%I #9 Oct 03 2023 15:48:00

%S 1,3,4,5,11,197,487,928,10995,19318

%N Numbers k such that k divides the sum of largest parts of all partitions of k.

%C Numbers k such that k divides the total number of parts in all partitions of k.

%C Next term, if it exists, is greater than 100000. - _Vaclav Kotesovec_, May 04 2018

%e 5 is in the sequence because we have [5], [4, 1], [3, 2], [3, 1, 1], [2, 2, 1], [2, 1, 1, 1], [1, 1, 1, 1, 1] and 5 + 4 + 3 + 3 + 2 + 2 + 1 = 20 is divisible by 5.

%t Select[Range[200],Mod[Total[IntegerPartitions[#][[;;,1]]],#]==0&] (* The program generates the first six terms of the sequence. *) (* _Harvey P. Dale_, Oct 03 2023 *)

%Y Cf. A006128, A051177.

%K nonn,more

%O 1,2

%A _Ilya Gutkovskiy_, Apr 12 2018