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    <p>The code shown below avoids several problems that are present in previously-presented code.</p> <p>A problem in the original post is counting divisors by testing all numbers less than the number <code>t</code> being processed. Various possible speedups exist: Don't test the numbers between <code>(t+2)/2</code> and <code>t-1</code>, because none of them can divide <code>t</code>. That could be extended (for example, numbers between <code>(t+3)/3</code> and <code>(t-1)/2</code> don't need to be tested) but returns diminish rapidly. A more important speedup is to use a multiplicative form of the <a href="https://en.wikipedia.org/wiki/Divisor_function" rel="nofollow">divisor function</a>; for example, if a number <code>t</code> equals <code>pᵃ·qᵇ·rᶜ</code> with <code>p, q, r</code> prime, then <code>σ₀(t) = τ(t) = (a+1)·(b+1)·(c+1)</code>. (The divisor-counting function σ₀ (sigma zero) often is indicated by τ (tau) instead of σ₀.)</p> <p>A problem in another answer is factoring <code>(i*(i+1)/2)</code> or <code>n*(n+1)/2</code> at each step. It is faster to just factor <code>i+1</code> or <code>n+1</code> at each step, and remember the factors from the previous step. This cuts execution time by a factor of at least 2.</p> <p>The code below computes <code>τ(k·(k+1)/2)</code> by <code>τ(k)·τ(k+1)·e/(e+1))</code>, where <code>e</code> represents the exponent of 2 in the product <code>τ(k)·τ(k+1)</code>. That is, because no prime factor of <code>k</code> can be a prime factor of <code>k+1</code>, and vice versa, <code>τ(k·(k+1)) = τ(k)·τ(k+1)</code>; and <code>τ(k·(k+1)/2)</code> is smaller than <code>τ(k·(k+1))</code> by the ratio <code>e/(e+1)</code>. </p> <p>The attached program runs in about 0.14 seconds on my older computer, when invoked by <code>findHiTau(120)</code>. The parameter to <code>findHiTau</code> is an upper limit on size of primes in the program's <code>oddprimes</code> table, and is the square root of the highest value of <code>k</code> tested; so triangular numbers as large as about <code>k⁴/2</code> will be processed.</p> <pre><code># Returns divisor count for integer v and power of 2 in v def tau(v): p2 = 0 while v%2 == 0: v, p2 = v/2, p2+1 t = p2+1 for p in oddprimes: if v%p == 0: e = 1 while v%p == 0: v, e = v/p, e+1 t *= e if p*p &gt; v: break if v&gt;1: # Rest of v is a prime t *= 2 return (t, p2) def findHiTau(primeLim): # Show record-high values of tau(k*(k+1)/2) up to k=m # Make array of small odd primes (expression is ok at least to 100K) oddprimes = [3,5,7,11,13]+[x for x in range(17,primeLim,2) if 1==pow(2,x-1,x)==pow(3,x-1,x)==pow(5,x-1,x)==pow(7,x-1,x)==pow(11,x-1,x)==pow(13,x-1,x)] taub, e2, hitau = 2, 1, 0 # e2 = exponent of 2 # In loop, taua = tau(k-1); taub = tau(k); e2 = power of 2 in (k-1)*k for k in range(3,primeLim*primeLim): taua = taub (taub, f2) = tau(k) if f2: e2 = f2 # If f is even, update e2 ttau = taua*taub*e2/(e2+1) if ttau &gt; hitau: hitau = ttau print '{:3} {:3} {:3}'.format(k-1, k, ttau) </code></pre>
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    1. POImproving my code for project euler p12. in Python
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    1. USJames Waldby - jwpat7
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