| a(1) = 7 = (10^0+1)^3 -(10^0)^3 , 2^3-1^3
a(2) = 331 =(10^1+1)^3 -(10^1)^3, 11^3-10^3
a(3) = 300030001 = (10^4+1)^3 - (10^4)^3, 10001^3-10000^3
These prime numbers (differences of consecutive cubes), for k>0, have only
three digits different from zero. The first is 3, in the middle 3 and the
last is 1. The other 2(k-1) digits have value 0 and are positioned,
in the same quantity, at the left and right, of the central digit.
If k=6*i or k=6*i-1, for each i positive integer, the number is always divisible for 7. [From Giacomo Fecondo (jackfertile(AT)alice.it), May 22 2010]
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