|
|
A285009
|
|
Subset sums (see Comments).
|
|
5
|
|
|
9, 17, 28, 42, 59, 79, 102, 128, 157, 189, 224, 262, 303, 347, 394, 444, 497, 553, 612, 674, 739, 807, 878, 952, 1029, 1109, 1192, 1278, 1367, 1459, 1554, 1652, 1753, 1857, 1964, 2074, 2187, 2303, 2422, 2544, 2669, 2797, 2928, 3062, 3199, 3339, 3482, 3628, 3777, 3929, 4084
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
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
|
|
|
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
|
|
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
|
|
|
STATUS
|
approved
|
|
|
|