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.

Showing 1-9 of 9 results.

A168266 A003557(A168264(n)).

Original entry on oeis.org

1, 1, 2, 1, 2, 4, 1, 2, 4, 6, 1, 2, 4, 6, 8, 1, 2, 4, 6, 8, 12, 1, 2, 4, 6, 8, 12, 1, 2, 4, 6, 8, 12, 1, 2, 4, 6, 8, 12, 1, 2, 4, 6, 8, 12, 24, 1, 2, 4, 6, 8, 12, 24, 1, 2, 4, 6, 8, 12, 24, 36, 1, 2, 4, 6, 8, 12, 24, 36, 1, 2, 4, 6, 8, 12, 24, 36, 1, 2, 4, 6, 8, 12, 24, 36, 1, 2, 4, 6, 8, 12, 24, 36, 48
Offset: 1

Views

Author

Matthew Vandermast, Nov 23 2009

Keywords

Crossrefs

For range of values, see A168267.

Formula

Also A111701(A168264(n)).

A025487 Least integer of each prime signature A124832; also products of primorial numbers A002110.

Original entry on oeis.org

1, 2, 4, 6, 8, 12, 16, 24, 30, 32, 36, 48, 60, 64, 72, 96, 120, 128, 144, 180, 192, 210, 216, 240, 256, 288, 360, 384, 420, 432, 480, 512, 576, 720, 768, 840, 864, 900, 960, 1024, 1080, 1152, 1260, 1296, 1440, 1536, 1680, 1728, 1800, 1920, 2048, 2160, 2304, 2310
Offset: 1

Views

Author

Keywords

Comments

All numbers of the form 2^k1*3^k2*...*p_n^k_n, where k1 >= k2 >= ... >= k_n, sorted.
A111059 is a subsequence. - Reinhard Zumkeller, Jul 05 2010
Choie et al. (2007) call these "Hardy-Ramanujan integers". - Jean-François Alcover, Aug 14 2014
The exponents k1, k2, ... can be read off Abramowitz & Stegun p. 831, column labeled "pi".
For all such sequences b for which it holds that b(n) = b(A046523(n)), the sequence which gives the indices of records in b is a subsequence of this sequence. For example, A002182 which gives the indices of records for A000005, A002110 which gives them for A001221 and A000079 which gives them for A001222. - Antti Karttunen, Jan 18 2019
The prime signature corresponding to a(n) is given in row n of A124832. - M. F. Hasler, Jul 17 2019

Examples

			The first few terms are 1, 2, 2^2, 2*3, 2^3, 2^2*3, 2^4, 2^3*3, 2*3*5, ...
		

Crossrefs

Subsequence of A055932, A191743, and A324583.
Cf. A085089, A101296 (left inverses).
Equals range of values taken by A046523.
Cf. A178799 (first differences), A247451 (squarefree kernel), A146288 (number of divisors).
Rearrangements of this sequence include A036035, A059901, A063008, A077569, A085988, A086141, A087443, A108951, A181821, A181822, A322827, A329886, A329887.
Cf. also array A124832 (row n = prime signature of a(n)) and A304886, A307056.

Programs

  • Haskell
    import Data.Set (singleton, fromList, deleteFindMin, union)
    a025487 n = a025487_list !! (n-1)
    a025487_list = 1 : h [b] (singleton b) bs where
       (_ : b : bs) = a002110_list
       h cs s xs'@(x:xs)
         | m <= x    = m : h (m:cs) (s' `union` fromList (map (* m) cs)) xs'
         | otherwise = x : h (x:cs) (s  `union` fromList (map (* x) (x:cs))) xs
         where (m, s') = deleteFindMin s
    -- Reinhard Zumkeller, Apr 06 2013
    
  • Maple
    isA025487 := proc(n)
        local pset,omega ;
        pset := sort(convert(numtheory[factorset](n),list)) ;
        omega := nops(pset) ;
        if op(-1,pset) <> ithprime(omega) then
            return false;
        end if;
        for i from 1 to omega-1 do
            if padic[ordp](n,ithprime(i)) < padic[ordp](n,ithprime(i+1)) then
                return false;
            end if;
        end do:
        true ;
    end proc:
    A025487 := proc(n)
        option remember ;
        local a;
        if n = 1 then
            1 ;
        else
            for a from procname(n-1)+1 do
                if isA025487(a) then
                    return a;
                end if;
            end do:
        end if;
    end proc:
    seq(A025487(n),n=1..100) ; # R. J. Mathar, May 25 2017
  • Mathematica
    PrimeExponents[n_] := Last /@ FactorInteger[n]; lpe = {}; ln = {1}; Do[pe = Sort@PrimeExponents@n; If[ FreeQ[lpe, pe], AppendTo[lpe, pe]; AppendTo[ln, n]], {n, 2, 2350}]; ln (* Robert G. Wilson v, Aug 14 2004 *)
    (* Second program: generate all terms m <= A002110(n): *)
    f[n_] := {{1}}~Join~
      Block[{lim = Product[Prime@ i, {i, n}],
       ww = NestList[Append[#, 1] &, {1}, n - 1], dec},
       dec[x_] := Apply[Times, MapIndexed[Prime[First@ #2]^#1 &, x]];
       Map[Block[{w = #, k = 1},
          Sort@ Prepend[If[Length@ # == 0, #, #[[1]]],
            Product[Prime@ i, {i, Length@ w}] ] &@ Reap[
             Do[
              If[# < lim,
                 Sow[#]; k = 1,
                 If[k >= Length@ w, Break[], k++]] &@ dec@ Set[w,
                 If[k == 1,
                   MapAt[# + 1 &, w, k],
                   PadLeft[#, Length@ w, First@ #] &@
                     Drop[MapAt[# + Boole[i > 1] &, w, k], k - 1] ]],
               {i, Infinity}] ][[-1]]
    ] &, ww]]; Sort[Join @@ f@ 13] (* Michael De Vlieger, May 19 2018 *)
  • PARI
    isA025487(n)=my(k=valuation(n,2),t);n>>=k;forprime(p=3,default(primelimit),t=valuation(n,p);if(t>k,return(0),k=t);if(k,n/=p^k,return(n==1))) \\ Charles R Greathouse IV, Jun 10 2011
    
  • PARI
    factfollow(n)={local(fm, np, n2);
      fm=factor(n); np=matsize(fm)[1];
      if(np==0,return([2]));
      n2=n*nextprime(fm[np,1]+1);
      if(np==1||fm[np,2]Franklin T. Adams-Watters, Dec 01 2011 */
    
  • PARI
    is(n) = {if(n==1, return(1)); my(f = factor(n));  f[#f~, 1] == prime(#f~) && vecsort(f[, 2],,4) == f[, 2]} \\ David A. Corneth, Feb 14 2019
    
  • PARI
    upto(Nmax)=vecsort(concat(vector(logint(Nmax,2),n,select(t->t<=Nmax,if(n>1,[factorback(primes(#p),Vecrev(p)) || p<-partitions(n)],[1,2]))))) \\ M. F. Hasler, Jul 17 2019
    
  • PARI
    \\ For fast generation of large number of terms, use this program:
    A283980(n) = {my(f=factor(n)); prod(i=1, #f~, my(p=f[i, 1], e=f[i, 2]); if(p==2, 6, nextprime(p+1))^e)}; \\ From A283980
    A025487list(e) = { my(lista = List([1, 2]), i=2, u = 2^e, t); while(lista[i] != u, if(2*lista[i] <= u, listput(lista,2*lista[i]); t = A283980(lista[i]); if(t <= u, listput(lista,t))); i++); vecsort(Vec(lista)); }; \\ Returns a list of terms up to the term 2^e.
    v025487 = A025487list(101);
    A025487(n) = v025487[n];
    for(n=1,#v025487,print1(A025487(n), ", ")); \\ Antti Karttunen, Dec 24 2019
    
  • Sage
    def sharp_primorial(n): return sloane.A002110(prime_pi(n))
    N = 2310
    nmax = 2^floor(log(N,2))
    sorted([j for j in (prod(sharp_primorial(t[0])^t[1] for k, t in enumerate(factor(n))) for n in (1..nmax)) if j <= N])
    # Giuseppe Coppoletta, Jan 26 2015

Formula

What can be said about the asymptotic behavior of this sequence? - Franklin T. Adams-Watters, Jan 06 2010
Hardy & Ramanujan prove that there are exp((2 Pi + o(1))/sqrt(3) * sqrt(log x/log log x)) members of this sequence up to x. - Charles R Greathouse IV, Dec 05 2012
From Antti Karttunen, Jan 18 & Dec 24 2019: (Start)
A085089(a(n)) = n.
A101296(a(n)) = n [which is the first occurrence of n in A101296, and thus also a record.]
A001221(a(n)) = A061395(a(n)) = A061394(n).
A007814(a(n)) = A051903(a(n)) = A051282(n).
a(A101296(n)) = A046523(n).
a(A306802(n)) = A002182(n).
a(n) = A108951(A181815(n)) = A329900(A181817(n)).
If A181815(n) is odd, a(n) = A283980(a(A329904(n))), otherwise a(n) = 2*a(A329904(n)).
(End)
Sum_{n>=1} 1/a(n) = Product_{n>=1} 1/(1 - 1/A002110(n)) = A161360. - Amiram Eldar, Oct 20 2020

Extensions

Offset corrected by Matthew Vandermast, Oct 19 2008
Minor correction by Charles R Greathouse IV, Sep 03 2010

A168263 For any m < n, and for all values of k, d(n^k) > d(m^k). (Let k, m, and n represent positive integers only.)

Original entry on oeis.org

1, 2, 4, 6, 12, 24, 60, 120, 180, 840, 1260, 1680, 27720
Offset: 1

Views

Author

Matthew Vandermast, Nov 23 2009

Keywords

Comments

d(n) is the number of divisors of n (A000005(n)).
All members must be highly composite numbers (A002182) with at least as many distinct prime factors as any smaller positive integer (A116998). (See Formula and Example sections.) It turns out that these two conditions are jointly sufficient.
Ramanujan proved that a) for any prime p, there exist a finite number of highly composite numbers with p as its largest prime factor; and b) in the canonical prime factorization of a highly composite number with largest prime factor p, the exponents for all primes > p are never smaller than they are in the factorization of A003418(p). (See formula 54 of the Ramanujan paper.)
It follows that, if the intersection of A003418 and A116998 is finite, so is the intersection of A002182 and A116998. For proof that the former intersection is finite, see A168262.
By using the given formula for the number of divisors, it is possible to define a canonical polynomial p_n(k) for every natural number n. For example, because 60 = (2^2)(3^1)(5^1), we define p_60(k) = (1+2k)(1+k)(1+k). The present sequence is defined only by examining whether p_n(k) achieves a record for natural numbers k, but the question could also be asked whether p_n(k) achieves a record for all k > 0. This stricter requirement does not hold for a(7)-a(13) at various positive values of k < 1, but it does hold for a(1)-a(6). The present sequence is "full", so a(1)-a(6) are the only numbers to satisfy the stronger property. - Hal M. Switkay, Aug 17 2025

Examples

			1) 1680 has more divisors than any smaller positive integer; thus for all m < n, d(1680^1) > d(m^1).
2) Since the exponents in 1680's prime factorization are (4,1,1,1), the k-th power of 1680 has (4k+1)(k+1)^3 = 4k^4 + 13k^3 + 15k^2 + 7k + 1 divisors. Comparison with the analogous formulas for all smaller members of A025487 shows the following:
a) No number smaller than 1680 has a positive coefficient in its "power formula" for any exponent larger than k^4.
b) The only power formula with a k^4 coefficient as high as 4 is that for 1260 (4k^4 + 12k^3 + 13k^2 + 6k + 1).
c) The k^3 coefficient for 1680 is higher than for 1260.
So for all sufficiently high values of k, d(1680^k) > d(m^k) for all m < 1680.
3) Careful comparison of 1680's "power formula" with the analogous formulas for smaller members of A025487 shows that no intermediate value of k can exist for which d(m^k) >= d(1680^k) if m < 1680.
		

References

  • S. Ramanujan, Highly composite numbers, Proc. Lond. Math. Soc. 14 (1915), 347-409; reprinted in Collected Papers, Ed. G. H. Hardy et al., Cambridge 1927; Chelsea, NY, 1962.

Crossrefs

Intersection of A002182 and A116998. Also, intersection of A002182 and A060735, and of A002182 and A168264. (A168264 is a subsequence of A060735, which is a subsequence of A116998.)

Formula

If the canonical factorization of n into prime powers is Product p^e(p), then the formula for the number of divisors of the k-th power of n is Product_p (ek + 1). (See also A146289, A146290.)
For two positive integers m and n with different prime signatures, let j be the largest exponent of k for which m and n have different coefficients, after the above formula for each integer is expanded as a polynomial. Let m_j and n_j denote the corresponding coefficients. d(n^k) > d(m^k) for all sufficiently high values of k if and only if n_j > m_j.

A168262 Intersection of A003418 and A116998.

Original entry on oeis.org

1, 2, 6, 12, 60, 420, 840, 27720, 360360, 5354228880
Offset: 1

Views

Author

Matthew Vandermast, Nov 23 2009

Keywords

Comments

If, for some prime p, A045948(p) > p^2, then all members of the sequence are less than A003418(p). (Let p_(n) be a prime for which the inequality is satisfied, and let p_(n+1) be the smallest prime > (p_(n))^2. No number smaller than A003418(p_(n+1)) can belong to this sequence. However, for any p_(n) that satisfies the inequality, so does p_(n+1), leading to an endless cycle.) This inequality is first satisfied at p=53, as A045948(53)=5040 > 53^2=2809.
Proof: It follows from the definitions of p_(n) and p_(n+1), and from Bertrand's Postulate, that 2(A045948(p_(n))) > 2((p_(n))^2) > p_(n+1). Therefore 2((A045948(p_(n)))^2 > (p_(n+1))^2.
Since any prime that divides A003418(p_(n)) divides A003418(p_(n+1)) at least twice as often, A045948(p_(n+1)) cannot be less than the product of (A045948(p_n))^2 and A034386(p_(n)). (The latter term greatly exceeds 2 for any actual p_(n).)
Therefore A045948(p_(n+1)) > 2((A045948(p_n))^2 > (p_(n+1))^2, and p_(n+1) satisfies the inequality, implying that no number smaller than A003418(p_(n+2)) can belong to this sequence.

Crossrefs

Also intersection of A003418 and A060735, and of A003418 and A168264. (A168264 is a subsequence of A060735, which is a subsequence of A116998.)
See also A001221, A168263.

A168265 a(n) = A003557(A060735(n)).

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 4, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
Offset: 1

Views

Author

Matthew Vandermast, Nov 23 2009

Keywords

Comments

A060735(n) belongs to A168264 if and only if a(n) belongs to A168267.
Looking at A060735 as an irregular triangle T(n,k) = k*A002110(n) with 1 <= k < prime(n+1), this sequence a(n) = k. - Michael De Vlieger, Jul 26 2016

Crossrefs

Programs

  • Mathematica
    Table[Range[Prime[n] - 1], {n, 9}] // Flatten (* or, per title definition: *)
    #/Times @@ (FactorInteger[#][[All, 1]]) & /@ Flatten@ Table[Range[Prime[n + 1] - 1] Apply[Times, Prime@ Range@ n], {n, 0, 8}] (* Michael De Vlieger, Jul 26 2016 *)

Formula

Integers 1 to A006093(1) inclusive, followed by integers 1 to A006093(2) inclusive, etc.
a(n) = A111701(A060735(n)).
T(n,k)=k for n >= 1 and 1 <= k < prime(n).

A168267 Range of values of A168266.

Original entry on oeis.org

1, 2, 4, 6, 8, 12, 24, 36, 48, 60, 72, 96, 120, 180, 240, 360, 480, 720, 840, 1080, 1260, 1440, 1680, 2160, 2520, 3360, 4320, 5040, 7560, 10080, 15120, 20160, 25200, 27720, 30240, 40320, 45360, 50400, 55440, 75600, 83160, 110880, 151200, 166320, 221760
Offset: 1

Views

Author

Matthew Vandermast, Nov 23 2009

Keywords

Crossrefs

Includes A002182. Subsequence of A140999.

A282472 Numbers k where records occur for d(k^2)/d(k), where d(k) is A000005(k).

Original entry on oeis.org

1, 2, 4, 6, 12, 24, 30, 60, 120, 180, 210, 420, 840, 1260, 2310, 4620, 9240, 13860, 27720, 30030, 60060, 120120, 180180, 360360, 510510, 1021020, 2042040, 3063060, 6126120, 9699690, 19399380, 38798760, 58198140, 116396280, 223092870, 446185740, 892371480
Offset: 1

Views

Author

Daniel Suteu, Feb 18 2017

Keywords

Comments

First 14 terms are similar, with A168264.
The quotients are (1, 3/2, 5/3, 9/4, 5/2, 21/8, 27/8, 15/4, 63/16, 25/6, 81/16, 45/8, 189/32, 25/4, 243/32, 135/16, 567/64, 75/8, 315/32, 729/64, 405/32, 1701/128, 225/16, 945/64, 2187/128, 1215/64, 5103/256, 675/32, 2835/128, 6561/256, 3645/128, 15309/512, 2025/64, 8505/256, 19683/512,...). - Lars Blomberg, Apr 10 2017

Crossrefs

Programs

  • Maple
    A282472 := proc(n)
        option remember;
        local a,a1,rec ;
        if n = 1 then
            1;
        else
            a1 := procname(n-1) ;
            rec := numtheory[tau](a1^2)/numtheory[tau](a1) ;
            for a from a1+1 do
                if numtheory[tau](a^2)/numtheory[tau](a) > rec then
                    return a;
                end if;
            end do:
        end if;
    end proc: # R. J. Mathar, Mar 03 2017
  • Mathematica
    s={}; rm=0; Do[r=DivisorSigma[0, n^2]/DivisorSigma[0, n]; If[r > rm, rm = r; AppendTo[s, n]], {n, 1, 10^4}]; s (* Amiram Eldar, Jul 17 2019 *)
  • PARI
    lista(nn) = {rec = 0; for (n=1, nn, if ((newrec = numdiv(n^2)/numdiv(n)) > rec, rec = newrec; print1(n, ", ")););} \\ Michel Marcus, Feb 20 2017
  • Perl
    use ntheory qw(:all);
    for (my ($n, $m) = (1, 0) ; ; ++$n) {
        my $d = divisors($n*$n) / divisors($n);
        if ($m < $d) {
            $m = $d;
            print "$n\n";
        }
    }
    

Extensions

a(32)-a(35) from Lars Blomberg, Apr 10 2017
a(36)-a(37) from Giovanni Resta, Apr 10 2017

A226959 a(0)=1 and a(n)=prime(n)*lcm(prime(n)-1, a(n-1)) for n > 0.

Original entry on oeis.org

1, 2, 6, 60, 420, 4620, 60060, 4084080, 232792560, 5354228880, 155272637520, 4813451763120, 178097715235440, 7302006324653040, 313986271960080720, 14757354782123793840, 782139803452561073520, 46146248403701103337680, 2814921152625767303598480
Offset: 0

Views

Author

Keywords

Comments

A226957(a(n)) = n.
Is this a subset of A168264? - Ralf Stephan, Jul 04 2013

Crossrefs

Programs

  • Mathematica
    a[0]=1; a[n_]:= a[n]=Prime[n]*LCM[a[n - 1],Prime[n] -1];
    Table[a[i], {i, 1, 18}]
  • PARI
    a(n)=if(n,my(v=primes(n),e=vector(n,i,1),f,j); forprime(p=2,v[n], f=factor(p-1); for(i=1,#f~,if(f[i,2]>1,j=primepi(f[i,1]);e[j]=max(e[j],f[i,2])))); prod(i=1,n,v[i]^e[i]),1) \\ Charles R Greathouse IV, Jun 28 2013

A384669 Positive integers setting a new record for the sum of the square roots of the prime exponents.

Original entry on oeis.org

1, 2, 4, 6, 12, 24, 30, 60, 120, 180, 210, 360, 420, 840, 1260, 1680, 2520, 3360, 4620, 6720, 7560, 9240, 13860, 18480, 27720, 36960, 55440, 73920, 83160, 110880, 120120, 180180, 221760, 240240, 360360, 480480, 720720, 960960, 1081080, 1441440, 2042040, 2882880, 4084080, 5765760, 6126120
Offset: 1

Views

Author

Hal M. Switkay, Jun 06 2025

Keywords

Comments

This sequence is the special case x = 1/2 of a class of sequences A_x indexed by real numbers x, defined as follows. Let k be a natural number with prime factorization k = Product_{i=1..r} (p_i)^(e_i), where p_i are distinct primes, and e_i are natural numbers. Define f_x(k) = Sum_{i=1..r} (e_i)^x. Then the sequence A_x consists of natural numbers k where f_x(k) sets a new record.
We may define g_k(x) = f_x(k), so that g_k consists of real functions indexed by natural numbers. Each g_k is a sum of exponential functions, whose bases are natural numbers.
We provide facts, conjectures, and questions about the class of sequences A_x.
Facts:
For all real x, A_x is an infinite, increasing sequence of numbers of least prime signature (A025487) starting with 1 and 2.
For all x <= 0, A_x coincides with the primorials (A002110). For all x >= 1, A_x coincides with the powers of 2 (A000079). Thus the interesting sequences A_x have 0 <= x <= 1.
For all x > 0, A_x(n+1) <= 2*A_x(n).
1, 2, 4, 6, 12, 24, and no other natural numbers, are terms of every A_x with 0 < x < 1.
Conjectures:
The sequences A_x appear to be distinct in the following sense. Given 0 <= x < y <= 1, not only are A_x and A_y distinct, their ranges appear to have finite intersection.
Define A_0+, respectively A_1- as the limit of A_x as x approaches 0 from the right, respectively as x approaches 1 from the left. Then A_0+ appears to coincide with A168264, and A_1- appears to coincide with A029744 without the term 3.
If 0 < x < 1 and d is a natural number, then d divides some term of A_x.
If 0 < x < 1 and d is a natural number, then d divides infinitely many terms of A_x.
If 0 < x < 1 and d is a natural number, then d divides all sufficiently large terms of A_x.
Define S_n to be the set of x for which A_x(n) achieves a minimum value as a function of x (well-defined because the values are integers). Then for fixed n >= 10, S_n appears to be a subset of the interval log(2)/log(3) <= x <= log(3)/log(4), although in general S_n itself is not an interval.
Questions:
Does every number of least prime signature appear in at least one A_x for some x? 216 is not a term in any A_x I have examined.
Are there any x for which A_x contains infinitely many highly composite numbers (A002182), respectively infinitely many deeply composite numbers (A095848)?
Does the sequence of sets S_n have any limit points x in [0,1]? If x is such a limit point, A_x would presumably grow more slowly than other A_x. Examining A_x for x rational with denominator 720, A_x appears to contain a maximum number of terms (80) less than 2 * 10! when x = 507/720, 511/720, 515/720, and 517/720, all between 0.70 and 0.72.
Answer to the first question above: No, 216 is not a term of A_x for any x. If it were, we would have g_216(x) > g_210(x) and g_216(x) > g_192(x), i.e., 2*3^x > 4 and 2*3^x > 6^x+1. The first inequality holds only if x > x0 = log(2)/log(3). But 2*3^x0 < 6^x0+1 and it is easily verified that the function 2*3^x - (6^x+1) is decreasing for x > x0, so the second equality cannot hold when x > x0. - Pontus von Brömssen, Jun 13 2025
Additional facts from Hal M. Switkay, Jun 29 2025: (Start)
If k1 and k2 have the same prime signature, then the functions g_k1(x) and g_k2(x) are identical. Hence g_k(x) is identical to g_A046523(k)(x), where A046523(k) is the smallest number with the same prime signature as k. This is why the terms of A_x are all numbers of least prime signature.
For all k, lim_{x->-oo} g_k(x) = A056169(k) = the number of unitary prime divisors of k.
For all k, g_k(0) = omega(k) = A001221(k) = the number of distinct prime divisors of k.
For all k, g_k(1) = bigomega(k) = A001222(k) = the number of prime divisors of k counted with multiplicity.
For all k > 1, lim_{x->oo} [log(g_k(x))/x] = log(A051903(k)), where A051903(k) = the maximum exponent in the prime factorization of k. When k has least prime signature, A051903(k) = A007814(k), the exponent of the largest power of 2 dividing k. (End)
Using the notation of A385722, this sequence is B_1. - Hal M. Switkay, Jul 27 2025

Examples

			f_(1/2)(24) = sqrt(3) + sqrt(1), because 24 = (2^3)(3^1). This is a record value for f_(1/2), so 24 is in the sequence. f_(1/2)(30) = sqrt(1) + sqrt(1) + sqrt(1) (because 30 = (2^1)(3^1)(5^1)), which is larger still, putting 30 in the sequence. However, f_(1/2)(32) = sqrt(5) (because 32 = 2^5), smaller than the previous value, so 32 is not in the sequence.
g_216(x) = 3^x + 3^x, because 216 = (2^3)(3^3).
		

Crossrefs

Programs

  • Mathematica
    s[n_] := Total[Sqrt[FactorInteger[n][[;; , 2]]]]; s[1] = 0; 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] (* Amiram Eldar, Jun 08 2025 *)
  • PARI
    s(n) = my(f=factor(n)); sum(k=1, #f~, sqrt(f[k,2]));
    lista(nn) = my(r=-oo, list=List()); for (n=1, nn, my(ss=s(n)); if (ss > r, r = ss; listput(list, n));); Vec(list); \\ Michel Marcus, Jun 15 2025
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