cp's OEIS Frontend

A037096 Periodic vertical binary vectors computed for powers of 3: a(n) = Sum_{k=0 .. (2^n)-1} (floor((3^k)/(2^n)) mod 2) * 2^k.

%I A037096 #28 Jan 14 2024 08:59:18
%S A037096 1,2,0,204,30840,3743473440,400814250895866480,
%T A037096 192435610587299441243182587501623263200,
%U A037096 2911899996313975217187797869354128351340558818020188112521784134070351919360
%N A037096 Periodic vertical binary vectors computed for powers of 3: a(n) = Sum_{k=0 .. (2^n)-1} (floor((3^k)/(2^n)) mod 2) * 2^k.
%C A037096 This sequence can be also computed with a recurrence that does not explicitly refer to 3^n. See the C program.
%C A037096 Conjecture: For n >= 3, each term a(n), when considered as a GF(2)[X] polynomial, is divisible by the GF(2)[X] polynomial (x + 1) ^ A055010(n-1). If this holds, then for n >= 3, a(n) = A048720(A136386(n), A048723(3,A055010(n-1))).
%D A037096 S. Wolfram, A New Kind of Science, Wolfram Media Inc., (2002), p. 119.
%H A037096 Antti Karttunen, <a href="/A037096/b037096.txt">Table of n, a(n) for n = 0..11</a>
%H A037096 Antti Karttunen, <a href="/A036284/a036284.c.txt">C program for computing this sequence</a>.
%H A037096 S. Wolfram, <a href="http://www.wolframscience.com/nksonline/page-119">A New Kind of Science, Wolfram Media Inc., (2002), p. 119</a>.
%F A037096 a(n) = Sum_{k=0 .. A000225(n)} (floor(A000244(k)/(2^n)) mod 2) * 2^k.
%F A037096 Other identities and observations:
%F A037096 For n >= 2, a(n) = A000215(n-1)*A037097(n) = A048720(A037097(n), A048723(3, A000079(n-1))).
%e A037096 When powers of 3 are written in binary (see A004656), under each other as:
%e A037096   000000000001 (1)
%e A037096   000000000011 (3)
%e A037096   000000001001 (9)
%e A037096   000000011011 (27)
%e A037096   000001010001 (81)
%e A037096   000011110011 (243)
%e A037096   001011011001 (729)
%e A037096   100010001011 (2187)
%e A037096 it can be seen that the bits in the n-th 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 11001100 = A007088(204), thus a(3) = 204.
%p A037096 a(n) := sum( 'bit_n(3^i, n)*(2^i)', 'i'=0..(2^(n))-1);
%p A037096 bit_n := (x, n) -> `mod`(floor(x/(2^n)), 2);
%Y A037096 Cf. A036284, A037095, A037097, A136386 for related sequences.
%Y A037096 Cf. A000079, A000215, A000225, A000244, A004656, A007088, A048720, A048723, A055010.
%Y A037096 Cf. also A004642, A265209, A265210 (for 2^n written in base 3).
%K A037096 nonn,base
%O A037096 0,2
%A A037096 _Antti Karttunen_, Jan 29 1999
%E A037096 Entry revised by _Antti Karttunen_, Dec 29 2007
%E A037096 Name changed and the example corrected by _Antti Karttunen_, Dec 05 2015