A323831 a(0) = 5; thereafter a(n) is obtained by doubling a(n-1) and repeatedly deleting any string of identical digits.

5, 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120, 10240, 20480, 40960, 81920, 163840, 327680, 6360, 12720, 250, 5, 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120, 10240, 20480, 40960, 81920, 163840, 327680, 6360, 12720, 250

Offset: 0

Author

N. J. A. Sloane, Feb 03 2019

Keywords

• nonn
• base
• easy

Comments

Periodic with period length 20.

Conjecture: If we start with any nonnegative number, and repeatedly double it and apply the "repeatedly delete any run of identical digits" operation described here, we eventually reach one of 0, 1, or 5.

In other words, the conjecture is that eventually we reach 0 or join the trajectory shown here or the trajectory shown in A323830.

The number of steps to reach 0, 1, or 5 is given in A323832.

Links

• Paolo Xausa, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1).

Crossrefs

Cf. A000079, A320487, A321801, A321802, A323830, A323832.

Programs

Mathematica

dad[n_]:=FromDigits[FixedPoint[Flatten[Select[Split[#],Length[#]==1&]]&,IntegerDigits[2n]]];NestList[dad,5,100] (* Paolo Xausa, Nov 14 2023 *)

Formula

a(n+1) = A321801(2*a(n)). For general numbers, the "repeatedly delete any run of identical digits" operation corresponds to repeatedly applying A321801. - Chai Wah Wu, Feb 11 2019