https://www.speedsolving.com/forum/threads/blindfold-accomplishment-thread.3582/page-103#post-178728
This happened in May 2009 and I thought it was time for a next
Pyraminx Crystal blind
49:49.17, memo 17:55 min

Did it with a QJ, which was the main challenge.
Haven't found any trace, that someone else has done it yet, so I claim place 2 worldwide for me

Memo was:
edges: 38,22 1,27 36,35 37,18 9,41 26,55 10,2 58,33 14,21 13,17 49,60 20,54 23,15 59,16 42,16
corners: 12,13 2,35 37,44 60,19 26,34 8,43 33,11 29,18 56,51 10/30
and I solved it with this system (colors as actual WCA megaminx-sheme, white on top, green in front):
edges:
buffer LB-U (yellow-white), target UF (white-green), helper RB-U (blue-white)
setup1 L R’ L’ R setup1’ setup2 R’ L R L’ setup2’
L is yellow, R is red
OK, the helper (it was flipped, too) needs a bit of thinking, but can be done with half of the alg. And flipping a edge goes with solving it first normal and then flipped.
corners:
Found probably the same as Istvan (8-mover repeated 7 times), but this seemed too long for a QJ.
Then I found a 8-mover, which seemed to be interesting:
R F’ R’ F RD F RD’ F’
it cycles 3 corners and 2*3 edges. My idea was to "eliminate" the edges-cycles after each corners-cycle. Before actual doing this, I found Stefan’s 14-mover in a video
RB L’ UBR’ DFL y L RB’ U’ RB L’ UBR DFL’ y' L RB’ U
which I used in my first attempt, but I couldn’t control the QJ enough when there were lock-ups and stuff.
So I thought again about my 8-mover, put it on top and searched for insertions for the undo-the-edge-cycles. Result:
F R F’ R’ U’ R’ L’ R L U FDR FDL R’ FDL’ FDR’ R
buffer is UFL, target is F-DR-DL, helper is URF
So it is: setup1 alg setup1’ setup2 alg’ setup2’
If it was easy, I put one corner of the next pair in URF and had to do the alg only once for the pair.