A159619 Slowest increasing sequence beginning with 4 such that n and a(n) are either both evil or both odious.

4, 7, 9, 11, 12, 15, 16, 19, 20, 23, 25, 27, 28, 31, 33, 35, 36, 39, 41, 43, 44, 47, 48, 51, 52, 55, 57, 59, 60, 63, 64, 67, 68, 71, 73, 75, 76, 79, 80, 83, 84, 87, 89, 91, 92, 95, 97, 99, 100, 103, 105, 107, 108, 111, 112, 115, 116, 119, 121, 123, 124, 127, 129, 131, 132, 135, 137

Offset: 1

Author

Vladimir Shevelev, Apr 17 2009, Apr 27 2009, May 04 2009

Keywords

• nonn
• base
• easy

Comments

(i) Theorem: For every initial value a(1) > 4, a minimum index n exists such that the a(n) obtained from that initial value coincides with this sequence here. Thus there exist essentially two slowest increasing sequences with this type of evil/odious congruence: A159615 and this one here.

(ii) In connection with this theorem, one can generalize to slowest increasing sequences a_m(n), a_m(1)=m, which let n and a(n) be at the same time in or not in some increasing sequence c(n). (This sequence here is c = A000069, m=4.)

We define a rank r of c as the minimum value a_r(1) such that for sufficiently large n (n depending on m) all sequences a_m(n), m>r, coincide with a_r(n).

In particular, c(n)=A004760(n+1) has rank r=2, and A000069 has rank r=3.

The problems are: 1) to find a sequence of rank r >= 4; 2) to find the rank of primes or to prove that it does not exist (in case of which it could be defined as infinity).

There is a conjecture arising in Sequence Machine that a(n) = A026491(2+n)-1. This appears to be true: Here we start from on odious or evil number and apply a minimum number of van-Eck-Transforms (of A171898) to reach a value larger than a(n-1). The Dekking formula in A026491 says that A026491 is essentially a partial sum of the backward van-Eck-Transforms, and in a (vague) manner this seems to match.

- R. J. Mathar, Jun 24 2021

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Hsien-Kuei Hwang, S. Janson and T.-H. Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint 2016.
• Hsien-Kuei Hwang, S. Janson and T.-H. Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585.
• Jon Maiga, Computer-generated formulas for A159619, Sequence Machine.
• Vladimir Shevelev, Several results on sequences which are similar to the positive integers, arXiv:0904.2101 [math.NT], 2009.

Crossrefs

Cf. A000069, A001969, A159615, A007814, A004760, A159559, A159560.

Programs

Maple

read("transforms") ; isA000069 := proc(n) option remember ; RETURN( type(wt(n), 'odd') ) ; end:
A159619 := proc(n) option remember; if n = 1 then 4; else for a from procname(n-1)+1 do if isA000069(a) = isA000069(n) then RETURN(a) ; fi; od: fi; end:
seq(A159619(n), n=1..120) ; # R. J. Mathar, Mar 25 2010

Mathematica

a[n_] := 2 * n + If[EvenQ[n] || EvenQ[IntegerExponent[n+1, 2]], 3, 2]; Array[a, 100] (* Amiram Eldar, Aug 30 2024 *)

PARI

a(n) = 2 * n + if(!(n % 2) || !(valuation(n+1, 2) % 2), 3, 2); \\ Amiram Eldar, Aug 30 2024

Formula

a(n) = 2n+3 if n*A007814(n+1) is even, and a(n) = 2n+2 otherwise.

Extensions

Edited and extended by R. J. Mathar, Mar 25 2010