A336363 -id:A336363 - cp's OEIS frontend

A331410 a(n) is the number of iterations needed to reach a power of 2 starting at n and using the map k -> k + k/p, where p is the largest prime factor of k.

0, 0, 1, 0, 2, 1, 1, 0, 2, 2, 2, 1, 2, 1, 3, 0, 3, 2, 3, 2, 2, 2, 2, 1, 4, 2, 3, 1, 4, 3, 1, 0, 3, 3, 3, 2, 4, 3, 3, 2, 3, 2, 3, 2, 4, 2, 2, 1, 2, 4, 4, 2, 4, 3, 4, 1, 4, 4, 4, 3, 2, 1, 3, 0, 4, 3, 4, 3, 3, 3, 3, 2, 5, 4, 5, 3, 3, 3, 3, 2, 4, 3, 3, 2, 5, 3, 5, 2, 5, 4, 3, 2, 2, 2, 5, 1, 3, 2, 4, 4, 5, 4, 3, 2, 4

Offset: 1

Ali Sada, Jan 16 2020

nonn

Let f(n) = A000265(n) be the odd part of n. Let p be the largest prime factor of k, and say k = p * m. Suppose that k is not a power of 2, i.e., p > 2, then f(k) = p * f(m). The iteration is k -> k + k/p = p*m + m = (p+1) * m. So, p * f(m) -> f(p+1) * f(m). Since for p > 2, f(p+1) < p, the odd part in each iteration decreases, until it becomes 1, i.e., until we reach a power of 2. - Amiram Eldar, Feb 19 2020
Any odd prime factor of k can be used at any step of the iteration, and the result will be same. Thus, like A329697, this is also fully additive sequence. - Antti Karttunen, Apr 29 2020
If and only if a(n) is equal to A005087(n), then sigma(2n) - sigma(n) is a power of 2. (See A336923, A046528). - Antti Karttunen, Mar 16 2021

			The trajectory of 15 is [15,18,24,32], taking 3 iterations to reach 32. So, a(15) = 3.

Antti Karttunen, Table of n, a(n) for n = 1..65537
Michael De Vlieger, Annotated fan style binary tree of a(n) labeling the index n and applying a color code where black represents a(n) = 0, red a(n) = 1, and magenta the largest value of a(n) for n = 1..16383.

Cf. A000265, A005087, A006530 (greatest prime factor), A052126, A078701, A087436, A329662 (positions of records and the first occurrences of each n), A334097, A334098, A334108, A334861, A336467, A336921, A336922, A336923 (A046528).
Cf. array A335430, and its rows A335431, A335882, and also A335874.
Cf. also A329697 (analogous sequence when using the map k -> k - k/p), A335878.
Cf. also A330437, A335884, A335885, A336362, A336363 for other similar iterations.

f:=func; g:=func; a:=[]; for n in [1..1000] do k:=n; s:=0; while not g(k) do  s:=s+1; k:=f(k); end while; Append(~a,s); end for; a; // Marius A. Burtea, Jan 19 2020

a[n_] := -1 + Length @ NestWhileList[# + #/FactorInteger[#][[-1, 1]] &, n, # / 2^IntegerExponent[#, 2] != 1 &]; Array[a, 100] (* Amiram Eldar, Jan 16 2020 *)

A331410(n) = if(!bitand(n,n-1),0,1+A331410(n+(n/vecmax(factor(n)[, 1])))); \\ Antti Karttunen, Apr 29 2020

A331410(n) = { my(k=0); while(bitand(n,n-1), k++; my(f=factor(n)[, 1]); n += (n/f[2-(n%2)])); (k); }; \\ Antti Karttunen, Apr 29 2020

A331410(n) = { my(f=factor(n)); sum(k=1,#f~,if(2==f[k,1],0,f[k,2]*(1+A331410(1+f[k,1])))); }; \\ Antti Karttunen, Apr 30 2020

From Antti Karttunen, Apr 29 2020: (Start)
This is a completely additive sequence: a(2) = 0, a(p) = 1+a(p+1) for odd primes p, a(m*n) = a(m)+a(n), if m,n > 1.
a(2n) = a(A000265(n)) = a(n).
If A209229(n) == 1, a(n) = 0, otherwise a(n) = 1 + a(n+A052126(n)), or equally, 1 + a(n+(n/A078701(n))).
a(n) = A334097(n) - A334098(n).
a(A122111(n)) = A334108(n).
(End)
a(n) = A334861(n) - A329697(n). - Antti Karttunen, May 14 2020
a(n) = a(A336467(n)) + A087436(n) = A336921(n) + A087436(n). - Antti Karttunen, Mar 16 2021

Data section extended up to a(105) by Antti Karttunen, Apr 29 2020

A336361 Number of iterations of A000593 (sum of divisors of odd part of n) needed to reach a power of 2, or -1 if never reached.

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 1, 0, 3, 2, 2, 1, 2, 1, 2, 0, 4, 3, 3, 2, 1, 2, 2, 1, 2, 2, 3, 1, 3, 2, 1, 0, 2, 4, 2, 3, 4, 3, 2, 2, 2, 1, 3, 2, 3, 2, 2, 1, 4, 2, 4, 2, 4, 3, 4, 1, 3, 3, 3, 2, 2, 1, 3, 0, 2, 2, 5, 4, 2, 2, 4, 3, 5, 4, 2, 3, 2, 2, 3, 2, 5, 2, 2, 1, 4, 3, 3, 2, 4, 3, 2, 2, 1, 2, 3, 1, 5, 4, 3, 2, 5, 4, 3, 2, 2

Offset: 1

Antti Karttunen, Jul 30 2020

nonn

Also, for n > 1, one less than the number of iterations of A000593 to reach 1.

If there exists any hypothetical odd perfect numbers w, then the iteration will get stuck into a fixed point after encountering them, and we will have a(w) = a(2^k * w) = -1 by the escape clause.

Antti Karttunen, Table of n, a(n) for n = 1..65537

Cf. A000265, A000593, A161942, A209229, A336362, A336363, A347249, A347250.
Cf. A054784 (positions of 0's and 1's in this sequence).
Cf. also A347240, A347241, A347242, A347243, A347244, A347245.

A336361(n) = if(!bitand(n,n-1),0,1+A336361(sigma(n>>valuation(n,2))));

If A209229(n) = 1 [when n is a power of 2], a(n) = 0, otherwise a(n) = 1+a(A000593(n)).

a(n) = a(2n) = a(A000265(n)).

A336362 Number of iterations of map k -> k*sigma(p^e)/p^e needed to reach a power of 2, where p is the smallest odd prime factor of k and e is its exponent, when starting from k = n. a(n) = -1 if number of the form 2^k is never reached.

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 1, 0, 3, 2, 2, 1, 2, 1, 3, 0, 4, 3, 3, 2, 2, 2, 2, 1, 2, 2, 3, 1, 4, 3, 1, 0, 3, 4, 3, 3, 4, 3, 3, 2, 3, 2, 3, 2, 5, 2, 2, 1, 5, 2, 5, 2, 4, 3, 4, 1, 4, 4, 4, 3, 2, 1, 4, 0, 4, 3, 5, 4, 3, 3, 4, 3, 5, 4, 3, 3, 3, 3, 3, 2, 6, 3, 3, 2, 6, 3, 5, 2, 6, 5, 3, 2, 2, 2, 5, 1, 6, 5, 5, 2, 6, 5, 3, 2, 4

Offset: 1

Antti Karttunen, Jul 30 2020

nonn

Informally: starting from k=n, keep on replacing p^e, the maximal power of the smallest odd prime factor in k, with (1 + p + p^2 + ... + p^e), until a power of 2 is reached. Sequence counts the steps needed.

			For n = 15 = 3*5, we obtain the following path, when starting from k = n, and when we always replace the maximal power of the lowest odd prime factor, p^e of k with sigma(p^e) = (1 + p + p^2 + ... + p^e) in the prime factorization k: 3^1 * 5^1 -> (1+3)*5 = 20 = 2^2 * 5 -> 4 * (1+5) = 24 = 2^3 * 3^1 -> 2^3 * 2^2 = 2^5, thus it took three iterations to reach a power of two, and a(15) = 3.

Antti Karttunen, Table of n, a(n) for n = 1..65537

Cf. A000265, A209229, A336650.

Cf. also A331410, A336361, A336363.

A336650(n) = if(!bitand(n,n-1),1,my(f=factor(n>>valuation(n,2))); f[1, 1]^f[1, 2]);
A336362(n) = if(!bitand(n,n-1),0,my(pe=A336650(n)); 1+A336362((n/pe)*sigma(pe)));

If A209229(n) = 1 [when n is a power of 2], a(n) = 0, otherwise a(n) = 1 + a(sigma(A336650(n))*(n/A336650(n))).

a(n) = a(2n) = a(A000265(n)).

cp's OEIS Frontend

A331410 a(n) is the number of iterations needed to reach a power of 2 starting at n and using the map k -> k + k/p, where p is the largest prime factor of k.

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A336361 Number of iterations of A000593 (sum of divisors of odd part of n) needed to reach a power of 2, or -1 if never reached.

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A336362 Number of iterations of map k -> k*sigma(p^e)/p^e needed to reach a power of 2, where p is the smallest odd prime factor of k and e is its exponent, when starting from k = n. a(n) = -1 if number of the form 2^k is never reached.

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