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A356631
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a(n) is the least number k such that the sum (with multiplicity) of prime factors of k*(k+1)*...*(k+n-1) is a perfect power.
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2
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1, 4, 2, 1, 4, 5, 2, 1, 11, 18, 8, 12, 8, 15, 4, 41, 10, 65, 10, 39, 21, 5, 54, 30, 25, 2, 1, 17, 43, 2, 1, 80, 12, 41, 206, 11, 70, 39, 81, 5, 289, 50, 18, 56, 24, 10, 49, 103, 146, 77, 53, 582, 31, 58, 37, 419, 140, 174, 77, 44, 100, 168, 44, 42, 99, 13, 11, 80, 60, 101, 71, 12, 24, 70, 11, 52, 671
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OFFSET
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1,2
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LINKS
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EXAMPLE
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a(5) = 4 because the sum of prime factors of 4*5*6*7*8 = 2^6 * 3 * 5 * 7 is 2*6 + 3 + 5 + 7 = 27 = 3^3 is a perfect power, and 4 is the least number that works.
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MAPLE
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spf:= proc(n) local t; add(t[1]*t[2], t=ifactors(n)[2]) end proc:
ispow:= proc(n) igcd(map(t -> t[2], ifactors(n)[2]))>1 end proc:
f:= proc(n) local S, t, i;
S:= Vector(n, spf); t:= convert(S, `+`);
for i from 1 do
if ispow(t) then return i fi;
t:= t-S[1];
S[1..n-1]:= S[2..n];
S[n]:= spf(i+n);
t:= t+S[n];
od
end proc:
f(1):= 1:
map(f, [$1..100]);
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MATHEMATICA
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sopfr[n_] := Plus @@ Times @@@ FactorInteger[n]; powQ[n_] := GCD @@ FactorInteger[n][[;; , 2]] > 1; a[n_] := Module[{k = 1}, While[! powQ[sopfr[Product[k + i, {i, 0, n - 1}]]], k++]; k]; a[1] = 1; Array[a, 100] (* Amiram Eldar, Aug 19 2022 *)
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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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