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 A319864 Exponents of the final nontrivial entry of the iterated Stern sequence; a(n) = log_2 min{s^k(n) : k > 0, s^k(n) > 1}, where s(n) = A002487(n). 0

%I

%S 1,1,2,1,1,1,3,2,1,1,1,1,1,2,4,1,2,1,1,3,1,1,1,1,1,3,1,1,2,1,5,1,1,2,

%T 2,1,1,1,1,1,3,1,1,1,1,2,1,2,1,1,1,1,3,1,1,1,1,1,2,2,1,1,6,1,1,1,1,1,

%U 2,1,2,2,1,2,1,1,1,1,1,1,1,1,3,3,1,2,1,1,1,1,1,4,2,1,1,1,2,4,1,1,1,1,1,2,1,3,3,1,1

%N Exponents of the final nontrivial entry of the iterated Stern sequence; a(n) = log_2 min{s^k(n) : k > 0, s^k(n) > 1}, where s(n) = A002487(n).

%C Let s(n) = A002487(n). Since s(n) < n for n > 1, iterating A002487 from any starting point eventually yields the fixed point 1 = s(1). Since s^-1(1) consists of the powers of 2, min{s^k(n) : k > 0, s^k(n) > 1} is a nontrivial power of 2 for any n > 1. Hence, the entries of this sequence are integers.

%C Since a(2^m) = m, every positive integer appears in this sequence.

%C Question: What is the asymptotic density of the number 1 in this sequence? Of the first 10^6 entries, more than 74% are 1.

%e Letting s(m) = A002487(m), we have s(7) = 3, s(3) = 2, and s(2) = 1. Hence, a(7) = log_2(2) = 1.

%t s[n_] := If[n<2, n, If[EvenQ[n], s[n/2], s[(n-1)/2] + s[(n+1)/2]]]; a[n_] := Module[{nn = s[n]}, If[nn==1, Log2[n], a[nn]]]; Array[a, 100, 2] (* _Amiram Eldar_, Nov 22 2018 *)

%o (Python)

%o from math import log

%o def s(n): return n if n<2 else s(n//2) if n%2==0 else s((n-1)//2)+s((n+1)//2)

%o def a(n): nn = s(n); return int(log(n,2)) if nn==1 else a(nn)

%o print([a(n) for n in range(2, 100)])

%o (PARI) s(n) = if( n<2, n>0, s(n\2) + if( n%2, s(n\2 + 1))); \\ A002487

%o a(n) = while((nn = s(n)) != 1, n = nn); valuation(n, 2); \\ _Michel Marcus_, Nov 23 2018

%Y Cf. A002487.

%K nonn

%O 2,3

%A _Oliver Pechenik_, Sep 29 2018

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Last modified July 2 09:21 EDT 2020. Contains 335398 sequences. (Running on oeis4.)