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User: Colm Bhandal

Colm Bhandal has authored 1 sequences.

A185102 a(n) is the recursion depth of repeatedly factoring n and then the exponents in its prime product factorization, until 1 is reached.

0, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 3, 2, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 3, 3, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1

Offset: 1

Colm Bhandal, Jan 23 2012

nonn

a(n) is the depth of a tree whose root is n and whose construction follows the rule: The children of n are the exponents in its prime factorization. If the number is 1, the convention is taken that the prime factorization is empty, and hence the set of children of 1 is the null set. The formulaic sequence definition given uses pattern matching on n as a product of primes, which is always possible by the fundamental theorem of arithmetic. The max{...} notation refers to the maximum value of the specified set.
The sequence differs from A096309 only at a(1) = 0. The justification for this choice is that the null product of primes has 0 levels.
For all numbers i less than or equal to 2^^x, i.e., 2 tetrated to the x, a(i) < x. Tetration is the next hyperoperation after exponentiation, and can be defined: x^^0 = 1 and x^^y = x^(x^^(y - 1)). Due to this bound the sequence can be seen to grow very slowly.

			a(156) = a(2^2*3*13) = 1 + max{a(2), a(1)} = 1 + max{1, 0} = 1 + 1 = 2

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Vitalii V. Iudelevich, On The Tree Structure of Natural Numbers, II, arXiv:2210.06055 [math.NT], 2022. See Figure 2 p. 2.
Wikipedia, Tetration

Cf. A096309.

f[n_Integer] := FactorInteger[n][[All, 2]]; a[1] = 0; a[n_] := Depth[f[n] //. k_Integer /; k > 1 :> f[k]] - 1; Table[a[n], {n, 1, 105}] (* Jean-François Alcover, Nov 20 2013 *)
a = {0}; Do[AppendTo[a, 1 + Max @@ a[[Union[FactorInteger[n][[All, 2]]]]]], {n, 2, 105}]; a (* Ray Chandler, Nov 22 2013 *)

a(n)=if(n<4,n>1,vecmax(apply(a,Set(factor(n)[,2])))+1) \\ Charles R Greathouse IV, Nov 21 2013

a(1) = 0. a(p1^k1*p2^k2*...*pn^kn) = 1 + max{a(k1), a(k2), ..., a(kn)} with all pi prime and distinct.

a(n) <= lg*(n), where lg*(x) = 0 if x < 2 and lg*(x) = lg*(log_2(x)) otherwise. - Charles R Greathouse IV, Nov 21 2013

Corrected (n=32, 36, 64, 72, 96, 100) by Jean-François Alcover, Nov 20 2013

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User: Colm Bhandal

A185102 a(n) is the recursion depth of repeatedly factoring n and then the exponents in its prime product factorization, until 1 is reached.

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