[seqfan] Re: Request for references

[Max Alekseyev]

These are just application of (power of) the forward difference operator:
http://mathworld.wolfram.com/ForwardDifference.html

The first identity is basically:
\Delta^m a(n) = \Delta^{m-1} n

The second one is:
\Delta^m (-1)^n*a(n) = \Delta^{m-1} (-1)^n*n^2

Regards,
Max


On Wed, Oct 6, 2010 at 11:26 AM, Charles Marion
<charliemath at optonline.net> wrote:
> Greetings,
>
>
>
> I've just submitted the following two comments for the triangular numbers in
> A000217:
>
>
>
> It is well known that a(n) - a(n-1) = n.  Less well known is that\Q
>
> a(n) - 2a(n-1) + a(n-2) = 1, a(n) - 3a(n-1) + 3a(n-2) - a(n-3) = 0 and\Q
>
> a(n) - 4a(n-1) + 6a(n-2) - 4(a-3) + a(n-4) = 0.\Q
>
> In general, for n>=m>2, sum_{k=0,...,m}(-1)^k*binomial(m,m-k)*a(n-k)=0.\Q
>
> For example, 1*28 - 5*21 + 10*15 - 10*10 + 5*6 - 1*3 = 0.\Q
>
>
>
> It is well known that a(n) + a(n-1) = n^2.  Less well known is that\Q
>
> a(n)+2a(n-1)+a(n-2) = n^2+(n-1)^2; e.g., 10+2*6+3=25=4^2+3^2  and\Q
>
> a(n)+3a(n-1)+3a(n-2)+a(n-3)= n^2+2*(n-1)^2+(n-2)^2;\Q
>
> e.g., 15+3*10+3*6+3=66=5^2+2*4^2+3^2.\Q
>
> In general, for n>=m>2,sum_{k=0,...,m}binomial(m,m-k)*a(n-k)=\Q
>
> sum_{k=0,...,m-1}binomial(m-1,m-1-k)*(n-k)^2  For example,\Q
>
> 1*28+5*21+10*15+10*10+5*6+1*3=416=1*7^2+4*6^2+6*5^2+4*4^2+1*3^2.\Q
>
>
>
> Can anyone supply a reference in the literature for these results?
>
>
>
> Thanks.
>
>
>
> Charlie Marion
>
> Yorktown Heights NY
>
>
>
> PS Any seqfans near Yorktown Heights?
>
>
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