[seqfan] Re: a question about triangular numbers

Tanya Khovanova mathoflove-seqfan at yahoo.com
Wed Dec 9 16:56:15 CET 2009

```I wonder if there is a limit on a(n)/n.

--- On Tue, 12/8/09, Max Alekseyev <maxale at gmail.com> wrote:

> From: Max Alekseyev <maxale at gmail.com>
> Subject: [seqfan] Re: a question about triangular numbers
> To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
> Date: Tuesday, December 8, 2009, 5:26 PM
> A trivial lower bound is a(n) >=
> n/3.
> Max
>
> On Tue, Dec 8, 2009 at 1:36 PM, Tanya Khovanova
> <mathoflove-seqfan at yahoo.com>
> wrote:
> > Right,
> >
> > Not any number is a sum of two triangular numbers.
> When n - (the largest triangular not exceeding n) can't be
> represented as a sum of two triangles, we get a drop out.
> >
> > Is there any lower bound for this sequence?
> >
> > Tanya
> >
> > --- On Tue, 12/8/09, Richard Mathar <mathar at strw.leidenuniv.nl>
> wrote:
> >
> >> From: Richard Mathar <mathar at strw.leidenuniv.nl>
> >> Subject: [seqfan] Re: a question about triangular
> numbers
> >> To: seqfan at seqfan.eu
> >> Date: Tuesday, December 8, 2009, 1:28 PM
> >>
> >> On behalf of http://list.seqfan.eu/pipermail/seqfan/2009-December/003182.html
> >> :
> >>
> >> If a(n)=the largest of the three triangular
> numbers x,y,z
> >> of partitioning n=x+y+z into any three triangulare
> numbers,
> >> I get the more
> >> irregular:
> >>
> >> 1, 1, 3, 3, 3, 6, 6, 6, 6, 10, 10, 10, 10, 10, 15,
> 15, 15,
> >> 15, 15, 10,
> >> 21, 21, 21, 21, 21, 15, 21, 28, 28, 28, 28, 28,
> 21, 28, 28,
> >> 36, 36, 36,
> >> 36, 36, 28, 36, 36, 28, 45, 45, 45, 45, 45,
> 28,...
> >> with "drop-outs"
> >>
> >> Maple:
> >> # The triangular numbers
> >> A000027 := proc(n)
> >>     option remember;
> >>     n*(n+1)/2 ;
> >> end proc:
> >> # test if the argument is a triangular number
> >> isA000027 := proc(n)
> >>     issqr(1+8*n) ;
> >> end proc:
> >> # calculate the maximum in the set Ta+Tb+Tc=n, any
> Ta, Tb,
> >> Tc of A000027
> >> ltn := proc(n)
> >>     local res, ai,bi,Ta,Tb,Tc ;
> >>     res := -1 ;
> >>     # loop Ta over all triangular numbers
> >>     for ai from 0 do
> >>         Ta := A000027(ai) ;
> >>         if Ta > n then
> >>
> >> break;
> >>         else
> >>             #
> >> loop Tb over all triangular numbers
> >>
> >> for bi from ai do
> >>
> >>     Tb := A000027(bi) ;
> >>
> >>     if Ta+Tb > n then
> >>
> >>         break;
> >>
> >>     else
> >>
> >>         # Tc the remainder to
> >> sum to n
> >>
> >>         Tc := n-Ta-Tb ;
> >>
> >>         if isA000027(Tc) then
> >>
> >>             res
> >> := max(res, Ta, Tb, Tc) ;
> >>
> >>             #
> >> printf("%d = %d + %d + %d\n",n,Ta,Tb,Tc) ;
> >>
> >>         end if;
> >>
> >>     end if ;
> >>
> >> end do ;
> >>         end if ;
> >>     end do;
> >>     return res;
> >> end proc:
> >>
> >> seq(ltn(n),n=1..50) ;
> >>
> >>
> >> _______________________________________________
> >>
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> >>
> >
> >
> > _______________________________________________
> >
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> >
>
>
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```