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A122723
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Primes which are the sum of three distinct positive cubes.
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8
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73, 197, 251, 281, 307, 349, 521, 547, 577, 701, 757, 853, 863, 881, 919, 953, 1009, 1091, 1217, 1249, 1483, 1559, 1637, 1861, 1907, 2069, 2087, 2267, 2269, 2287, 2339, 2477, 2521, 2729, 2753, 2843, 2927, 2953, 2969, 3067, 3257, 3413, 3457, 3527, 3529
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| Considering parity, a prime sum of three cubes cannot be the sum of three evens nor two odds and an even, but must be the sum of three odds (such as 1^3 + 3^3 + 9^3 = 757 or 3^3 + 5^3 + 9^3 = 881) or two evens and an odd (such as 1^3 + 2^3 + 10^3 = 1009). Without "distinct" we have solutions such as 1^3 + 1^3 + 3^3 = 29; 2^3 + 2^3 + 3^3 = 43; 1^3 + 1^3 + 5^3 = 127. A subset of the three odds subset is primes which are the sum of the cubes of three distinct primes, such as 3^3 + 5^3 + 11^3 = 1483; or 3^3 + 7^3 + 19^3 = 7229; or 7^3 + 11^3 + 23^3 = 13841; or 3^3 + 5^3 + 41^3 = 69073.
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LINKS
| T. D. Noe, Table of n, a(n) for n=1..1000
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FORMULA
| Primes in A024975.
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EXAMPLE
| a(1) = 73 = 1^3 + 2^3 + 4^3.
a(7) = 1^3 + 2^3 + 8^3 = 521.
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MATHEMATICA
| lst={}; Do[Do[Do[p=n^3+m^3+k^3; If[PrimeQ[p], AppendTo[lst, p]], {n, m+1, 4!}], {m, k+1, 4!}], {k, 4!}]; Take[Union[lst], 30] [From Vladimir Orlovsky (4vladimir(AT)gmail.com), May 23 2009]
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CROSSREFS
| Cf. A000040, A024975.
Sequence in context: A142741 A088199 A140010 * A089786 A142894 A141909
Adjacent sequences: A122720 A122721 A122722 * A122724 A122725 A122726
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KEYWORD
| easy,nonn
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AUTHOR
| Jonathan Vos Post (jvospost3(AT)gmail.com), Sep 23 2006
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EXTENSIONS
| Corrected and extended by Vladimir Orlovsky (4vladimir(AT)gmail.com) and T. D. Noe (noe(AT)sspectra.com), Jul 16 2010
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