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 1-3 of 3 results.

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

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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

A168264 For all sufficiently high values of k, d(n^k) > d(m^k) for all m < n. (Let k, m, and n represent positive integers only.)

Original entry on oeis.org

1, 2, 4, 6, 12, 24, 30, 60, 120, 180, 210, 420, 840, 1260, 1680, 2310, 4620, 9240, 13860, 18480, 27720, 30030, 60060, 120120, 180180, 240240, 360360, 510510, 1021020, 2042040, 3063060, 4084080, 6126120, 9699690, 19399380, 38798760, 58198140
Offset: 1

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Author

Matthew Vandermast, Nov 23 2009

Keywords

Comments

d(n) is the number of divisors of n (A000005(n)).

Examples

			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.
		

Crossrefs

Subsequence of A025487, A060735, A116998. Includes A002110, A168262, A168263.
See also A168265, A168266, A168267.

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.

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.
Showing 1-3 of 3 results.