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A317628
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Take a number z of x digits and consider any concatenation z = concat(y_1, y_2, ..., y_i) where y_1, y_2, ..., y_i have the same number of digits. Then be g(z) the product of the sums y_1 + y_2 + ... + y_i for all those concatenations. Sequence lists numbers z such that g(g(z)) = z. (See example.)
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1
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1782, 109000, 208000, 307000, 406000, 505000, 604000, 703000, 802000, 901000, 18802160, 33534424, 67452850, 71272872, 2496688768, 100000000000
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OFFSET
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1,1
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COMMENTS
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The number of concatenations is d(x)-1, where d(x) is the number of divisors of x (z = concat(z) is excluded).
So far only fixed points of the transform g(z).
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LINKS
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EXAMPLE
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z = 1782, x = 4. The concatenations are (1,7,8,2), (17,82) and g(z) = (1+7+8+2) * (17+82) = 1782.
z = 18802160, x = 8. The concatenations are (1,8,8,0,2,1,6,0), (18,80,21,60), (1880,2160) and g(z) = (1+8+8+0+2+1+6+0)*(18+80+21+60)*(1880+2160) = 18802160.
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MAPLE
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with(numtheory): T:=proc(w) local i, k, x, y, z; z:=1; i:=sort([op(divisors(ilog10(w)+1))]); for k from 1 to nops(i)-1 do x:=0; y:=w; while y>0 do x:=x+(y mod 10^i[k]); y:=trunc(y/10^i[k]); od; z:=z*x; od; z; end: P:=proc(q) local c, d, j, n; for j from 1 to q do if not isprime(j+1) then for n from 10^j to 10^(j+1)-1 do c:=T(n); if c>0 then d:=T(c); fi; if d=n and d<>c then print(n); fi; od; fi; od; end: P(10^9); # Paolo P. Lava, Aug 02 2018
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MATHEMATICA
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g[n_] := Block[{d = IntegerDigits[n], nd}, nd = Length@d; Times @@ (Plus @@ FromDigits /@ Partition[d, #] & /@ Most[Divisors@ nd])]; Select[Range[2, 10^6], g[g[#]] == # &] (* Giovanni Resta, Aug 04 2018 *)
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CROSSREFS
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KEYWORD
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nonn,base,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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