1,1

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.

a(1)= 3t(t+1)+1 with t=10^0 a(2)= 3t(t+1)+1 with t=10^1 a(3)= 3t(t+1)+1 with t=10^4

A002407, A003215

Sequence in context: A002000 A092588 A049686 this_sequence A009587 A054325 A161582

Adjacent sequences: A160429 A160430 A160431 this_sequence A160433 A160434 A160435

nonn

Giacomo Fecondo (jackfertile(AT)alice.it), May 13 2009