A323830 a(0) = 1; thereafter a(n) is obtained by doubling a(n-1) and repeatedly deleting any string of identical digits.

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 636, 1272, 25, 50, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 636, 1272, 25, 50, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 636, 1272, 25, 50

Offset: 0

Author

N. J. A. Sloane, Feb 03 2019, following a suggestion from Yukun Yao

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 A323831.

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

Examples

			After a(15) = 32768 we get 65536 which becomes 636 after deleting "55". Then doubling 636 we get 1272, then 2544 which becomes 25 after deleting "44", then 50, then 100 which becomes 1 after deleting "00", and now the sequence repeats.
		

Links

• Colin Barker, 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, A323831, A323832.

See A035615 for a classic related base-2 sequence.

Programs

Mathematica

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

PARI

Vec((1 + 2*x)*(1 + 4*x^2 + 16*x^4 + 64*x^6 + 256*x^8 + 1024*x^10 + 4096*x^12 + 16384*x^14 + 636*x^16 + 25*x^18) / (1 - x^20) + O(x^40)) \\ Colin Barker, Feb 03 2019

Formula

From Colin Barker, Feb 03 2019: (Start)

G.f.: (1 + 2*x)*(1 + 4*x^2 + 16*x^4 + 64*x^6 + 256*x^8 + 1024*x^10 + 4096*x^12 + 16384*x^14 + 636*x^16 + 25*x^18) / (1 - x^20).

a(n) = a(n-20) for n>19.

(End)

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