A153013 - OEIS

%I #44 Apr 05 2020 21:19:07

%S 0,1,2,3,4,5,6,9,10,15,16,11,12,25,50,147,220,6125,1968750,

%T 89142864525,84252896510182189218,

%U 34892570216750728458698250328871491829901861750593684043

%N Starting with input 0, find the binary value of the input. Then interpret resulting string of 1's and 0's as prime-based numbers, as follows: 0's are separators, uninterrupted strings of 1's are interpreted from right to left as exponents of the prime numbers. Output is returned as input for the next number in sequence.

%C From _Antti Karttunen_, Oct 15 2016: (Start)

%C Iterates of map f : n -> A005940(1+n), (Doudna-sequence, but with starting offset zero) starting from the initial value 0. Conversely, the unique infinite sequence such that a(n) = A156552(a(n+1)) and a(0) = 0.

%C Note that map f can also form cycles, like 7 <-> 8 (A005940(1+7) = 8, A005940(1+8) = 7).

%C On the other hand, this sequence cannot ever fall into a loop because 0 is not in the range of map f, for n=0.., while f is injective on [1..]. Thus the values obtained by this sequence are not bounded, although there might be more nonmonotonic positions like for example there is from a(10) = 16 to a(11) = 11.

%C The formula A008966(a(n+1)) = A085357(a(n)) tells that the squarefreeness of the next term a(n+1) is determined by whether the previous term a(n) is a fibbinary number (A003714) or not. Numerous other such correspondences hold, and they hold also for any other trajectories outside of this sequence.

%C Even and odd terms alternate. No two squares can occur in succession because A106737 obtains even values for all squares > 1 and A000005 is odd for all squares. More directly this is seen from the fact that the rightmost 1-bit in the binary expansion of any square is always alone.

%C (End)

%H Yang Haoran, <a href="/A153013/b153013.txt">Table of n, a(n) for n = 0..23</a>

%F From _Antti Karttunen_, Oct 15 2016: (Start)

%F a(0) = 0; for n >= 1, a(n) = A005940(1+a(n-1)).

%F A008966(a(n+1)) = A085357(a(n)). [See the comment.]

%F A181819(a(1+n)) = A246029(a(n)).

%F A000005(a(n+1)) = A106737(a(n)).

%F (End)

%e 101 is interpreted as 3^1 * 2^1 = 6. 1110011 is interpreted as 5^3 * 2^2 = 500.

%t NestList[Times @@ Flatten@ MapIndexed[Prime[#2]^#1 &, #] &@ Flatten@ MapIndexed[If[Total@ #1 == 0, ConstantArray[0, Boole[First@ #2 == 1] + Length@ #1 - 1], Length@ #1] &, Reverse@ Split@ IntegerDigits[#, 2]] &, 0, 21] (* _Michael De Vlieger_, Oct 17 2016 *)

%o (PARI) step(n)=my(t=1,v); forprime(p=2,, v=valuation(n+1,2); t*=p^v; n>>=v+1; if(!n, return(t)))

%o t=0; concat(0,vector(20,n, t=step(t))) \\ _Charles R Greathouse IV_, Sep 01 2015

%o (Scheme, with memoization-macro definec)

%o (definec (A153013 n) (if (zero? n) n (A005940 (+ 1 (A153013 (- n 1))))))

%o ;; _Antti Karttunen_, Oct 15 2016

%Y Cf. A000005, A003714, A005940, A008966, A085357, A106737, A156552, A181819, A246029.

%K nonn

%O 0,3

%A Mark Zegarelli (mtzmtz(AT)gmail.com), Dec 16 2008

%E a(20)-a(22) from _Yang Haoran_, Aug 31 2015