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A134611
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Nonprime numbers such that the root mean cube of their prime factors is an integer (where the root mean cube of c and d is ((c^3+d^3)/2)^(1/3)).
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13
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1, 4, 8, 9, 16, 25, 27, 32, 49, 64, 81, 121, 125, 128, 169, 243, 256, 289, 343, 361, 512, 529, 625, 729, 841, 961, 1024, 1331, 1369, 1512, 1681, 1849, 2048, 2187, 2197, 2209, 2401, 2809, 3125, 3481, 3721, 4096, 4489, 4913, 5041, 5329, 6241, 6561, 6859, 6889
(list;
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refs;
listen;
history;
text;
internal format)
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OFFSET
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1,2
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COMMENTS
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The prime factors are taken with multiplicity.
Most terms have a last digit of 1 or 9 (i.e., 8326 out of 9000 terms). Mainly, this comes from the fact that all squares of primes are included. Since each prime > 10 has a last digit of 1, 3, 7 or 9, its square has a last digit of 1 or 9. In addition, m-th powers of primes have a last digit of 1, if m == 0 (mod 4), and have a last digit of 1 or 9 if m == 2 (mod 4), and have a 50% chance, roughly, for a last digit of 1 or 9, if m == 1 (mod 4) or m == 3 (mod 4). Since the number of terms <= N which are squares of primes is PrimePi(sqrt(N)) = A000720(sqrt(N)), it follows that the number of terms <= N which have a last digit of 1 or 9 is greater than PrimePi(sqrt(N)). This can be estimated as 2*N^(1/2)/log(N), approximately.
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LINKS
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EXAMPLE
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a(6) = 25, since 25 = 5*5 and ((5^3+5^3)/2)^(1/3) = 5.
a(30) = 1512, since 1512 = 2*2*2*3*3*3*7 and ((3*2^3+3*3^3+7^3)/7)^(1/3) = 64^(1/3) = 4.
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PROG
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(PARI) lista(m) = {for (i=2, m, if (! isprime(i), f = factor(i); s = sum (j=1, length(f~), f[j, 1]^3*f[j, 2]); s /= bigomega(i); if (type(s) == "t_INT" && ispower(s, 3), print1(i, ", ")); ); ); } \\ Michel Marcus, Apr 14 2013
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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