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%I A201221 #29 Dec 07 2018 19:16:29
%S A201221 4,6,8,10,12,14,16,16,19,16,18,17,19,18,28,34,72,49,57,35,51,36,97,65,
%T A201221 56,56,57,69,64,65,219,136,96,72,79,75,78,74,78,70,72,73
%N A201221 The smallest number of previous terms of sequence A008829 which are required to sum to A008829(n).
%e A201221 For n=2, the corresponding term of A008829 is 4, and the only preceding term is 1, therefore four 1s are needed and term a(2)=4.
%e A201221 Likewise, for n=3, the corresponding term is 15, which is three 4s and three 1s, so term a(3)=6.
%Y A201221 Cf. A008829.
%K A201221 nonn,base,more
%O A201221 2,1
%A A201221 _Alastair Stanley_, Nov 30 2011
%E A201221 More terms from _Alastair Stanley_, Nov 29 2018

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