A026166 - OEIS

%I #35 Jul 03 2020 07:13:59

%S 1,2,4,3,8,10,5,6,16,7,20,22,9,26,28,11,12,34,13,14,40,15,44,46,17,18,

%T 52,19,56,58,21,62,64,23,24,70,25,74,76,27,80,82,29,30,88,31,32,94,33,

%U 98,100,35,36,106,37,38,112,39,116,118,41,42

%N For n >= 2, let h=floor((n-1)/2), L=n-h, R=n+h; then a(L)=n if a(L) not yet defined, otherwise a(R)=n; thus |a(n)-n| = floor((1/2)*(a(n)-1)).

%C Every positive integer occurs exactly once. The inverse permutation of the positive integers is given by A026167. - _Clark Kimberling_, Oct 20 2019

%H Clark Kimberling, <a href="/A026166/b026166.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..1502 from R. J. Mathar)

%H F. M. Dekking, <a href="https://arxiv.org/abs/2001.08915">Permutations of N generated by left-right filling algorithms</a>, arXiv:2001.08915 [math.CO], 2020.

%F |a(n)-n| = floor((1/2)*(a(n)-1)).

%F This formula does not permit us to calculate the n-th term of the sequence, since the equation |x-n| = floor((1/2)*(x-1)) has at least 2 integer solutions for all n. - _Michel Dekking_, Nov 26 2019

%t a[1] = 1; z = 300;

%t Do[{L, R} = {n - #, n + #} &[Floor[(n - 1)/2]];

%t   If[! Head[a[L]] === Integer, a[L] = n, a[R] = n], {n, 2, z}];

%t a026166 = Most[Last[

%t    Last[Reap[NestWhile[# + 1 &, 1, Head[Sow[a[#]]] === Integer &]]]]];

%t ListPlot[a026166]  (* _Peter J. C. Moses_, Oct 20 2019 *)

%o (Python)

%o A026166 = {1: 1}

%o for n in range(2, 1000):

%o     h=(n-1)//2

%o     L=n-h

%o     R=n+h

%o     if not L in A026166 :

%o         A026166[L]=n

%o     else :

%o         A026166[R]=n

%o for n in range(1,2000):

%o     if n in A026166:

%o        print(A026166[n], end=',')

%o     else:

%o         break

%o # _R. J. Mathar_, Aug 26 2019

%o (PARI) seq(n)={my(a=vector(n)); a[1]=1; for(i=1, 2*n-1, my(h=(i-1)\2); if(!a[i-h], a[i-h]=i, if(i+h<=n, a[i+h]=i))); a} \\ _Andrew Howroyd_, Oct 15 2019

%Y Cf. A026167, A026182, A026198, A026202.

%K nonn

%O 1,2

%A _Clark Kimberling_

%E Edited by _N. J. A. Sloane_, Jan 31 2020