

A153013


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


3



0, 1, 2, 3, 4, 5, 6, 9, 10, 15, 16, 11, 12, 25, 50, 147, 220, 6125, 1968750, 89142864525, 84252896510182189218, 34892570216750728458698250328871491829901861750593684043
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,3


COMMENTS

From Antti Karttunen, Oct 15 2016: (Start)
Iterates of map f : n > A005940(1+n), (Doudnasequence, 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 1bit in the binary expansion of any square is always alone.
(End)


LINKS

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


FORMULA

From Antti Karttunen, Oct 15 2016: (Start)
a(0) = 0; for n >= 1, a(n) = A005940(1+a(n1)).
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)


EXAMPLE

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


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 *)


PROG

(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 memoizationmacro definec)
(definec (A153013 n) (if (zero? n) n (A005940 (+ 1 (A153013 ( n 1))))))
;; Antti Karttunen, Oct 15 2016


CROSSREFS

Cf. A000005, A003714, A005940, A008966, A085357, A106737, A156552, A181819, A246029.
Sequence in context: A082642 A217348 A319450 * A052492 A162228 A085714
Adjacent sequences: A153010 A153011 A153012 * A153014 A153015 A153016


KEYWORD

nonn


AUTHOR

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


EXTENSIONS

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


STATUS

approved



