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Numbers c such that the difference of consecutive cubes (c+1)^3 - c^3 is the sum of two positive cubes.
4

%I #20 Jan 11 2016 20:58:35

%S 5,8,18,40,53,70,102,114,188,197,213,248,255,297,306,453,460,477,487,

%T 491,495,564,632,671,684,768,909,958,989,1190,1290,1324,1331,1346,

%U 1744,1745,1779,2068,2130,2178,2208,2262,2448,2790,2813,3320,3327,3402,3414

%N Numbers c such that the difference of consecutive cubes (c+1)^3 - c^3 is the sum of two positive cubes.

%C The numbers c in A225909.

%C The sequence is infinite, because A226903 is a parametrized infinite subsequence.

%H Chai Wah Wu, <a href="/A226902/b226902.txt">Table of n, a(n) for n = 1..5000</a>

%F a(n) = (-3 + sqrt(9 + 12*(A225909(n) - 1)))/6

%e (5+1)^3 - 5^3 = 3^3 + 4^3, so 5 is a member.

%Y Cf. A225909, A226903.

%K nonn

%O 1,1

%A _Jonathan Sondow_, Jun 21 2013