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Primes between consecutive squares

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  • Kermit Rose
    By prime number theorem, the approximate number of primes less than N**2 is N**2/(2 ln(N)). I considered using this relationship to estimate whether or not
    Message 1 of 3 , Dec 7, 2010
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      By prime number theorem, the approximate number of primes less

      than N**2 is N**2/(2 ln(N)).


      I considered using this relationship to estimate whether or not
      there always exist a prime between a square and the next larger
      square.

      I derive the conjecture that there exist infinitely many
      integers, N, such that there are no primes between N**2 and
      (N+1)**2.

      I "reasoned" as follows:

      Suppose we had a graph of the function,

      f(x) = number of primes < x**2.

      For large x, this function is approximated by

      f(x) = x**2/ln(x**2) = x**2/(2 ln(x))


      I calculate the first and second derivatives of this
      approximation.


      y = x**2/(2 ln(x))

      2 ln(x) y = x**2

      2 ln(x) y' + 2y/x = 2x

      ln(x) y' + y/x = x

      x ln(x) y' + y = x**2

      (2 ln(x)) x ln(x) y' + (2 ln(x)) y = (2 ln(x)) x**2

      2 ln(x) y = x**2

      (2 ln(x)) x ln(x) y' + x**2 = (2 ln(x)) x**2

      2x [ln(x)]**2 y' = (2 ln(x) - 1) x**2


      2 [ln(x)]**2 y' = (2 ln(x) - 1) x


      2 [ln(x)]**2 y'' + 4 ([ln(x)] /x) y' = (2/x)x + (2 ln(x) - 1)

      2x [ln(x)]**2 y'' + 4 [ln(x)] y' = x(2 ln(x) + 1)


      2x [ln(x)]**2 y' = (2 ln(x) - 1) x**2


      4x [ln(x)]**2 y' = 2 (2 ln(x) - 1) x**2


      2x [ln(x)]**2 y'' + 4 [ln(x)] y' = x(2 ln(x) + 1)


      2x [ln(x)]**3 y'' + 4 [ln(x)]**2 y' = x(2 ln(x) + 1)[ln(x)]

      4x [ln(x)]**2 y' = 2 (2 ln(x) - 1) x**2

      2x [ln(x)]**3 y'' + 2 (2 ln(x) - 1) x**2 = x(2 ln(x) + 1)[ln(x)]

      2x [ln(x)]**3 y'' =x(2 ln(x) + 1)[ln(x)] -2(2 ln(x) - 1) x**2

      2[ln(x)]**3 y'' =(2 ln(x) + 1)[ln(x)] -2(2 ln(x) - 1) x


      showing that the second derivative is negative for all x.



      This suggests that the rate of increase of the number of primes
      between consecutive squares will slow, and eventually, on the
      average, decrease.

      As a test, I calculated the actual number of primes between
      consecutive squares, for the first thousand squares.


      Here is the result for the last 100 of those calculated:

      First Column: N
      Second column: Number of primes < N**2
      Third column: Number of primes between N**2 and (N+1)**2
      Fourth Column: Number of primes between (N+1)**2 and (N+2)**2
      minus the number of primes between N**2 and (N+1)
      squared.

      These results indicate that the second difference of the actual
      prime count is often negative.


      Perhaps someone can calculate, by finding when the first
      derivative of x**2/(2 ln(x)), becomes less than 1, the order of
      magnitude of the smallest value of N, such that there is no prime
      between N**2 and (N+1)**2.

      900 64683 137 -7
      901 64820 130 -3
      902 64950 127 11
      903 65077 138 -18
      904 65215 120 24
      905 65335 144 -7
      906 65479 137 2
      907 65616 139 1
      908 65755 140 -3
      909 65895 137 -13
      910 66032 124 16
      911 66156 140 0
      912 66296 140 -19
      913 66436 121 14
      914 66557 135 -3
      915 66692 132 -6
      916 66824 126 14
      917 66950 140 0
      918 67090 140 -17
      919 67230 123 12
      920 67353 135 3
      921 67488 138 -3
      922 67626 135 -10
      923 67761 125 7
      924 67886 132 13
      925 68018 145 -14
      926 68163 131 11
      927 68294 142 0
      928 68436 142 -13
      929 68578 129 12
      930 68707 141 2
      931 68848 143 0
      932 68991 143 -13
      933 69134 130 -5
      934 69264 125 23
      935 69389 148 -16
      936 69537 132 12
      937 69669 144 0
      938 69813 144 -12
      939 69957 132 0
      940 70089 132 8
      941 70221 140 2
      942 70361 142 -12
      943 70503 130 11
      944 70633 141 -2
      945 70774 139 0
      946 70913 139 3
      947 71052 142 -17
      948 71194 125 10
      949 71319 135 -6
      950 71454 129 15
      951 71583 144 8
      952 71727 152 4
      953 71879 156 -21
      954 72035 135 -14
      955 72170 121 14
      956 72291 135 7
      957 72426 142 -1
      958 72568 141 -2
      959 72709 139 5
      960 72848 144 10
      961 72992 154 -16
      962 73146 138 -1
      963 73284 137 0
      964 73421 137 1
      965 73558 138 9
      966 73696 147 -11
      967 73843 136 -1
      968 73979 135 4
      969 74114 139 7
      970 74253 146 2
      971 74399 148 -19
      972 74547 129 7
      973 74676 136 5
      974 74812 141 2
      975 74953 143 -3
      976 75096 140 -2
      977 75236 138 -6
      978 75374 132 9
      979 75506 141 7
      980 75647 148 0
      981 75795 148 -9
      982 75943 139 2
      983 76082 141 2
      984 76223 143 11
      985 76366 154 -18
      986 76520 136 14
      987 76656 150 -18
      988 76806 132 2
      989 76938 134 8
      990 77072 142 -2
      991 77214 140 3
      992 77354 143 -3
      993 77497 140 3
      994 77637 143 5
      995 77780 148 -16
      996 77928 132 16
      997 78060 148 3
    • Peter Kosinar
      Hi Kermit, ... Unless I ve made a mistake, the first and second derivative look like this: f (x) = x/log(x) - 1/(2*log^2(x)) f (x) = 1/log(x) -
      Message 2 of 3 , Dec 7, 2010
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        Hi Kermit,

        > f(x) = x**2/ln(x**2) = x**2/(2 ln(x))
        >
        > I calculate the first and second derivatives of this
        > approximation.
        >
        > [ long calculation snipped ]
        >
        > showing that the second derivative is negative for all x.

        Unless I've made a mistake, the first and second derivative
        look like this:

        f'(x) = x/log(x) - 1/(2*log^2(x))
        f''(x) = 1/log(x) - (3/2)*(1/log^2(x)) + 1/log^3(x)

        Unfortunately, they're both positive for x > sqrt(e) :-)

        Peter
      • Kermit Rose
        ... Thank you. I trust your calculations more than my own. Kermit.
        Message 3 of 3 , Dec 7, 2010
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          On 12/7/2010 11:34 PM, Peter Kosinar wrote:
          >
          > Hi Kermit,
          >
          >> f(x) = x**2/ln(x**2) = x**2/(2 ln(x))
          >>
          >> I calculate the first and second derivatives of this
          >> approximation.
          >>
          >> [ long calculation snipped ]
          >>
          >> showing that the second derivative is negative for all x.
          >
          > Unless I've made a mistake, the first and second derivative look like this:
          >
          > f'(x) = x/log(x) - 1/(2*log^2(x))
          > f''(x) = 1/log(x) - (3/2)*(1/log^2(x)) + 1/log^3(x)
          >
          > Unfortunately, they're both positive for x > sqrt(e) :-)


          Thank you.

          I trust your calculations more than my own.

          Kermit.
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