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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. 2
1, 0, 4, 2, 0, 32, 8, 1, 0, 64, 4, 0, 1024, 32, 0, 32768, 512, 4, 0, 16384, 64, 0, 1048576, 2048, 2, 0, 131072, 64, 0, 16777216, 4096, 0, 4294967296, 524288, 32, 0, 134217728, 4096, 0, 68719476736, 1048576, 8, 0, 536870912, 2048, 0, 549755813888, 1048576, 1, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
Write the Mersenne numbers 2^n-1, n=1,2,..., in binary in form of triangle T_0(M) consisting of all 1's:
1
11
111
1111
11111
......
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, ...
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.
The sequence lists numbers obtained by reading rows of T_inf(M) in binary and converting them to decimal.
The n-th entry t_n of T_inf(M) equals 1, if n is perfect square, and 0 otherwise (A010052, for n>=1).
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.
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.
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..3453 (all terms less than 10^1000)
FORMULA
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.
Thus there are n/sqrt(2) + O(1) positive terms among the first n.
EXAMPLE
T_inf(M) begins
1
00
100
0010
00000
100000
0001000
00000001
.........
Let n=4. Then the interval in the formula is [sqrt(7), sqrt(10)], so x=3 and a(4) = 2^(10-9) = 2.
PROG
(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
CROSSREFS
Sequence in context: A371076 A019155 A107724 * A177715 A164789 A203986
KEYWORD
nonn,base,easy
AUTHOR
EXTENSIONS
a(25)-a(50) from Charles R Greathouse IV, Nov 19 2014
STATUS
approved

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Last modified July 19 06:32 EDT 2024. Contains 374389 sequences. (Running on oeis4.)