|
%I #8 Mar 07 2021 14:48:00
%S 1,2,2,2,3,3,4,3,4,5,3,5,6,5,5,6,7,3,7,8,7,6,8,9,7,8,8,9,10,8,10,11,8,
%T 10,8,11,12,10,8,11,10,12,13,8,12,11,13,14,11,13,6,14,15,13,10,14,9,
%U 15,16,11,14,14,15,15,16,17,13,17,18,10,16,9,17,15,18,19,16,15,17,15,18,13
%N a(1) = 1, a(2) = 2; for n > 2, a(n) = the number of terms in the maximal length sum of previous consecutive terms that equals n.
%C The equivalent sequence for a minimal length sum is given by A003059.
%H Scott R. Shannon, <a href="/A342219/a342219.png">Scatterplot of the first 50000 terms</a>.
%e a(3) = 2 as the only way to sum previous consecutive terms to make 3 is 1 + 2 = 3, which contains two terms.
%e a(7) = 4 as the previous consecutive terms 1 + 2 + 2 + 2 = 7, which contains four terms. Note that 7 can also be made by consecutive terms 2 + 2 + 3 = 7, but the sequence is the maximal sum length.
%e a(10) = 5 as the previous consecutive terms 1 + 2 + 2 + 2 + 3 = 10, which contains five terms. Three other consecutive term sums also exist that sum to 10 but they contain fewer terms.
%Y Cf. A163169, A003059, A064816, A118139
%K nonn
%O 1,2
%A _Scott R. Shannon_, Mar 05 2021
|
