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

On Mon, Jun 1, 2009 at 7:33 PM, <rwg@sdf.lonestar.org> wrote:

...

Matrix products efficiently subsume nearly all sums, continued fractions, and higher recurrences. ....

If only more people understood this! I have often wondered whether Ramanujan's success was due to an implicit understanding of this basic truth, even though the use of matrices wasn't all that common during his lifetime. ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos

...

On Mon, Jun 1, 2009 at 7:33 PM, <rwg@sdf.lonestar.org> wrote:

...

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.

Link? -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com

http://gosper.org/stanfordn1.dvi http://gosper.org/stanfordn2.dvi http://gosper.org/stanfordn3.dvi http://gosper.org/stanfordn4.dvi

plus innumerable posts to this list. --rwg

I seem to recall that for the older kid I once path-transformed sum(1/n^2) into the series for 6 asin(1/2)^2, which should have greatly pleased old Bernoulli. Damned if I can find it though. I'll rederive it if there's interest. (Bleeping Yahoo won't let me search oldish mail.) --rwg

Yi-Wen Liu, a former student of Julius Smith (Stanford Dept of Music) noticed, in effect, binom(4*(n+1)^2,(n+1)*(2*n+1))/binom(2*(2*n^2+4*n+1),n*(2*n+3)) = 4 4(n+1)^2 ( ) (n+1) (2n+1) ---------------- = 4 . 2 (2n^2+4n+1) ( ) n (2n+3) I just found binom((2*n+1)^2,(n+1)*(2*n+1))/binom(4*n^2+4*n-1,n*(2*n+3)) = 4 (2n+1)^2 ( ) (n+1) (2n+1) --------------- = 4 . 4n^2+4n-1 ( ) n (2n+3) Both of these are trivial to prove, but perhaps not to find, and I think at least one of them was discussed here a few years ago. There are more. Have they been characterized? Less obvious: binom(fib(2*n+4)-fib(2*n)-2,fib(2*n+1)-1)/binom(fib(2*n+4)-fib(2*n)-4,fib(2*n+1)-2) fib(2n+4)-fib(2n)-2 ( ) fib(2n+1)-1 --------------------- = 5 , fib(2n+4)-fib(2n)-4 ( ) fib(2n+1)-2 which is equivalent to the identity (fib(2*(n+2))-fib(2*n)-3)*(fib(2*(n+2))-fib(2*n)-2)/((fib(2*n+1)-1)*(fib(2*n+3)-1))=5 (fib(2 (n + 2)) - fib(2 n) - 3) (fib(2 (n + 2)) - fib(2 n) - 2) --------------------------------------------------------------- = 5 . (fib(2 n + 1) - 1) (fib(2 n + 3) - 1) Known? --rwg ISOMETRIC EROTICISM ISOTONIC COITIONS

Oh, foo, it's basically just Pell's equation.

binom(-(sqrt(3)-sqrt(2))^n*(sqrt(6)+1)/4+(sqrt(3)+sqrt(2))^n*(sqrt(6)-1)/4+1/2,(sqrt(3)-sqrt(2))^n*(sqrt(6)+6)/24+(sqrt(3)+sqrt(2))^n*(6-sqrt(6))/24+1/2)/binom(-(sqrt(3)-sqrt(2))^n*(sqrt(6)+1)/4+(sqrt(3)+sqrt(2))^n*(sqrt(6)-1)/4-3/2,(sqrt(3)-sqrt(2))^n*(sqrt(6)+6)/24+(sqrt(3)+sqrt(2))^n*(6-sqrt(6))/24-3/2) = 6 n n (sqrt(3) - sqrt(2)) (sqrt(6) + 1) (sqrt(3) + sqrt(2)) (sqrt(6) - 1) 1 - ---------------------------------- + ---------------------------------- + - 4 4 2 ( ) n n (sqrt(3) - sqrt(2)) (sqrt(6) + 6) (sqrt(3) + sqrt(2)) (6 - sqrt(6)) 1 ---------------------------------- + ---------------------------------- + - 24 24 2 ------------------------------------------------------------------------------- = 6, n n (sqrt(3) - sqrt(2)) (sqrt(6) + 1) (sqrt(3) + sqrt(2)) (sqrt(6) - 1) 3 - ---------------------------------- + ---------------------------------- - - 4 4 2 ( ) n n (sqrt(3) - sqrt(2)) (sqrt(6) + 6) (sqrt(3) + sqrt(2)) (6 - sqrt(6)) 3 ---------------------------------- + ---------------------------------- - - 24 24 2 n = floor(n) /= 0, likewise for binom(-(sqrt(2)-1)^n*(3*sqrt(2)+2)/8+(sqrt(2)+1)^n*(3*sqrt(2)-2)/8+1/2,(3-sqrt(2))*(sqrt(2)+1)^n/12+(sqrt(2)-1)^n*(sqrt(2)+3)/12+1/2)/binom(-(sqrt(2)-1)^n*(3*sqrt(2)+2)/8+(sqrt(2)+1)^n*(3*sqrt(2)-2)/8-3/2,(3-sqrt(2))*(sqrt(2)+1)^n/12+(sqrt(2)-1)^n*(sqrt(2)+3)/12-3/2) = 9/2 n n (sqrt(2) - 1) (3 sqrt(2) + 2) (sqrt(2) + 1) (3 sqrt(2) - 2) 1 - ------------------------------ + ------------------------------ + - 8 8 2 (---------------------------------------------------------------------) n n (3 - sqrt(2)) (sqrt(2) + 1) (sqrt(2) - 1) (sqrt(2) + 3) 1 ---------------------------- + ---------------------------- + - 12 12 2 9 ----------------------------------------------------------------------- = -. n n 2 (sqrt(2) - 1) (3 sqrt(2) + 2) (sqrt(2) + 1) (3 sqrt(2) - 2) 3 - ------------------------------ + ------------------------------ - - 8 8 2 (---------------------------------------------------------------------) n n (3 - sqrt(2)) (sqrt(2) + 1) (sqrt(2) - 1) (sqrt(2) + 3) 3 ---------------------------- + ---------------------------- - - 12 12 2 Note that for n /=0 mod 4, the latter binomials are exotic. Likewise for odd n in the former. Actually, with binom(a,b) := Gamma(a+1)/(Gamma(-b+a+1)*Gamma(b+1)) a Gamma(a + 1) ( ) := -------------------------------, b Gamma(- b + a + 1) Gamma(b + 1) both identities hold for complex n /= 0. --rwg PS, the original binomial coincidence appeared during the derivation of cos(x)^odd = sum_k a_k cos(k*x). In case anyone wants them, I went ahead and derived several expansions of cos(x)^complex, and sin^complex, some of which are Fourier series valid only on an interval. One expansion for 1 - cos(pi*x)^0 gave Assuming[Element[x, Reals], Sum[Sin[2*Pi*k*x]/k, {k, 1, Infinity}]/Pi + x - 1/2] which should be (Floor[x]+Ceiling[x]-1)/2, but Mma 7.0 gives -(1/2) + x + (I*(Log[1 - E^(2*I*Pi*x)] - Log[(-1 + E^(2*I*Pi*x))/E^(2*I*Pi*x)]))/(2*Pi) which is, unlike the sum, undefined at the integers, and erroneously mimics Floor[x] if you take directional limits at the integers. An actual "analytic" Floor[x] appears to be Assuming[Element[x, Reals], FullSimplify[ArcCot[Tan[Pi*x]]/Pi + x + 1/2 == Floor[x]]] but again, no True, even Assuming[Element[x, Reals] && Not[Element[x + 1/2, Integers]],...]

...

Yi-Wen Liu, a former student of Julius Smith (Stanford Dept of Music) noticed, in effect,

binom(4*(n+1)^2,(n+1)*(2*n+1))/binom(2*(2*n^2+4*n+1),n*(2*n+3)) = 4

4(n+1)^2 ( ) (n+1) (2n+1) ---------------- = 4 . 2 (2n^2+4n+1) ( ) n (2n+3)

I just found

binom((2*n+1)^2,(n+1)*(2*n+1))/binom(4*n^2+4*n-1,n*(2*n+3)) = 4

(2n+1)^2 ( ) (n+1) (2n+1) --------------- = 4 . 4n^2+4n-1 ( ) n (2n+3)

Both of these are trivial to prove, but perhaps not to find, and I think at least one of them was discussed here a few years ago. There are more. Have they been characterized?

Less obvious:

binom(fib(2*n+4)-fib(2*n)-2,fib(2*n+1)-1)/binom(fib(2*n+4)-fib(2*n)-4,fib(2*n+1)-2)

fib(2n+4)-fib(2n)-2 ( ) fib(2n+1)-1 --------------------- = 5 , fib(2n+4)-fib(2n)-4 ( ) fib(2n+1)-2

which is equivalent to the identity

(fib(2*(n+2))-fib(2*n)-3)*(fib(2*(n+2))-fib(2*n)-2)/((fib(2*n+1)-1)*(fib(2*n+3)-1))=5

(fib(2 (n + 2)) - fib(2 n) - 3) (fib(2 (n + 2)) - fib(2 n) - 2) --------------------------------------------------------------- = 5 . (fib(2 n + 1) - 1) (fib(2 n + 3) - 1)

Known? --rwg ISOMETRIC EROTICISM ISOTONIC COITIONS

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