A116536 Numbers that can be expressed as the ratio of the product and the sum of consecutive prime numbers starting from 2.
1, 3, 125970, 1278362451795, 305565807424800745258151050335, 2099072522743338791053378243660769678400212601239922213271230, 330455532167461882998265688366895823334392289157931734871641555
Offset: 1
Examples
a(1) = 1 because 2/2 = 1. a(2) = 3 because (2*3*5)/(2+3+5) = 30/10 = 3. a(3) = 125970 because (2*3*5*7*11*13*17*19)/(2+3+5+7+11+13+17+19) = 9699690/77 = 125790.
References
- G. Balzarotti and P. P. Lava, Le sequenze di numeri interi, Hoepli, 2008, p. 158.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..81 (terms 1..42 from Vincenzo Librandi)
Crossrefs
Programs
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Haskell
import Data.Maybe (catMaybes) a116536 n = a116536_list !! (n-1) a116536_list = catMaybes $ zipWith div' a002110_list a007504_list where div' x y | m == 0 = Just x' | otherwise = Nothing where (x',m) = divMod x y -- Reinhard Zumkeller, Oct 03 2011
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Magma
[p/s: n in [1..40] | IsDivisibleBy(p,s) where p is &*[NthPrime(i): i in [1..n]] where s is &+[NthPrime(i): i in [1..n]]]; // Bruno Berselli, Sep 30 2011
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Maple
P:=proc(n) local i,j, pp,sp; pp:=1; sp:=0; for i from 1 by 1 to n do pp:=pp*ithprime(i); sp:=sp+ithprime(i); j:=pp/sp; if j=trunc(j) then print(j); fi; od; end: P(100);
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Mathematica
seq = {}; sum = 0; prod = 1; p = 1; Do[p = NextPrime[p]; prod *= p; sum += p; If[Divisible[prod, sum], AppendTo[seq, prod/sum]], {50}]; seq (* Amiram Eldar, Nov 02 2020 *)
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