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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

%I #12 May 06 2023 09:01:01

%S 1,2,4,7,11,16,24,33,46,62,78,106,140,184,235,302,391,487,612,768,943,

%T 1155,1367,1698,2075,2513,3023,3649,4386,5286,6306,7544,8936,10408,

%U 12255,14102,16590,19426,22902,26902,31345,36634,42707,49486,57463

%N 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.

%H Sean A. Irvine, <a href="/A063676/b063676.txt">Table of n, a(n) for n = 1..500</a>

%H Sean A. Irvine, <a href="https://github.com/archmageirvine/joeis/blob/master/src/irvine/oeis/a063/A063676.java">Java program</a> (github)

%e 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.

%t 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 *)

%Y Cf. A063677, A063678.

%K nice,nonn

%O 1,2

%A _Reinhard Zumkeller_, Jul 28 2001

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Last modified April 24 07:44 EDT 2024. Contains 371922 sequences. (Running on oeis4.)