I think the 4-triangle-partitions-equals-sigma-2n+1identity (which RWG introduced me to long ago, thanks!) is due to Legendre. (I've used it as the basis for a prime-testing programming puzzle: “1 point: what does this program do? 10 points: how does it do it? 100 points: why does it work?” One guy fiddled with it for 8 years and then sent me a 31 page PDF about it!)
On Jul 8, 2016, at 5:30 AM, Bill Gosper <billgosper@gmail.com> wrote:
I haven't been following this discussion, but psi^4 is https://oeis.org/A008438 ("sum of divisors of 2n+1"), which is quite a long entry with many references and comments.
If there's something new to add there, please add it (or tell me what to add)!
Best regards Neil
Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com <http://neilsloane.com/> Email: njasloane@gmail.com
Nothing much, except that it's the secret behind this obscure prime test: primeq[n_Integer]:=1==SeriesCoefficient[\[Integral](3 q^2+1/16 EllipticTheta[2,0,q]^4)\[DifferentialD]q/q,{q,0,n}]
In[16]:= Select[Range@99,primeq] Out[16]= {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97} --rwg
On Wed, Jul 6, 2016 at 11:37 PM, Mike Hirschhorn <m.hirschhorn@unsw.edu.au> wrote:
Dear all,
The right side is \psi(q)^4, where \psi(q) generates the triangular numbers, so the identity gives the number of reps of a number as the sum of 4 triangles, which must be well-known.
Mike Hirschhorn.
________________________________ From: mathfuneavesdroppers@googlegroups.com < mathfuneavesdroppers@googlegroups.com> on behalf of Bill Gosper < billgosper@gmail.com> Sent: 07 July 2016 13:08:14 To: math-fun@mailman.xmission.com Subject: Re: How I miss Mizan Rahman, or, what the heck is this?
George Andrews got it from Bailey's ??? transformation (rather than finding it scrawled on the wall of an ancient tomb.) Merely with elementary manipulations, starting with partial fractions, I reduced it to n q Sum[---------------, {n, -?, ?}] == 1 + 2 n 2 (1 - q )
2 2 4 QPochhammer[q , q ] -------------------- 2 4 QPochhammer[q, q ]
which is presumably known, since it's a simple q-extension of
In[1653]:= Sum[(2 n - 1)^-2, {n, -?, ?}]
Out[1653]= ?^2/4.
Exercises: Evaluate
3 n + 1 q Sum[---------------, {n, -?, ?}] 1 + 2 n 2 (1 - q )
and n q Sum[------------, {n, -?, ?}] 1 + 2 n 1 - q
Spoilers: 2 2 4 QPochhammer[q , q ] -------------------- 2 4 QPochhammer[q, q ]
and 0 (off the unit circle).
Problem: n integer, t real, L=Limit[sin nt,n->?]. Then L=0 or it doesn't exist? --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun