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URL: https://oeis.org/A075254

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A075254
a(n) = n + (sum of prime factors of n taken with repetition).
36
1, 4, 6, 8, 10, 11, 14, 14, 15, 17, 22, 19, 26, 23, 23, 24, 34, 26, 38, 29, 31, 35, 46, 33, 35, 41, 36, 39, 58, 40, 62, 42, 47, 53, 47, 46, 74, 59, 55, 51, 82, 54, 86, 59, 56, 71, 94, 59, 63, 62, 71, 69, 106, 65, 71, 69, 79, 89, 118, 72, 122, 95, 76, 76, 83, 82, 134, 89, 95, 84, 142
OFFSET
1,2
COMMENTS
a(n) = n + A001414(n).
Product of prime factors plus sum of prime factors of n. For minus instead of plus we have A075255, zeros A175787. - Gus Wiseman, Jan 26 2025
LINKS
FORMULA
From Gus Wiseman, Jan 26 2025: (Start)
First differences are 1 - A090340(n).
a(n) = 2*n - A075255(n).
a(n) = 2*A001414(n) + A075255(n).
(End)
EXAMPLE
a(6)=11 because 6=2*3, sopfr(6)=2+3=5 and 6+5=11.
MAPLE
A075254 := proc(n)
n+A001414(n) ;
end proc: # R. J. Mathar, Jul 27 2015
MATHEMATICA
Table[If[n==1, 1, n +Plus@@Times@@@FactorInteger@n], {n, 80}] (* G. C. Greubel, Jan 10 2019 *)
PROG
(Haskell)
a075254 n = n + a001414 n -- Reinhard Zumkeller, Feb 27 2012
(PARI) a(n) = my(f = factor(n)); n + sum(k=1, #f~, f[k, 1]*f[k, 2]); \\ Michel Marcus, Feb 22 2017
(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
(SageMath) [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
CROSSREFS
A000027 gives product of prime factors, indices A003963.
A000040 lists the primes, differences A001223.
A001414 gives sum of prime factors, indices A056239.
A027746 lists prime factors, indices A112798, count A001222.
A075255 gives product of prime factors minus sum of prime factors.
Sequence in context: A379842 A272475 A184016 * A284913 A139203 A179372
KEYWORD
nonn,look
AUTHOR
Zak Seidov, Sep 10 2002
STATUS
approved