sum of 1/A007504(n)
N. J. A. Sloane
njas at research.att.com
Mon May 14 19:40:18 CEST 2007
Hello, I am trying a first calculation with
100,000 terms to see if the 1/2 + Pi/6 makes sense.
SImon plouffe
At 1:40 PM -0400 5/14/07, N. J. A. Sloane wrote:
>Dear Seqfans, A correspondent,
>"fabio mercurio" <mercurio.fabio at gmail.com>
>writes to say that this number
>
>%I A122989
>%S A122989 1,0,2,3,4,7,6,3,2
>%N A122989 Decimal expansion of Sum_{n >= 1} 1/A007504(n), where
>A007504(n) is the sum of the f\
>irst n primes.
>%e A122989 1/2+1/5+1/10+1/17+1/28+1/41+1/58+1/77+1/100+... = 1.02347632...
>%Y A122989 Cf. A007504.
>%Y A122989 Adjacent sequences: A122986 A122987 A122988 this_sequence
>A122990 A122991 A122992
>%Y A122989 Sequence in context: A026237 A125150 A072275 this_sequence
>A077223 A055265 A117922
>%K A122989 cons,nonn,more
>%O A122989 1,3
>%A A122989 Pierre CAMI (pierrecami(AT)tele2.fr), Oct 28 2006
>
>is really equal to 1/2 + Pi/6. But the agreement is
>not very close. Can someone compute more decimal places?
Is this sum finite? A007504(n) grows roughly as 0.5 n^2 log(n). However,
the sum of the reciprocals 2/(n^2 log(n)) diverges.
Tony
At 1:40 PM -0400 5/14/07, N. J. A. Sloane wrote:
>Dear Seqfans, A correspondent,
>"fabio mercurio" <mercurio.fabio at gmail.com>
>writes to say that this number
>
>%I A122989
>%S A122989 1,0,2,3,4,7,6,3,2
>%N A122989 Decimal expansion of Sum_{n >= 1} 1/A007504(n), where
>A007504(n) is the sum of the f\
>irst n primes.
>%e A122989 1/2+1/5+1/10+1/17+1/28+1/41+1/58+1/77+1/100+... = 1.02347632...
>%Y A122989 Cf. A007504.
>%Y A122989 Adjacent sequences: A122986 A122987 A122988 this_sequence
>A122990 A122991 A122992
>%Y A122989 Sequence in context: A026237 A125150 A072275 this_sequence
>A077223 A055265 A117922
>%K A122989 cons,nonn,more
>%O A122989 1,3
>%A A122989 Pierre CAMI (pierrecami(AT)tele2.fr), Oct 28 2006
>
>is really equal to 1/2 + Pi/6. But the agreement is
>not very close. Can someone compute more decimal places?
Is this sum finite? A007504(n) grows roughly as 0.5 n^2 log(n). However,
the sum of the reciprocals 2/(n^2 log(n)) diverges.
Ignore that last message!
Tony
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