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%I #18 Jul 22 2017 20:49:22
%S 1,2,5,10,16,19,26,29,34,37,43,46,58,64,65,67,73,82,86,94,101,109,122,
%T 130,134,142,145,146,149,157,163,190,193,197,199,202,206,211,229,257,
%U 262,281,283,290,298,302,310,334,337,347,349,367,401,409,421,430
%N Numbers n that cannot be obtained as a partial sum of the divisors (taken in descending order, from m down to 1) of any m < n.
%C Numbers such that A243970(n) = n.
%C Is this sequence infinite? - _Franklin T. Adams-Watters_, Jul 20 2017
%H Alois P. Heinz, <a href="/A243971/b243971.txt">Table of n, a(n) for n = 1..10000</a>
%e From n=1 to 4, these partial sums are: 1; 2, 3; 3, 4; 4, 6, 7. So it is not possible to obtain 5 with any partial sum of divisors of numbers that are less than 5. And indeed A243970(5) is equal to 5. Hence 5 is in this sequence.
%t Module[{nn = 432, s}, s = Array[Function[d, Array[Total@ Take[d, -#] &, Length@ d]]@ Divisors@ # &, nn - 1]; Select[Range@ nn, ! MemberQ[Flatten@ Take[s, # - 1], #] &]] (* _Michael De Vlieger_, Jul 22 2017 *)
%Y Cf. A243970.
%K nonn
%O 1,2
%A _Michel Marcus_, Jun 16 2014
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