dude just use a six-deep for loop and generate your dice that way, in order. It's extremely fast even for N=30

[QUOTE=voidme;294168]dude just use a six-deep for loop and generate your dice that way, in order. It's extremely fast even for N=30[/QUOTE]

I was talking of 377. I did that for 376 and with all my edits I only cut it from a 112 digit number of years down to a 32 digit number.

like most problems 377 likely will not require you to actually figure out all the partitions but rather partial sums

there are patterns in the partial sums but they're messy imo

[QUOTE=voidme;294170]like most problems 377 likely will not require you to actually figure out all the partitions but rather partial sums

there are patterns in the partial sums but they're messy imo[/QUOTE]

I think the reason it runs out of memory is I take 13 with the i being an exponent.

[QUOTE=voidme;294147]377 is hilarious

there is an OEIS sequence but it's incorrect[/QUOTE]
The OEIS sequence is correct up to 9. It diverges starting at 10. In fact it's the comment in the OEIS page that is misleading.

[QUOTE=ldesnogu;294174]The OEIS sequence is correct up to 9. It diverges starting at 10. In fact it's the comment in the OEIS page that is misleading.[/QUOTE]

right, this is correct; it is still nevertheless apparent that the OEIS sequence is pretty much useless for the purposes of 377

[QUOTE=ldesnogu;294174]The OEIS sequence is correct up to 9. It diverges starting at 10. In fact it's the comment in the OEIS page that is misleading.[/QUOTE]

Does the comment need to be corrected?

[QUOTE=CRGreathouse;294194]Does the comment need to be corrected?[/QUOTE]

the comment is wrong because the sequence does not describe what the comment says in the strictest sense

I wonder if 377 can be solved with a closed formula. I was able to get one for [TEX]1\leq n \leq 19[/TEX], but not yet for larger values. More thoughts needed :smile:

[QUOTE=ldesnogu;294243]I wonder if 377 can be solved with a closed formula. I was able to get one for [TEX]1\leq n \leq 19[/TEX], but not yet for larger values. More thoughts needed :smile:[/QUOTE]

basically what I found is 212+221+122 = 555 and all these are in the list to add up to 5 so the answer for 5 can be drawn from: 5,41,23,113,122,1112, and 11111

under 11111 there are 1,2,2,1, amount of unique arrangements each with 1,2,2,3,3,4, possibilities of rearrangement 1*5+2*55+2*555+1*5555+11111 = 17891

I am pretty close to one but rearranging is a challenge; almost positive it is possible

EDIT: lol just derived the OEIS equation purely by accident as an approximation to my data. The multiplicative difference between each result is about 11 (minus some change as you go on).