[seqfan] Re: Need proof that Kimberling's A26185 = A198173

[Neil Sloane]

Don, Thanks.  I merged the two entries (into A026185).



On Sat, Feb 1, 2020 at 8:45 PM Don Reble via SeqFan <seqfan at list.seqfan.eu>
wrote:

> > A026185 =? A198473 is an open question. ... Could someone find a proof?
>
> ---
>
> Starting with A026136 (where a(1)=1),
>
> > %N A026136
> >   for n >= 2, let L=n-[ n/2 ], R=n+[ n/2 ];
> >   then a(L)=n if a(L) not yet defined, else a(R)=n
>
> If n = 2m-1: L = (2m-1)-(m-1) = m.
> If n = 2m  : L =  2m   - m    = m; R = 2m+m = 3m.
> So 2m never goes to position a(L)=a(m): 2m-1 or some smaller number
> fills that slot. In fact 2m -> a(3m), since no smaller number has that
> R value.
>
> Modulo 6:
>     6m+1 -> a(3m+1) or a(9m+1)
>     6m+3 -> a(3m+2) or a(9m+4)
>     6m+5 -> a(3m+3) or a(9m+7)
> We see that 3m+2 is never a R-index, so 6m+3 -> a(3m+2).
>
> That resolves the multiples of 3: 6m -> a(9m), 6m+3 -> a(3m+2).
>
> ---
>
> Those multiples of 3 define A026184:
>
> > %N A026184 a(n) = (1/3)*s(n),
> >    where s(n) is the n-th multiple of 3 in A026136.
>
> In A026136, the triples are at positions a(9m+2), a(9m+5), a(9m+8),
> a(9m+9).
> (There is no a(0), so I changed 9m to 9m+9.)
> The corresponding A026136 values are 18m+3, 18m+9, 18m+15, 6m+6,
> and the (1/3) values (A026184) are 6m+1, 6m+3, 6m+5, 2m+2.
>
> So for A026184, a(4m+1) = 6m+1, a(4m+2) = 6m+3, a(4m+3) = 6m+5;
> a(4m+4) = 2m+2 (and so a(4m) = 2m).
>
> ---
>
> A026185 is the inverse permutation to A026184.
> Is A198473 also that inverse permutation?
>
> > %N A198473 If n even, then 2n. If n odd, then nearest integer to 2n/3.
> > %F A198473 a(2n)=4n ; a(6n+1)=4n+1 ; a(6n+3)=4n+2 ; a(6n+5)=4n+3.
>
> That formula suggests yes, but let's prove it.
>
> If n is even (n=2m):
>     A026184[A198473(n)] = A026184[2n] = A026184[4m] = 2m = n.
> If n is odd (n=6m+{1,3,5}):
>     The nearest integer to 2n/3 = 4m+{1,2,3}.
>     A026184[A198473(n)] = A026184[4m+{1,2,3}] = 6m+{1,3,5} = n.
> QED
>
>
> --
> Don Reble  djr at nk.ca
>
>
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>