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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A000688 Number of Abelian groups of order n; number of factorizations of n into prime powers.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 5, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 7, 1, 1, 1, 4, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 5, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 11, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 2, 2, 1, 1, 1, 5, 5, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 7, 1, 2, 2, 4, 1, 1, 1, 3, 1, 1, 1
Offset: 1

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Author

Keywords

Comments

Equivalently, number of Abelian groups with n conjugacy classes. - Michael Somos, Aug 10 2010
a(n) depends only on prime signature of n (cf. A025487). So a(24) = a(375) since 24 = 2^3*3 and 375 = 3*5^3 both have prime signature (3, 1).
Also number of rings with n elements that are the direct product of fields; these are the commutative rings with n elements having no nilpotents; likewise the commutative rings where for every element x there is a k > 0 such that x^(k+1) = x. - Franklin T. Adams-Watters, Oct 20 2006
Range is A033637.
a(n) = 1 if and only if n is from A005117 (squarefree numbers). See the Ahmed Fares comment there, and the formula for n>=2 below. - Wolfdieter Lang, Sep 09 2012
Also, from a theorem of Molnár (see [Molnár]), the number of (non-isomorphic) abelian groups of order 2*n + 1 is equal to the number of non-congruent lattice Z-tilings of R^n by crosses, where a "cross" is a unit cube in R^n for which at each facet is attached another unit cube (Z, R are the integers and reals, respectively). (Cf. [Horak].) - L. Edson Jeffery, Nov 29 2012
Zeta(k*s) is the Dirichlet generating function of the characteristic function of numbers which are k-th powers (k=1 in A000012, k=2 in A010052, k=3 in A010057, see arXiv:1106.4038 Section 3.1). The infinite product over k (here) is the number of representations n=product_i (b_i)^(e_i) where all exponents e_i are distinct and >=1. Examples: a(n=4)=2: 4^1 = 2^2. a(n=8)=3: 8^1 = 2^1*2^2 = 2^3. a(n=9)=2: 9^1 = 3^2. a(n=12)=2: 12^1 = 3*2^2. a(n=16)=5: 16^1 = 2*2^3 = 4^2 = 2^2*4^1 = 2^4. If the e_i are the set {1,2} we get A046951, the number of representations as a product of a number and a square. - R. J. Mathar, Nov 05 2016
See A060689 for the number of non-abelian groups of order n. - M. F. Hasler, Oct 24 2017
Kendall & Rankin prove that the density of {n: a(n) = m} exists for each m. - Charles R Greathouse IV, Jul 14 2024

Examples

			a(1) = 1 since the trivial group {e} is the only group of order 1, and it is Abelian; alternatively, since the only factorization of 1 into prime powers is the empty product.
a(p) = 1 for any prime p, since the only factorization into prime powers is p = p^1, and (in view of Lagrange's theorem) there is only one group of prime order p; it is isomorphic to (Z/pZ,+) and thus Abelian.
From _Wolfdieter Lang_, Jul 22 2011: (Start)
a(8) = 3 because 8 = 2^3, hence a(8) = pa(3) = A000041(3) = 3 from the partitions (3), (2, 1) and (1, 1, 1), leading to the 3 factorizations of 8: 8, 4*2 and 2*2*2.
a(36) = 4 because 36 = 2^2*3^2, hence a(36) = pa(2)*pa(2) = 4 from the partitions (2) and (1, 1), leading to the 4 factorizations of 36: 2^2*3^2, 2^2*3^1*3^1, 2^1*2^1*3^2 and 2^1*2^1*3^1*3^1.
(End)
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 274-278.
  • D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Section XIII.12, p. 468.
  • J. S. Rose, A Course on Group Theory, Camb. Univ. Press, 1978, see p. 7.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • A. Speiser, Die Theorie der Gruppen von endlicher Ordnung, 4. Auflage, Birkhäuser, 1956.

Crossrefs

Cf. A080729 (Dgf at s=2), A369634 (Dgf at s=3).

Programs

  • Haskell
    a000688 = product . map a000041 . a124010_row
    -- Reinhard Zumkeller, Aug 28 2014
    
  • Maple
    with(combinat): readlib(ifactors): for n from 1 to 120 do ans := 1: for i from 1 to nops(ifactors(n)[2]) do ans := ans*numbpart(ifactors(n)[2][i][2]) od: printf(`%d,`,ans): od: # James Sellers, Dec 07 2000
  • Mathematica
    f[n_] := Times @@ PartitionsP /@ Last /@ FactorInteger@n; Array[f, 107] (* Robert G. Wilson v, Sep 22 2006 *)
    Table[FiniteAbelianGroupCount[n], {n, 200}] (* Requires version 7.0 or later. - Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)
  • PARI
    A000688(n)=local(f);f=factor(n);prod(i=1,matsize(f)[1],numbpart(f[i,2])) \\ Michael B. Porter, Feb 08 2010
    
  • PARI
    a(n)=my(f=factor(n)[,2]); prod(i=1,#f,numbpart(f[i])) \\ Charles R Greathouse IV, Apr 16 2015
    
  • Python
    from sympy import factorint, npartitions
    from math import prod
    def A000688(n): return prod(map(npartitions,factorint(n).values())) # Chai Wah Wu, Jan 14 2022
  • Sage
    def a(n):
        F=factor(n)
        return prod([number_of_partitions(F[i][1]) for i in range(len(F))])
    # Ralf Stephan, Jun 21 2014
    

Formula

Multiplicative with a(p^k) = number of partitions of k = A000041(k); a(mn) = a(m)a(n) if (m, n) = 1.
a(2n) = A101872(n).
a(n) = Product_{j = 1..N(n)} A000041(e(j)), n >= 2, if
n = Product_{j = 1..N(n)} prime(j)^e(j), N(n) = A001221(n). See the Richert reference, quoting A. Speiser's book on finite groups (in German, p. 51 in words). - Wolfdieter Lang, Jul 23 2011
In terms of the cycle index of the symmetric group: Product_{q=1..m} [z^{v_q}] Z(S_v) 1/(1-z) where v is the maximum exponent of any prime in the prime factorization of n, v_q are the exponents of the prime factors, and Z(S_v) is the cycle index of the symmetric group on v elements. - Marko Riedel, Oct 03 2014
Dirichlet g.f.: Sum_{n >= 1} a(n)/n^s = Product_{k >= 1} zeta(ks) [Kendall]. - Álvar Ibeas, Nov 05 2014
a(n)=2 for all n in A054753 and for all n in A085987. a(n)=3 for all n in A030078 and for all n in A065036. a(n)=4 for all n in A085986. a(n)=5 for all n in A030514 and for all n in A178739. a(n)=6 for all n in A143610. - R. J. Mathar, Nov 05 2016
A050360(n) = a(A025487(n)). a(n) = A050360(A101296(n)). - R. J. Mathar, May 26 2017
a(n) = A000001(n) - A060689(n). - M. F. Hasler, Oct 24 2017
From Amiram Eldar, Nov 01 2020: (Start)
a(n) = a(A057521(n)).
Asymptotic mean: lim_{n->oo} (1/n) * Sum_{k=1..n} a(k) = A021002. (End)
a(n) = A005361(n) except when n is a term of A046101, since A000041(x) = x for x <= 3. - Miles Englezou, Feb 17 2024
Inverse Moebius transform of A188585: a(n) = Sum_{d|n} A188585(d). - Amiram Eldar, Jun 10 2025

A384912 The number of unordered factorizations of n into exponentially squarefree prime powers (A384419).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 6, 1, 1, 1, 4, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 4, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 9, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 2, 2, 1, 1, 1, 4, 4, 1, 1, 2, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Jun 12 2025

Keywords

Comments

First differs from A384913 at n = 64.

Examples

			a(4) = 2 since 4 has 2 factorizations: 2^1 * 2^1 and 2^2, with squarefree exponents 1 and 2.
		

Crossrefs

Programs

  • Mathematica
    s[n_] := s[n] = If[n == 0, 1, Sum[Sum[d * Abs[MoebiusMu[d]], {d, Divisors[j]}] * s[n-j], {j, 1, n}] / n]; (* Jean-François Alcover at A073576 *)
    f[p_, e_] := s[e]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    s(n) = if(n < 1, 1, sum(j = 1, n, sumdiv(j, d, d*issquarefree(d)) * s(n-j))/n);
    a(n) = vecprod(apply(s, factor(n)[, 2]));

Formula

Multiplicative with a(p^e) = A073576(e).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} f(1/p) = 2.1069024289184419840496..., where f(x) = (1-x) / Product_{k>=1} (1-x^A005117(k)).

A384913 The number of unordered factorizations of n into exponentially Fibonacci powers of primes (A115975).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 1, 1, 3, 2, 1, 3, 2, 1, 1, 1, 6, 1, 1, 1, 4, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 4, 2, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 1, 1, 2, 8, 1, 1, 1, 2, 1, 1, 1, 6, 1, 1, 2, 2, 1, 1, 1, 4, 4, 1, 1, 2, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Jun 12 2025

Keywords

Comments

First differs from A384912 at n = 64.

Examples

			a(4) = 2 since 4 has 2 factorizations: 2^1 * 2^1 and 2^2, with exponents 1 and 2 that are Fibonacci numbers.
		

Crossrefs

Programs

  • Mathematica
    fib[n_] := Boole[Or @@ IntegerQ /@ Sqrt[5*n^2 + {-4, 4}]];
    s[n_] := s[n] = If[n == 0, 1, Sum[Sum[d * fib[d], {d, Divisors[j]}] * s[n-j], {j, 1, n}] / n];
    f[p_, e_] := s[e]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    isfib(n) = issquare(5*n^2 - 4) || issquare(5*n^2 + 4);
    s(n) = if(n < 1, 1, sum(j = 1, n, sumdiv(j, d, d*isfib(d)) * s(n-j))/n);
    a(n) = vecprod(apply(s, factor(n)[, 2]));

Formula

Multiplicative with a(p^e) = A003107(e).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} f(1/p) = 2.05893526314055968638..., where f(x) = (1-x) / Product_{k>=2} (1-x^A000045(k)).

A384914 The number of unordered factorizations of n into numbers of the form p^(k^2) where p is prime and k >= 0 (A323520).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Jun 12 2025

Keywords

Comments

First differs from A203640, A295658 and A365333 at n = 64, from A043289 and A053164 at n = 81, and from A063775 at n = 512.

Examples

			a(16) = 2 since 4 has 2 factorizations: 2^1 * 2^1 * 2^1 * 2^1 and 2^4, with exponents 1 and 4 that are squares.
		

Crossrefs

Programs

  • Mathematica
    s[n_] := s[n] = If[n == 0, 1, Sum[Sum[d * Boole[IntegerQ[Sqrt[d]]], {d, Divisors[j]}] * s[n-j], {j, 1, n}] / n];
    f[p_, e_] := s[e]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    s(n) = if(n < 1, 1, sum(j = 1, n, sumdiv(j, d, d*issquare(d)) * s(n-j))/n);
    a(n) = vecprod(apply(s, factor(n)[, 2]));

Formula

Multiplicative with a(p^e) = A001156(e).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} f(1/p) = 1.08451356983124311685..., where f(x) = (1-x) / Product_{k>=1} (1-x^(k^2)).

A384915 The number of unordered factorizations of n into powers of primes of the form p^e where p is prime and 0 <= e <= p (A074583).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 3, 2, 1, 1, 1, 3, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 3, 2, 2, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 2, 2, 1, 1, 1, 3, 4, 1, 1, 2, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Jun 12 2025

Keywords

Examples

			a(4) = 2 since 4 has 2 factorizations: 2^1 * 2^1 and 2^2, with exponents 1 and 2 that are <= 2.
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := Length[IntegerPartitions[e, p]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    T(n, k)=my(s); forpart(v=n, s++, , k); s \\ Charles R Greathouse IV at A026820
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, T(f[i,2], f[i,1]));}

Formula

Multiplicative with a(p^e) = A026820(e, p).
a(n) >= A384916(n), with equality if and only if n is in A048103.

A384916 The number of unordered factorizations of n into powers of primes of the form p^e where p is prime and 0 <= e < p.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Jun 12 2025

Keywords

Comments

First differs from A298735 at n = 125.

Examples

			a(9) = 2 since 9 has 2 factorizations: 3^1 * 3^1 and 3^2, with exponents 1 and 2 that are < 3.
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := Length[IntegerPartitions[e, p-1]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    T(n, k)=my(s); forpart(v=n, s++, , k); s \\ Charles R Greathouse IV at A026820
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, T(f[i,2], f[i,1]-1));}

Formula

Multiplicative with a(p^e) = A026820(e, p-1).
a(n) <= A384915(n), with equality if and only if n is in A048103.

A259362 a(1) = 1, for n > 1: a(n) is the number of ways to write n as a nontrivial perfect power.

Original entry on oeis.org

1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0
Offset: 1

Views

Author

Doug Bell, Jun 24 2015

Keywords

Comments

a(n) = number of integer pairs (i,j) for distinct values of i where i > 0, j > 1 and n = i^j. Since 1 = 1^r for all real values of r, the requirement for a distinct i causes a(1) = 1 instead of a(1) = infinity.
Alternatively, the sequence can be defined as: a(1) = 1, for n > 1: a(n) = number of pairs (i,j) such that i > 0, j > 1 and n = i^j.
A007916 = n, where a(n) = 0.
A001597 = n, where a(n) > 0.
A175082 = n, where n = 1 or a(n) = 0.
A117453 = n, where n = 1 or a(n) > 1.
A175065 = n, where n > 1 and a(n) > 0 and this is the first occurrence in this sequence of a(n).
A072103 = n repeated a(n) times where n > 1.
A075802 = min(1, a(n)).
A175066 = a(n), where n = 1 or a(n) > 1. This sequence is an expansion of A175066.
A253642 = 0 followed by a(n), where n > 1 and a(n) > 0.
A175064 = a(1) followed by a(n) + 1, where n > 1 and a(n) > 0.
Where n > 1, A001597(x) = n (which implies a(n) > 0), i = A025478(x) and j = A253641(n), then a(n) = A000005(j) - 1, which is the number of factors of j greater than 1. The integer pair (i,j) comprises the smallest value i and the largest value j where i > 0, j > 1 and n = i^j. The a(n) pairs of (a,b) where a > 0, b > 1 and n = a^b are formed with b = each of the a(n) factors of j greater than 1. Examples for n = {8,4096}:
a(8) = 1, A001597(3) = 8, A025478(3) = 2, A253641(8) = 3, 8 = 2^3 and A000005(3) - 1 = 1 because there is one factor of 3 greater than 1 [3]. The set of pairs (a,b) is {(2,3)}.
a(4096) = 5, A001597(82) = 4096, A025478(82) = 2, A253641(4096) = 12, 4096 = 2^12 and A000005(12) - 1 = 5 because there are five factors of 12 greater than 1 [2,3,4,6,12]. The set of pairs (a,b) is {(64,2),(16,3),(8,4),(4,6),(2,12)}.
A023055 = the ordered list of x+1 with duplicates removed, where x is the number of consecutive zeros appearing in this sequence between any two nonzero terms.
A070428(x) = number of terms a(n) > 0 where n <= 10^x.
a(n) <= A188585(n).

Examples

			a(6) = 0 because there is no way to write 6 as a nontrivial perfect power.
a(9) = 1 because there is one way to write 9 as a nontrivial perfect power: 3^2.
a(16) = 2 because there are two ways to write 16 as a nontrivial perfect power: 2^4, 4^2.
From _Friedjof Tellkamp_, Jun 14 2025: (Start)
n:       1, 2, 3, 4, 5, 6, 7, 8, 9, ...
Squares: 1, 0, 0, 1, 0, 0, 0, 0, 1, ... (A010052)
Cubes:   1, 0, 0, 0, 0, 0, 0, 1, 0, ... (A010057)
...
Sum:    oo, 0, 0, 1, 0, 0, 0, 1, 1, ...
a(1)=1:  1, 0, 0, 1, 0, 0, 0, 1, 1, ... (= this sequence). (End)
		

Crossrefs

Programs

  • Mathematica
    a[n_] := If[n == 1, 1, Sum[Boole[IntegerQ[n^(1/k)]], {k, 2, Floor[Log[2, n]]}]]; Array[a, 100] (* Friedjof Tellkamp, Jun 14 2025 *)
    a[n_] := If[n == 1, 1, DivisorSigma[0, Apply[GCD, Transpose[FactorInteger[n]][[2]]]] - 1]; Array[a, 100] (* Michael Shamos, Jul 06 2025 *)
  • PARI
    a(n) = if (n==1, 1, sum(i=2, logint(n, 2), ispower(n, i))); \\ Michel Marcus, Apr 11 2025

Formula

a(1) = 1, for n > 1: a(n) = A000005(A253641(n)) - 1.
If n not in A001597, then a(n) = 0, otherwise a(n) = A175064(x) - 1 where A001597(x) = n.
From Friedjof Tellkamp, Jun 14 2025: (Start)
a(n) = A089723(n) - 1, for n > 1.
a(n) = A010052(n) + A010057(n) + A374016(n) + (...), for n > 1.
Sum_{k>=2..n} a(k) = A089361(n), for n > 1.
G.f.: x + Sum_{j>=2, k>=2} x^(j^k).
Dirichlet g.f.: 1 + Sum_{k>=2} zeta(k*s)-1. (End)

A384783 The number of unordered factorizations of the n-th powerful number into 1 and prime powers p^e where p is prime and e >= 2 (A025475).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 4, 1, 2, 1, 1, 1, 1, 4, 2, 1, 1, 1, 1, 1, 2, 7, 2, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 8, 1, 4, 2, 2, 1, 1, 4, 2, 2, 1, 2, 1, 1, 1, 2, 1, 12, 1, 1, 4, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 4, 1, 4, 1, 1, 1, 2, 2, 2, 2, 14, 1, 4, 1, 1, 7, 1, 2
Offset: 1

Views

Author

Amiram Eldar, Jun 10 2025

Keywords

Comments

The positive terms in A188585.

Examples

			a(5) = 2 since the 5th powerful number, A001694(5) = 16, has 2 factorizations: 2^2 * 2^2 and 2^4.
a(11) = 4 since the 11th powerful number, A001694(11) = 64, has 4 factorizations: 2^2 * 2^2 * 2^2, 2^2 * 2^4, 2^3 * 2^3, and 2^6.
		

Crossrefs

Cf. A001694, A025475, A188585, A384784, A384785 (cubefull analog).

Programs

  • Mathematica
    f[p_, e_] := PartitionsP[e] - PartitionsP[e-1]; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; seq[lim_] := Module[{pow = Union[Flatten[Table[i^2*j^3, {j, 1, Surd[lim, 3]}, {i, 1, Sqrt[lim/j^3]}]]]}, Select[s /@ pow, # > 0 &]]; seq[10^4]
  • PARI
    s(n) = vecprod(apply(x -> numbpart(x)-numbpart(x-1), factor(n)[, 2]));
    pows(lim) = {my(p = List()); for(j = 1, sqrtnint(lim, 3), for(i = 1, sqrtint(lim \ j^3), listput(p, i^2 * j^3))); Set(p); }
    list(lim) = {my(p = pows(lim), v = List(), s1); for(k = 1, #p, s1 = s(p[k]); if(s1 > 0, listput(v, s1))); Vec(v);}

Formula

a(n) = A188585(A001694(n)).

A384784 Numbers with a record number of unordered factorizations into 1 and prime powers p^e where p is prime and e >= 2 (A025475).

Original entry on oeis.org

1, 16, 64, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576, 2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728, 268435456, 536870912, 1073741824, 2147483648, 4294967296, 8589934592, 17179869184, 34359738368
Offset: 1

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Author

Amiram Eldar, Jun 10 2025

Keywords

Comments

The least term that is not a power of 2 is a(47) = 2^35 * 3^10.
Indices of records in A188585.
All the terms are powerful numbers since A188585(1) = 1 and A188585(n) = 0 if n is a nonpowerful number.
The corresponding record values are 1, 2, 4, 7, 8, 12, 14, 21, 24, 34, 41, 55, 66, 88, ... (see the link for more values).

Crossrefs

Subsequence of A001694 and A025487 (i.e., of A181800).
Cf. A025475, A046055, A188585, A384783, A384786 (cubefull analog).

Programs

  • Mathematica
    f[p_, e_] := PartitionsP[e] - PartitionsP[e-1]; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; With[{lps = Cases[Import["https://oeis.org/A025487/b025487.txt", "Table"], {, }][[;; , 2]]}, sm = -1; seq = {}; Do[s1 = s[lps[[i]]]; If[s1 > sm, sm = s1; AppendTo[seq, lps[[i]]]], {i, 1, Length[lps]}]; seq]

A365550 The number of square coreful divisors of n.

Original entry on oeis.org

1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0
Offset: 1

Views

Author

Amiram Eldar, Sep 08 2023

Keywords

Comments

First differs from A188585 at n = 64.
A coreful divisor d of a number n is a divisor with the same set of distinct prime factors as n.

Examples

			a(16) = 2 since the coreful divisors of 16 are {2, 4, 8, 16}, and 2 of them, 4 and 16, are squares.
		

Crossrefs

Cf. A001694, A005361 (number of coreful divisors), A046951 (number of square divisors), A325837.

Programs

  • Mathematica
    f[p_, e_] := Floor[e/2]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = vecprod(apply(x -> x\2, factor(n)[, 2]));

Formula

Multiplicative with a(p^e) = floor(e/2).
a(n) > 0 if and only if n is a powerful number (A001694).
Dirichlet g.f.: zeta(s) * zeta(2*s) * Product_{p prime} (1 - 1/p^s + 1/p^(3*s)).
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