Help find reference

[Max Alekseyev]

On Dec 15, 2007 11:00 AM, Charles Marion <charliemath at optonline.net> wrote:

> A045944
> Rhombic matchstick numbers: n*(3*n+2).
> 0, 5, 16, 33, 56, 85, 120, 161, 208, 261, 320, 385, 456, 533, 616, 705, 800,
> 901, 1008,

[...]

> For n>0,
>
> a(n)^3+(a(n)+1)^3 +...+(a(n)+n)^3 +2*A000217(n)^2=
> (a(n)+n+1)^3+...+(a(n)+2n)^3;

This identity can be rewritten as

S_3(a(n)+n) - S_3(a(n)-1) + 2*A000217(n)^2 = S_3(a(n)+2n) - S_3(a(n)+n)

where S_3(k) = (k^4 + 2k^3 + k^2) / 4.
See http://mathworld.wolfram.com/FaulhabersFormula.html

Plugging the formula for a(n) in, we have
2*A000217(n)^2 = S_3(3*n^2+4n) + S_3(3*n^2+2n-1) - 2*S_3(3*n^2+3*n)
= 1/2*n^4 + n^3 + 1/2*n^2 = (n^2 + n)^2/2
that obviously holds.

> A033954 n*(4*n+3). Also, second 10-gonal (or decagonal) numbers.
> 0, 7, 22, 45, 76, 115, 162, 217, 280, 351, 430, 517, 612, 715, 826, 945,
> 1072, 1207,

[...]

> For n>0,
>
> a(n)^4+(a(n)+1)^4 +...+(a(n)+n)^4 +(4*A000217(n))^3 =
> (a(n)+n+1)^4+...+(a(n)+2n)^4;

Similarly, this one is equivalent

S_4(a(n)+n) - S_4(a(n)-1) + (4*A000217(n))^3 = S_4(a(n)+2n) - S_4(a(n)+n)

or

(4*A000217(n))^3 = S_4(4*n^2+5*n) + S_4(4*n^2+3*n-1) - 2*S_4(4*n^2+4*n)

where S_4(k) = (6*k^5 + 15*k^4 + 10*k^3 - k) / 30. The last identity
trivially holds since its r.h.s. is (2*(n^2+n))^3.

Regards,
Max