

A063676


a(1) = 1 and for n > 1 let a(n) = a(n1) + 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
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OFFSET

1,2


LINKS



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 = n1, k >= 1, k, s = Select[Mean /@ Subsets[Array[a, n1], {k}], IntegerQ] // Sort; If[s != {}, Return[a[n1] + Last[s]]]]]; Table[a[n], {n, 1, 45}] (* JeanFrançois Alcover, Dec 11 2014 *)


CROSSREFS



KEYWORD

nice,nonn


AUTHOR



STATUS

approved



