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.

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A384185 Integers in A375491 in order of their first appearance.

Original entry on oeis.org

1, 2, 4, 6, 12, 5, 8, 24, 3, 7, 18, 16, 30, 36, 48, 10, 32, 14, 72, 9, 60, 96, 120, 19, 64, 13, 40, 144, 21, 35, 56, 38, 28, 90, 26, 240, 15, 192, 384, 44, 76, 360, 88, 80, 180, 168, 114, 54, 288, 112, 264, 25, 41, 33, 256, 98, 20, 55, 312, 128
Offset: 1

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Author

Robin Jones, May 21 2025

Keywords

Comments

Does every positive integer appear in this sequence?

Examples

			a(1)=1 since the first squarefree integer (1) has 1 group of that order.
a(2)=2 since the first squarefree integer not to have a(1)=1 group of its order (6) has 2 groups of its order.
a(3)=4 since the first squarefree integer not to have a(1)=1 or a(2)=2 groups of its order (30) has 4 groups of its order.
		

Crossrefs

Cf. A375491.

A375483 Number of nonabelian groups of order m where m is the n-th squarefree number.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 3, 0, 0, 1, 0, 0, 1, 1, 0, 5, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 3, 0, 0, 3, 0, 0, 1, 0, 5, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 3, 0, 1, 1, 0, 0, 5, 1, 0, 5, 0, 1, 0, 1, 0, 0, 1, 3, 0, 0, 1, 0, 3, 0, 0
Offset: 1

Views

Author

Chai Wah Wu, Aug 17 2024

Keywords

References

  • O. Hölder. Die Gruppen mit quadratfreier Ordnungszahl. Nachrichten von der Gesellschaft der Wissenschaften zu Göttingen. Mathematisch-Physikalische Klasse, pages 211-219 (1895).

Crossrefs

Programs

  • Python
    from math import isqrt, prod
    from itertools import combinations
    from sympy import mobius, primefactors, npartitions, factorint
    def A375483(n):
        def f(x): return n+x-sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1))
        m, k = n, f(n)
        while m != k:
            m, k = k, f(k)
        a = set(primefactors(m))
        return sum(prod((p**sum(1 for q in b if q%p==1)-1)//(p-1) for p in a-set(b)) for l in range(0,len(a)+1) for b in combinations(a,l))-prod(map(npartitions, factorint(m).values()))

Formula

a(n) = A060689(A005117(n)) = A375491(n) - A000688(A005117(n)).
Showing 1-2 of 2 results.