cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 25 results. Next

A068993 Numbers k such that A062799(k) = 4.

Original entry on oeis.org

6, 10, 14, 15, 16, 21, 22, 26, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58, 62, 65, 69, 74, 77, 81, 82, 85, 86, 87, 91, 93, 94, 95, 106, 111, 115, 118, 119, 122, 123, 129, 133, 134, 141, 142, 143, 145, 146, 155, 158, 159, 161, 166, 177, 178, 183, 185, 187, 194, 201
Offset: 1

Views

Author

Benoit Cloitre, Apr 06 2002

Keywords

Comments

4*a(n)^2 are the solutions to A048272(x) = -Sum_{d|x} (-1)^d = -9. - Benoit Cloitre, Apr 14 2002

Crossrefs

Union of A006881 and A030514.

Programs

  • Mathematica
    f[n_] := DivisorSum[n, PrimeNu[#] &]; Select[Range[201], f[#] == 4 &] (* Amiram Eldar, Jul 25 2020 *)
  • PARI
    for(n=1,100,if(sumdiv(n,d,omega(d))==4,print1(n,",")))
    
  • PARI
    is(n)=my(f=factor(n)[,2]~); f==[1,1] || f==[4] \\ Charles R Greathouse IV, Oct 15 2015

Formula

A113901(a(n)) = 4. - Reinhard Zumkeller, Mar 13 2011

A069551 Numbers k such that the squarefree part of k equals A062799(k).

Original entry on oeis.org

27, 28, 40, 63, 175, 208, 250, 847, 1053, 1183, 1216, 1408, 1656, 2023, 2527, 3125, 3703, 4600, 5887, 6727, 8125, 8496, 9016, 9583, 11767, 12943, 13851, 15463, 19116, 19652, 19663, 22264, 23600, 24367, 26047, 26568, 31096, 31213, 31423, 31744
Offset: 1

Views

Author

Benoit Cloitre, Apr 17 2002

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[OddQ[e], p, 1]; sqf[n_] := Times @@ (f @@@ FactorInteger[n]); imomega[n_] := DivisorSum[n, PrimeNu[#] &]; Select[Range[32000], sqf[#] == imomega[#] &] (* Amiram Eldar, Apr 06 2020 *)
  • PARI
    for(n=1,50000,if(core(n)==sumdiv(n,d,omega(d)),print1(n,",")))

Formula

k such that A007913(k) = A062799(k).

A001248 Squares of primes.

Original entry on oeis.org

4, 9, 25, 49, 121, 169, 289, 361, 529, 841, 961, 1369, 1681, 1849, 2209, 2809, 3481, 3721, 4489, 5041, 5329, 6241, 6889, 7921, 9409, 10201, 10609, 11449, 11881, 12769, 16129, 17161, 18769, 19321, 22201, 22801, 24649, 26569, 27889, 29929, 32041, 32761, 36481
Offset: 1

Views

Author

Keywords

Comments

Also 4, together with numbers n such that Sum_{d|n}(-1)^d = -A048272(n) = -3. - Benoit Cloitre, Apr 14 2002
Also, all solutions to the equation sigma(x) + phi(x) = 2x + 1. - Farideh Firoozbakht, Feb 02 2005
Unique numbers having 3 divisors (1, their square root, themselves). - Alexandre Wajnberg, Jan 15 2006
Smallest (or first) new number deleted at the n-th step in an Eratosthenes sieve. - Lekraj Beedassy, Aug 17 2006
Subsequence of semiprimes A001358. - Lekraj Beedassy, Sep 06 2006
Integers having only 1 factor other than 1 and the number itself. Every number in the sequence is a multiple of 1 factor other than 1 and the number itself. 4 : 2 is the only factor other than 1 and 4; 9 : 3 is the only factor other than 1 and 9; and so on. - Rachit Agrawal (rachit_agrawal(AT)daiict.ac.in), Oct 23 2007
The n-th number with p divisors is equal to the n-th prime raised to power p-1, where p is prime. - Omar E. Pol, May 06 2008
There are 2 Abelian groups of order p^2 (C_p^2 and C_p x C_p) and no non-Abelian group. - Franz Vrabec, Sep 11 2008
Also numbers n such that phi(n) = n - sqrt(n). - Michel Lagneau, May 25 2012
For n > 1, n is the sum of numbers from A006254(n-1) to A168565(n-1). - Vicente Izquierdo Gomez, Dec 01 2012
A078898(a(n)) = 2. - Reinhard Zumkeller, Apr 06 2015
Let r(n) = (a(n) - 1)/(a(n) + 1); then Product_{n>=1} r(n) = (3/5) * (4/5) * (12/13) * (24/25) * (60/61) * ... = 2/5. - Dimitris Valianatos, Feb 26 2019
Numbers k such that A051709(k) = 1. - Jianing Song, Jun 27 2021

Crossrefs

Programs

Formula

n such that A062799(n) = 2. - Benoit Cloitre, Apr 06 2002
A000005(a(n)^(k-1)) = A005408(k) for all k>0. - Reinhard Zumkeller, Mar 04 2007
a(n) = A000040(n)^(3-1)=A000040(n)^2, where 3 is the number of divisors of a(n). - Omar E. Pol, May 06 2008
A000005(a(n)) = 3 or A002033(a(n)) = 2. - Juri-Stepan Gerasimov, Oct 10 2009
A033273(a(n)) = 3. - Juri-Stepan Gerasimov, Dec 07 2009
For n > 2: (a(n) + 17) mod 12 = 6. - Reinhard Zumkeller, May 12 2010
A192134(A095874(a(n))) = A005722(n) + 1. - Reinhard Zumkeller, Jun 26 2011
For n > 2: a(n) = 1 (mod 24). - Zak Seidov, Dec 07 2011
A211110(a(n)) = 2. - Reinhard Zumkeller, Apr 02 2012
a(n) = A087112(n,n). - Reinhard Zumkeller, Nov 25 2012
a(n) = prime(n)^2. - Jon E. Schoenfield, Mar 29 2015
Product_{n>=1} a(n)/(a(n)-1) = Pi^2/6. - Daniel Suteu, Feb 06 2017
Sum_{n>=1} 1/a(n) = P(2) = 0.4522474200... (A085548). - Amiram Eldar, Jul 27 2020
From Amiram Eldar, Jan 23 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = zeta(2)/zeta(4) = 15/Pi^2 (A082020).
Product_{n>=1} (1 - 1/a(n)) = 1/zeta(2) = 6/Pi^2 (A059956). (End)

A030078 Cubes of primes.

Original entry on oeis.org

8, 27, 125, 343, 1331, 2197, 4913, 6859, 12167, 24389, 29791, 50653, 68921, 79507, 103823, 148877, 205379, 226981, 300763, 357911, 389017, 493039, 571787, 704969, 912673, 1030301, 1092727, 1225043, 1295029, 1442897, 2048383, 2248091, 2571353, 2685619, 3307949
Offset: 1

Views

Author

Keywords

Comments

Numbers with exactly three factorizations: A001055(a(n)) = 3 (e.g., a(4) = 1*343 = 7*49 = 7*7*7). - Reinhard Zumkeller, Dec 29 2001
Intersection of A014612 and A000578. Intersection of A014612 and A030513. - Wesley Ivan Hurt, Sep 10 2013
Let r(n) = (a(n)-1)/(a(n)+1) if a(n) mod 4 = 1, (a(n)+1)/(a(n)-1) otherwise; then Product_{n>=1} r(n) = (9/7) * (28/26) * (124/126) * (344/342) * (1332/1330) * ... = 48/35. - Dimitris Valianatos, Mar 06 2020
There exist 5 groups of order p^3, when p prime, so this is a subsequence of A054397. Three of them are abelian: C_p^3, C_p^2 X C_p and C_p X C_p X C_p = (C_p)^3. For 8 = 2^3, the 2 nonabelian groups are D_8 and Q_8; for odd prime p, the 2 nonabelian groups are (C_p x C_p) : C_p, and C_p^2 : C_p (remark, for p = 2, these two semi-direct products are isomorphic to D_8). Here C, D, Q mean Cyclic, Dihedral, Quaternion groups of the stated order; the symbols X and : mean direct and semidirect products respectively. - Bernard Schott, Dec 11 2021

Examples

			a(3) = 125; since the 3rd prime is 5, a(3) = 5^3 = 125.
		

References

  • Edmund Landau, Elementary Number Theory, translation by Jacob E. Goodman of Elementare Zahlentheorie (Vol. I_1 (1927) of Vorlesungen über Zahlentheorie), by Edmund Landau, with added exercises by Paul T. Bateman and E. E. Kohlbecker, Chelsea Publishing Co., New York, 1958, pp. 31-32.

Crossrefs

Other sequences that are k-th powers of primes are: A000040 (k=1), A001248 (k=2), this sequence (k=3), A030514 (k=4), A050997 (k=5), A030516 (k=6), A092759 (k=7), A179645 (k=8), A179665 (k=9), A030629 (k=10), A079395 (k=11), A030631 (k=12), A138031 (k=13), A030635 (k=16), A138032 (k=17), A030637 (k=18).
Cf. A060800, A131991, A000578, subsequence of A046099.
Subsequence of A007422 and of A054397.

Programs

Formula

n such that A062799(n) = 3. - Benoit Cloitre, Apr 06 2002
a(n) = A000040(n)^3. - Omar E. Pol, Jul 27 2009
A064380(a(n)) = A000010(a(n)). - Vladimir Shevelev, Apr 19 2010
A003415(a(n)) = A079705(n). - Reinhard Zumkeller, Jun 26 2011
A056595(a(n)) = 2. - Reinhard Zumkeller, Aug 15 2011
A000005(a(n)) = 4. - Wesley Ivan Hurt, Sep 10 2013
a(n) = A119959(n) * A008864(n) -1.- R. J. Mathar, Aug 13 2019
Sum_{n>=1} 1/a(n) = P(3) = 0.1747626392... (A085541). - Amiram Eldar, Jul 27 2020
From Amiram Eldar, Jan 23 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = zeta(3)/zeta(6) (A157289).
Product_{n>=1} (1 - 1/a(n)) = 1/zeta(3) (A088453). (End)

A050997 Fifth powers of primes.

Original entry on oeis.org

32, 243, 3125, 16807, 161051, 371293, 1419857, 2476099, 6436343, 20511149, 28629151, 69343957, 115856201, 147008443, 229345007, 418195493, 714924299, 844596301, 1350125107, 1804229351, 2073071593, 3077056399, 3939040643, 5584059449, 8587340257, 10510100501
Offset: 1

Views

Author

Keywords

Comments

Numbers k such that A062799(k) = 5.
Let r(n) = (a(n)+1)/(a(n)-1) if a(n) mod 4 = 3, (a(n)-1)/(a(n)+1) otherwise; then Product_{n>=1} r(n) = (31/33) * (244/242) * (3124/3126) * (16808/16806) * ... = 246016/259875. - Dimitris Valianatos, Mar 09 2020

Crossrefs

Programs

Formula

A056595(a(n)) = 3. - Reinhard Zumkeller, Aug 15 2011
Sum_{n>=1} 1/a(n) = P(5) = 0.0357550174... (A085965). - Amiram Eldar, Jul 27 2020
From Amiram Eldar, Jan 23 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = zeta(5)/zeta(10) (A157291).
Product_{n>=1} (1 - 1/a(n)) = 1/zeta(5) = 1/A013663. (End)

A063647 Number of ways to write 1/n as a difference of exactly 2 unit fractions.

Original entry on oeis.org

0, 1, 1, 2, 1, 4, 1, 3, 2, 4, 1, 7, 1, 4, 4, 4, 1, 7, 1, 7, 4, 4, 1, 10, 2, 4, 3, 7, 1, 13, 1, 5, 4, 4, 4, 12, 1, 4, 4, 10, 1, 13, 1, 7, 7, 4, 1, 13, 2, 7, 4, 7, 1, 10, 4, 10, 4, 4, 1, 22, 1, 4, 7, 6, 4, 13, 1, 7, 4, 13, 1, 17, 1, 4, 7, 7, 4, 13, 1, 13, 4, 4, 1, 22, 4, 4, 4, 10, 1, 22, 4, 7, 4, 4, 4
Offset: 1

Views

Author

Henry Bottomley, Jul 23 2001

Keywords

Comments

Also number of ways to write 1/n as sum of exactly two distinct unit fractions. - Thomas L. York, Jan 11 2014
Also number of positive integers m such that 1/n + 1/m is a unit fraction. - Jon E. Schoenfield, Apr 17 2018
If 1/n = 1/b - 1/c then n = bc/(c-b) and 1/n = 1/(2n-b) + 1/(c+2n) (though it is also the case that 1/n = 1/(2n) + 1/(2n) equivalent to b = c = 0).
Also number of divisors of n^2 less than n. - Vladeta Jovovic, Aug 13 2001
Number of elements in the set {(x,y): x|n, y|n, xVladeta Jovovic, May 03 2002
Also number of positive integers of the form k*n/(k+n). - Benoit Cloitre, Jan 04 2002
This is similar to A062799, having the same first 29 terms. But they are different sequences.
If A001221(n) = omega(n) <= 2, then a(n) = A062799(n); if A001221(n) > 2, then a(n) > A062799(n). - Matthew Vandermast, Aug 25 2004
Number of r X s integer-sided rectangles such that r + s = 4n, r < s and (s - r) | (s * r). - Wesley Ivan Hurt, Apr 24 2020
Also number of integer-sided right triangles with 2n as a leg. Equivalent to the even indices of A046079. - Nathaniel C Beckman, May 14 2020; Jun 26 2020
a(n) is the number of positive integers k such that k+n divides k*n. - Thomas Ordowski, Dec 02 2024

Examples

			a(10) = 4 since 1/10 = 1/5 - 1/10 = 1/6 - 1/15 = 1/8 - 1/40 = 1/9 - 1/90.
a(12) = 7: the divisors of 12 are 1, 2, 3, 4, 6 and 12 and the decompositions are (1, 2), (1, 3), (1, 4), (1, 6), (1, 12), (2, 3), (3, 4).
		

Crossrefs

First twenty-nine terms identical to those of A062799.

Programs

  • Magma
    [(NumberOfDivisors(n^2)-1)/2 : n in [1..100]]; // Vincenzo Librandi, Apr 18 2018
  • Mathematica
    Table[(Length[Divisors[n^2]] - 1)/2, {n, 1, 100}]
    (DivisorSigma[0,Range[100]^2]-1)/2 (* Harvey P. Dale, Apr 15 2013 *)
  • PARI
    for(n=1,100,print1(sum(i=1,n^2,if((n*i)%(i+n),0,1)),","))
    
  • PARI
    a(n)=numdiv(n^2)\2 \\ Charles R Greathouse IV, Oct 03 2016
    

Formula

a(n) = (tau(n^2)-1)/2.
a(n) = A018892(n)-1. If n = (p1^a1)(p2^a2)...(pt^at), a(n) = ((2*a1+1)(2*a2+1)...(2*at+1)-1)/2.
If n is prime a(n)=1. Conjecture: (1/n)*Sum_{i=1..n} a(i) = C*log(n)*log(log(n)) + o(log(n)) with C=0.7... [The conjecture is false. See the plot and the asymptotic formula below. - Amiram Eldar, Oct 03 2024]
Bisection of A046079. - Lekraj Beedassy, Jul 09 2004
a(n) = Sum_{i=1..2*n-1} (1 - ceiling(i*(4*n-i)/(4*n-2*i)) + floor(i*(4*n-i)/(4*n-2*i))). - Wesley Ivan Hurt, Apr 24 2020
Sum_{k=1..n} a(k) ~ (n/(2*zeta(2)))*(log(n)^2/2 + log(n)*(3*gamma - 1) + 1 - 3*gamma + 3*gamma^2 - 3*gamma_1 - zeta(2) + (2 - 6*gamma - 2*log(n))*zeta'(2)/zeta(2) + (2*zeta'(2)/zeta(2))^2 - 2*zeta''(2)/zeta(2)), where gamma is Euler's constant (A001620) and gamma_1 is the first Stieltjes constant (A082633). - Amiram Eldar, Oct 03 2024

A000977 Numbers that are divisible by at least three different primes.

Original entry on oeis.org

30, 42, 60, 66, 70, 78, 84, 90, 102, 105, 110, 114, 120, 126, 130, 132, 138, 140, 150, 154, 156, 165, 168, 170, 174, 180, 182, 186, 190, 195, 198, 204, 210, 220, 222, 228, 230, 231, 234, 238, 240, 246, 252, 255, 258, 260, 264, 266, 270, 273, 276, 280, 282, 285
Offset: 1

Views

Author

Keywords

Comments

a(n+1)-a(n) seems bounded and sequence appears to give n such that the number of integers of the form nk/(n+k) k>=1 is not equal to Sum_{ d | n} omega(d) (i.e., n such that A062799(n) is not equal to A063647(n)). - Benoit Cloitre, Aug 27 2002
The first differences are bounded: clearly a(n+1) - a(n) <= 30. - Charles R Greathouse IV, Dec 19 2011

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 844.

Crossrefs

Complement of A070915.

Programs

  • Haskell
    a000977 n = a000977_list !! (n-1)
    a000977_list = filter ((> 2) . a001221) [1..]
    -- Reinhard Zumkeller, May 03 2013
  • Maple
    A000977 := proc(n)
    if (nops(numtheory[factorset](n)) >= 3) then
       RETURN(n)
    fi: end:  seq(A000977(n), n=1..500); # Jani Melik, Feb 24 2011
  • Mathematica
    DeleteCases[Table[If[Count[PrimeQ[Divisors[i]], True] >= 3, i, 0], {i, 1, 274}], 0]
    Select[Range[300], PrimeNu[#] >= 3 &] (* Paolo Xausa, Mar 28 2024 *)
  • PARI
    is(n)=omega(n)>2 \\ Charles R Greathouse IV, Dec 19 2011
    

Formula

a(n) = n + O(n log log n / log n). - Charles R Greathouse IV, Dec 19 2011 A001221(a(n)) > 2. - Reinhard Zumkeller, May 03 2013
A033992 UNION A033993 UNION A051270 UNION A074969 UNION A176655 UNION ... - R. J. Mathar, Dec 05 2016

Extensions

More terms from Vit Planocka (planocka(AT)mistral.cz), Sep 17 2002

A346009 a(n) is the numerator of the average number of distinct prime factors of the divisors of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 7, 1, 1, 1, 4, 1, 7, 1, 7, 1, 1, 1, 5, 2, 1, 3, 7, 1, 3, 1, 5, 1, 1, 1, 4, 1, 1, 1, 5, 1, 3, 1, 7, 7, 1, 1, 13, 2, 7, 1, 7, 1, 5, 1, 5, 1, 1, 1, 5, 1, 1, 7, 6, 1, 3, 1, 7, 1, 3, 1, 17, 1, 1, 7, 7, 1, 3, 1, 13, 4, 1, 1, 5, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Jul 01 2021

Keywords

Examples

			The fractions begin with 0, 1/2, 1/2, 2/3, 1/2, 1, 1/2, 3/4, 2/3, 1, 1/2, 7/6, ...
f(2) = 1/2 since 2 has 2 divisors, 1 and 2, and (omega(1) + omega(2))/2 = (0 + 1)/2 = 1/2.
f(6) = 1 since 6 has 4 divisors, 1, 2, 3 and 6 and (omega(1) + omega(2) + omega(3) + omega(6))/4 = (0 + 1 + 1 + 2)/4 = 1.
		

References

  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Exercise 3.3.21 on page 100.

Crossrefs

Programs

  • Mathematica
    a[n_] := Numerator[DivisorSum[n, PrimeNu[#] &]/DivisorSigma[0, n]]; Array[a, 100]
    (* or *)
    f[p_, e_] := e/(e+1); a[1] = 0; a[n_] := Numerator[Plus @@ f @@@ FactorInteger[n]]; Array[a, 100]

Formula

Let f(n) = a(n)/A346010(n) be the sequence of fractions. Then:
f(n) = A062799(n)/A000005(n).
f(n) = (Sum_{p prime, p|n} d(n/p))/d(n), where d(n) is the number of divisors of n (A000005).
f(n) depends only on the prime signature of n: If n = Product_{i} p_i^e_i, then a(n) = Sum_{i} e_i/(e_i + 1).
f(p) = 1/2 for prime p.
f(n) = 1 for squarefree semiprimes n (A006881).
Sum_{k=1..n} f(k) ~ (1/2) * A013939(n) + C*n + O(n/log(n)) ~ n*log(log(n))/2 + (B/2 + C)*n + O(n/log(n)), where B is Mertens's constant (A077761) and C = A346011 (Duncan, 1961).

A319131 a(n) = Sum_{d|n} Sum_{p|d, p prime} p.

Original entry on oeis.org

0, 2, 3, 4, 5, 10, 7, 6, 6, 14, 11, 17, 13, 18, 16, 8, 17, 18, 19, 23, 20, 26, 23, 24, 10, 30, 9, 29, 29, 40, 31, 10, 28, 38, 24, 30, 37, 42, 32, 32, 41, 48, 43, 41, 27, 50, 47, 31, 14, 26, 40, 47, 53, 26, 32, 40, 44, 62, 59, 64, 61, 66, 33, 12, 36, 64, 67, 59, 52, 56
Offset: 1

Views

Author

Ilya Gutkovskiy, Sep 11 2018

Keywords

Comments

Inverse Möbius transform of A008472.

Examples

			a(12) = 13 as 12 has 6 divisors and 2 * 6 * (2/3) + 3 * 6 * (1/2) = 17. - _David A. Corneth_, Oct 08 2019
		

Crossrefs

Programs

  • Magma
    [0] cat  [&+[&+[PrimeDivisors(d)[i]:i in [1..#PrimeDivisors(d)]]:d in Set(Divisors(n)) diff {1}]:n in [2..70]]; // Marius A. Burtea, Oct 08 2019
    
  • Magma
    [0] cat [&+[p*#Divisors(n div p):p in PrimeDivisors(n)]:n in [2..70]]; // Marius A. Burtea, Oct 08 2019 (According to the formula given by Ridouane Oudra)
  • Maple
    with(numtheory): seq(add(p*tau(n/p), p in factorset(n)), n=1..80); # Ridouane Oudra, Oct 08 2019
  • Mathematica
    Table[Sum[Total[Select[Divisors[d], PrimeQ]], {d, Divisors[n]}], {n, 70}]
    nmax = 70; Rest[CoefficientList[Series[Sum[DivisorSum[k, # &, PrimeQ[#] &] x^k/(1 - x^k), {k, 1, nmax}], {x, 0, nmax}], x]]
    nmax = 70; Rest[CoefficientList[Series[-Log[Product[(1 - x^k)^(DivisorSum[k, # &, PrimeQ[#] &]/k), {k, 1, nmax}]], {x, 0, nmax}], x] Range[0, nmax]]
  • PARI
    a(n) = sumdiv(n, d, my(f=factor(d)); vecsum(f[,1])); \\ Michel Marcus, Oct 08 2019
    
  • PARI
    a(n) = my(f = factor(n), nd = numdiv(f)); sum(i = 1, #f~, f[i, 1] * nd / (f[i, 2] + 1) * f[i, 2]) \\ David A. Corneth, Oct 08 2019
    

Formula

G.f.: Sum_{k>=1} A008472(k)*x^k/(1 - x^k).
L.g.f.: -log(Product_{k>=1} (1 - x^k)^(A008472(k)/k)) = Sum_{n>=1} a(n)*x^n/n.
a(p^k) = p*k, where p is a prime.
a(n) = Sum_{p|n} p*tau(n/p), where p is a prime and tau(n) = A000005(n). - Ridouane Oudra, Oct 08 2019
a(n) = Sum_{p|n} p*tau(n)*(e_p-1)/(e_p) where e_p is the exponent of p in the factorization of n. - David A. Corneth, Oct 08 2019
a(n) = Sum_{d|n} sopf(d). - Wesley Ivan Hurt, May 23 2021

A329354 a(n) = Sum_{d|n} d*omega(d).

Original entry on oeis.org

0, 2, 3, 6, 5, 17, 7, 14, 12, 27, 11, 45, 13, 37, 38, 30, 17, 62, 19, 71, 52, 57, 23, 101, 30, 67, 39, 97, 29, 162, 31, 62, 80, 87, 82, 162, 37, 97, 94, 159, 41, 220, 43, 149, 137, 117, 47, 213, 56, 152, 122, 175, 53, 197, 126, 217, 136, 147, 59, 410, 61, 157, 187, 126, 148, 336, 67, 227, 164, 342, 71, 362, 73, 187, 213, 253, 172, 394, 79
Offset: 1

Views

Author

Antti Karttunen, Nov 15 2019

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[d*PrimeNu[d], {d, Divisors[n]}], {n, 1, 100}] (* Vaclav Kotesovec, Aug 18 2021 *)
  • PARI
    A329354(n) = sumdiv(n,d,omega(d)*d);

Formula

a(n) = Sum_{d|n} d*A001221(d).
a(n) = A180253(n) - A323599(n).
a(n) = A328260(n) + A329375(n).
a(n) = Sum_{d|n} (n/d) * sopf(d). - Wesley Ivan Hurt, May 24 2021
Dirichlet g.f.: primezeta(s-1) * zeta(s-1) * zeta(s). - Ilya Gutkovskiy, Aug 18 2021
Conjecture: Sum_{k=1..n} a(k) ~ Pi^2 * n^2 * (log(log(n)) + A077761) / 12. - Vaclav Kotesovec, Mar 03 2023
Showing 1-10 of 25 results. Next