A261416 Let b(k) denote A260273(k). It appears that for k >= 200, whenever b(k) just passes a power of 2, 2^m say, the successive differences b(k)-2^m converge to this sequence.

2, 5, 8, 11, 17, 20, 23, 29, 38, 43, 49, 54, 61, 70, 75, 81, 84, 87, 93, 102, 107, 114, 119, 128, 131, 136, 139, 145, 148, 151, 157, 167, 173, 180, 187, 196, 201, 206, 211, 218, 225, 230, 235, 244, 253, 262, 267, 273, 276, 279, 285, 294, 299, 305, 310, 317, 327, 333, 340, 343, 349, 358, 365, 372, 381

Offset: 0

Author

N. J. A. Sloane, Aug 25 2015

Keywords

• nonn

Comments

It would be nice to have an independent characterization of this sequence.

A partial answer: set a(0)=2, and for n>0, a(n) = A261281(a(n-1)). - N. J. A. Sloane, Sep 17 2015

Examples

			At k=200, b(k)=b(200)=1026 has just passed 2^10. The successive differences b(200+i)-2^10 (i>=0) beyond this point are 2, 5, 8, 11, 17, 20, 23, 29, 38, 43, 49, 54, 61, 70, 75, 81, 84, 87, 93, 102, 107, 114, 119, 128, 131, 136, 139, 145, 148, 151, 157, [165, ...], which are the first 31 terms of the present sequence.
At k=371, b(371)=2050, and the successive differences b(371+i)-2^11 are 2, 5, ..., 279, 285, ... giving the first 51 terms of the present sequence.
		

Crossrefs

Cf. A260273, A261281. For when A260273 just passes a power of 2, see A261396.