A082091 a(n) = one more than the number of iterations of A005361 needed to reach 1 from the starting value n.
1, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 3, 2, 2, 2, 4, 2, 3, 2, 3, 2, 2, 2, 3, 3, 2, 3, 3, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 4, 3, 3, 2, 3, 2, 3, 2, 3, 2, 2, 2, 3, 2, 2, 3, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 3, 3, 2, 2, 2, 4, 4, 2, 2, 3, 2, 2, 2, 3, 2, 3, 2, 3, 2, 2, 2, 3, 2, 3, 3, 4, 2, 2, 2, 3, 2, 2
Offset: 1
Keywords
Examples
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.
Links
Programs
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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 *) -
PARI
A005361(n) = factorback(factor(n)[, 2]); \\ This function from Charles R Greathouse IV, Nov 07 2014 A082091(n) = if(1==n,1,1+A082091(A005361(n))); \\ Antti Karttunen, Jul 28 2017
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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
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Scheme
(define (A082091 n) (if (= 1 n) n (+ 1 (A082091 (A005361 n))))) ;; Antti Karttunen, Jul 28 2017
Formula
a(1) = 1, and for n > 1, a(n) = 1 + a(A005361(n)).
Extensions
Term a(1)=1 prepended, Name and Example sections edited by Antti Karttunen, Jul 28 2017