login
A063676
a(1) = 1 and for n > 1 let a(n) = a(n-1) + m, where m is the arithmetic mean of the largest subset of all predecessors such that m is an integer and m is maximal.
3
1, 2, 4, 7, 11, 16, 24, 33, 46, 62, 78, 106, 140, 184, 235, 302, 391, 487, 612, 768, 943, 1155, 1367, 1698, 2075, 2513, 3023, 3649, 4386, 5286, 6306, 7544, 8936, 10408, 12255, 14102, 16590, 19426, 22902, 26902, 31345, 36634, 42707, 49486, 57463
OFFSET
1,2
LINKS
Sean A. Irvine, Java program (github)
EXAMPLE
The arithmetic mean of the first 7 elements is (1 + 2 + 4 + 7 + 11 + 16 + 24) / 7 = 65 / 7; this is not an integer. But skipping a(5) = 11 yields m = 54 / 6 = 9, so we have a(8) = a(7) + m = 24 + 9 = 33.
MATHEMATICA
a[1] = 1; a[n_] := a[n] = Module[{k, s}, For[k = n-1, k >= 1, k--, s = Select[Mean /@ Subsets[Array[a, n-1], {k}], IntegerQ] // Sort; If[s != {}, Return[a[n-1] + Last[s]]]]]; Table[a[n], {n, 1, 45}] (* Jean-François Alcover, Dec 11 2014 *)
CROSSREFS
Sequence in context: A129339 A196719 A011912 * A099385 A331387 A357308
KEYWORD
nice,nonn
AUTHOR
Reinhard Zumkeller, Jul 28 2001
STATUS
approved