login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A059377
Jordan function J_4(n).
31
1, 15, 80, 240, 624, 1200, 2400, 3840, 6480, 9360, 14640, 19200, 28560, 36000, 49920, 61440, 83520, 97200, 130320, 149760, 192000, 219600, 279840, 307200, 390000, 428400, 524880, 576000, 707280, 748800, 923520, 983040, 1171200, 1252800, 1497600, 1555200, 1874160
OFFSET
1,2
COMMENTS
This sequence is multiplicative. - Mitch Harris, Apr 19 2005
For n = 4 or n >= 6, a(n) is divisible by 240. - Jianing Song, Apr 06 2019
REFERENCES
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 199, #3.
R. Sivaramakrishnan, "The many facets of Euler's totient. II. Generalizations and analogues", Nieuw Arch. Wisk. (4) 8 (1990), no. 2, 169-187.
LINKS
D. H. Lehmer, On a theorem of von Sterneck, Bull. Amer. Math. Soc. 37(10): 723-726 (1931)
Michael Lugo, A little number theory problem (2008)
László Tóth, Multiplicative arithmetic functions of several variables: a survey, arXiv preprint arXiv:1310.7053 [math.NT], 2013.
FORMULA
a(n) = Sum_{d|n} d^4*mu(n/d). - Benoit Cloitre, Apr 05 2002
Multiplicative with a(p^e) = p^(4e)-p^(4(e-1)).
Dirichlet generating function: zeta(s-4)/zeta(s). - Franklin T. Adams-Watters, Sep 11 2005
a(n) = Sum_{k=1..n} gcd(k,n)^4 * cos(2*Pi*k/n). - Enrique Pérez Herrero, Jan 18 2013
a(n) = n^4*Product_{distinct primes p dividing n} (1 - 1/p^4). - Tom Edgar, Jan 09 2015
G.f.: Sum_{n>=1} a(n)*x^n/(1 - x^n) = x*(1 + 11*x + 11*x^2 + x^3)/(1 - x)^5. - Ilya Gutkovskiy, Apr 25 2017
Sum_{k=1..n} a(k) ~ n^5 / (5*zeta(5)). - Vaclav Kotesovec, Feb 07 2019
From Amiram Eldar, Oct 12 2020: (Start)
lim_{n->oo} (1/n) * Sum_{k=1..n} a(k)/k^4 = 1/zeta(5).
Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + p^4/(p^4-1)^2) = 1.0870036174... (End)
O.g.f.: Sum_{n >= 1} mu(n)*x^n*(1 + 11*x^n + 11*x^(2*n) + x^(3*n))/(1 - x^n)^5 = x + 15*x^2 + 80*x^3 + 240*x^4 + 624*x^5 + .... - Peter Bala, Jan 31 2022
From Peter Bala, Jan 01 2024: (Start)
a(n) = Sum_{d divides n} d * J_3(d) * J_1(n/d) = Sum_{d divides n} d^2 * J_2(d) * J_2(n/d) = Sum_{d divides n} d^3 * J_1(d) * J_3(n/d), where J_1(n) = phi(n) = A000010(n), J_2(n) = A007434(n) and J(3,n) = A059376(n).
a(n) = Sum_{k = 1..n} gcd(k, n) * J_3(gcd(k, n)) = Sum_{1 <= j, k <= n} gcd(j, k, n)^2 * J_2(gcd(j, k, n)) = Sum_{1 <= i, j, k <= n} gcd(i, j, k, n)^3 * J_1(gcd(i, j, k, n)). (End)
a(n) = Sum_{1 <= i, j <= n, lcm(i, j) = n} J_2(i) * J_2(j) = Sum_{1 <= i, j <= n, lcm(i, j) = n} phi(i) * J_3(j) (apply Lehmer, Theorem 1). - Peter Bala, Jan 29 2024
MAPLE
J := proc(n, k) local i, p, t1, t2; t1 := n^k; for p from 1 to n do if isprime(p) and n mod p = 0 then t1 := t1*(1-p^(-k)); fi; od; t1; end:
seq(J(n, 4), n=1..40);
MATHEMATICA
JordanJ[n_, k_: 1] := DivisorSum[n, #^k*MoebiusMu[n/#] &]; f[n_] := JordanJ[n, 4]; Array[f, 38]
f[p_, e_] := p^(4*e) - p^(4*(e-1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 12 2020 *)
PROG
(PARI) for(n=1, 100, print1(sumdiv(n, d, d^4*moebius(n/d)), ", "))
(PARI) a(n)=if(n<1, 0, sumdiv(n, d, d^4*moebius(n/d)))
(PARI) a(n)=if(n<1, 0, dirdiv(vector(n, k, k^4), vector(n, k, 1))[n])
(PARI) { for (n = 1, 1000, write("b059377.txt", n, " ", sumdiv(n, d, d^4*moebius(n/d))); ) } \\ Harry J. Smith, Jun 26 2009
CROSSREFS
See A059379 and A059380 (triangle of values of J_k(n)), A000010 (J_1), A007434 (J_2), A059376 (J_3), A059378 (J_5), A069091 - A069095 (J_6 through J_10).
Cf. A013663.
Sequence in context: A085808 A180577 A033594 * A370533 A123865 A024002
KEYWORD
nonn,mult,easy
AUTHOR
N. J. A. Sloane, Jan 28 2001
STATUS
approved