0,1

This is to Lucas numbers as A130150 is to Fibonacci numbers.

a(n) = A003617(A000032(n)) = (MIN{p(i) such that p(i) is prime and number of digits of decimal expansion of p(i) = L(n) for Lucas number L(n)} = MIN{A000040(i) such that A055642(A000040(i)) = A000032(n)}.

a(0) = 11 because L(0) = A000032(0) = 2 and 11 is the smallest 2 digit prime.

a(1) = 2 because L(1) = A000032(1) = 1 and 2 is the smallest 1 digit prime.

a(2) = 101 because L(2) = A000032(2) = 3 and 101 is the smallest 3 digit prime.

Cf. A000032, A000040, A003617, A130150.

Sequence in context: A095157 A110767 A089365 this_sequence A096044 A160464 A038316

Adjacent sequences: A130214 A130215 A130216 this_sequence A130218 A130219 A130220

easy,nonn,base,less

Jonathan Vos Post (jvospost3(AT)gmail.com), Aug 04 2007