Mistake? A038853 and A038860

[Maximilian Hasler]

Independently from Robert, I made my little calculation confirming
that these sequences (without much interest, IMHO) are given by
(k+5j)^3-k^3 and nothing else (any j>0 resp. any odd j>0).
Here's a little PARI program (KISS principle) that should give the
correct sequence:

{A038853(Nmax)=local(t,j5,a=[]);for(j=1,Nmax^(1/3)/5, j5=5*j;
for(k=1,sqrt((Nmax/j5-j5^2-3*j5)/3),
if(Nmax < t=(k+j5)^3-k^3, next);a=concat(a,t)));vecsort(a)}

> A038853(10000)
%688 = [215, 335, 485, 665, 875, 1115, 1330, 1385, 1685, 1720, 2015,
2170, 2375, 2680, 2765, 3185, 3250, 3635, 3880, 4095, 4115, 4570,
4625, 4905, 5165, 5320, 5735, 5805, 6130, 6335, 6795, 6965, 7000,
7625, 7875, 7930, 8315, 8920, 9035, 9045, 9260, 9785, 9970]


{A038860(Nmax)=local(t,j5,a=[]);forstep(j=1,Nmax^(1/3)/5,2, j5=5*j;
for(k=1,sqrt((Nmax/j5-j5^2-3*j5)/3),
if(Nmax < t=(k+j5)^3-k^3, next);a=concat(a,t)));vecsort(a)}

> A038860(20000)
%690 = [215, 335, 485, 665, 875, 1115, 1385, 1685, 2015, 2375, 2765,
3185, 3635, 4095, 4115, 4625, 4905, 5165, 5735, 5805, 6335, 6795,
6965, 7625, 7875, 8315, 9035, 9045, 9785, 10305, 10565, 11375, 11655,
12215, 13085, 13095, 13985, 14625, 14915, 15875, 16245, 16865, 17575,
17885, 17955, 18935, 19675, 19755]

If I have the time later in the day, I'll submit all of that in a
properly formatted %I A038860 ... mail to NJAS.

Maximilian

On 5/31/07, Robert Israel <israel at math.ubc.ca> wrote:
> Yes:
> x^3 - y^3 == 0 mod 5 iff x - y == 0 mod 5 (so you're right about A038853).
> Also
> x^3 - y^3 == 1 mod 2 iff x - y == 1 mod 2
> so x^3 - y^3 == 5 mod 10 iff x - y == 5 mod 10 (so you're right about
> A038860).
> Perhaps it should be mentioned that here "cubes" means "cubes of positive
> integers", e.g. 65 = 4^3 - (-1)^3 and 125 = 5^3 - 0^3 are not included.
>
> Cheers,
> Robert Israel
>
> On Thu, 31 May 2007, Maximilian Hasler wrote:
>
> > at first sight it seems to me as if
> > A038860 = { b(k,j) ; k=1,2,3..., j=1,3,5,7... }
> > A038853 = { b(k,j) ; k=1,2,3..., j=1,2,3,4... }
> > with b(k,j)=(k+5j)^3-k^3
> > (and k=0 might be allowed depending on personal idea of cubes)
> > but I don't have time to think about "completeness" and since I tend
> > to make small errors when I don't have time...
> > M.H.
> >
> > On 5/31/07, Maximilian Hasler <maximilian.hasler at gmail.com> wrote:
> >> Definitely, any number of the form 5(3k(k+5)+25) = (k+5)^3-k^3
> >> should be in that sequence, and 1385 is of that form.
> >> M.H.
> >>
> >> On 5/31/07, Tanya Khovanova <tanyakh at tanyakhovanova.com> wrote:
> >> > Hello Seqfans,
> >> >
> >> > Looking at the definitions, A038860 should be a subsequence of A038853;
> >> but 1385 is missing in A038853.
> >> >
> >> > A038853                  Numbers that are divisible by 5 and are
> >> differences between two cubes in at least one way.
> >> >         215, 335, 485, 665, 875, 1115, 1330, 1685, 2015, 2170, 2375,
> >> 2680, 2765, 3150, 3185, 3635, 3880, 4095, 4115, 4570, 4625, 4905, 5165,
> >> 5320, 5735, 5805, 6130, 6335, 6795, 6965, 7000, 7625, 7875, 7930, 8315,
> >> 8920, 9035, 9045, 9260, 9785, 9970, 10305
> >> >
> >> > A038860                  Numbers n such that n ends with '5' and is
> >> difference between two cubes in at least one way.
> >> >         215, 335, 485, 665, 875, 1115, 1385, 1685, 2015, 2375, 2765,
> >> 3185, 3635, 4095, 4115, 4625, 4905, 5165, 5735, 5805, 6335, 6795, 6965,
> >> 7625, 7875, 8315, 9035, 9045, 9785, 10305, 10565, 11375, 11655, 12215,
> >> 13085, 13095, 13985, 14625, 14915, 15875
> >> >
> >> > Tanya
> >> >
> >> >
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