A107281 a(0) = 1, a(1) = 1, a(2) = 2 and for n >= 1: a(n+1) = SORT[a(n) + a(n-1) + a(n-2)] where SORT places digits in ascending order and deletes 0's.

1, 1, 2, 4, 7, 13, 24, 44, 18, 68, 13, 99, 18, 13, 13, 44, 7, 46, 79, 123, 248, 45, 146, 349, 45, 45, 349, 349, 347, 145, 148, 46, 339, 335, 27, 17, 379, 234, 36, 469, 379, 488, 1336, 223, 247, 168, 368, 378, 149, 589, 1116, 1458, 1336, 139, 2339, 1348, 2368, 556, 2247

Offset: 0

Author

Jonathan Vos Post, Jun 08 2005

Keywords

• base
• easy
• nonn

Comments

The maximum value is 56899, which first occurs at a(275). The maximum next occurs at a(977). T. D. Noe verified that the terms around a(275) and a(977) are the same. Hence the period is 977 - 275 = 702. The actual period starts at a(24) with the interesting terms 349, 45, 45, 349, 349. For some different initial conditions, the period is different. The point of the SORT operation here is that it "mixes" the sequence and the questions are, considering cycles as orbits, all about ergodicity. To turn this into the sorted Fibonacci sequence (A069638), use a(0)=0, a(1)=1, a(2)=1. This is a "base" sequence, but has analogs in other bases; for instance, SORT(base 2)[n] means count the 1's in the binary, call that k and output 2^(k-1). How does this sequence depend on SORT(base M)[n] for various M? Are there any initial values such that the sequence us unbounded? If not, how does cycle length depend upon initial values?

Examples

			a(8) = 18 because a(5) + a(6) + a(7) = 13 + 24 + 44 = 81 and SORT(81) = 18.
		

Links

• Richard I. Hess, Problem 920: sorted Fibonacci sequence, Pi Mu Epsilon Journal, Vol. 10 (Fall 1998) No. 9, pp. 754-755.

Crossrefs

Cf. A000073, A069638.

Programs

Mathematica

nxt[{a_,b_,c_}]:=Module[{d=FromDigits[Sort[IntegerDigits[a+b+c]]]}, {b,c,d}]; Transpose[NestList[nxt,{1,1,2},65]][[1]]  (* Harvey P. Dale, Feb 07 2011 *)

Formula

a(0) = 1, a(1) = 1, a(2) = 2 and for n>1: a(n+1) = SORT[a(n) + a(n-1) + a(n-2)] where SORT places digits in ascending order and deletes 0.