[seqfan] Re: a question about triangular numbers

Max Alekseyev maxale at gmail.com
Tue Dec 8 23:26:55 CET 2009


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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