|
|
A094662
|
|
Prime numerators of the sums of the ratios of consecutive primes.
|
|
1
|
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Sum of reciprocals = 0.3941031881461705902079511494...
The next term, A094661(53) ~ 1.497*10^100, is too large to include.
It might be preferable to record the index of these primes in A094661: a(n)=A094661(b(n)) with b=(1,2,3,4,7,32,53,55,94,183,189,...). - M. F. Hasler, Mar 06 2009
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
3/2 + 5/3 + 7/5 + 11/7 + 13/11 + 17/13 + 19/17 = 4975049/510510. 4975049 is prime, the fifth entry in the sequence.
|
|
MATHEMATICA
|
Select[Numerator[Accumulate[#[[2]]/#[[1]]&/@Partition[Prime[Range[ 100]], 2, 1]]], PrimeQ] (* Harvey P. Dale, Jun 01 2018 *)
|
|
PROG
|
(PARI) consecpr(n) = { s=0; y=0; forprime(x=2, n, y+=nextprime(x+1)/x; z=numerator(y); s+=1./z; if(isprime(z), print1(z", ")) ); print(); print(s) }
(PARI) s=0; for( i=1, 999, isprime(numerator(s+=prime(i+1)/prime(i))) & print1(numerator(s)", ")) \\ M. F. Hasler, Mar 06 2009
|
|
CROSSREFS
|
|
|
KEYWORD
|
frac,nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|