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User talk:Gary Detlefs
Hi Gary
See my comments and discussion notes in the (draft) version of A024166.
Did you see my formulas there and in other OEIS sequences (I provide the list in my links), which already covered squares and cubes (but not fourth degrees which is due to you)?
Also see "k-fold nested sum of integer powers" section of my OEIS wiki page, which is
https://oeis.org/wiki/User:Alexander_R._Povolotsky
and
http://math.stackexchange.com/questions/128763/k-fold-nested-sum-of-integer-powers
Two questions:
1) How this could be related and/or derived from the Faulhaber's formula ?
2) Any takers for generalizing the formula to any degree of power ? That is, how this could be extended to the formula for the m-fold nested sum of integer powers of "m-th" degree ?
Best Regards, Alex
Hi Alex,
I think it would be easier for me to communicate through regular e-mail than the Wiki page but I could not get this email through. Also, I could attach word documents which would be in a better format. I have looked over your Wiki page and see that you have some unique insights into mathematics.
The world of mathematics is so vast and there are so many people out there who are way smarter than we are. On just the topic of the sums of powers, one could fill a small book with just the references to all of the highly advanced papers on the subject. .. way over my head. Still it is fun to explore and as John Lennon said..we all shine on.
Is it possible to create a Wiki page on the OEIS and then link to it from a given sequence? If so it would be neat to put together a page of links to all of the sequences associated with sums of powers, nested sums of powers, alternating sums of powers..etc. There is an index in the OEIS but it is pretty bare bones.
I had been studying primality criterion and that is what had led me to revisit the sums of powers topic. There is an interesting relationship to the sums of powers and the rough numbers..numbers with no prime factors less that k. For example , the sum of the fourth powers is divisible by n only if n is prime or not divisible by {2,3,5}, n>5. Similarly the sum of the 12 powers is divisible by n only if n is prime not divisible by {2,3,5,7,13},n>7. I made a chart of all the even power summations from 4 to 40.
Anyway, I just like to look at a lot of topics and collect formulas. I have attached an index of Fibonaaci formulas I had put together if you find this topic of interest.(Was not able to send this)
Best regards Gary Detlefs
(Alexander R. Povolotsky) Alexander R. Povolotsky 22:09, 4 March 2013 (UTC)
=========================================
Thank you Alex...I admire your technology skills. Mine are next to nil..I can't even seem to post a document on the cloud. I have revisited this topic of sums of powers many times and finally found what I considered to be one of the best kept secrets in mathematics. Since you mention Falhauber, I assume you are familiar with Knuth's paper on him. In a sense, this paper does give a general formula for the r-th fold summation of any power. Another site I had been looking at was at http://www.noticingnumbers.net/ . Somehow I figured out that there is a double summation formula involving the Eulerian triangle which I had posted at A008292
The Eulerian triangle is an element of the formula for the r-th successive summation of sum(k^j,k=1..n)which appears to be sum(T(j,k-1) * binomial(j-k+n+r,j+r), k=1..n). [From Gary Detlefs, Nov 11 2001]
Don't ask me to prove it but it works. The formula itself just states the recursion in a different way but.. Maple is amazing in its ability to express it in algebraic terms. I could submit a formula for the r-th successive summation of any power at all but they get large quickly. Speaking of technology, lets see if I can paste some Maple output.
Powesum template...the primary formula. >
> > Eulerian := proc (n, k) options operator, arrow; sum((-1)^j*(k-j+1)^n*binomial(n+1, j), j = 0 .. k+1) end proc;
k + 1
-----
\
) j n
(n, k) -> / (-1) (k - j + 1) binomial(n + 1, j)
-----
j = 0
> s2 := proc (n, j, r) options operator, arrow; sum(Eulerian(j, k-1)*binomial(j-k+n+r, j+r), k = 1 .. n) end proc;
n
-----
\
)
(n, j, r) -> / Eulerian(j, k - 1) binomial(j - k + n + r, j + r)
-----
k = 1
> seq(s2(n, 3, 2), n = 1 .. 20); 1, 10, 46, 146, 371, 812, 1596, 2892, 4917, 7942, 12298, 18382, 26663, 37688,
52088, 70584, 93993, 123234, 159334, 203434
> > > > for n to 5 do print(simplify(s2(x, n, r))) end do;
GAMMA(1 + x + r)
---------------------
GAMMA(2 + r) GAMMA(x)
(r + 2 x) GAMMA(1 + x + r)
--------------------------
GAMMA(3 + r) GAMMA(x)
/ 2 2\
\r + 6 x r - r + 6 x / GAMMA(1 + x + r)
----------------------------------------
GAMMA(4 + r) GAMMA(x)
/ 2 2\
(r + 2 x) \r + 12 x r - 5 r + 12 x / GAMMA(1 + x + r)
------------------------------------------------------
GAMMA(5 + r) GAMMA(x)
1 // 4 3 3 2 2 2 2
--------------------- \\r + 30 x r - 16 r - 90 x r + 11 r + 150 x r
GAMMA(6 + r) GAMMA(x)
3 2 4\ \ + 240 x r - 90 x r + 4 r + 120 x / GAMMA(1 + x + r)/
>
well that sort of worked. Recall that the Gamma(r+k) = (r+k-1)! If you have Maple, I can send you a worksheet.
Although there is nothing attributable to me regarding these formulas, there is one observation I made many years ago which I have never seen in the literature regarding this topic. If we denote the r-th fold summation of n^k as [n,k,r] then [n,k,r] = n*[n,k-1,r]-r*[n-1,k-1,r+1]... again, don't ask me to prove this but I know it to be true. I recall spending a sun up to sundown August day on a deserted Lake Michigan beach pouring over some printouts of the three dimensional array.I recall looking at a binary tree data structure regarding the topic....confusing
Anyway that's all for now...nice meeting you. I see you live in Mass.
Gary
=======================
Likewise, it was nice to communicate with you too, Gary. It was purely my luck then ( ;-)) that you took some time between posting your Eulerian triangle comment at A008292 on Nov 11 2001 and starting posting r-th fold summation formulas only now - this delay did give me chance to stick into OEIS few formulas of my own without even knowing about existence of your comment and without knowing about relationship between r-th fold summation and Eulerian triangle ... ;-).
I do not have Maple and/or Mathematica (speaking of technology) - so I am using free WolframAlpha (poor emigrant choice ;-)).
I am retired (just got 65 few days ago) and do live in Brookline (next to Boston) MA - you could find me on LinkedIn to get the idea of my sufferings in the area of embedded software ;-).
But, seriously speaking, I am impressed with your knowledge in the area being discussed.
I would be delighted if one nice day (sunny or not) you would go through my wiki page and share with me your insights on my small ideas. I would be happy if any cooperation and/or synergy would arise from that.
Cheers, Alex
PS I noticed that you play instruments while I write poetry - is it good or bad for synergy ? ;-)
