My guess (I could be wrong of course) is that he rediscovered some version of the van Staudt or Kummer congruences (or the generalization which says, that, suitably normalized, the Bernoulli numbers are p-adically continuous). It's an interesting sea change -- before the advent of the web -- especially excellent Wikipedia articles -- it might have been very difficult for someone from the "hinterlands" to find things in the literature. Just think of the initial letters that Ramanujan sent to GH Hardy, in which he discovered for himself some things that were already "well-known". Nevertheless, it sounds like a significant accomplishment by a 16 year old -- if nothing else it shows his ability and concentration. Victor
On Mon, Jun 1, 2009 at 3:14 PM, N. J. A. Sloane <njas@research.att.com>wrote:
A later news item reports that the formula is well-known.
http://www.uu.se/news/news_item.php?typ=artikel&id=693
Neil
Here's what I sent a couple of young friends: -------------------------------------- Original Message -------------------------------------- Subject: Swedish-Iraqi teen "solves" Bernoulli numbers From: rwg@sdf.lonestar.org Date: Sat, May 30, 2009 12:16 pm To: Neil, mitchell riley Cc: cjbickford rwg@sdf.lonestar.org ---------------------------------------------------------------------------------------------- Hi guys, this is all over the net: http://www.thelocal.se/19710/20090528/ Unless some idiot cropped more than the outer right paren, that formula bern(n)=sum(1/(i+1)*sum((-1)^p*binom(p,n)*p^n,p,0,i),i,0,n) i ==== \ n p > p (- 1) binomial(p, n) n / ==== ==== \ p = 0 B = > ------------------------------ n / i + 1 ==== i = 0 simply doesn't work, but it's close. One of many such that works is bern(n) = -'sum(-(-1)^i*binomial(n+1,i+1)*('sum(p^n,p,0,i))/(i+1),i,0,n) i ==== i \ n (- 1) binomial(n + 1, i + 1) > p n / ==== ==== \ p = 0 B = - > (- --------------------------------------) n / i + 1 ==== i = 0 Test: (c218) makelist(apply_nouns(apply_nouns(bern(n) = -'sum(-(-1)^i*binomial(n+1,i+1)*('sum(p^n,p,0,i))/(i+1),i,0,n))),n,0,9) 1 1 1 1 1 1 1 1 (d218) [1 = 1, - - = - -, - = -, 0 = 0, - -- = - --, 0 = 0, -- = --, 2 2 6 6 30 30 42 42 1 1 0 = 0, - -- = - --, 0 = 0] 30 30 (Quick, Neil, what's the next one?-) But contrary to custom, I claim there remain two stupidities in this working formula: 1) You should almost never use Bernoulli numbers, but rather Bernoulli polynomials in a totally new unknown. Amazingly, this greatly generalizes most formulae while complicating them hardly at all. 2) Summation notation is stoopid! Look what that formula says: sum up n things, each of which is a product of about n things times a sum of about n/2 things. That's at least quadratic complexity in n--doubling n quadruples the work. Here is how I'd write the general case: [0,0,bernpoly(x,n);0,0,harmonic(m);0,0,m]/m= prod([(k-m)/(k+1),(k-m)*(x+k-1)^n/(k+1),(x+k-1)^n/k;0,(k-m)/(k+1),1/k;0,0,1],k,1,m) [ B (x) ] [ n ] [ n n ] [ 0 0 ----- ] m [ k - m (k - m) (x + k - 1) (x + k - 1) ] [ m ] /===\ [ ----- -------------------- ------------ ] [ ] | | [ k + 1 k + 1 k ] [ H ] = | | [ ] [ m ] | | [ k - m 1 ] [ 0 0 -- ] k = 1 [ 0 ----- - ] [ m ] [ k + 1 k ] [ ] [ ] [ 0 0 1 ] [ 0 0 1 ] i.e., a product of m consecutive matrices, where m is any integer > n. B[n] := B[n](0), which is also a mistake--it should have been B[n](1)! Another motivation to junk Bernoulli numbers. (Although it only matters for n=1.) Matrix products efficiently subsume nearly all sums, continued fractions, and higher recurrences. My path-invariant calculus of them is my one great public relations failure. I hope I can turn you both onto it. --Bill