Higher-order composite numbers

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A050435 $a(n)$ = composite(composite($n$)), where composite = A002808, composite numbers.

{9, 12, 15, 16, 18, 21, 24, 25, 26, 28, 32, 33, 34, 36, 38, 39, 40, 42, 45, 48, 49, 50, 51, 52, 55, 56, 57, 60, 63, 64, 65, 68, 69, 70, 72, 74, 76, 77, 78, 80, 81, 84, 86, 87, ...}

Order of compositeness

Let $c_{k}=k$-th composite number, let $S(c)=S(c_{k})=k$, the subscript of $c$; a(n) is order of compositeness of $c(n)=1+m$ where $m$ is largest number such that $S(S(..S(c(n))...))$ with $m$ $S$'s is not a composite.

A059981 Order of compositeness of $n$ (it is actually: of $n$-th composite): $a(c(n))=a(n)+1$ for $n$-th composite $c(n)$.

{1, 1, 1, 2, 1, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 4, 1, 3, 1, 2, 4, 2, 1, 3, 4, 5, 2, 4, 1, 2, 1, 3, 5, 3, 2, 4, 1, 5, 6, 3, 1, 5, 1, 2, 3, 2, 1, 4, 6, 4, 3, 5, 1, 2, 6, 7, 4, 2, ...}

A?????? Order of compositeness of $n$: $a(c(n))=a(n)+1$ for $n$-th composite $c(n)$; $a(n)=0$ if $n$ is not composite.

{0, 0, 0, 1, 0, 1, 0, 1, 2, 1, 0, 2, 0, 1, 2, 3, 0, 2, 0, 1, 3, 1, 0, 2, 3, 4, 1, 3, 0, 1, 0, 2, ...}

See also

• Higher-order prime numbers
• Order of primeness