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 On Sat, Jul 2, 2016 at 11:19 AM, Bill Gosper <billgosper@gmail.com<mailto:billgosper@gmail.com>> wrote: Out[1131]= k q Sum -------------------------------, 1 - 2 k 2 1 + 2 k 2 (1 - q ) (1 - q ) k,-?,? = 1 2 4 (1 + q) QFactorial[-(-), q ] 2 ----------------------------- 2 4 (1 - q ) q-Pochhammers in the sum but not the summand? "q-Mittag Leffler"? It q-retracts to 16 Sum[-----------, {k,-?,?}] == 2 2 (1 - 4 k ) 2 2 ? Mathematica is dumbstruck. For minutes on end. Are these actually common? Test for q=1/2: In[1132]:= N[Activate[%1131/.q->.5]] Out[1132]= True Term ratio: In[1180]:= DiscreteRatio[%1131[[1,1]],k] Out[1180]= (q (1-q^(1-2 k))^2 (1-q^(1+2 k))^2)/((1-q^(-1-2 k))^2 (1-q^(3+2 k))^2) Nine terms of the sum give 27 term agreement with the product, despite the lack of q^k^2: In[1189]:= Series[%1131/.\[Infinity]->9/.-\[Infinity]->-9//Activate,{q,0,29}] Out[1189]= 1+5 q+12 q^2+24 q^3+44 q^4+68 q^5+102 q^6+150 q^7+202 q^8+270 q^9+354 q^10+446 q^11+561 q^12+693 q^13+838 q^14+1002 q^15+1195 q^16+1407 q^17+1638 q^18+1906 q^19+2179 q^20+2491 q^21+2842 q^22+3202 q^23+3610 q^24+4042 q^25+4504 q^26+5008 q^27+5549 q^28+6114 q^29+O[q]^30==1+5 q+12 q^2+24 q^3+44 q^4+68 q^5+102 q^6+150 q^7+202 q^8+270 q^9+354 q^10+446 q^11+561 q^12+693 q^13+838 q^14+1002 q^15+1195 q^16+1407 q^17+1638 q^18+1906 q^19+2179 q^20+2491 q^21+2842 q^22+3202 q^23+3610 q^24+4042 q^25+4504 q^26+5008 q^27+5550 q^28+6114 q^29+O[q]^30 In[1190]:= Subtract@@% Out[1190]= -q^28+O[q]^30 I got it by mildly abusing the (bilateral) "q-Dixon's theorem" in http://math.sun.ac.za/~hproding/pdffiles/Triples-long.pdf which is not the (unilateral) q-Dixon's I get from path invariant matrices. --rwg -- You received this message because you are subscribed to the Google Groups "mathfuneavesdroppers" group. To unsubscribe from this group and stop receiving emails from it, send an email to mathfuneavesdroppers+unsubscribe@googlegroups.com<mailto:mathfuneavesdroppers+unsubscribe@googlegroups.com>. For more options, visit https://groups.google.com/d/optout.