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A365850
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Numbers k for which k^2 + (k')^2 is a square, where k' is the arithmetic derivative of k (A003415).
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0
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0, 1, 12, 15, 35, 81, 143, 323, 400, 441, 899, 1540, 1763, 2700, 3599, 4641, 5183, 5929, 9375, 10395, 10403, 11663, 13585, 18225, 19043, 21952, 22499, 30576, 32399, 35581, 36863, 39203, 48841, 51983, 57599, 72899, 79523, 97343, 121103, 148176, 166375, 175692, 176399
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OFFSET
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1,3
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COMMENTS
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If p and p + 2 are twin primes (A001359) then m = p*(p + 2) is a term. Indeed, m' = p + (p + 2) = 2*p + 2 and m^2 + (m')^2 = p^2*(p + 2)^2 + (2*p + 2)^2 = (p^2 + 2*p + 2)^2.
More generally, if p and p + 2*k, k >= 1, are prime numbers, then m = p^k*(p + 2*k)^k is a term. Indeed, m' = k*p^(k - 1)*(p + 2*k)^k+ k*p^k*(p + 2*k)^(k - 1) = k*p^(k - 1)*(p + 2*k)^(k-1)*(2*p + 2*k). Thus, m^2 + (m')^2 = p^(2*k)*(p + 2*k)^(2*k) + (k^2)*p^(2*k - 2)*(p + 2*k)^(2*k - 2)*(2*p + 2*k)^2 = p^(2*k - 2)*(p + 2*k)^(2*k - 2)*(p^2*(p + 2*k)^2 + k^2*(2*p + 2*k)^2) = p^(2*k - 2)*(p + 2*k)^(2*k - 2)*(2*k^2 + 2*k*p + p^2)^2.
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LINKS
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EXAMPLE
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For k = 12, k' = 16 and 12^2 + 16^2 = 144 + 256 = 400 = 20^2, so 12 is a term.
For k = 15, k' = 8 and 15^2 + 8^2 = 225 + 64 = 289 = 17^2, so 15 is a term.
For k = 143, k' = 24 and 143^2 + 24^2 = 144 + 256 = 21025 = 145^2, so 143 is a term.
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MAPLE
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ader:= proc(n) local f; n*add(f[2]/f[1], f=ifactors(n)[2]) end proc:
select(t -> issqr(t^2 + ader(t)^2), [$0..10^6]; # Robert Israel, Oct 17 2023
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MATHEMATICA
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d[0] = d[1] = 0; d[n_] := n*Plus @@ ((Last[#]/First[#]) & /@ FactorInteger[n]); Select[Range[0, 180000], IntegerQ[Sqrt[#^2 + d[#]^2]] &] (* Amiram Eldar, Oct 11 2023 *)
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PROG
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(Magma) f:=func<n |n le 1 select 0 else n*(&+[Factorisation(n)[i][2] / Factorisation(n)[i][1]: i in [1..#Factorisation(n)]])>; [n:n in [0..200000] |IsSquare( n^2+( Floor(f(n))^2))];
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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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