OFFSET
1,1
COMMENTS
With this constant f(1) and using the formula f(n+1) = (floor(f(n))*(f(n))) - ((floor(f(n)))^2 - floor(f(n))) it is possible to obtain the prime numbers repeated exactly a number of times corresponding to the position of the prime number. That is, 2 once, 3 twice, 5 thrice, etc.
FORMULA
Equals 2 + (3-2)/(2) + (5-3)/(2*3^2) + (7-5)/(2*3^2*5^3) + (11-7)/(2*3^2*5^3*7^4) + ...
EXAMPLE
2.61200074043...
MATHEMATICA
imax:=87; First[RealDigits[N[2+Sum[(Prime[i]-Prime[i-1])/Product[Prime[j-1]^(j-1), {j, 2, i}], {i, 2, imax}], imax]]] (* Stefano Spezia, Dec 16 2020 *)
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Davide Rotondo, Dec 16 2020
STATUS
approved
