[seqfan] Comments regarding "Number of Garden of Eden partitions of n in Bulgarian Solitaire"

Thomas Baruchel baruchel at gmx.com
Sat May 12 15:04:19 CEST 2018


some months ago, I released the tool https://github.com/baruchel/oeis for searching
linear relations between sequences; then I went to holidays and forgot a little
about it. Some days ago, I decided to make it run again and began to gather some
results, but rather than copying the whole dump-output of the software, I decided
to chose some of these results in order to submit it here in a well formatted way.

One result is:

A064173 A101198 A123975    -->    1 -1 -1    (3)
A064173 Number of partitions of n with positive rank.
A101198 Number of partitions of n with rank 1 (the rank of a partition is the largest part minus the number of parts).
A123975 Number of Garden of Eden partitions of n in Bulgarian Solitaire.

meaning that A064173(n) =  A101198(n) + A123975(n)

The interesting result here is that A123975 is not a large entry; maybe it would
have some interest to add some more properties.

Would the definition "Number of partitions of n with rank at least 2" be the right
definition for A123975 (as long as the empirically detected linear relation is true)?

Furthermore, A101198 and A123975 have a G.f.; thus it would be easy to build a G.f.
for A064173 (no g.f. yet). With Maxima, I get:

taylor( product( 1/(1-q^i), i, 1, infty) * sum( (-1)^(r-1)*q^((3*r^2+3*r)/2), r, 1, infty)     + sum( (-1)^k*q^(k)*(q^((3*k^2+k)/2)-q^((3*k^2-k)/2)), k, 1, infty)/product(1-q^k, k, 1, infty), q, 0, 60 );

(%o18) q^2+q^3+2*q^4+3*q^5+5*q^6+6*q^7+10*q^8+13*q^9+19*q^10+25*q^11+35*q^12

which seems to match:
A064173          Number of partitions of n with positive rank.
     0, 1, 1, 2, 3, 5, 6, 10, 13, 19, 25, 35, 45, 62, 80, 106, 136, 178, 225, 291,
     366, 466, 583, 735, 912, 1140, 1407, 1743, 2140, 2634, 3214, 3932, 4776, 5807,
     7022, 8495, 10225, 12313, 14762, 17696, 21136, 25236, 30030, 35722, 42367,
     50216, 59368, 70138

Could someone simplify the g.f. above (in the Maxima expression)?

Best regards,

Thomas Baruchel

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