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

A101296 n has the a(n)-th distinct prime signature.

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 7, 2, 6, 2, 6, 4, 4, 2, 8, 3, 4, 5, 6, 2, 9, 2, 10, 4, 4, 4, 11, 2, 4, 4, 8, 2, 9, 2, 6, 6, 4, 2, 12, 3, 6, 4, 6, 2, 8, 4, 8, 4, 4, 2, 13, 2, 4, 6, 14, 4, 9, 2, 6, 4, 9, 2, 15, 2, 4, 6, 6, 4, 9, 2, 12, 7, 4, 2, 13, 4, 4, 4, 8, 2, 13, 4, 6, 4, 4, 4, 16, 2, 6, 6, 11, 2, 9, 2, 8, 9, 4, 2, 15, 2, 9, 4, 12, 2, 9, 4, 6, 6, 4, 4, 17
Offset: 1

Views

Author

David Wasserman, Dec 21 2004

Keywords

Comments

From Antti Karttunen, May 12 2017: (Start)
Restricted growth sequence transform of A046523, the least representative of each prime signature. Thus this partitions the natural numbers to the same equivalence classes as A046523, i.e., for all i, j: a(i) = a(j) <=> A046523(i) = A046523(j), and for that reason satisfies in that respect all the same conditions as A046523. For example, we have, for all i, j: if a(i) = a(j), then:
A000005(i) = A000005(j), A008683(i) = A008683(j), A286605(i) = A286605(j).
So, this sequence (instead of A046523) can be used for finding sequences where a(n)'s value is dependent only on the prime signature of n, that is, only on the multiset of prime exponents in the factorization of n. (End)
This is also the restricted growth sequence transform of many other sequences, for example, that of A181819. See further comments there. - Antti Karttunen, Apr 30 2022

Examples

			From _David A. Corneth_, May 12 2017: (Start)
1 has prime signature (), the first distinct prime signature. Therefore, a(1) = 1.
2 has prime signature (1), the second distinct prime signature after (1). Therefore, a(2) = 2.
3 has prime signature (1), as does 2. Therefore, a(3) = a(2) = 2.
4 has prime signature (2), the third distinct prime signature after () and (1). Therefore, a(4) = 3. (End)
From _Antti Karttunen_, May 12 2017: (Start)
Construction of restricted growth sequences: In this case we start with a(1) = 1 for A046523(1) = 1, and thereafter, for all n > 1, we use the least so far unused natural number k for a(n) if A046523(n) has not been encountered before, otherwise [whenever A046523(n) = A046523(m), for some m < n], we set a(n) = a(m).
For n = 2, A046523(2) = 2, which has not been encountered before (first prime), thus we allot for a(2) the least so far unused number, which is 2, thus a(2) = 2.
For n = 3, A046523(2) = 2, which was already encountered as A046523(1), thus we set a(3) = a(2) = 2.
For n = 4, A046523(4) = 4, not encountered before (first square of prime), thus we allot for a(4) the least so far unused number, which is 3, thus a(4) = 3.
For n = 5, A046523(5) = 2, as for the first time encountered at n = 2, thus we set a(5) = a(2) = 2.
For n = 6, A046523(6) = 6, not encountered before (first semiprime pq with distinct p and q), thus we allot for a(6) the least so far unused number, which is 4, thus a(6) = 4.
For n = 8, A046523(8) = 8, not encountered before (first cube of a prime), thus we allot for a(8) the least so far unused number, which is 5, thus a(8) = 5.
For n = 9, A046523(9) = 4, as for the first time encountered at n = 4, thus a(9) = 3.
(End)
From _David A. Corneth_, May 12 2017: (Start)
(Rough) description of an algorithm of computing the sequence:
Suppose we want to compute a(n) for n in [1..20].
We set up a vector of 20 elements, values 0, and a number m = 1, the minimum number we haven't checked and c = 0, the number of distinct prime signatures we've found so far.
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
We check the prime signature of m and see that it's (). We increase c with 1 and set all elements up to 20 with prime signature () to 1. In the process, we adjust m. This gives:
[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]. The least number we haven't checked is m = 2. 2 has prime signature (1). We increase c with 1 and set all elements up to 20 with prime signature (1) to 2. In the process, we adjust m. This gives:
[1, 2, 2, 0, 2, 0, 2, 0, 0, 0, 2, 0, 2, 0, 0, 0, 2, 0, 2, 0]
We check the prime signature of m = 4 and see that its prime signature is (2). We increase c with 1 and set all numbers up to 20 with prime signature (2) to 3. This gives:
[1, 2, 2, 3, 2, 0, 2, 0, 3, 0, 2, 0, 2, 0, 0, 0, 2, 0, 2, 0]
Similarily, after m = 6, we get
[1, 2, 2, 3, 2, 4, 2, 0, 3, 4, 2, 0, 2, 4, 4, 0, 2, 0, 2, 0], after m = 8 we get:
[1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 0, 2, 4, 4, 0, 2, 0, 2, 0], after m = 12 we get:
[1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 0, 2, 6, 2, 0], after m = 16 we get:
[1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 7, 2, 6, 2, 0], after m = 20 we get:
[1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 7, 2, 6, 2, 8]. Now, m > 20 so we stop. (End)
The above method is inefficient, because the step "set all elements a(n) up to n = Nmax with prime signature s(n) = S[c] to c" requires factoring all integers up to Nmax (or at least comparing their signature, once computed, with S[c]) again and again. It is much more efficient to run only once over each m = 1..Nmax, compute its prime signature s(m), add it to an ordered list in case it did not occur earlier, together with its "rank" (= new size of the list), and assign that rank to a(m). The list of prime signatures is much shorter than [1..Nmax]. One can also use m'(m) := the smallest n with the prime signature of m (which is faster to compute than to search for the signature) as representative for s(m), and set a(m) := a(m'(m)). Then it is sufficient to have just one counter (number of prime signatures seen so far) as auxiliary variable, in addition to the sequence to be computed. - _M. F. Hasler_, Jul 18 2019
		

Crossrefs

Cf. A025487, A046523, A064839 (ordinal transform of this sequence), A181819, and arrays A095904, A179216.
Sequences that are unions of finite number (>= 2) of equivalence classes determined by the values that this sequence obtains (i.e., sequences mentioned in David A. Corneth's May 12 2017 formula): A001358 (A001248 U A006881, values 3 & 4), A007422 (values 1, 4, 5), A007964 (2, 3, 4, 5), A014612 (5, 6, 9), A030513 (4, 5), A037143 (1, 2, 3, 4), A037144 (1, 2, 3, 4, 5, 6, 9), A080258 (6, 7), A084116 (2, 4, 5), A167171 (2, 4), A217856 (6, 9).
Cf. also A077462, A305897 (stricter variants, with finer partitioning) and A254524, A286603, A286605, A286610, A286619, A286621, A286622, A286626, A286378 for other similarly constructed sequences.

Programs

  • Maple
    A101296 := proc(n)
        local a046523, a;
        a046523 := A046523(n) ;
        for a from 1 do
            if A025487(a) = a046523 then
                return a;
            elif A025487(a) > a046523 then
                return -1 ;
            end if;
        end do:
    end proc: # R. J. Mathar, May 26 2017
  • Mathematica
    With[{nn = 120}, Function[s, Table[Position[Keys@s, k_ /; MemberQ[k, n]][[1, 1]], {n, nn}]]@ Map[#1 -> #2 & @@ # &, Transpose@ {Values@ #, Keys@ #}] &@ PositionIndex@ Table[Times @@ MapIndexed[Prime[First@ #2]^#1 &, Sort[FactorInteger[n][[All, -1]], Greater]] - Boole[n == 1], {n, nn}] ] (* Michael De Vlieger, May 12 2017, Version 10 *)
  • PARI
    find(ps, vps) = {for (k=1, #vps, if (vps[k] == ps, return(k)););}
    lisps(nn) = {vps = []; for (n=1, nn, ps = vecsort(factor(n)[,2]); ips = find(ps, vps); if (! ips, vps = concat(vps, ps); ips = #vps); print1(ips, ", "););} \\ Michel Marcus, Nov 15 2015; edited by M. F. Hasler, Jul 16 2019
    
  • PARI
    rgs_transform(invec) = { my(occurrences = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(occurrences,invec[i]), my(pp = mapget(occurrences, invec[i])); outvec[i] = outvec[pp] , mapput(occurrences,invec[i],i); outvec[i] = u; u++ )); outvec; };
    write_to_bfile(start_offset,vec,bfilename) = { for(n=1, length(vec), write(bfilename, (n+start_offset)-1, " ", vec[n])); }
    write_to_bfile(1,rgs_transform(vector(100000,n,A046523(n))),"b101296.txt");
    \\ Antti Karttunen, May 12 2017

Formula

A025487(a(n)) = A046523(n).
Indices of records give A025487. - Michel Marcus, Nov 16 2015
From David A. Corneth, May 12 2017: (Start) [Corresponding characteristic function in brackets]
a(A000012(n)) = 1 (sig.: ()). [A063524]
a(A000040(n)) = 2 (sig.: (1)). [A010051]
a(A001248(n)) = 3 (sig.: (2)). [A302048]
a(A006881(n)) = 4 (sig.: (1,1)). [A280710]
a(A030078(n)) = 5 (sig.: (3)).
a(A054753(n)) = 6 (sig.: (1,2)). [A353472]
a(A030514(n)) = 7 (sig.: (4)).
a(A065036(n)) = 8 (sig.: (1,3)).
a(A007304(n)) = 9 (sig.: (1,1,1)). [A354926]
a(A050997(n)) = 10 (sig.: (5)).
a(A085986(n)) = 11 (sig.: (2,2)).
a(A178739(n)) = 12 (sig.: (1,4)).
a(A085987(n)) = 13 (sig.: (1,1,2)).
a(A030516(n)) = 14 (sig.: (6)).
a(A143610(n)) = 15 (sig.: (2,3)).
a(A178740(n)) = 16 (sig.: (1,5)).
a(A189975(n)) = 17 (sig.: (1,1,3)).
a(A092759(n)) = 18 (sig.: (7)).
a(A189988(n)) = 19 (sig.: (2,4)).
a(A179643(n)) = 20 (sig.: (1,2,2)).
a(A189987(n)) = 21 (sig.: (1,6)).
a(A046386(n)) = 22 (sig.: (1,1,1,1)).
a(A162142(n)) = 23 (sig.: (2,2,2)).
a(A179644(n)) = 24 (sig.: (1,1,4)).
a(A179645(n)) = 25 (sig.: (8)).
a(A179646(n)) = 26 (sig.: (2,5)).
a(A163569(n)) = 27 (sig.: (1,2,3)).
a(A179664(n)) = 28 (sig.: (1,7)).
a(A189982(n)) = 29 (sig.: (1,1,1,2)).
a(A179666(n)) = 30 (sig.: (3,4)).
a(A179667(n)) = 31 (sig.: (1,1,5)).
a(A179665(n)) = 32 (sig.: (9)).
a(A189990(n)) = 33 (sig.: (2,6)).
a(A179669(n)) = 34 (sig.: (1,2,4)).
a(A179668(n)) = 35 (sig.: (1,8)).
a(A179670(n)) = 36 (sig.: (1,1,1,3)).
a(A179671(n)) = 37 (sig.: (3,5)).
a(A162143(n)) = 38 (sig.: (2,2,2)).
a(A179672(n)) = 39 (sig.: (1,1,6)).
a(A030629(n)) = 40 (sig.: (10)).
a(A179688(n)) = 41 (sig.: (1,3,3)).
a(A179689(n)) = 42 (sig.: (2,7)).
a(A179690(n)) = 43 (sig.: (1,1,2,2)).
a(A189991(n)) = 44 (sig.: (4,4)).
a(A179691(n)) = 45 (sig.: (1,2,5)).
a(A179692(n)) = 46 (sig.: (1,9)).
a(A179693(n)) = 47 (sig.: (1,1,1,4)).
a(A179694(n)) = 48 (sig.: (3,6)).
a(A179695(n)) = 49 (sig.: (2,2,3)).
a(A179696(n)) = 50 (sig.: (1,1,7)).
(End)

Extensions

Data section extended to 120 terms by Antti Karttunen, May 12 2017
Minor edits/corrections by M. F. Hasler, Jul 18 2019

A179643 Products of exactly 2 distinct squares of primes and a different prime (p^2 * q^2 * r).

Original entry on oeis.org

180, 252, 300, 396, 450, 468, 588, 612, 684, 700, 828, 882, 980, 1044, 1100, 1116, 1300, 1332, 1452, 1476, 1548, 1575, 1692, 1700, 1900, 1908, 2028, 2124, 2156, 2178, 2196, 2205, 2300, 2412, 2420, 2450, 2475, 2548, 2556, 2628, 2844, 2900, 2925, 2988
Offset: 1

Views

Author

Keywords

Comments

A050326(a(n)) = 5, subsequence of A225228. - Reinhard Zumkeller, May 03 2013

Examples

			180 = 2^2 * 3^2 * 5, 252 = 2^2 * 3^2 * 7, 300 = 2^2 * 3 * 5^2, ...
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=Sort[Last/@FactorInteger[n]]=={1,2,2}; Select[Range[3000], f]
  • PARI
    list(lim)=my(v=List(),t);forprime(p=2,sqrt(lim\12),forprime(q=p+1,sqrt(lim\p^2\2),t=(p*q)^2;forprime(r=2,lim\t,if(p==r||q==r,next);listput(v,t*r))));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jul 19 2011
    
  • Python
    from math import isqrt
    from sympy import primepi, primerange, integer_nthroot
    def A179643(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x): return n+x+sum((t:=primepi(s:=isqrt(y:=isqrt(x//r))))+(t*(t-1)>>1)-sum(primepi(y//k) for k in primerange(1, s+1)) for r in primerange(x+1))+sum(primepi(isqrt(x//p**3)) for p in primerange(integer_nthroot(x,3)[0]+1))-primepi(integer_nthroot(x,5)[0])
        return bisection(f,n,n) # Chai Wah Wu, Mar 27 2025

A030628 1 together with numbers of the form p*q^4 and p^9, where p and q are distinct primes.

Original entry on oeis.org

1, 48, 80, 112, 162, 176, 208, 272, 304, 368, 405, 464, 496, 512, 567, 592, 656, 688, 752, 848, 891, 944, 976, 1053, 1072, 1136, 1168, 1250, 1264, 1328, 1377, 1424, 1539, 1552, 1616, 1648, 1712, 1744, 1808, 1863, 1875, 2032, 2096, 2192, 2224, 2349, 2384
Offset: 1

Views

Author

Keywords

Comments

Also 1 together with numbers with 10 divisors. Also numbers n such that product of all proper divisors of n equals n^4.
If M(n) denotes the product of all divisors of n, then n is said to be k-multiplicatively perfect if M(n)=n^k. All such numbers are of the form p*q^(k-1) or p^(2k-1). The sequence A030628 is therefore 5-multiplicatively perfect. See the Links for A007422. - Walter Kehowski, Sep 13 2005

References

  • D. M. Burton, Elementary Number Theory, Allyn and Bacon Inc., Boston, MA, 1976. p. 119.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, entry for 48, page 106, 1997.

Crossrefs

Programs

  • Maple
    with(numtheory): k:=5: MPL:=[]: for z from 1 to 1 do for n from 1 to 5000 do if convert(divisors(n),`*`) = n^k then MPL:=[op(MPL),n] fi od; od; MPL; # Walter Kehowski, Sep 13 2005
  • Mathematica
    Join[{1},Select[Range[6000],DivisorSigma[0,#]==10&]] (* Vladimir Joseph Stephan Orlovsky, May 05 2011 *)
    Select[Range[2500],Times@@Most[Divisors[#]]==#^4&] (* Harvey P. Dale, Nov 04 2024 *)
  • PARI
    {v=[]; for(n=1,500,v=concat(v, if(numdiv(n)==10,n,",")); ); v} \\ Jason Earls, Jun 18 2001
    
  • PARI
    list(lim)=my(v=List([1]), t); forprime(p=2, (lim\2+.5)^(1/4), t=p^4; forprime(q=2, lim\t, if(p==q, next); listput(v, t*q))); forprime(p=2,(lim+.5)^(1/9),listput(v,p^9)); vecsort(Vec(v)) \\ Charles R Greathouse IV, Apr 26 2012
    
  • Python
    from sympy import primepi, primerange, integer_nthroot
    def A030628(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x): return n-1+x-sum(primepi(x//p**4) for p in primerange(integer_nthroot(x,4)[0]+1))+primepi(integer_nthroot(x,5)[0])-primepi(integer_nthroot(x,9)[0])
        return bisection(f,n,n) # Chai Wah Wu, Feb 21 2025

Formula

Union A178739 U A179665 {1}. - R. J. Mathar, Apr 03 2011

Extensions

Better description from Sharon Sela (sharonsela(AT)hotmail.com), Dec 23 2001
More terms from Walter Kehowski, Sep 13 2005

A179644 Product of the 4th power of a prime and 2 different distinct primes (p^4*q*r).

Original entry on oeis.org

240, 336, 528, 560, 624, 810, 816, 880, 912, 1040, 1104, 1134, 1232, 1360, 1392, 1456, 1488, 1520, 1776, 1782, 1840, 1904, 1968, 2064, 2106, 2128, 2256, 2288, 2320, 2480, 2544, 2576, 2754, 2832, 2835, 2928, 2960, 2992, 3078, 3216, 3248, 3280, 3344, 3408
Offset: 1

Views

Author

Keywords

Examples

			240=2^4*3*5,336=2^4*3*7,..810=2^3^4*5,..
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=Sort[Last/@FactorInteger[n]]=={1,1,4}; Select[Range[4000], f]
    Take[Union[#[[1]]^4 #[[2]]#[[3]]&/@(Flatten[Permutations/@ Subsets[ Prime[ Range[ 20]],{3}],1])],50] (* Harvey P. Dale, Feb 07 2013 *)
  • PARI
    list(lim)=my(v=List(),t);forprime(p=2,(lim\6)^(1/4),forprime(q=2,sqrt(lim\p^4),if(p==q,next);t=p^4*q;forprime(r=q+1,lim\t,if(p==r,next);listput(v,t*r))));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jul 19 2011
    
  • Python
    from math import isqrt
    from sympy import primepi, primerange, integer_nthroot
    def A179644(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x): return n+x+sum((t:=primepi(s:=isqrt(y:=x//r**4)))+(t*(t-1)>>1)-sum(primepi(y//k) for k in primerange(1, s+1)) for r in primerange(integer_nthroot(x,4)[0]+1))+sum(primepi(x//p**5) for p in primerange(integer_nthroot(x,5)[0]+1))-primepi(integer_nthroot(x,6)[0])
        return bisection(f,n,n) # Chai Wah Wu, Mar 27 2025

A178740 Product of the 5th power of a prime (A050997) and a different prime (p^5*q).

Original entry on oeis.org

96, 160, 224, 352, 416, 486, 544, 608, 736, 928, 992, 1184, 1215, 1312, 1376, 1504, 1696, 1701, 1888, 1952, 2144, 2272, 2336, 2528, 2656, 2673, 2848, 3104, 3159, 3232, 3296, 3424, 3488, 3616, 4064, 4131, 4192, 4384, 4448, 4617, 4768, 4832, 5024, 5216
Offset: 1

Views

Author

Will Nicholes, Jun 08 2010

Keywords

Comments

Subsequence of A030630, integers whose number of divisors is 12. - Michel Marcus, Nov 11 2015

Crossrefs

Programs

  • Mathematica
    f[n_]:=Sort[Last/@FactorInteger[n]]=={1,5};Select[Range[6000],f] (* Vladimir Joseph Stephan Orlovsky, May 03 2011 *)
    With[{nn=50},Take[Union[Flatten[{#[[1]]^5 #[[2]],#[[1]]#[[2]]^5}&/@Subsets[ Prime[ Range[nn]],{2}]]],nn]] (* Harvey P. Dale, Mar 18 2013 *)
  • PARI
    list(lim)=my(v=List(), t); forprime(p=2, (lim\2)^(1/5), t=p^5; forprime(q=2, lim\t, if(p==q, next); listput(v, t*q))); vecsort(Vec(v)) \\ Altug Alkan, Nov 11 2015
    
  • PARI
    isok(n)=my(f=factor(n)[, 2]); f==[5, 1]~||f==[1, 5]~
    for(n=1, 1e4, if(isok(n), print1(n,", "))) \\ Altug Alkan, Nov 11 2015
    
  • Python
    from sympy import primepi, primerange, integer_nthroot
    def A178740(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x): return n+x-sum(primepi(x//p**5) for p in primerange(integer_nthroot(x,5)[0]+1))+primepi(integer_nthroot(x,6)[0])
        return bisection(f,n,n) # Chai Wah Wu, Mar 27 2025

A179668 Products of the 8th power of a prime and a distinct prime (p^8*q).

Original entry on oeis.org

768, 1280, 1792, 2816, 3328, 4352, 4864, 5888, 7424, 7936, 9472, 10496, 11008, 12032, 13122, 13568, 15104, 15616, 17152, 18176, 18688, 20224, 21248, 22784, 24832, 25856, 26368, 27392, 27904, 28928, 32512, 32805, 33536, 35072, 35584, 38144, 38656, 40192
Offset: 1

Views

Author

Keywords

Crossrefs

Programs

  • Mathematica
    f[n_]:=Sort[Last/@FactorInteger[n]]=={1,8}; Select[Range[40000], f]
    With[{nn=40},Take[Union[#[[1]]^8 #[[2]]&/@Flatten[Permutations/@Subsets[ Prime[Range[nn]],{2}],1]],nn]] (* Harvey P. Dale, Jan 20 2016 *)
  • PARI
    list(lim)=my(v=List(),t);forprime(p=2,(lim\2)^(1/8),t=p^8;forprime(q=2,lim\t,if(p==q,next);listput(v,t*q)));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jul 20 2011
    
  • Python
    from sympy import primepi, primerange, integer_nthroot
    def A179668(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x): return n+x-sum(primepi(x//p**8) for p in primerange(integer_nthroot(x,8)[0]+1))+primepi(integer_nthroot(x,9)[0])
        return bisection(f,n,n) # Chai Wah Wu, Feb 21 2025

A179692 Numbers of the form p^9*q where p and q are distinct primes.

Original entry on oeis.org

1536, 2560, 3584, 5632, 6656, 8704, 9728, 11776, 14848, 15872, 18944, 20992, 22016, 24064, 27136, 30208, 31232, 34304, 36352, 37376, 39366, 40448, 42496, 45568, 49664, 51712, 52736, 54784, 55808, 57856, 65024, 67072, 70144, 71168, 76288, 77312, 80384, 83456
Offset: 1

Views

Author

Keywords

Crossrefs

Programs

  • Mathematica
    f[n_]:=Sort[Last/@FactorInteger[n]]=={1,9}; Select[Range[90000], f]
  • PARI
    list(lim)=my(v=List(),t);forprime(p=2, (lim\2)^(1/9), t=p^9;forprime(q=2, lim\t, if(p==q, next);listput(v,t*q))); vecsort(Vec(v)) \\ Charles R Greathouse IV, Jul 24 2011
    
  • Python
    from sympy import primepi, integer_nthroot, primerange
    def A179692(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x): return n+x-sum(primepi(x//p**9) for p in primerange(integer_nthroot(x,9)[0]+1))+primepi(integer_nthroot(x,10)[0])
        return bisection(f,n,n) # Chai Wah Wu, Feb 21 2025

A189975 Numbers with prime factorization pqr^3 for distinct p, q, r.

Original entry on oeis.org

120, 168, 264, 270, 280, 312, 378, 408, 440, 456, 520, 552, 594, 616, 680, 696, 702, 728, 744, 750, 760, 888, 918, 920, 945, 952, 984, 1026, 1032, 1064, 1128, 1144, 1160, 1240, 1242, 1272, 1288, 1416, 1464, 1480, 1485, 1496, 1566, 1608, 1624, 1640, 1672
Offset: 1

Views

Author

Keywords

Crossrefs

Programs

  • Mathematica
    f[n_]:=Sort[Last/@FactorInteger[n]]=={1,1,3}; Select[Range[2000],f]
  • PARI
    list(lim)=my(v=List(),t);forprime(p=2,(lim\6)^(1/3),forprime(q=2,sqrt(lim\p^3),if(p==q,next);t=p^3*q;forprime(r=q+1,lim\t,if(p==r,next);listput(v,t*r))));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jul 19 2011
    
  • Python
    from math import isqrt
    from sympy import primepi, primerange, integer_nthroot
    def A189975(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x): return n+x+sum((t:=primepi(s:=isqrt(y:=x//r**3)))+(t*(t-1)>>1)-sum(primepi(y//k) for k in primerange(1, s+1)) for r in primerange(integer_nthroot(x,3)[0]+1))+sum(primepi(x//p**4) for p in primerange(integer_nthroot(x,4)[0]+1))-primepi(integer_nthroot(x,5)[0])
        return bisection(f,n,n) # Chai Wah Wu, Mar 27 2025

A058061 Number of prime factors (counted with multiplicity) of d(n), the number of divisors of n.

Original entry on oeis.org

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

Views

Author

Labos Elemer, Nov 23 2000

Keywords

Comments

From Bernard Schott, Mar 24 2020: (Start)
a(n) = 1 iff n = p^(q-1) with p, q primes (A009087).
a(n) = 2 if n=p*q with p, q primes (A006881), or if n=p^2*q with p, q primes (A054753), or if n=p^4*q with p, q primes (A178739), or if n=p^6*q with p, q primes (A189987), or if n=p^2*q^4 with p, q primes (A189988), or if n=p^(m-1) with p prime and m is semiprime in A001358 (not exhaustive). (End)

Examples

			For n=120, d(120)=16, a(120)=4.
		

Crossrefs

Cf. A001222, A000005, A058060, A079057 (partial sums).

Programs

  • Mathematica
    Table[PrimeOmega@ DivisorSigma[0, n], {n, 120}] (* Michael De Vlieger, Feb 18 2017 *)
  • PARI
    a(n) = bigomega(numdiv(n)); \\ Michel Marcus, Dec 14 2013

Formula

a(n) = A001222(A000005(n)).
Additive with a(p^e) = A001222(e+1). - Amiram Eldar, Jan 15 2024
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