A007948 -id:A007948 - cp's OEIS frontend

A384555 The largest infinitary divisor of n that is cubefree.

1, 2, 3, 4, 5, 6, 7, 4, 9, 10, 11, 12, 13, 14, 15, 1, 17, 18, 19, 20, 21, 22, 23, 12, 25, 26, 9, 28, 29, 30, 31, 2, 33, 34, 35, 36, 37, 38, 39, 20, 41, 42, 43, 44, 45, 46, 47, 3, 49, 50, 51, 52, 53, 18, 55, 28, 57, 58, 59, 60, 61, 62, 63, 4, 65, 66, 67, 68, 69

Offset: 1

Amiram Eldar, Jun 03 2025

The number of these divisors is A368883(n), and their sum is A384554(n).

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A004709, A007948, A013666, A077609, A368883, A384554.

f[p_, e_] := Switch[Mod[e, 4], 0, 1, 1, p, 2, p^2, 3, p^2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

a(n) = {my(f = factor(n)); prod(i = 1, #f~, p = f[i,1]; e = f[i,2]; [1, p, p^2, p^2][e%4+1]);}

Multiplicative with a(p^e) = 1 if e == 0 (mod 4), p if e == 1 (mod 4), p^2 if e == 2 or 3 (mod 4).
a(n) = n if and only if n is cubefree (A004709).
Dirichlet g.f.: zeta(4*s) * Product_{p prime} (1 + 1/p^(s-1) + 1/p^(2*s-2) + 1/p^(3*s-2)).
Sum_{k=1..n} a(k) = c * n^2 / 2, where c = zeta(8) * Product_{p prime} (1 - 1/p^3 + 1/p^4 - 1/p^5) = 0.87406992849637563411... .

A128251 n^4 - 1 divided by its largest fourth power divisor.

Original entry on oeis.org

15, 5, 255, 39, 1295, 150, 4095, 410, 9999, 915, 20735, 1785, 38415, 3164, 65535, 5220, 104975, 8145, 159999, 12155, 234255, 17490, 331775, 24414, 456975, 33215, 614655, 44205, 809999, 57720, 1048575, 74120, 1336335, 93789, 1679615, 117135

Offset: 2

Jonathan Vos Post, May 03 2007

In other words, biquadratefree part of n^4-1, or biquadratefree kernel of n^4-1. Fourth power analog of what A128972 is to cubes and A068310 is to squares. A046100 Biquadratefree numbers. A008835 Largest 4th power dividing n.

			a(3) = 5 because (3^4 - 1)/16 = 80/16 = (2^4 * 5)/(2^4) = 5.
a(5) = 39 because (5^4 - 1)/16 = 624/16 = (2^4 * 3 * 13)/(2^4) = 39.
a(7) = 150 because (7^4 - 1)/16 = 2400/16 = (2^5 * 3 * 5^2)/(2^4) = 150.
a(9) = 410 because (9^4 - 1)/16 = 6560/16 = (2^5 * 5 * 41)/(2^4) = 410.
a(63) = 61535 because (63^4 - 1)/256 = 15752960/256 = (2^8 * 5 * 31 * 397)/(2^8) = 61535.

Eric Weisstein's World of Mathematics, Biquadratefree.

Cf. A000188, A000583, A002350, A004709, A007948, A008835, A062378, A067872, A033314, A068310, A128972.

a(n) = (n^4 - 1)/A008835(n^4 - 1) = (A000583(n)-1)/A008835((A000583(n)-1)).

A309875 Cubefree colossally superabundant numbers: cubefree numbers (A004709) k for which there is a positive exponent epsilon such that sigma(k)/k^{1 + epsilon} >= sigma(j)/j^{1 + epsilon} for all cubefree j > 1, so that k attains the maximum value of sigma(k)/k^{1 + epsilon} over the cubefree numbers.

Original entry on oeis.org

2, 6, 12, 60, 180, 1260, 13860, 180180, 900900, 15315300, 290990700, 6692786100, 194090796900, 6016814703900, 42117702927300, 1558355008310100, 63892555340714100, 2747379879650706300, 129126854343583196100, 6843723280209909393300, 403779673532384654204700

Offset: 1

Amiram Eldar, Aug 21 2019

nonn

This sequence is formed by the largest cubefree divisors (A007948) of the colossally superabundant numbers (A004490).

Paul Erdős and Jean-Louis Nicolas, Répartition des nombres superabondants", Bulletin de la Société Mathématique de France, Vol. 103 (1975), pp. 65-90. See section 5, p. 83.

Cf. A000203, A004490, A004709, A007948.

Subsequence of A025487, A220423 and A308618.

cp's OEIS Frontend

A384555 The largest infinitary divisor of n that is cubefree.

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A128251 n^4 - 1 divided by its largest fourth power divisor.

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