A246553 Limiting sequence obtained by taking the sequence 0, 2, 3, 5, 7, 11, 13, ... and applying an infinite process which is described in the comments.

1, 2, 7, 7, 7, 43, 5, 16, 19, 87, 25, 31, 1061, 9, 43, 32815, 565, 63, 61, 16451, 7, 73, 1048655, 2131, 91, 97, 131173, 39, 107, 16777325, 4209, 127, 4294967427, 524425, 171, 149, 134217879, 4253, 163, 68719476903, 1048749, 187, 181, 536871103, 2241, 197, 549755814087

Offset: 1

Author

Vladimir Shevelev and Peter J. C. Moses, Nov 16 2014

Keywords

• nonn

Comments

Consider the (0,1)-triangle T_0(P) with entries numbered 1,2,3,4,..., the rows of which, read in binary and converted to decimal, give the sequence 0 followed by consecutive primes, 0,2,3,5,7,...

Let the operator A_k map every k-th entry to its binary opposite (1->0, 0->1), for k=1,2,... .

Put T_inf(P) = ...*A_3*A_2*A_1(T_(0)P), with successive applications of the operators A_1, A_2, A_3, ...

Note that the (0,1)-triangle T_inf(P) 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(P) in binary and converting them to decimal.

Examples

			T_0(P) begins:
0
10
11
101
111
1011
1101
10001
........
T_inf(P) begins:
1
10
111
0111
00111
101011
0000101
00010000
000010011
0001010111
00000011001
000000011111
0010000100101
.............
		

Links

• Peter J. C. Moses, Table of n, a(n) for n = 1..500

Crossrefs

Cf. A000040, A000225, A247092.

Programs

Mathematica

seq=Apply[BitXor,{Map[If[IntegerQ[Sqrt[#]],1,0]&,Range[Length[#]]],#}&[Flatten[Join[{{0}},Map[IntegerDigits[Prime[#],2,#+1]&,Range[50]]]]]];
Map[FromDigits[#,2]&,MapThread[seq[[#1;;#2]]&,({Join[{0},Most[#1]]+1,#1}&)[#/2(#+1)&[Range[NestWhile[#+1&,1,((1+#1) (2+#1)<=2Length[seq])&]]]]]] (* Peter J. C. Moses, Nov 18 2014 *)

Formula

If we take the initial triangle T_0(O) to consist of all 0's, then in T_inf(O) the 1's are only on positions of squares of all positive numbers, i.e., 1,4,9,16,... . Indeed, in order to get an entry in the n-th position of T_inf(O), we should use all considered operators A_d, d|n. The number of these operators is the number of divisors of n which is odd iff n is a perfect square. Thus only in this case, we obtain that entry in the n-th position is flipped, beginning with 0, an odd number of times, such that in the n-th position of T_inf(O) we have 1, while, if n is nonsquare, in the n-th position we have 0.

T_inf(O) begins:

1

00

100

0010

00000

100000

0001000

00000001

.........

Now we have T_inf(P) = XNOR(T_0(P), T_inf(O)).