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A317593
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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 let f(z) be the sum of the products y_1*y_2*...*y_i for all those concatenations. Sequence lists numbers z such that f(f(z)) = z. (see example.)
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1
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1488, 2992, 5692, 101875, 165864, 207712, 425830, 569953, 872539, 20988160, 25768720, 26645280, 48218580, 48418542, 50519902, 1055195775, 1175196540, 1661395240, 1971592669, 2056199802, 2601999369, 3016199710, 3068196150, 3545196660, 3835794250, 3896098816, 5520199708
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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).
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LINKS
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EXAMPLE
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z = 1488, x = 4. The concatenations are (1,4,8,8), (14,88) and f(z) = 1*4*8*8 + 14*88 = 1488 (fixed point).
z = 165864, x = 6. The concatenations are (1,6,5,8,6,4), (16,58,64), (165,864) and f(z) = 1*6*5*8*6*4 + 16*58*64 + 165*864 = 207712.
Now, z = 207712, x = 6. The concatenations are (2,0,7,7,1,2), (20,77,12), (207,712) and f(z) = 2*0*7*7*1*2 + 20*77*12 + 207*712 = 165864.
Therefore both 165864 and 207712 are terms of the sequence.
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MAPLE
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with(numtheory): T:=proc(w) local i, k, x, y, z; z:=0; i:=sort([op(divisors(ilog10(w)+1))]); for k from 1 to nops(i)-1 do x:=1; 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 then print(n); fi; od; fi; od; end: P(10^9);
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MATHEMATICA
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f[n_] := With[{w = IntegerDigits@ n}, Total@ Flatten@ Map[Times @@ Map[FromDigits, Partition[w, #, #]] &, Most@ Divisors@ Length@ w]]; Select[Range[10^6], f@ f@ # == # &] (* Michael De Vlieger, Aug 05 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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