0,2

This sequence can be also computed with a recurrence that does not explicitly refer to 3^n. See the given C-program.

Conjecture: For n>=3, each term a(n), when considered as a GF(2)[X]-polynomial, is divisible by GF(2)[X] -polynomial (x + 1) ^ A055010(n-1). If this holds, then for n>=3, a(n) = A048720bi(A136386(n),A048723bi(3,A055010(n-1))).

S. Wolfram, A New Kind of Science, Wolfram Media Inc., (2002), p. 119.

A. Karttunen, Table of n, a(n) for n = 0..12

A. Karttunen, C program for computing this sequence

S. Wolfram, A New Kind of Science, Wolfram Media Inc., (2002), p. 119.

a(n) = Sum_{k=0..A000225(n)} ([A000244(k)/(2^n)] mod 2) * 2^k, where [] stands for floor function.

When powers of 3 are written in binary (see A004656), under each other as:

000000000001 (1)

000000000011 (3)

000000001001 (9)

000000011011 (27)

000001010001 (81)

000011110011 (243)

001011011001 (729)

100010001011 (2187)

it can be seen that the bits in the nth column from the right can be arranged in periods of 2^n: 1, 2, 4, 8, ... This sequence is formed from those bits: 1, is binary for 1, thus a(0) = 1. 01, reversed is 10, which is binary for 2, thus a(1) = 2, 0000 is binary for 0, thus a(2)=0, 000110011, reversed is 111001100, which is binary for 204, thus a(3) = 204.

a(n) := sum( 'bit_n(3^i, n)*(2^i)', 'i'=0..(2^(n))-1);

bit_n := (x, n) -> `mod`(floor(x/(2^n)), 2);

For n>=2, a(n) = A000215(n-1)*A037097(n) = A048720bi(A037097(n),A048723bi(3,A000079(n-1))).

Cf. A036284, A037095, A037097, A136386.

Sequence in context: A013555 A012335 A012331 this_sequence A111814 A036938 A012330

Adjacent sequences: A037093 A037094 A037095 this_sequence A037097 A037098 A037099

nonn,base

Antti Karttunen (His_Firstname.His_Surname(AT)gmail.com), Jan 29 1999. Entry revised Dec 29 2007.