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a(n) = n + (sum of prime factors of n taken with repetition).
18

%I #26 Mar 31 2022 21:59:53

%S 1,4,6,8,10,11,14,14,15,17,22,19,26,23,23,24,34,26,38,29,31,35,46,33,

%T 35,41,36,39,58,40,62,42,47,53,47,46,74,59,55,51,82,54,86,59,56,71,94,

%U 59,63,62,71,69,106,65,71,69,79,89,118,72,122,95,76,76,83,82,134,89,95,84,142

%N a(n) = n + (sum of prime factors of n taken with repetition).

%C a(n) = n + A001414(n).

%H Reinhard Zumkeller, <a href="/A075254/b075254.txt">Table of n, a(n) for n = 1..10000</a>

%e a(6)=11 because 6=2*3, sopfr(6)=2+3=5 and 6+5=11.

%p A075254 := proc(n)

%p n+A001414(n) ;

%p end proc: # _R. J. Mathar_, Jul 27 2015

%t Table[If[n==1,1, n +Plus@@Times@@@FactorInteger@n], {n, 80}] (* _G. C. Greubel_, Jan 10 2019 *)

%o (Haskell)

%o a075254 n = n + a001414 n -- _Reinhard Zumkeller_, Feb 27 2012

%o (PARI) a(n) = my(f = factor(n)); n += sum(k=1, #f~, f[k,1]*f[k,2]); \\ _Michel Marcus_, Feb 22 2017

%o (Magma) [n eq 1 select 1 else (&+[p[1]*p[2]: p in Factorization(n)]) + n: n in [1..80]]; // _G. C. Greubel_, Jan 10 2019

%o (Sage) [n + sum(factor(n)[j][0]*factor(n)[j][1] for j in range(0, len(factor(n)))) for n in range(1, 80)] # _G. C. Greubel_, Jan 10 2019

%Y Cf. A001414, A008472, A075255.

%Y Cf. A096461 (iteration).

%K nonn

%O 1,2

%A _Zak Seidov_, Sep 10 2002