62 lines
1.7 KiB
Ruby
62 lines
1.7 KiB
Ruby
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descr=%{
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Pentagonal numbers are generated by the formula, P_(n)=n(3n−1)/2. The first ten pentagonal numbers are:
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1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
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It can be seen that P_(4) + P_(7) = 22 + 70 = 92 = P_(8). However, their difference, 70 − 22 = 48, is not pentagonal.
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Find the pair of pentagonal numbers, P_(j) and P_(k), for which their sum and difference is pentagonal and D = |P_(k) − P_(j)| is minimised; what is the value of D?
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}
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thoughts=%{
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We can make a fast test to know whether a number is pentagonal.
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We can also remark that this function is strictly growing.
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If P(k+1) - P(k) > P(j) then we should no more try to use the index j.
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}
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def P(n)
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return n*(3*n - 1)/2
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end
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def is_pentagonal(k)
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# k = n(3n - 1)/2
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# 2k = n(3n - 1)
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# 3n^2 - n - 2k = 0
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#
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a=3.0; b=-1; c=-2.0*k
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# delta = sqrt( b^2 - 4ac )
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delta=Math.sqrt( b**2 - 4*a*c )
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sol1=( -b + delta )/(2*a)
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sol2=( -b - delta )/(2*a)
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# renvoie vrai s'il existe une solution entiere positive
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return true if sol1 == sol1.floor && sol1 > 0
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return true if sol2 == sol2.floor && sol2 > 0
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return false
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end
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# (1..117).each do |x|
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# puts x if is_pentagonal(x)
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# end
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j=1
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k=2
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while true
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pk=P(k)
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pj=P(j)
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diff=pk-pj
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sum=pk+pj
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# puts %{P(#{k})=#{pk}, P(#{j})=#{pj}, Pk-Pj=#{diff}#{"*" if is_pentagonal(diff)}, Pk+Pj=#{sum}#{"*" if is_pentagonal(sum)}}
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if is_pentagonal( sum ) && is_pentagonal( diff )
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puts %{P(#{k})=#{pk}, P(#{j})=#{pj}, Pk-Pj=#{diff}#{"*" if is_pentagonal(diff)}, Pk+Pj=#{sum}#{"*" if is_pentagonal(sum)}}
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puts diff
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break
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end
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if P(k+1) - pk > pj
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k += 1
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j = k-1
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else
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j -= 1
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end
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end
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