A235400 a(1)=1. For n>1, assume a(n-1) has decimal expansion qrstuvwxyz (with at most ten digits, where some of q, r, s, ... may be zero). Then a(n) = sum 99...9 (with z 9's) + 88...8 (with y 8's) + 77...7 (with x 7's) + ... For example, if a(n-1) were 243, we would sum 77 + 8888 + 999 and get 9964.
1, 9, 999999999, 4999999995, 4000099995, 2888988885, 800099995, 2900099996, 3000999996, 3334333329, 1000006195, 889655560, 368199996, 2923444555, 111394795, 1452329335, 666792195, 1449655625, 334781195, 948992565, 1145513395, 889070335, 372323885, 168990195
Offset: 1
Examples
1 leads to one 9, then 9 leads to 999999999, then we have to sum 111111111 + 222222222 + ... + 999999999 = 4999999995, then we have to sum 1111 + 222222222 + ... + 888888888 + 99999 = 4000099995, and so on.
References
- Rodolfo Kurchan, Problems with repdigits, to appear in Proceedings of Gathering for Gardner 2014 conference
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..1000
Crossrefs
Cf. A235460.
Programs
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Mathematica
nxt[n_]:=Total[FromDigits/@(Table[#[[1]],{#[[2]]}]&/@Thread[{ Range[ 9,10-IntegerLength[n],-1],Reverse[IntegerDigits[n]]}])]; NestList[ nxt,1,30] (* Harvey P. Dale, Apr 24 2015 *)
Comments