Totally mind-boggling Gary. Tracing through On-Line Dictionary of Integer Sequences I figured out how to do it recursively, which might not be as much fun. For one thing, it takes forever to run. I despair of ever reaching p(200) this way. Major MacMahon rocks 'n rules (Hardy and Ramanujan too).
This song about Ramanujan's life is both funny and kinda sad: http://www.archive.org/details/Ramanujan (play control in upper right) Kirby >>> gen = a000041() >>> [next(gen) for i in range(60)] [1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, 6842, 8349, 10143, 12310, 14883, 17977, 21637, 26015, 31185, 37338, 44583, 53174, 63261, 75175, 89134, 105558, 124754, 147273, 173525, 204226, 239943, 281589, 329931, 386155, 451276, 526823, 614154, 715220, 831820] Here's the code: def a000041(): gen = a137683() while True: yield sum(next(gen)) def pascal(): row = [1] while True: yield row row = [i+j for i,j in zip(row + [0], [0] + row)] def invpascal(): gen = pascal() while True: row = next(gen) if len(row)%2: yield [row[i]*pow(-1,i) for i in range(len(row))] else: yield [row[i]*pow(-1,i+1) for i in range(len(row))] def a137683(): gen1 = invpascal() gen2 = a026794() n = 1 ipmatrix = [] # inverse pascal triangular matrix while True: ipmatrix.append(next(gen1)) result = [] row = next(gen2) for i in range(n): col = [0 for j in range(n)] for k in range(i,n): col[k] = ipmatrix[k][i] result.append(sum([i * j for i,j in zip(row, col)])) yield result n += 1 def a026794(): n = 1 while True: row = [] for k in range(1, n+1): row.append(T(n,k)) yield row n += 1 def T(n, k): """ T(n, k)=if(k<1|k>n, 0, if(n==k, 1, sum(i=k, n-k, T(n-k, i)))) """ if k < 1 or k > n: return 0 if n == k: return 1 else: return sum([T(n-k, i) for i in range(k, n-k+1)]) # bonus, not used def a010054(): n = 0 incr = 0 tri = incr while True: if n == tri: yield 1 incr += 1 tri = tri + incr else: yield 0 n += 1 On Wed, Jun 30, 2010 at 6:49 PM, Litvin <lit...@skylit.com> wrote: > At 07:01 PM 6/30/2010, kirby urner wrote: > > I did something similar with a Ramanujan series converging to pi... > > Since you mentioned Ramanujan... > > He and Hardy worked quite a bit on the partition function: p(n) = the number > of different ways n can be represented as a sum of positive integers. The > sequence p(n) starts like Fibonacci numbers, 1, 1, 2, 3, 5... but the next > one is 7. Nice try! > > At some point around 1916, an amature mathematician Major MacMahon, who was > "good with numbers" calculated _by hand_ for Hardy and Ramanujan p(n) for n > = 1 through 200. p(200) happens to be 3,972,999,029,388. An interesting > story. > > It is possible, of course, to write a recursive function to calculate p(n), > but a more fun way is to use the generating function (1 + x + x^2 > +x^3+...x^n)*(1+x^2+x^4+x^6+...)(1+x^3+x^6+...)*...*(1+x^n) (due to Euler, > as most everything else...) When you distribute, the coefficients at x, > x^2, ... of the resulting polynomial give you p(1), p(2), ... So this could > be a basis for a meaningful project on lists and polynomials. > > If you plug a large enough number into the generating function, say, x = > 10000, you get a number whose groups of five digits (from the right) give > you p(n) as long as p(n) < 10000. So for a relatively small n instead of > multiplying polynomials you can multiply "periodic" integers > 100010001...0001*100000001...00000001*... > It is easy, of course, to generate these numbers using strings. Then this > becomes a meaningful exercise on strings. > > I am currently working on a "test package" for the Math and Python book and > I've just added a question based on a small peace of this, so I felt like > sharing. Not sure whether this is for the left brain or for the right > brain; these periodic strings are kinda cute. :) > > Gary Litvin > www.skylit.com > _______________________________________________ Edu-sig mailing list Edu-sig@python.org http://mail.python.org/mailman/listinfo/edu-sig
