A246685 Run Length Transform of sequence 1, 3, 5, 17, 257, 65537, ... (1 followed by Fermat numbers).

1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 1, 3, 3, 3, 5, 17, 1, 1, 1, 3, 1, 1, 3, 5, 3, 3, 3, 9, 5, 5, 17, 257, 1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 1, 3, 3, 3, 5, 17, 3, 3, 3, 9, 3, 3, 9, 15, 5, 5, 5, 15, 17, 17, 257, 65537, 1, 1, 1, 3, 1, 1, 3, 5, 1, 1, 1, 3, 3, 3, 5, 17, 1, 1, 1, 3, 1, 1, 3, 5, 3, 3, 3, 9, 5, 5, 17, 257

Offset: 0

Author

Antti Karttunen, Sep 22 2014

Keywords

• nonn

Comments

The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g. 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product).

This sequence is obtained by applying Run Length Transform to sequence b = 1, 3, 5, 17, 257, 65537, ... (1 followed by Fermat numbers, with b(1) = 1, b(2) = 3, b(3) = 5, ..., b(n) = 2^(2^(n-2)) + 1 for n >= 2).

Examples

			115 is '1110011' in binary. The run lengths of 1-runs are 2 and 3, thus we multiply the second and the third elements of the sequence 1, 3, 5, 17, 257, 65537, ... to get a(115) = 3*5 = 15.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..1024

Crossrefs

Cf. A003714 (gives the positions of ones).

Cf. A000215.

A001316 is obtained when the same transformation is applied to A000079, the powers of two. Cf. also A001317.

Run Length Transforms of other sequences: A071053, A227349, A246588, A246595, A246596, A246660, A246661, A246674, A247282.

Programs

Mathematica

f[n_] := Switch[n, 0|1, 1, , 2^(2^(n-2))+1]; Table[Times @@ (f[Length[#]] &) /@ Select[s = Split[IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 95}] (* _Jean-François Alcover, Jul 11 2017 *)

Python

# use RLT function from A278159
def A246685(n): return RLT(n,lambda m: 1 if m <= 1 else 2**(2**(m-2))+1) # Chai Wah Wu, Feb 04 2022