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A061148
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Smallest positive integer for which the number of divisors is a product of 2 distinct primes: Min{x; d[x]=pq}.
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5
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12, 48, 192, 144, 576, 3072, 12288, 9216, 196608, 5184, 786432, 36864, 12582912, 589824, 82944, 2359296, 805306368, 3221225472, 331776, 37748736, 206158430208, 746496, 3298534883328, 5308416, 13194139533312, 2415919104
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OFFSET
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1,1
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COMMENTS
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Least solutions of d(x)=A000005(x)=pq equation, where 1<p<q, primes: x=(2^q)*(3^p) where the exponent of smaller base[=2] is the larger prime factor[=q].
Since 2^(pq-1)>2^(q-1)*3^(p-1) -> [(2^q)/3]^(p-1) holds for q>1, p>1, therefore these solutions are in fact minimal.
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LINKS
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FORMULA
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EXAMPLE
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If d(x)=253=11*23, then 1<p=11<q=23,q-1=22,p-1=10, 2^22=4194304, 3^10=59049 so the smallest number x, which has 253 divisors is 4194304*59049=247669456896.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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