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A263534
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Consider the 10's complements mod 10 of the digits of a number k. Take their sum and repeat the process deleting the first addend and adding the previous sum. The sequence lists the numbers that after some iterations reach a sum equal to k.
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9
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29, 76, 157, 174, 191, 475, 713, 1129, 1961, 3286, 4424, 7812, 8973, 19978, 24317, 35845, 37041, 51712, 68022, 166838, 443275, 444247, 445219, 509439, 706317, 1189312, 1933197, 2686010, 10809303, 55558901, 58338037, 257990335, 504050156, 839186880
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OFFSET
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1,1
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COMMENTS
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Like Keith numbers but using the ten's complements of their digits.
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LINKS
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EXAMPLE
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For 29, the 10's complements of its digits are 8, 1. Then:
8 + 1 = 9;
1 + 9 = 10;
9 + 10 = 19;
10 + 19 = 29.
For 475, the 10's complements of its digits are 6, 3, 5. Then:
6 + 3 + 5 = 14;
3 + 5 + 14 = 22;
5 + 14 + 22 = 41;
14 + 22 + 41 = 77;
22 + 41 + 77 = 140;
41 + 77 + 140 = 258;
77 + 140 + 258 = 475.
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MAPLE
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with(numtheory): P:=proc(q, h) local a, b, c, k, n, t, v; v:=array(1..h);
for n from 10 to q do b:=ilog10(n)+1; c:=n; a:=[];
for k from 1 to b do a:=[(10-c) mod 10, op(a)]; c:=trunc(c/10); od;
for k from 1 to b do v[k]:=a[k]; od; t:=b+1; v[t]:=add(v[k], k=1..b);
while v[t]<n do t:=t+1; v[t]:=add(v[k], k=t-b..t-1); od;
if v[t]=n then print(n); fi; od; end: P(10^9, 1000);
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MATHEMATICA
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Select[Range[10^5], Function[{m, n}, Last@ NestWhile[Append[#, Total@ Take[#, -m]] &, Flatten[{#, Total@ #}] &[IntegerDigits[n] /. d_?Positive :> 10 - d], Last@ # < n &, 1, 10^2] == n] @@ {IntegerLength@#, #} &] (* Michael De Vlieger, Mar 09 2018 *)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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Name clarified, some terms and Maple code corrected by Paolo P. Lava, Mar 08 2018
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STATUS
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approved
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