Eric (and Seqfans),

please have a look there, if you have a couple of minutes: http://www.cetteadressecomportecinquantesignes.com/Kimberlike01.htm Best, É. Your generalization is curious. Perhaps a deeper look at the hidden structure beneath Clark Kimberling's sequences will shed light on yours.

Consider Kimberling's sequences A026242 and A026249 in particular. In these, every n occurs twice, at 'distance factors' 1 (A026242) and 2 (A026249). In A026242, the distance between positions where the 2 n's occur is 1*n; in A026249, the distance between positions where the 2 n's occur is 2*n. As you know, these can be constructed from complementary Beatty sequences. The distance factor, if it exists, is determined by the parameter used in the Beatty sequences. Here is how to determine the parameter r of the Beatty sequences from the distance factor d. The r is a root of quadratic equation 1 - d*x - x^2 = 0, or r=(sqrt(d^2+4)-d)/2. distance; r=root ; equation ; Sequence# d=1; r=(sqrt(5)-1)/2 ; 1-x-x^2 = 0 ; A026242 d=2; r=sqrt(2)-1 ; 1-2*x-x^2 = 0 ; A026249 d=3; r=(sqrt(13)-3)/2; 1-3*x-x^2 = 0 ; NOT IN OEIS d=4; r=sqrt(5)-2 ; 1-4*x-x^2 = 0 ; NOT IN OEIS

From r, we generate the Lower and Upper complementary Beatty sequences. Then the integer n will occur at postions L(n) and U(n) in the sequence. r=(sqrt(d^2+4)-d)/2 L(n)=floor(n*(1+r)) /* Lower Beatty seq */ U(n)=floor(n*(1+1/r)) /* Upper Beatty seq */ A=vector(U(100)); for(n=1,100,A[L(n)]=n;A[U(n)]=n)

Here are some examples for d=3, and d=4. Distance factor d=3; r=(sqrt(13)-3)/2; NOT IN OEIS 1,2,3,1,4,5,6,2,7,8,9,3,10,11,12,13,4,14,15,16,5,17,18,19,6, 20,21,22,23,7,24,25,26,8,27,28,29,9,30,31,32,33,10,34,35,36, 11,37,38,39,12,40,41,42,13,43,44,45,46,14,47,48,49,15,50,51, 52,16,53,54,55,56,17,57,58,59,18,60,61,62,19,63,64,65,66,20, 67,68,69,21,70,71,72,22,73,74,75,23,76,77,78,79,24,80,81,82, 25,83,84,85,26,86,87,88,89,27,90,91,92,28, Distance factor d=4; r=sqrt(5)-2; NOT IN OEIS 1,2,3,4,1,5,6,7,8,2,9,10,11,12,3,13,14,15,16,4,17,18,19,20, 21,5,22,23,24,25,6,26,27,28,29,7,30,31,32,33,8,34,35,36,37, 38,9,39,40,41,42,10,43,44,45,46,11,47,48,49,50,12,51,52,53, 54,55,13,56,57,58,59,14,60,61,62,63,15,64,65,66,67,16,68,69, 70,71,72,17,73,74,75,76,18,77,78,79,80,19,81,82,83,84,20,85, 86,87,88,21,89,90,91,92,93,22,94,95,96,97,23, Now, for your case, where you choose every integer n to occur 3 times, perhaps the sequence could be generated using something akin to Beatty sequences. But there would have to be 3 complementary sequences generated from irrational parameter(s). And maybe these parameters would be roots to some cubic equation? These are just surmisings, but what I am attempting to do is find a way of easily generating your sequences similar to the Beatty sequence method. Unfortunately, I do not know of an analogue to Beatty Theorem that will generate 3 complementary sequences! That would be nice ... 'Nough Said, Paul