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A287637 a(n) = A249125(n)/concatenation of prime factors of A249125(n).

%I A287637 #7 May 31 2017 13:15:51
%S A287637 2,4,3,8,5,9,16,7,2,32,27,4,11,25,64,13,8,81,10,128,17,49,19,16,20,
%T A287637 256,23,125,243,32,29,31,40,512,50,121,37,64,41,43,80,1024,729,169,47,
%U A287637 343,100,53,625,128,59,61,160,2048,67,289,200,71,73,79,250,256
%N A287637 a(n) = A249125(n)/concatenation of prime factors of A249125(n).
%C A287637 The squares of the sequence are, in increasing order: 4, 9, 16, 25, 49, 64, 81, 100, 121, 169, 256, 289, 361, 400, 625, 729, 1024, 4096,... including the squares of the prime numbers.
%C A287637 The numbers p^n, p prime and n = 1, 2, 3, 4,... are in the sequence.
%C A287637 The twin primes (a(m), a(m+1)) of the sequence are (29, 31), (41, 43), (59, 61), (71, 73), (101, 103), (107, 109), (137, 139),...
%C A287637 The numbers whose prime factors are 2 and 5 (A033846) are in the sequence.
%e A287637 a(9)=2 because A249125(9) = 50 and the concatenation of the prime factors of 50 is 25. Hence, 50/25 = 2.
%p A287637 with(numtheory):
%p A287637 for n from 2 to 10000 do:
%p A287637   if type(n,prime)=false
%p A287637    then
%p A287637     x:=factorset(n):n0:=nops(x):
%p A287637     d:=sum('length(x[i])', 'i'=1..n0):
%p A287637     l:=sum('x[i]*10^sum('length(x[j])', 'j'=i+1..n0)', 'i'=1..n0):
%p A287637     z:=n/l:
%p A287637      if floor(z)=z
%p A287637       then
%p A287637       printf(`%d, `,z):
%p A287637       else
%p A287637      fi:
%p A287637    fi:
%p A287637 od:
%t A287637 cf[n_] := FromDigits@ Flatten[ IntegerDigits /@ First /@ FactorInteger@n]; Reap[ Do[If[ CompositeQ[n] && IntegerQ[rz = n/cf[n]], Sow[rz]], {n, 6400}]][[2, 1]] (* _Giovanni Resta_, May 29 2017 *)
%Y A287637 Cf. A033846, A077800, A084317, A249125.
%K A287637 nonn,base
%O A287637 1,1
%A A287637 _Michel Lagneau_, May 28 2017