OFFSET
1,2
LINKS
FORMULA
a(1) = 1, and for n > 1, a(n) = 1 + a(A005361(n)).
EXAMPLE
For n = 2 = 2^1, A005361(2) = 1, so we reach 1 in one step, and thus a(2) = 1+1 = 2.
For n = 4 = 2^2, A005361(4) = 2; A005361(2) = 1, so we reach 1 in two steps, and thus a(4) = 2+1 = 3.
For n = 6 = 2^1 * 3^1, A005361(6) = 1*1 = 1, so we reach 1 in one step, and thus a(6) = 1+1 = 2.
For n = 64 = 2^6, A005361(64) = 6, thus a(64) = 1 + a(6) = 3.
For n = 10! = 3628800 = 2^8 * 3^4 * 5^2 * 7*1, A005361(3628800) = 64, thus a(3628800) = 1 + a(64) = 4.
MATHEMATICA
ffi[x_] := Flatten[FactorInteger[x]] lf[x_] := Length[FactorInteger[x]] ep[x_] := Table[Part[ffi[x], 2*w], {w, 1, lf[x]}] expr[x_] := Apply[Times, ep[x]] Table[Length[FixedPointList[expr, w]]-1, {w, 2, 128}]
(* Second program: *)
Table[Length@ NestWhileList[Apply[Times, FactorInteger[#][[All, -1]]] &, n, # != 1 &], {n, 105}] (* Michael De Vlieger, Jul 29 2017 *)
PROG
(PARI)
A005361(n) = factorback(factor(n)[, 2]); \\ This function from Charles R Greathouse IV, Nov 07 2014
(PARI) first(n) = my(v = vector(n)); v[1] = 1; for(i=2, n, v[i] = v[factorback(factor(i)[, 2])] + 1); v \\ David A. Corneth, Jul 28 2017
(Scheme) (define (A082091 n) (if (= 1 n) n (+ 1 (A082091 (A005361 n))))) ;; Antti Karttunen, Jul 28 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Labos Elemer, Apr 09 2003
EXTENSIONS
Term a(1)=1 prepended, Name and Example sections edited by Antti Karttunen, Jul 28 2017
STATUS
approved