

A159619


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


10



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
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OFFSET

1,1


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 vanEckTransforms (of A171898) to reach a value larger than a(n1). The Dekking formula in A026491 says that A026491 is essentially a partial sum of the backward vanEckTransforms, 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
HsienKuei 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.
HsienKuei Hwang, S. Janson and T.H. Tsai, Exact and Asymptotic Solutions of a DivideandConquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585.
Jon Maiga, Computergenerated formulas for A159619, Sequence Machine.
Vladimir Shevelev, Several results on sequences which are similar to the positive integers, arXiv:0904.2101 [math.NT], 2009.


FORMULA

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


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(n1)+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


CROSSREFS

Cf. A000069, A001969, A159615, A007814, A004760, A159559, A159560.
Sequence in context: A273916 A053169 A007656 * A272454 A207017 A174724
Adjacent sequences: A159616 A159617 A159618 * A159620 A159621 A159622


KEYWORD

nonn,base,easy


AUTHOR

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


EXTENSIONS

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


STATUS

approved



