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A180082
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Semiprime centered cube numbers: m^3 + (m+1)^3.
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3
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9, 35, 91, 341, 559, 1241, 6119, 7471, 17261, 19909, 75241, 143009, 257651, 323839, 671509, 860851, 967591, 1433969, 1482571, 1970299, 2348641, 2772559, 3413159, 4548059, 5313691, 5666509, 7233841, 7520291, 9568441
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OFFSET
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1,1
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COMMENTS
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There are no prime centered cube numbers because m^3 + (m+1)^3 = (2m+1)*(m^2+m+1). - Zak Seidov, Feb 08 2011
Products of two primes p and q = (p^2+3)/4 with p's in A118939. - Zak Seidov, Feb 08 2011
Also numbers which are products of two primes whose sum and difference are both promic (A002378).
Subsequence of A217843 (sums of consecutive nonnegative cubes) limited to the terms that have only two prime factors (multiplicity counted).
As stated in A217843, any number that is the sum of consecutive nonnegative cubes can also be expressed as the product of two integers whose sum and difference are both promic. (Therefore, it cannot be prime.) It can also be expressed as the difference of the squares of two triangular numbers (A000217); thus, the two primes that are the factors of any term of this sequence are respectively the sum and difference of two triangular numbers. (End)
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LINKS
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FORMULA
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EXAMPLE
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a(1) = 1^3 + (1+1)^3 = 9 = 3^2 is semiprime.
a(2) = 2^3 + (2+1)^3 = 35 = 5 * 7.
a(3) = 3^3 + (3+1)^3 = 91 = 7 * 13.
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MATHEMATICA
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Select[Total/@Partition[Range[200]^3, 2, 1], PrimeOmega[#]==2&] (* Harvey P. Dale, Feb 02 2019 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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