cp's OEIS Frontend

A247092 Limiting sequence obtained by taking the sequence of Mersenne numbers 2^n-1, n=1,2,...(A000225) and applying an infinite process which is described in the comments.

%I A247092 #26 Dec 11 2015 12:11:15
%S A247092 1,0,4,2,0,32,8,1,0,64,4,0,1024,32,0,32768,512,4,0,16384,64,0,1048576,
%T A247092 2048,2,0,131072,64,0,16777216,4096,0,4294967296,524288,32,0,
%U A247092 134217728,4096,0,68719476736,1048576,8,0,536870912,2048,0,549755813888,1048576,1,0
%N A247092 Limiting sequence obtained by taking the sequence of Mersenne numbers 2^n-1, n=1,2,...(A000225) and applying an infinite process which is described in the comments.
%C A247092 Write the Mersenne numbers 2^n-1, n=1,2,..., in binary in form of triangle T_0(M) consisting of all 1's:
%C A247092 1
%C A247092 11
%C A247092 111
%C A247092 1111
%C A247092 11111
%C A247092 ......
%C A247092 Let the operator A_k map every k-th entry to its binary opposite (1->0, 0->1), for k=2,3,... . Put T_inf(M) = ...*A_4*A_3*A_2(T_(0)M), with successive applications of the operators A_2, A_3, A_4, ...
%C A247092 Note that the (0,1)-triangle T_inf(M) is well-defined, since the operator T_n does not affect entries in the first floor((sqrt(8*n-7) - 1)/2) rows.
%C A247092 The sequence lists numbers obtained by reading rows of T_inf(M) in binary and converting them to decimal.
%C A247092 The n-th entry t_n of T_inf(M) equals 1, if n is perfect square, and 0 otherwise (A010052, for n>=1).
%C A247092 Indeed, in order to get an entry t_n of T_inf(M), we should use all considered operators A_d, d|n, d>1. The number of these operators is the diminished on 1 the number of divisors of n which is even iff n is a perfect square. Thus only in this case we obtain that entry in the n-th position is flipped, beginning with 1, an even number of times, such that t_n=1, while, if n is nonsquare, t_n=0.
%C A247092 Note that, since (n+1)^2 - n^2 > n, then in every row of T_inf(M) there exists at most one 1. So every term is either 0 or a power of 2.
%H A247092 Charles R Greathouse IV, <a href="/A247092/b247092.txt">Table of n, a(n) for n = 1..3453</a> (all terms less than 10^1000)
%F A247092 If there is an integer x in [sqrt((n-1)*n/2 +1), sqrt(n*(n+1)/2)] then it is unique and a(n) = 2^(n(n+1)/2-x^2); otherwise, a(n)=0.
%F A247092 Thus there are n/sqrt(2) + O(1) positive terms among the first n.
%e A247092 T_inf(M) begins
%e A247092 1
%e A247092 00
%e A247092 100
%e A247092 0010
%e A247092 00000
%e A247092 100000
%e A247092 0001000
%e A247092 00000001
%e A247092 .........
%e A247092 Let n=4. Then the interval in the formula is [sqrt(7), sqrt(10)], so x=3 and a(4) = 2^(10-9) = 2.
%o A247092 (PARI) a(n)=my(mx=n*(n+1)/2,s=sqrtint(mx)^2); if(s<=mx-n, 0, 2^(mx-s)) \\ _Charles R Greathouse IV_, Nov 19 2014
%Y A247092 Cf. A000225, A010052, A246553.
%K A247092 nonn,base,easy
%O A247092 1,3
%A A247092 _Vladimir Shevelev_ and _Peter J. C. Moses_, Nov 18 2014
%E A247092 a(25)-a(50) from _Charles R Greathouse IV_, Nov 19 2014