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%I #22 Aug 06 2018 09:03:53
%S 1488,2992,5692,101875,165864,207712,425830,569953,872539,20988160,
%T 25768720,26645280,48218580,48418542,50519902,1055195775,1175196540,
%U 1661395240,1971592669,2056199802,2601999369,3016199710,3068196150,3545196660,3835794250,3896098816,5520199708
%N 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.)
%C The number of concatenations is d(x)-1, where d(x) is the number of divisors of x (z = concat(z) is excluded).
%e 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).
%e 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.
%e 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.
%e Therefore both 165864 and 207712 are terms of the sequence.
%p 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);
%t 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 *)
%Y Cf. A000005, A317628.
%K nonn,base,more
%O 1,1
%A _Paolo P. Lava_, Aug 01 2018
%E a(16)-a(27) from _Giovanni Resta_, Aug 01 2018
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