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A286353
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Zeroless numbers n = concat(s,t) such that s * t is the 10’s complement of the digits of n.
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1
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75, 92, 696, 794, 921, 6946, 9211, 9418, 67365, 69446, 73515, 92111, 92592, 584799, 653597, 694446, 921111, 5793295, 6693466, 6944446, 7145554, 7694443, 9211111, 58788989, 61728398, 66733665, 69444446, 72175395, 76445374, 88183426, 89245679, 91721145, 92111111, 92592592
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OFFSET
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1,1
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COMMENTS
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Numbers of the form concat(92,(10^k - 1)/9) and concat(69,4*(10^k - 1)/9,6) for k>0 are part of the sequence.
The first digit of a term must be 5 or higher. Numbers of the form 925925...92592 are terms, i.e. concatenate 925 m times for m >= 0 and append 92. Numbers of the form concat(2*(10^k-1)/3,9,(10^(k-1)-1)/3,4,2*(10^k-1)/3) for k > 1 are also terms. It also works for the case k=1 if the middle 0 that results is removed. If a term n is even, then the digits of n besides the least significant digit must contain either 4 or 9. If a term n is odd and not divisible by 5, then the digits of n besides the least significant digit must contain 9. - Chai Wah Wu, May 10 2017
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LINKS
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EXAMPLE
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75 = concat(7,5) and 7 * 5 = 35 is the 10’s complement of 75.
584799 = concat(58479,9) and 58479 * 9 = 526311 is the 10’s complement of 584799.
7694443 = concat(769,4443) and 769 * 4443 = 3416667 is the 10’s complement of 7694443.
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MAPLE
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P:=proc(q) local a, b, c, d, k, ok, n; for n from 1 to q do b:=ilog10(n)+1; c:=n; ok:=1;
d:=0; for k from 1 to b do if (c mod 10)>0 then a:=(10-(c mod 10)); d:=d+10^(k-1)*a; c:=trunc(c/10); else ok:=0; break; fi; od; if ok=1 then for k from 1 to b do
if d=(n mod 10^k)*trunc(n/10^k) then print(n); fi; od; fi; od; end: P(10^7);
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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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STATUS
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approved
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