1,2

a(n) = b(n) for n<=2, a(n) = b(a(n-1)) for n>2, where b(n) = A000384(n) = n*(2*n-1), the hexagonal numbers.

Agrees with A097419 for n>1.

J. V. Post, "Iterated Polygonal Numbers", preprint.

J. V. Post, "Iterated Triangular Numbers", preprint.

Eric Weisstein's World of Mathematics, "Hexagonal Number."

a(1) = 1, a(2) = 6, a(n) = 2*a(n-1)^2 - a(n-1) for n>2.

Let H(n) = n*(2*n-1) = the n-th hexagonal number. Define A(n, k) recursively by A(1, k) = H(k), A(n, k) = A(1, A(n-1, k)) for n>1. Then a(1) = A(1, 1), a(n) = A(n-1, 2) for n>1.

a(4) = b(a(3)) = b(b(a(2))) = b(b(b(2))) = b(b(6)) = b(66) = 8646, where b(n) = A000384(n).

(PARI) {hexagonal(n) = n*(2*n-1)} {a(n) = if(n<=2, hexagonal(n), hexagonal(a(n-1)))} - Klaus Brockhaus, Jan 10 2007

Cf. A000384, A097419.

Sequence in context: A068966 A063039 A082781 this_sequence A073326 A024203 A073562

Adjacent sequences: A099144 A099145 A099146 this_sequence A099148 A099149 A099150

nonn

Jonathan Vos Post (jvospost2(AT)yahoo.com), Nov 14 2004

Edited by Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Jan 10 2007