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A260981 Primes p that are equal to the sum of the first k primes where p=prime(prime(k)).

5, 17, 41

Offset: 1

Waleed Mohammed, Ali Adams, Aug 06 2015

Terms listed are the only three primes p found to satisfy the condition that p = prime(m) = Sum_{i=1..k} prime(i) where m=prime(k).
From Jon E. Schoenfield, Aug 19 2015: (Start)
Let S(k) be the sum of the first k primes, and let PP(k) = prime(prime(k)); then the terms of the sequence are the values of prime(prime(k)) at those values of k at which S(k) = PP(k). (This occurs at k = 2, 4 and 6.)
Given the behavior of the ratio S(k)/PP(k) over the range of values of k shown in the table below, it seems very unlikely that this ratio will return to 1 for any k beyond the values that have been tested, and thus very likely that a(3) = 41 = PP(6) is the final term in the sequence:
     k        S(k)         PP(k)     S(k)/PP(k)
  ======  ===========    ========  ==============
       1            2           3     0.666666...
       2            5  =        5     1
       3           10          11     0.909090...
       4           17  =       17     1
       5           28          31     0.903225...
       6           41  =       41     1
       7           58          59     0.983050...
       8           77          67     1.149253...
       9          100          83     1.204819...
      10          129         109     1.183486...
     ...
     100        24133        3911     6.170544...
    1000      3682913       80917    45.514700...
   10000    496165411     1366661   363.049367...
  100000  62260698721    20491057  3038.432752... (End)

			k=3: prime(3) = 5 = 2+3 = prime(1) + prime(2).
k=7: prime(7) = 17 = 2+3+5+7 = prime(1) + prime(2) + prime(3) + prime(4).
k=13: prime(13) = 41 = 2+3+5+7+11+13 = prime(1) + prime(2) + prime(3) + prime(4) + prime(5) + prime(6).

Cf. A006450, A007504, A013918.

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User: Waleed Mohammed

A260981 Primes p that are equal to the sum of the first k primes where p=prime(prime(k)).

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