92 reputation
3
bio website
location
age
visits member for 1 year, 11 months
seen Dec 13 '13 at 15:43

Mar
13
comment How to create a generating function / closed form from this recurrence?
Can you explain the last step at all, how you get to 7x + ... etc
Mar
13
accepted How to create a generating function / closed form from this recurrence?
Mar
13
asked How to create a generating function / closed form from this recurrence?
Feb
5
accepted Concatenated number mod m
Feb
5
comment Concatenated number mod m
sorry, I mean each variable is a digit 0-9 (first digit 1-9). otherwise yes
Feb
5
awarded  Commentator
Feb
5
comment Concatenated number mod m
integers yes. I assumed I could just add the values all mod m but am unsure
Feb
5
comment Concatenated number mod m
Basic yes, but having trouble here
Feb
5
asked Concatenated number mod m
Jan
7
awarded  Scholar
Jan
7
accepted Is n! mod p doable in sub O(n) time?
Jan
7
comment Is n! mod p doable in sub O(n) time?
forgive my ignorance but explain p(x) please?
Jan
7
comment Is n! mod p doable in sub O(n) time?
Well i can solve it iteratively by multiplying by n-k+1 all the way up modulo p every step. rationale: (n!/((n-k)!)) /(n!/((n-(k-1))!)) = n-k+1
Jan
7
comment Is n! mod p doable in sub O(n) time?
technically time complexity of min(n,p) i think, right?
Jan
7
comment Is n! mod p doable in sub O(n) time?
i also know about the zero condition but i am mainly concerned about nontrivial case k < p
Jan
7
comment Is n! mod p doable in sub O(n) time?
arbitrary. i am writing a function
Jan
7
comment Is n! mod p doable in sub O(n) time?
I mean this may be an XY question or whatever. I just want n choose k mod p in fast execution. I just thought the factorial difference was what i needed. mywiki.wooledge.org/XyProblem
Jan
7
comment Is n! mod p doable in sub O(n) time?
well i know how to get combinations quickly and i know the only difference between combinations and permutations is a denominator factor of k!
Jan
7
asked Is n! mod p doable in sub O(n) time?
Dec
23
comment Division into $x(x-1)$
Right but this is maximal x given g