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.

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

Offset: 1

Author

Vladimir Shevelev and Peter J. C. Moses, Nov 18 2014

Keywords

• nonn
• base
• easy

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.

Examples

			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.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..3453 (all terms less than 10^1000)

Crossrefs

Cf. A000225, A010052, A246553.

Programs

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

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.

Extensions

a(25)-a(50) from Charles R Greathouse IV, Nov 19 2014