* Bill Gosper <billgosper@gmail.com> [Oct 10. 2012 10:13]:
Warren, you have GMail. Neil just taught me to click on▼, then "Message text garbled?" That should fix everything but the overwrapped lines, which you can fix by copying and pasting into an editor and deleting the bogus linebreaks.
I too hate ASCII art. Why can't math-fun allow html with in-line images?
From the path-invariant 3x3s {{j, {{0, q^(j + 2*n), 1 - q^n}, {1, q^n, 0}, {0, 0, 1}}}, {n, {{q^(j + 2*n)/(1 - q^(1 + n)), 0, 1 - q^n}, {0, q^(1 + j + 2*n)/(1 - q^(1 + n)),1}, {0, 0, 1}}}}
Neil and I just found [ inf ] [ ==== n n (n + 1) ] [ \ a q ] [ ? a > -------------- ] inf [ / qpoch(q, q, n) ] /===\ [ ==== ] | | [ j ] [ n = 0 ] | | [ 0 a q ] = [ ] | | [ ] [ inf 2 ] j = 0 [ 1 1 ] [ ==== n n ] [ \ a q ] [ ? > -------------- ] [ / qpoch(q, q, n) ] [ ==== ] [ n = 0 ] Maybe Corey&Julian can help me determine the "?"s. --rwg
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Let G(q)=sum(n=0,S,q^(n^2)/prod(k=1,n,1-q^k)); /* g.f. of A003114 */ \\Vec(G(q)) \\ A003114 partitions of n into parts 5k+1 or 5k+4. H(q)=sum(n=0,S,q^(n^2+n)/prod(k=1,n,1-q^k)); /* g.f. of A003106 */ \\Vec(H(q)) \\ A003106 partitions of n into parts 5k+2 or 5k+3. P(q) = sum(n=0,S,q^(n^2-n)/prod(k=1,n,1-q^k)); /* g.f. of A003113 */ \\Vec(P(q)) \\ A003113 Coefficients in expansion of permanent ... B(q) = q * sum(n=0,S,q^(n^2+2*n)/prod(k=1,n,1-q^k)); /* g.f. of A006141 */ \\Vec(B(q)) \\ A006141 balanced partitions: smallest part equals the number of parts. A specialization (a==1, starting product with k=1): inf /===\ | | [ k ] [ B B ] | | [ 0 q ] = [ ] | | [ ] [ H H ] k = 1 [ 1 1 ] And (a==1, starting with k=0): inf /===\ | | [ k ] [ H H ] [ H H ] | | [ 0 q ] = [ ] = [ ] | | [ ] [ G G ] [ H+B H+B ] k = 0 [ 1 1 ] So I added ( https://oeis.org/draft/A003106 ): R. W. Gosper and _N. J. A. Sloane_ give (message to math-fun, Oct 10 2012) prod(k>=0, [0 , a*q^k; 1, 1]) = [?, U(a,q); ?, V(a,q)] where U(a,q) = a * sum(n>=0, a^n*q^(n^2+n) / prod(k=1..n, 1-q^n) ) and V(a,q) = sum(n>=0, a^n*q^(n^2) / prod(k=1..n, 1-q^n) ). Set a=1 to obtain prod(k>=0, [0 , q^k; 1, 1]) = [H(q), H(q); G(q), G(q)]. Now we really like to know the two '?' ...