A199745 Numbers such that the sum of the largest and the smallest prime divisor equals the sum of the other distinct prime divisors.
2145, 2730, 4641, 4845, 5005, 5460, 5610, 6435, 7410, 8190, 8778, 9177, 10725, 10920, 11220, 11305, 11730, 13485, 13585, 13650, 13923, 14535, 14820, 16380, 16830, 17017, 17556, 19110, 19305, 20010, 20930, 21489, 21505, 21840, 22230, 22440, 23460, 23529, 23595
Offset: 1
Keywords
Examples
22440 is in the sequence because the distinct prime divisors are {2, 3, 5, 11, 17} and 17+2 = 3+5+11.
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Programs
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Haskell
a199745 n = a199745_list !! (n-1) a199745_list = filter (\x -> 2 * (a074320 x) == a008472 x) [1..] -- Reinhard Zumkeller, Nov 10 2011
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Maple
isA199745 := proc(n) local p; p := sort(convert(numtheory[factorset](n),list)) ; if nops(p) >= 3 then return ( op(1,p) + op(-1,p) = add(op(i,p),i=2..nops(p)-1) ) ; else false; end if; end proc: for n from 2 to 20000 do if isA199745(n) then printf("%d,",n) ; end if ; end do: # R. J. Mathar, Nov 10 2011
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Mathematica
Select[Range[25000],Plus@@(pl=First/@FactorInteger[#])/2==pl[[1]]+pl[[-1]]&] (* Ray Chandler, Nov 10 2011 *)
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Sage
def isA199745(n) : p = factor(n) return len(p) > 2 and p[0][0] + p[-1][0] == add(p[i][0] for i in (1..len(p)-2)) [n for n in (2..20000) if isA199745(n)] # Peter Luschny, Nov 10 2011
Comments