A153013 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.

0, 1, 2, 3, 4, 5, 6, 9, 10, 15, 16, 11, 12, 25, 50, 147, 220, 6125, 1968750, 89142864525, 84252896510182189218, 34892570216750728458698250328871491829901861750593684043

Offset: 0

Author

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

Keywords

• nonn

Comments

From Antti Karttunen, Oct 15 2016: (Start)

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.

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

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.

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.

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.

(End)

Examples

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

Links

• Yang Haoran, Table of n, a(n) for n = 0..23

Crossrefs

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

Cf. also A109162, A328316 for similar iteration sequences.

Programs

Mathematica

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 *)

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)))
t=0; concat(0,vector(20,n, t=step(t))) \\ Charles R Greathouse IV, Sep 01 2015

Scheme

;; With memoization-macro definec.
(definec (A153013 n) (if (zero? n) n (A005940 (+ 1 (A153013 (- n 1))))))
;; Antti Karttunen, Oct 15 2016

Formula

From Antti Karttunen, Oct 15 2016: (Start)

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

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

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

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

(End)

Extensions

a(20)-a(22) from Yang Haoran, Aug 31 2015