A069359 -id:A069359 - cp's OEIS frontend

A276076 Factorial base exp-function: digits in factorial base representation of n become the exponents of successive prime factors whose product a(n) is.

1, 2, 3, 6, 9, 18, 5, 10, 15, 30, 45, 90, 25, 50, 75, 150, 225, 450, 125, 250, 375, 750, 1125, 2250, 7, 14, 21, 42, 63, 126, 35, 70, 105, 210, 315, 630, 175, 350, 525, 1050, 1575, 3150, 875, 1750, 2625, 5250, 7875, 15750, 49, 98, 147, 294, 441, 882, 245, 490, 735, 1470, 2205, 4410, 1225, 2450, 3675, 7350, 11025, 22050, 6125, 12250, 18375, 36750, 55125, 110250, 343

Offset: 0

Antti Karttunen, Aug 18 2016

These are prime-factorization representations of single-variable polynomials where the coefficient of term x^(k-1) (encoded as the exponent of prime(k) in the factorization of n) is equal to the digit in one-based position k of the factorial base representation of n. See the examples.

			   n  A007623   polynomial     encoded as             a(n)
   -------------------------------------------------------
   0       0    0-polynomial   (empty product)        = 1
   1       1    1*x^0          prime(1)^1             = 2
   2      10    1*x^1          prime(2)^1             = 3
   3      11    1*x^1 + 1*x^0  prime(2) * prime(1)    = 6
   4      20    2*x^1          prime(2)^2             = 9
   5      21    2*x^1 + 1*x^0  prime(2)^2 * prime(1)  = 18
   6     100    1*x^2          prime(3)^1             = 5
   7     101    1*x^2 + 1*x^0  prime(3) * prime(1)    = 10
and:
  23     321  3*x^2 + 2*x + 1  prime(3)^3 * prime(2)^2 * prime(1)
                                      = 5^3 * 3^2 * 2 = 2250.

Antti Karttunen, Table of n, a(n) for n = 0..5040
Indranil Ghosh, Python program for computing this sequence.
Index entries for sequences related to factorial base representation.

Cf. A000040, A007623, A084558, A099563, A257687, A276009.
Cf. A276075 (a left inverse).
Cf. A276078 (same terms in ascending order).
Cf. also A000142, A001221, A001222, A002110, A007489, A008836, A019565, A033312, A034968, A048675, A051903, A059590, A060130, A076954, A246359, A248663, A262725, A276073, A276074, A351576, A351577, A351950, A351951, A351952, A351954.
Cf. also A275733, A275734, A275735, A275725 for other such encodings of factorial base related polynomials, and A276086 for a primorial base analog.

a[n_] := Module[{k = n, m = 2, r, p = 2, q = 1}, While[{k, r} = QuotientRemainder[k, m]; k != 0|| r != 0, q *= p^r; p = NextPrime[p]; m++]; q]; Array[a, 100, 0] (* Amiram Eldar, Feb 07 2024 *)

a(0) = 1, for n >= 1, a(n) = A275733(n) * a(A276009(n)).
Or: for n >= 1, a(n) = a(A257687(n)) * A000040(A084558(n))^A099563(n).
Other identities.
For all n >= 0:
A276075(a(n)) = n.
A001221(a(n)) = A060130(n).
A001222(a(n)) = A034968(n).
A051903(a(n)) = A246359(n).
A048675(a(n)) = A276073(n).
A248663(a(n)) = A276074(n).
a(A007489(n)) = A002110(n).
a(A059590(n)) = A019565(n).
For all n >= 1:
a(A000142(n)) = A000040(n).
a(A033312(n)) = A076954(n-1).
From Antti Karttunen, Apr 18 2022: (Start)
a(n) = A276086(A351576(n)).
A276085(a(n)) = A351576(n)
A003557(a(n)) = A351577(n).
A003415(a(n)) = A351950(n).
A069359(a(n)) = A351951(n).
A083345(a(n)) = A342001(a(n)) = A351952(n).
A351945(a(n)) = A351954(n).
A181819(a(n)) = A275735(n).
(End)
lambda(a(n)) = A262725(n+1), where lambda is Liouville's function, A008836. - Antti Karttunen and Peter Munn, Aug 09 2024

Name changed by Antti Karttunen, Apr 18 2022

A054377 Primary pseudoperfect numbers: numbers n > 1 such that 1/n + sum 1/p = 1, where the sum is over the primes p | n.

Original entry on oeis.org

2, 6, 42, 1806, 47058, 2214502422, 52495396602, 8490421583559688410706771261086

Offset: 1

Eric W. Weisstein

Primary pseudoperfect numbers are the solutions of the "differential equation" n' = n-1, where n' is the arithmetic derivative of n. - Paolo P. Lava, Nov 16 2009
Same as n > 1 such that 1 + sum n/p = n (and the only known numbers n > 1 satisfying the weaker condition that 1 + sum n/p is divisible by n). Hence a(n) is squarefree, and is pseudoperfect if n > 1. Remarkably, a(n) has exactly n (distinct) prime factors for n < 9. - Jonathan Sondow, Apr 21 2013
From the Wikipedia article: it is unknown whether there are infinitely many primary pseudoperfect numbers, or whether there are any odd primary pseudoperfect numbers. - Daniel Forgues, May 27 2013
Since the arithmetic derivative of a prime p is p' = 1, 2 is obviously the only prime in the sequence. - Daniel Forgues, May 29 2013
Just as 1 is not a prime number, 1 is also not a primary pseudoperfect number, according to the original definition by Butske, Jaje, and Mayernik, as well as Wikipedia and MathWorld. - Jonathan Sondow, Dec 01 2013
Is it always true that if a primary pseudoperfect number N > 2 is adjacent to a prime N-1 or N+1, then in fact N lies between twin primes N-1, N+1? See A235139. - Jonathan Sondow, Jan 05 2014
Also, integers n > 1 such that A069359(n) = n - 1. - Jonathan Sondow, Apr 16 2014

			From _Daniel Forgues_, May 24 2013: (Start)
With a(1) = 2, we have 1/2 + 1/2 = (1 + 1)/2 = 1;
with a(2) = 6 = 2 * 3, we have
  1/2 + 1/3 + 1/6 = (3 + 2 + 1)/6 = (1*3 + 3)/(2*3) = (1 + 1)/2 = 1;
with a(3) = 42 = 6 * 7, we have
  1/2 + 1/3 + 1/7 + 1/42 = (21 + 14 + 6 + 1)/42 =
  (3*7 + 2*7 + 7)/(6*7) = (3 + 2 + 1)/6 = 1;
with a(4) = 1806 = 42 * 43, we have
  1/2 + 1/3 + 1/7 + 1/43 + 1/1806 = (903 + 602 + 258 + 42 + 1)/1806 =
  (21*43 + 14*43 + 6*43 + 43)/(42*43) = (21 + 14 + 6 + 1)/42 = 1;
with a(5) = 47058 (not oblong number), we have
  1/2 + 1/3 + 1/11 + 1/23 + 1/31 + 1/47058 =
  (23529 + 15686 + 4278 + 2046 + 1518 + 1)/47058 = 1.
For n = 1 to 8, a(n) has n prime factors:
  a(1) = 2
  a(2) = 2 * 3
  a(3) = 2 * 3 *  7
  a(4) = 2 * 3 *  7 * 43
  a(5) = 2 * 3 * 11 * 23 *  31
  a(6) = 2 * 3 * 11 * 23 *  31 * 47059
  a(7) = 2 * 3 * 11 * 17 * 101 *   149 *       3109
  a(8) = 2 * 3 * 11 * 23 *  31 * 47059 * 2217342227 * 1729101023519
If a(n)+1 is prime, then a(n)*[a(n)+1] is also primary pseudoperfect. We have the chains: a(1) -> a(2) -> a(3) -> a(4); a(5) -> a(6). (End)
A primary pseudoperfect number (greater than 2) is oblong if and only if it is not the initial member of a chain. - _Daniel Forgues_, May 29 2013
If a(n)-1 is prime, then a(n)*(a(n)-1) is a Giuga number (A007850). This occurs for a(2), a(3), and a(5). See A235139 and the link "The p-adic order . . .", Theorem 8 and Example 1. - _Jonathan Sondow_, Jan 06 2014

M. A. Alekseyev, J. M. Grau, A. M. Oller-Marcen. Computing solutions to the congruence 1^n + 2^n + ... + n^n == p (mod n). Discrete Applied Mathematics, 2018. doi:10.1016/j.dam.2018.05.022 arXiv:1602.02407 [math.NT]
W. Butske, L. M. Jaje, and D. R. Mayernik, On the Equation Sum_{p|N} 1/p + 1/N = 1, Pseudoperfect numbers and partially weighted graphs, Math. Comput., 69 (1999), 407-420. [Title corrected by _Jonathan Sondow_, Apr 11 2012]
J. M. Grau, A. M. Oller-Marcen, and J. Sondow, On the congruence 1^m + 2^m + ... + m^m == n (mod m) with n|m, arXiv:1309.7941 [math.NT], 2013.
J. M. Grau, A. M. Oller-Marcén, and D. Sadornil, On µ-Sondow Numbers, arXiv:2111.14211 [math.NT], 2021.
John Machacek, Egyptian Fractions and Prime Power Divisors, arXiv:1706.01008 [math.NT], 2017.
J. Sondow and K. MacMillan, Reducing the Erdos-Moser equation 1^n + 2^n + . . . + k^n = (k+1)^n modulo k and k^2, Integers 11 (2011), #A34.
J. Sondow and K. MacMillan, Primary pseudoperfect numbers, arithmetic progressions, and the Erdos-Moser equation, Amer. Math. Monthly, 124 (2017) 232-240; arXiv:math/1812.06566 [math.NT], 2018.
J. Sondow and E. Tsukerman, The p-adic order of power sums, the Erdos-Moser equation, and Bernoulli numbers, arXiv:1401.0322 [math.NT], 2014; see section 4.
Eric Weisstein's World of Mathematics, Primary pseudoperfect number.
Wikipedia, Primary pseudoperfect number.
OEIS Wiki, Primary pseudoperfect numbers.

Cf. A005835, A007850, A069359, A168036, A190272, A191975, A203618, A216825, A216826, A230311, A235137, A235138, A235139, A236433.

pQ[n_] := (f = FactorInteger[n]; 1/n + Sum[1/f[[i]][[1]], {i, Length[f]}] == 1)
Select[Range[2, 10^6], pQ[#] &] (* Robert Price, Mar 14 2020 *)

isok(n) = if (n > 1, my(f=factor(n)[,1]); 1/n + sum(k=1, #f, 1/f[k]) == 1); \\ Michel Marcus, Oct 05 2017

from sympy import primefactors
A054377 = [n for n in range(2,10**5) if sum([n/p for p in primefactors(n)]) +1 == n] # Chai Wah Wu, Aug 20 2014

A031971(a(n)) (mod a(n)) = A233045(n). - Jonathan Sondow, Dec 11 2013
A069359(a(n)) = a(n) - 1. - Jonathan Sondow, Apr 16 2014
a(n) == 36*(n-2) + 6 (mod 288) for n = 2,3,..,8. - Kieren MacMillan and Jonathan Sondow, Sep 20 2017

A349394 a(p^e) = p^(e-1) for prime powers, a(n) = 0 for all other n; Dirichlet convolution of A003415 (arithmetic derivative of n) with A055615 (Dirichlet inverse of n).

Original entry on oeis.org

0, 1, 1, 2, 1, 0, 1, 4, 3, 0, 1, 0, 1, 0, 0, 8, 1, 0, 1, 0, 0, 0, 1, 0, 5, 0, 9, 0, 1, 0, 1, 16, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 7, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 32, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 27, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0

Offset: 1

Antti Karttunen, Nov 18 2021

nonn

Dirichlet convolution of this sequence with Euler phi (A000010) is A300251.
Convolving this sequence with sigma (A000203) produces A319684.
With a(1) = 1 instead of 0, this would be the Dirichlet convolution of A129283 (A003415(n)+n) with A055615. Thus when we subtract A063524 from that convolution, we get this sequence. (See also A349434). Compare also to the convolution of A069359 (sequence agreeing with A003415 on squarefree numbers) with A055615, which is the characteristic function of primes, A010051. - Antti Karttunen, Nov 20 2021

Antti Karttunen, Table of n, a(n) for n = 1..20000
P. Haukkanen, J. K. Merikoski and T. Tossavainen, Asymptotics of partial sums of the Dirichlet series of the arithmetic derivative, Mathematical Communications 25 (2020), 107-115.

Cf. A003415, A003557, A055615, A063524, A069513, A100995, A120007, A129283, A246655.

Cf. also A000010, A000203, A069359, A300251, A319684, A327564, A349340, A349396, A349434, A349618, A349619, A349620, A349621.

import Math.NumberTheory.Primes
a n = case factorise n of
    [(p,e)] -> unPrime p^(e-1) :: Int
     -> 0 -- _Sebastian Karlsson, Nov 19 2021

f[p_, e_] := e/p; d[1] = 0; d[n_] := n * Plus @@ f @@@ FactorInteger[n]; a[n_] := DivisorSum[n, # * MoebiusMu[#] * d[n/#] &]; Array[a, 100] (* Amiram Eldar, Nov 19 2021 *)

A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
A055615(n) = (n*moebius(n));
A349394(n) = sumdiv(n,d,A003415(n/d)*A055615(d));

A349394(n) = { my(p=0,e); if((e=isprimepower(n,&p)),p^(e-1),0); }; \\ (After Sebastian Karlsson's new formula) - Antti Karttunen, Nov 20 2021

a(n) = Sum_{d|n} A003415(n/d) * A055615(d).
a(n) = 0 unless n is a prime power (A246655), in which case a(p^e) = p^(e-1). - Sebastian Karlsson, Nov 19 2021
a(n) = A003557(n) * A069513(n). [From above] - Antti Karttunen, Nov 20 2021
Dirichlet g.f.: Sum_{p prime} 1/(p^s-p) [Follows from the D.g.f. of A003415 proved by Haukkanen et al.]. - Sebastian Karlsson, Nov 25 2021
Sum_{k=1..n} a(k) has an average value c*n, where c = A137245 = Sum_{primes p} 1/(p*log(p)) = 1.63661632335... - Vaclav Kotesovec, Mar 03 2023

Added Sebastian Karlsson's formula as the new primary definition - Antti Karttunen, Nov 20 2021

A323599 Dirichlet convolution of the identity function with omega.

Original entry on oeis.org

0, 1, 1, 3, 1, 7, 1, 7, 4, 9, 1, 19, 1, 11, 10, 15, 1, 25, 1, 25, 12, 15, 1, 43, 6, 17, 13, 31, 1, 54, 1, 31, 16, 21, 14, 67, 1, 23, 18, 57, 1, 68, 1, 43, 37, 27, 1, 91, 8, 49, 22, 49, 1, 79, 18, 71, 24, 33, 1, 142, 1, 35, 45, 63, 20, 96, 1, 61, 28, 90, 1, 151, 1, 41, 55

Offset: 1

Torlach Rush, Jan 18 2019

nonn

a(n) = omega(n) = 1 iff n is prime.
a(n) = A323600(n) = 1 iff n is prime.
a(n) = A323600(n) - 1 = 1 iff n is the square of a prime.
a(n) = A323600(n) - 2 = 2 iff n is a squarefree semiprime.
a(n) = A323600(n) - (p + 2) if n is the cube of a prime p.

Antti Karttunen, Table of n, a(n) for n = 1..16384
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537

Cf. A001221, A180253, A276085, A319684, A323600, A329347, A329380.

Inverse Möbius transform of A069359.

with(numtheory):
a:= n-> add(d*nops(factorset(n/d)), d=divisors(n)):
seq(a(n), n=1..100);  # Alois P. Heinz, Jan 28 2019

Table[DivisorSum[n, # PrimeNu[n/#] &], {n, 75}] (* Michael De Vlieger, Jan 27 2019 *)

a(n) = sumdiv(n, d, d*omega(n/d)); \\ Michel Marcus, Jan 22 2019

a(n) = Sum_{d|n} d * A001221(n/d).
a(n) = Sum_{p|n} sigma(n/p) where p is prime and sigma(n) = A000203(n). - Ridouane Oudra, Apr 28 2019
a(n) = Sum_{d|n} A069359(d), a(n) = A276085(A329380(n)). - Antti Karttunen, Nov 12 2019
From Torlach Rush, Mar 23 2024: (Start)
For p in primes: (Start)
a(p^(k+1)) = a(p^k) + p^k, k >= 0.
a(p^2) = p + 1.
(End)
a(2^k) = 2^k - 1, k >= 0.
(End)

A326690 Denominator of the fraction (Sum_{prime p | n} 1/p - 1/n).

Original entry on oeis.org

1, 1, 1, 4, 1, 3, 1, 8, 9, 5, 1, 4, 1, 7, 15, 16, 1, 9, 1, 20, 7, 11, 1, 24, 25, 13, 27, 28, 1, 1, 1, 32, 33, 17, 35, 36, 1, 19, 13, 40, 1, 21, 1, 44, 45, 23, 1, 16, 49, 25, 51, 52, 1, 27, 11, 8, 19, 29, 1, 60, 1, 31, 63, 64, 65, 11, 1, 68, 69, 35, 1, 72

Offset: 1

Jonathan Sondow, Jul 18 2019

Theorem. If n is a prime or a Carmichael number, then a(n) = A309132(n) = denominator of (N(n-1)/n + D(n-1)/n^2), where B(k) = N(k)/D(k) is the k-th Bernoulli number. This is a generalization of Theorem 1 in A309132 that A309132(p) = 1 if p is a prime. The proof generalizes that in A309132. As an application of Theorem, for n a prime or a Carmichael number one can compute A309132(n) without calculating Bernoulli numbers; see A309268.
A composite number n is a Giuga number A007850 if and only if a(n) = 1. (In fact, Sum_{prime p | n} 1/p - 1/n = 1 for all known Giuga numbers n.)
Semiprimes m = pq such that 1/p + 1/q - 1/m = p/q are exactly A190275. - Amiram Eldar and Thomas Ordowski, Jul 22 2019
The preceding comment may be rephrased as "Semiprimes m = pq such that A326689(m) = p and a(m) = q are exactly A190275." - Jonathan Sondow, Jul 22 2019
More generally, semiprimes m = pq such that 1/p + 1/q - 1/m = P/Q are exactly A190273, where P <> Q are primes. In other words, semiprimes m such that A326689(m) is prime and a(m) is prime are exactly A190273. - Amiram Eldar and Thomas Ordowski, Jul 25 2019

			-1/1, 0/1, 0/1, 1/4, 0/1, 2/3, 0/1, 3/8, 2/9, 3/5, 0/1, 3/4, 0/1, 4/7, 7/15, 7/16, 0/1, 7/9, 0/1, 13/20, 3/7, 6/11, 0/1, 19/24, 4/25, 7/13, 8/27, 17/28, 0/1, 1/1
a(12) = denominator of (Sum_{prime p | 12} 1/p - 1/12) = denominator of (1/2 + 1/3 - 1/12) = denominator of 3/4 = 4.
Computing A309132(561) involves numerator(B(560)) which has 865 digits. But 561 is a Carmichael number, so Theorem implies A309132(561) = a(561) = denominator(1/3 + 1/11 + 1/17 - 1/561) = denominator(90/187) = 187.

Antti Karttunen, Table of n, a(n) for n = 1..16384
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537
Wikipedia, Bernoulli number
Wikipedia, Carmichael number
Wikipedia, Giuga number

Numerators are A326689. Quotients n/a(n) are A326691.
Cf. A069359, A007947 (denominator of Sum_{prime p | n} 1/p).
Cf. A002997, A007850, A190275, A309132, A309235, A309268, A309378, A326692.

[1] cat [Denominator(&+[1/p:p in PrimeDivisors(k)]-1/k):k in [2..72]]; // Marius A. Burtea, Jul 27 2019

A326690 := n -> denom((A069359(n)-1)/n):
seq(A326690(n), n=1..72); # Peter Luschny, Jul 22 2019

PrimeFactors[n_] := Select[Divisors[n], PrimeQ];
f[n_] := Denominator[Sum[1/p, {p, PrimeFactors[n]}] - 1/n];
Table[ f[n], {n, 100}]

a(n) = denominator(sumdiv(n, d, isprime(d)/d) - 1/n); \\ Michel Marcus, Jul 19 2019

p = lambda n: [n//f[0] for f in factor(n)]
A326690 = lambda n: ((sum(p(n)) - 1)/n).denominator()
[A326690(n) for n in (1..72)] # Peter Luschny, Jul 22 2019

a(n) = 1 if n is a prime or a Giuga number A007850.
a(n) = denominator of (N(n-1)/n + D(n-1)/n^2) if n is a Carmichael number A002997.
a(n) = denominator((A069359(n) - 1)/n). - Peter Luschny, Jul 22 2019

A117494 a(n) is the number of m's, 1 <= m <= n, where gcd(m,n) is prime.

Original entry on oeis.org

0, 1, 1, 1, 1, 3, 1, 2, 2, 5, 1, 4, 1, 7, 6, 4, 1, 8, 1, 6, 8, 11, 1, 8, 4, 13, 6, 8, 1, 14, 1, 8, 12, 17, 10, 10, 1, 19, 14, 12, 1, 20, 1, 12, 14, 23, 1, 16, 6, 24, 18, 14, 1, 24, 14, 16, 20, 29, 1, 20, 1, 31, 18, 16, 16, 32, 1, 18, 24, 34, 1, 20, 1, 37, 28, 20, 16, 38, 1, 24, 18, 41, 1, 28

Offset: 1

Leroy Quet, Mar 22 2006

nonn

Dirichlet convolution of A000010 (Euler phi) and A010051 (characteristic function of primes), therefore also Möbius transform of A069359. - Antti Karttunen, Nov 17 2021

			Of the positive integers <= 12, exactly four (2, 3, 9 and 10) have a GCD with 12 that is prime. (gcd(2,12) = 2, gcd(3,12) = 3, gcd(9,12) = 3, gcd(10,12) = 2.)
So a(12) = 4.

Antti Karttunen, Table of n, a(n) for n = 1..65537

Cf. A000010, A008683, A010051, A069359, A085548, A122410, A122411, A175787, A349338.

Coincides with A300251 on squarefree numbers, A005117.

a:=proc(n) local c,m: c:=0: for m from 1 to n do if isprime(gcd(m,n))=true then c:=c+1 else c:=c fi od: end: seq(a(n),n=1..100); # Emeric Deutsch, Apr 01 2006

f[n_] := Length@ Select[GCD[n, Range@n], PrimeQ@ # &]; Array[f, 84] (* Robert G. Wilson v, Apr 06 2006 *)
Table[Count[Range@ n, ?(PrimeQ@ GCD[#, n] &)], {n, 84}] (* _Michael De Vlieger, Feb 25 2018 *)
a[n_] := Module[{f = FactorInteger[n], p, e}, p = f[[;; , 1]]; e = f[[;; , 2]]; n * Times @@ (1-1/p) * Total[1/(p - (Boole[# == 1] & /@ e))]]; a[1] = 0; Array[a, 100] (* Amiram Eldar, Jun 21 2025 *)

A117494(n) = sum(k=1,n,isprime(gcd(n,k))); \\ Antti Karttunen, Feb 25 2018

a(n) = my(f=factor(n)[, 1]); sum(k=1, #f, eulerphi(n/f[k])); \\ Daniel Suteu, Jun 23 2018

A117494(n) = sumdiv(n,d,eulerphi(n/d)*isprime(d)); \\ Antti Karttunen, Nov 17 2021

Dirichlet g.f: P(s)*Z(s-1)/Z(s) with P(s) the prime zeta function and Z(s) the Riemann zeta function. - Pierre-Louis Giscard, Jul 16 2014
a(n) = Sum_{distinct primes p dividing n} phi(n/p), where phi(k) is the Euler totient function. - Daniel Suteu, Jun 23 2018
From Antti Karttunen, Nov 17 2021: (Start)
a(n) = Sum_{d|n} A008683(n/d) * A069359(d).
a(n) = Sum_{d|n} A000010(n/d) * A010051(d).
a(n) = A349338(n) - A000010(n).
a(A005117(n)) = A300251(A005117(n)) for all n >= 1. (End)
a(n) = 1 iff n = 4 or n is prime (A175787). - Bernard Schott, Nov 18 2021
Sum_{k=1..n} a(k) ~ 3 * A085548 * n^2 / Pi^2. - Vaclav Kotesovec, Nov 20 2021

More terms from Emeric Deutsch, Apr 01 2006

A322078 a(n) = n^2 * Sum_{p|n, p prime} 1/p^2.

Original entry on oeis.org

0, 1, 1, 4, 1, 13, 1, 16, 9, 29, 1, 52, 1, 53, 34, 64, 1, 117, 1, 116, 58, 125, 1, 208, 25, 173, 81, 212, 1, 361, 1, 256, 130, 293, 74, 468, 1, 365, 178, 464, 1, 673, 1, 500, 306, 533, 1, 832, 49, 725, 298, 692, 1, 1053, 146, 848, 370, 845, 1, 1444, 1, 965, 522

Offset: 1

Daniel Suteu, Nov 25 2018

nonn

Generalized formula is f(n,m) = n^m * Sum_{p|n} 1/p^m, where f(n,0) = A001221(n) and f(n,1) = A069359(n).

Dirichlet convolution of A010051(n) and n^2. - Wesley Ivan Hurt, Jul 15 2025

			a(40) = 464 because the prime factors of 40 are 2 and 5, so we have 40^2 * (1/2^2 + 1/5^2) = 464.

Daniel Suteu, Table of n, a(n) for n = 1..10000

Cf. A000040, A000330, A001221, A007434, A010051, A069359, A085541.

Sequences of the form n^k * Sum_{p|n, p prime} 1/p^k for k = 0..10: A001221 (k=0), A069359 (k=1), this sequence (k=2), A351242 (k=3), A351244 (k=4), A351245 (k=5), A351246 (k=6), A351247 (k=7), A351248 (k=8), A351249 (k=9), A351262 (k=10).

[0] cat [n^2*&+[1/p^2:p in PrimeDivisors(n)]:n in [2..70]]; // Marius A. Burtea, Oct 10 2019

a:= n-> n^2*add(1/i[1]^2, i=ifactors(n)[2]):
seq(a(n), n=1..70);  # Alois P. Heinz, Oct 11 2019

f[p_, e_] := 1/p^2; a[n_] := If[n==1, 0, n^2*Plus@@f@@@FactorInteger[n]]; Array[a, 60] (* Amiram Eldar, Nov 26 2018 *)

a(n) = my(f=factor(n)[,1]~); sum(k=1, #f, n^2\f[k]^2);

Sum_{k=1..n} a(k) ~ A085541 * A000330(n).
G.f.: Sum_{k>=1} x^prime(k) * (1 + x^prime(k)) / (1 - x^prime(k))^3. - Ilya Gutkovskiy, Oct 10 2019
a(n) = 1 <=> n is prime.  _Alois P. Heinz, Oct 11 2019
Dirichlet g.f.: zeta(s-2)*primezeta(s).  This follows because Sum_{n>=1} a(n)/n^s = Sum_{n>=1} (n^2/n^s) Sum_{p|n} 1/p^2. Since n = p*j, rewrite the sum as Sum_{p} Sum_{j>=1} 1/(p^2*(p*j)^(s-2)) = Sum_{p} 1/p^s Sum_{j>=1} 1/j^(s-2) = zeta(s-2)*primezeta(s). The result generalizes to higher powers of p. - Michael Shamos, Mar 02 2023
a(n) = Sum_{d|n} A007434(d)*A001221(n/d). - Ridouane Oudra, Jul 13 2025
From Wesley Ivan Hurt, Jul 15 2025: (Start)
a(n) = Sum_{d|n} c(d) * (n/d)^2, where c = A010051.
a(p^k) = p^(2*k-2) for p prime and k>=1. (End)

A028235 If n = Product (p_j^k_j), a(n) = numerator of Sum 1/p_j (the denominator of this sum is A007947).

Original entry on oeis.org

0, 1, 1, 1, 1, 5, 1, 1, 1, 7, 1, 5, 1, 9, 8, 1, 1, 5, 1, 7, 10, 13, 1, 5, 1, 15, 1, 9, 1, 31, 1, 1, 14, 19, 12, 5, 1, 21, 16, 7, 1, 41, 1, 13, 8, 25, 1, 5, 1, 7, 20, 15, 1, 5, 16, 9, 22, 31, 1, 31, 1, 33, 10, 1, 18, 61, 1, 19, 26, 59, 1, 5, 1, 39, 8, 21, 18, 71, 1, 7, 1, 43, 1, 41, 22, 45, 32

Offset: 1

N. J. A. Sloane

For n=1, the empty sum = 0 = 0/1 = a(1)/A007947(1), thus a(1) should be 0. - Antti Karttunen, Mar 04 2018

			Fractions begin with 0, 1/2, 1/3, 1/2, 1/5, 5/6, 1/7, 1/2, 1/3, 7/10, 1/11, 5/6, ...

Antti Karttunen, Table of n, a(n) for n = 1..65537

Cf. A007947 (denominators), A003415, A069359, A085548, A379967.

a[1] = 0; a[n_] := 1/FactorInteger[n][[All, 1]] // Total // Numerator;
Array[a, 100] (* Jean-François Alcover, May 08 2019 *)

A028235(n) = numerator(vecsum(apply(p->(1/p), factor(n)[, 1]))); \\ Antti Karttunen, Mar 04 2018

Fraction is additive with a(p^e) = 1/p.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k)/A007947(k) = Sum_{p prime} 1/p^2 = 0.452247... (A085548). - Amiram Eldar, Sep 29 2023
a(n) = A003415(A007947(n)) = A069359(A007947(n)). - Antti Karttunen, Jan 22 2025

More terms from Erich Friedman.

Term a(1) changed to 0 by Antti Karttunen, Mar 04 2018

A351242 a(n) = n^3 * Sum_{p|n, p prime} 1/p^3.

Original entry on oeis.org

0, 1, 1, 8, 1, 35, 1, 64, 27, 133, 1, 280, 1, 351, 152, 512, 1, 945, 1, 1064, 370, 1339, 1, 2240, 125, 2205, 729, 2808, 1, 4591, 1, 4096, 1358, 4921, 468, 7560, 1, 6867, 2224, 8512, 1, 12221, 1, 10712, 4104, 12175, 1, 17920, 343, 16625, 4940, 17640, 1, 25515, 1456, 22464

Offset: 1

Wesley Ivan Hurt, Feb 05 2022

Dirichlet convolution of A010051(n) and n^3. - Wesley Ivan Hurt, Jul 15 2025

			a(6) = 35; a(6) = 6^3 * Sum_{p|6, p prime} 1/p^3 = 216 * (1/2^3 + 1/3^3) = 35.

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Sequences of the form n^k * Sum_{p|n, p prime} 1/p^k for k = 0..10: A001221 (k=0), A069359 (k=1), A322078 (k=2), this sequence (k=3), A351244 (k=4), A351245 (k=5), A351246 (k=6), A351247 (k=7), A351248 (k=8), A351249 (k=9), A351262 (k=10).

Cf. A000040, A001221, A010051, A059376, A085964.

a[n_]:= n^3 * Sum[1/p^3,{p,Select[Divisors[n],PrimeQ]}]; Array[a,56] (* Stefano Spezia, Jul 14 2025 *)

a(n) = my(f = factor(n)); sum(i = 1, #f~, (n/f[i,1])^3) \\ David A. Corneth, Jul 14 2025

a(A000040(n)) = 1.
G.f.: Sum_{k>=1} x^prime(k) * (1 + 4*x^prime(k) + x^(2*prime(k))) / (1 - x^prime(k))^4. - Ilya Gutkovskiy, Feb 05 2022
Dirichlet g.f. = zeta(s-3)*primezeta(s).  This follows because Sum_{n>=1} a(n)/n^s = Sum_{n>=1} (n^3/n^s) Sum_{p|n} 1/p^3. Since n = p*j, rewrite the sum as Sum_{p} Sum_{j>=1} 1/(p^3*(p*j)^(s-3)) = Sum_{p} 1/p^s Sum_{j>=1} 1/j^(s-3) = zeta(s-3)*primezeta(s).  The result generalizes to higher powers of p in a(n). - Michael Shamos, Mar 01 2023
Sum_{k=1..n} a(k) ~ A085964 * n^4/4. - Vaclav Kotesovec, Mar 03 2023
a(n) = Sum_{d|n} A059376(d)*A001221(n/d). - Ridouane Oudra, Jul 14 2025
From Wesley Ivan Hurt, Jul 15 2025: (Start)
a(n) = Sum_{d|n} c(d) * (n/d)^3, where c = A010051.
a(p^k) = p^(3*k-3) for p prime and k>=1. (End)

A351244 a(n) = n^4 * Sum_{p|n, p prime} 1/p^4.

Original entry on oeis.org

0, 1, 1, 16, 1, 97, 1, 256, 81, 641, 1, 1552, 1, 2417, 706, 4096, 1, 7857, 1, 10256, 2482, 14657, 1, 24832, 625, 28577, 6561, 38672, 1, 61921, 1, 65536, 14722, 83537, 3026, 125712, 1, 130337, 28642, 164096, 1, 234193, 1, 234512, 57186, 279857, 1, 397312, 2401, 400625, 83602

Offset: 1

Wesley Ivan Hurt, Feb 05 2022

nonn

Dirichlet convolution of A010051(n) and n^4. - Wesley Ivan Hurt, Jul 15 2025

			a(6) = 97; a(6) = 6^4 * Sum_{p|6, p prime} 1/p^4 = 1296 * (1/2^4 + 1/3^4) = 97.

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Sequences of the form n^k * Sum_{p|n, p prime} 1/p^k for k = 0..10: A001221 (k=0), A069359 (k=1), A322078 (k=2), A351242 (k=3), this sequence (k=4), A351245 (k=5), A351246 (k=6), A351247 (k=7), A351248 (k=8), A351249 (k=9), A351262 (k=10).

Cf. A000040, A001221, A010051, A059377, A085965.

a[n_]:= n^4 * Sum[1/p^4, {p, Select[Divisors[n], PrimeQ]}]; Array[a, 51] (* Stefano Spezia, Jul 14 2025 *)

a(A000040(n)) = 1.
G.f.: Sum_{k>=1} x^prime(k) * (1 + 11*x^prime(k) + 11*x^(2*prime(k)) + x^(3*prime(k))) / (1 - x^prime(k))^5. - Ilya Gutkovskiy, Feb 05 2022
Dirichlet g.f.: zeta(s-4)*primezeta(s). This follows because Sum_{n>=1} a(n)/n^s = Sum_{n>=1} (n^4/n^s) Sum_{p|n} 1/p^4. Since n = p*j, rewrite the sum as Sum_{p} Sum_{j>=1} 1/(p^4*(p*j)^(s-4)) = Sum_{p} 1/p^s Sum_{j>=1} 1/j^(s-4) = zeta(s-4)*primezeta(s). The result generalizes to higher powers of p. - Michael Shamos, Mar 02 2023
Sum_{k=1..n} a(k) ~ A085965 * n^5/5. - Vaclav Kotesovec, Mar 03 2023
a(n) = Sum_{d|n} A059377(d)*A001221(n/d). - Ridouane Oudra, Jul 14 2025
From Wesley Ivan Hurt, Jul 15 2025: (Start)
a(n) = Sum_{d|n} c(d) * (n/d)^4, where c = A010051.
a(p^k) = p^(4*k-4) for p prime and k>=1. (End)

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