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%I #22 Feb 19 2023 09:20:02
%S 4,16,22,27,33,55,63,77,143,187,209,222,248,253,256,319,341,407,451,
%T 473,484,517,555,583,649,656,671,737,777,781,803,837,869,913,979,1067,
%U 1111,1133,1152,1177,1199,1221,1243,1397,1441,1443,1507,1529,1639,1661,1727
%N Composite numbers n divisible by their concatenated exponents in prime factorization.
%C The numbers 2^(2^m), m = 1, 2,... are in the sequence. A majority of semiprimes of the form 11*p where is p prime different from 11 are in the sequence. The numbers of the form p*111 = p*3*37 where p is prime different from 3 or 37 are in the sequence. In the general case, the numbers of the form n = p_1*p_2*...*p_m*R where the p_k are prime numbers and R is a repunit number (A002275) with q digits "1" and (q-m) prime divisors are in the sequence, for example the numbers of the form n = p_1*p_2*p_3*p_4*11111111 = p_1*p_2*p_3*p_4*11*73*101*137 are in the sequence if the primes p_k are different from 11, 73, 101 or 137. So, 11111111 divides n.
%H Harvey P. Dale, <a href="/A227815/b227815.txt">Table of n, a(n) for n = 1..1000</a>
%e 248 = 2^3*31 => 31 is the concatenate exponents 3 and 1, so 31 divides 248.
%p with(numtheory):for n from 1 to 10000 do:x:=ifactors(n):y:=x[2];n1:=nops(y):s:=0:for i from 1 to n1 do:z:=y[i][2]:s:=s+z*10^(n1-i):od:if type(n,prime)=false and irem(n,s)=0 then printf(`%d, `, n):else fi:od:
%t With[{predicate = And[CompositeQ[#], Divisible[#, FromDigits[Join @@ IntegerDigits@(Last /@ FactorInteger[#])]]] &},
%t Select[Range[10000], predicate]] (* _Sidney Cadot_, Feb 19 2023 *)
%o (Python)
%o from sympy import isprime, factorint
%o def ok(n): return n > 1 and not isprime(n) and n%int("".join(str(e) for e in factorint(n).values())) == 0
%o print([k for k in range(1728) if ok(k)]) # _Michael S. Branicky_, Feb 19 2023
%Y Cf. A002275, A037916, A070529.
%K nonn,base
%O 1,1
%A _Michel Lagneau_, Jul 31 2013
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