain Teasers", Sourcebooks, 2010, p. 55.
LINKS
Colin Barker, Table of n, a(n) for n = 3..1000
Terrel Trotter, Normal Magic Triangles of Order n, Journal of Recreational Mathematics Vol. 5, No. 1, 1972, pp. 28-32.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), for n > 5.
a(n) = (8 + (n-2)*(3*n+1))/2, for n > 2.
G.f.: x^3*(9 - 10*x + 4*x^2) / (1 - x)^3. - Colin Barker, Apr 08 2017
E.g.f.: (1/2)*exp(x)*(3*x^2 - 2*x + 6) - 2*x*(x + 1) - 3. - Indranil Ghosh, Apr 08 2017; corrected by Ilya Gutkovskiy, Apr 10 2017
a(n) = A005449(n-1) + 2. - Hugo Pfoertner, Feb 18 2021
EXAMPLE
For n = 3, the set is S = {1,2,3,4,5,6} and the subsets are S1 = {1,2,6}, S2 = {1,3,5} and S3 = {2,3,4}. Therefore, a(3) = 9.
MATHEMATICA
Table[(8+(n-2)*(3 *n+1))/2, {n, 3, 53}]
Drop[CoefficientList[Series[x^3*(9 - 10*x + 4*x^2) / (1 - x)^3 , {x, 0, 60}], x], 3] (* Indranil Ghosh, Apr 08 2017 *)
PROG
(PARI) Vec(x^3*(9 - 10*x + 4*x^2) / (1 - x)^3 + O(x^60)) \\ Colin Barker, Apr 08 2017
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Ivan N. Ianakiev, Apr 07 2017
STATUS
approved