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 A320021 Numbers equal to the sum of the aliquot parts of the previous k numbers, for some k. 1
 5, 6, 7, 8, 35, 40, 51, 237, 263, 264, 280, 387, 899, 1300, 7300, 8363, 8364, 11764, 26740, 26939, 46595, 59004, 80877, 131580, 5244549, 5462385, 17062317, 75097524, 127838820, 323987589, 1162300835, 1381439877, 4943600220 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS So far 2 <= k <= 4 (k = 2 for 7, 35, 51, 237, 263, 387, 899, 8363, 26939, 46595, 80877, ...; k = 3 for 5, 8, 40, 264, 280, 1300, 7300, 8364, 11764, 26740, 59004, 131580, ...; k = 4 for 6). Are there terms with k = 5, 6, 7, ...? a(34) > 10^12. - Giovanni Resta, Oct 09 2018 If we were looking at numbers equal to the sum of the aliquot parts of the previous k numbers and of the following k, for some k, the first terms would be 2263024 and 128508838576. LINKS FORMULA a(n) = Sum_{i = 1..k} A001065(a(n)-i), for some k. EXAMPLE 5 is in the sequence because aliquot parts of 4 are 1, 2, of 3 is 1, of 2 is 1: 1 + 2 + 1 + 1 = 5. 6 is in the sequence because aliquot parts of 5 is 1, of 4 are 1, 2, of 3 is 1, of 2 is 1: 1 + 1 + 2 + 1 + 1 = 6. 7 is in the sequence because aliquot parts of 6 are 1, 2, 3, of 5 is 1: 1 + 2 + 3 + 1 = 7. MAPLE with(numtheory): P:=proc(q) local a, j, k, n; for n from 1 to q do a:=0; k:=0; while a0 && s

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Last modified July 4 16:24 EDT 2020. Contains 335448 sequences. (Running on oeis4.)