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%I #13 Jun 05 2015 03:31:33
%S 0,1,1,3,1,5,6,5,1,6,7,12,10,12,11,12,1,8,16,14,17,18,18,23,13,21,18,
%T 22,23,24,19,14,1,22,20,23,24,31,27,25,26,36,28,37,29,30,42,37,22,32,
%U 37,38,35,41,36,37,43,42,37,44,44,34,33,47,1,48,49,43,53
%N Summative Fission - For a positive integer n, find the greatest number of consecutive positive integers (at least 2) which add to n. For each of these do the same ... iterate to completion. a(n) = the total number of integers (including n itself) defined.
%C The iteration that leads to this sequence is worthy of consideration for the grade 2 classroom learning addition.
%C a(2^k)=1 for all nonnegative integers k as can be seen from A138591.
%H Martin Büttner, <a href="/A256504/b256504.txt">Table of n, a(n) for n = 0..10000</a>
%e a(23) = 23 because there are 23 numbers generated by the iteration:
%e 23
%e /\
%e / \
%e / \
%e / \
%e / \
%e / \
%e / \
%e 11 12
%e /\ /|\
%e / \ / | \
%e / \ / | \
%e / \ 3 4 5
%e / \ / \ / \
%e 5 6 1 2 2 3
%e / \ /|\ / \
%e 2 3 / | \ 1 2
%e / \ / | \
%e 1 2 1 2 3
%e / \
%e 1 2
%e a(24) = 13 because there are 13 numbers generated by the iteration:
%e 24
%e /|\
%e / | \
%e / | \
%e 7 8 9
%e / \ /|\
%e 3 4 / | \
%e / \ / | \
%e 1 2 2 3 4
%e / \
%e 1 2
%t fission[0] = 0;
%t fission[n_] := fission@n = Module[{div = SelectFirst[Reverse@Divisors[2 n], (OddQ@# == IntegerQ[n/#] && n/# > (# - 1)/2) &]}, If[div == 1, 1, 1 + Total[fission /@ (Range@div + n/div - (div + 1)/2)]]];
%t fission /@ Range[0, 100] (* _Martin Büttner_, Jun 04 2015 *)
%Y Cf. A138591.
%K nonn,easy
%O 0,4
%A _Gordon Hamilton_, Mar 31 2015
%E Corrected and extended by _Martin Büttner_, Jun 04 2015