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A098699
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Anti-derivative of n: or the first occurrence of n in A003415, or zero if impossible.
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9
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1, 2, 0, 0, 4, 6, 9, 10, 15, 14, 21, 0, 8, 22, 33, 26, 12, 0, 65, 34, 51, 18, 57, 0, 20, 46, 69, 27, 115, 0, 161, 30, 16, 62, 93, 0, 155, 0, 217, 45, 111, 42, 185, 82, 24, 50, 129, 0, 44, 94, 141, 63, 235, 0, 329, 75, 52, 0, 265, 70, 36, 66, 177, 122, 183, 0, 305, 0, 40, 134
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OFFSET
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0,2
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COMMENTS
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With Goldbach's conjecture, any even integer n = 2k > 2 can be written as sum of two primes, n = p + q, and therefore admits N = pq as (not necessarily smallest) anti-derivative, so a(2k) > 0, and a(2k) <= pq <= k^2. [Remark inspired by L. Polidori.] - M. F. Hasler, Apr 09 2015
a(n) <= n^2/4 for n > 1. This is because if A003415(x) = n > 1, x = a*b for some a,b > 1, and then n = A003415(x) = a*A003415(b) + A003415(a)*b >= a + x/a >= 2*sqrt(x), i.e. x <= (n/2)^2. - Robert Israel, May 29 2023
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LINKS
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FORMULA
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a(n) = n for { 4, 27, 3125, 823543, ... } = { p^p; p prime } = A051674.
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MAPLE
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ader:= proc(n) local t;
n * add(t[2]/t[1], t = ifactors(n)[2])
end proc:
N:= 100: # for a(0) .. a(N)
V:= Array(0..N): count:= 0:
for x from 1 to N^2/4 while count < 100 do
v:= ader(x);
if v > 0 and v <= 100 and V[v] = 0 then
count:= count+1; V[v]:= x;
fi;
od:
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MATHEMATICA
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a[1] = 0; a[n_] := Block[{f = Transpose[ FactorInteger[ n]]}, If[ PrimeQ[n], 1, Plus @@ (n*f[[2]]/f[[1]])]]; b = Table[0, {70}]; b[[1]] = 1; Do[c = a[n]; If[c < 70 && b[[c + 1]] == 0, b[[c + 1]] = n], {n, 10^3}]; b
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PROG
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(Python)
from sympy import factorint
if n < 2:
return n+1
for m in range(1, (n**2>>2)+1):
if sum((m*e//p for p, e in factorint(m).items())) == n:
return m
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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