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.

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

A157704 G.f.s of the z^p coefficients of the polynomials in the GF3 denominators of A156927.

Original entry on oeis.org

1, 1, 5, 32, 186, 132, 10, 56, 2814, 17834, 27324, 11364, 1078, 10, 48, 17988, 494720, 3324209, 7526484, 6382271, 2004296, 203799, 4580, 5, 16, 72210, 7108338, 146595355, 1025458635, 2957655028, 3828236468
Offset: 0

Views

Author

Johannes W. Meijer, Mar 07 2009

Keywords

Comments

The formula for the PDGF3(z;n) polynomials in the GF3 denominators of A156927 can be found below.
The general structure of the GFKT3(z;p) that generate the z^p coefficients of the PDGF3(z; n) polynomials can also be found below. The KT3(z;p) polynomials in the numerators of the GFKT3(z; p) have a nice symmetrical structure.
The sequence of the number of terms of the first few KT3(z;p) polynomials is 1, 2, 4, 7, 10, 13, 14, 17, 20, 23, 26, 29, 32, 34, 36, 39, 42. The differences of this sequence and that of the number of terms of the KT4(z;p), see A157705, follow a simple pattern.
A Maple algorithm that generates relevant GFKT3(z;p) information can be found below.

Examples

			Some PDGF3 (z;n) are:
  PDGF3(z;n=3) = (1-z)*(1-2*z)^4*(1-3*z)^7*(1-4*z)^10
  PDGF3(z;n=4) = (1-z)*(1-2*z)^4*(1-3*z)^7*(1-4*z)^10*(1-5*z)^13
The first few GFKT3's are:
  GFKT3(z;p=0) = 1/(1-z)
  GFKT3(z;p=1) = -(5*z+1)/(1-z)^4
  GFKT3(z;p=2) = z*(32+186*z+132*z^2+10*z^3)/(1-z)^7
Some KT3(z,p) polynomials are:
  KT3(z;p=2) = 32+186*z+132*z^2+10*z^3
  KT3(z;p=3) = 56+2814*z+17834*z^2+27324*z^3+11364*z^4+1078*z^5+10*z^6
		

Crossrefs

Originator sequence A156927.
See A002414 for the z^1 coefficients and A157707 for the z^2 coefficients divided by 2.
Row sums equal A064350 and those of A157705.

Programs

  • Maple
    p:=2; fn:=sum((-1)^(n1+1)*binomial(3*p+1,n1) *a(n-n1),n1=1..3*p+1): fk:=rsolve(a(n) = fn,a(k)): for n2 from 0 to 3*p+1 do fz(n2):=product((1-(k+1)*z)^(1+3*k), k=0..n2): a(n2):= coeff(fz(n2),z,p); end do: b:=n-> a(n): seq(b(n), n=0..3*p+1); a(n)=fn; a(k)=sort(simplify(fk)); GFKT3(p):=sum((fk)*z^k, k=0..infinity); q3:=ldegree((numer(GFKT3(p)))): KT3(p):=sort((-1)^(p)*simplify((GFKT3(p)*(1-z)^(3*p+1))/z^q3),z, ascending);

Formula

PDGF3(z;n) = Product_{k=0..n} (1-(k+1)*z)^(1+3*k) with n = 1, 2, 3, ...
GFKT3(z;p) = (-1)^(p)*(z^q3)*KT3(z, p)/(1-z)^(3*p+1) with p = 0, 1, 2, ...
The recurrence relation for the z^p coefficients a(n) is a(n) = Sum_{k=1..3*p+1} (-1)^(k+1)*binomial(3*p + 1, k)*a(n-k) with p = 0, 1, 2, ... .

A157705 G.f.s of the z^p coefficients of the polynomials in the GF4 denominators of A156933.

Original entry on oeis.org

1, 1, 3, 2, 18, 128, 171, 42, 1, 22, 1348, 11738, 26293, 17693, 3271, 115, 13, 6122, 228986, 2070813, 6324083, 7397855, 3361536, 544247, 24590, 155, 3, 17248, 2413434, 67035224, 612026240, 2274148882
Offset: 0

Views

Author

Johannes W. Meijer, Mar 07 2009

Keywords

Comments

The formula for the PDGF4(z;n) polynomials in the GF4 denominators of A156933 can be found below.
The general structure of the GFKT4(z;p) that generate the z^p coefficients of the PDGF4(z;n) polynomials can also be found below. The KT4(z;p) polynomials in the numerators of the GFKT4(z;p) have a nice symmetrical structure.
The sequence of the number of terms of the first few KT4(z;p) polynomials is 1, 3, 5, 7, 10, 13, 15, 18, 20, 23, 26, 29, 32, 34, 37, 40, 42. The differences of this sequence and that of the number of terms of the KT3(z;p), see A157704, follow a simple pattern.
A Maple algorithm that generates relevant GFKT4(z;p) information can be found below.

Examples

			Some PDGF4 (z;n) are:
  PDGF4(z; n=3) = (1-7*z)*(1-5*z)^4*(1-3*z)^7*(1-z)^10
  PDGF4(z; n=4) = (1-9*z)*(1-7*z)^4*(1-5*z)^7*(1-3*z)^10*(1-z)^13
The first few GFKT4's are:
  GFKT4(z;p=0) = 1/(1-z)
  GFKT4(z;p=1) = -(1+3*z+2*z^2)/(1-z)^4
  GFKT4(z;p=2) = z*(18+128*z+171*z^2+42*z^3+z^4)/(1-z)^7
Some KT4(z,p) polynomials are:
  KT4(z;p=2) = 18+128*z+171*z^2+42*z^3+z^4
  KT4(z;p=3) = 22+1348*z+11738*z^2+26293*z^3+17693*z^4+3271*z^5+115*z^6
		

Crossrefs

Originator sequence A156933.
See A081436 for the z^1 coefficients and A157708 for the z^2 coefficients.
Row sums equal A064350 and those of A157704.

Programs

  • Maple
    p:=2; fn:=sum((-1)^(n1+1)*binomial(3*p+1,n1) *a(n-n1),n1=1..3*p+1): fk:=rsolve(a(n) = fn,a(k)): for n2 from 0 to 3*p+1 do fz(n2):=product((1-(2*n2+1-(2*k))*z)^(3*k+1), k=0..n2): a(n2):= coeff(fz(n2),z,p): end do: b:=n-> a(n): seq(b(n), n=0..3*p+1); a(n)=fn; a(k)= sort (simplify(fk)); GFKT4(p):=sum((fk)*z^k,k=0..infinity); q4:=ldegree((numer (GFKT4(p)))): KT4(p):=sort((-1)^(p)*simplify((GFKT4(p)*(1-z)^(3*p+1))/z^q4),z, ascending);

Formula

PDGF4(z;n) = Product_{k=0..n} (1-(2*n+1-2*k)*z)^(3*k+1) with n = 1, 2, 3, ...
GFKT4(z;p) = (-1)^(p)*(z^q4)*KT4(z, p)/(1-z)^(3*p+1) with p = 0, 1, 2, ...
The recurrence relation for the z^p coefficients a(n) is a(n) = Sum_{k=1..3*p+1} (-1)^(k+1)*binomial(3*p + 1, k)*a(n-k) with p = 0, 1, 2, ... .

A250270 Products of terms of A003418.

Original entry on oeis.org

1, 2, 4, 6, 8, 12, 16, 24, 32, 36, 48, 60, 64, 72, 96, 120, 128, 144, 192, 216, 240, 256, 288, 360, 384, 420, 432, 480, 512, 576, 720, 768, 840, 864, 960, 1024, 1152, 1296, 1440, 1536, 1680, 1728, 1920, 2048, 2160, 2304, 2520, 2592, 2880, 3072, 3360, 3456
Offset: 1

Views

Author

Matthew Vandermast, Dec 16 2014

Keywords

Comments

Includes all factorials and Jordan-Polya numbers, since n! = Product_{i = 1..n} A003418(floor(n/i)) for positive n.

Examples

			720 = 2*6*60 = 12*60. Since 2, 6, 12 and 60 are all terms of A003418, 720 is a term of this sequence.
		

Crossrefs

Range of values of A253139. Subsequences include A000142, A001013, A001813, A025527, A064350, A166338, A250569.
Subsequence of A025487.

Programs

  • PARI
    f(n) = lcm(vector(n, i, i)); \\ A003418
    mul(x,y) = x*y;
    lista(nn) = {my(v = vector(nn, k, f(k)), lim = f(nn+1), ok = 0, nv); while (!ok,  nv = select(x->(xMichel Marcus, May 09 2021

A202946 a(n+1) = 6*A060544(n)*a(n).

Original entry on oeis.org

3, 18, 1080, 181440, 59875200, 32691859200, 26676557107200, 30411275102208000, 46164315605151744000, 90020415430045900800000, 219289731987591814348800000, 652606242395073239502028800000
Offset: 1

Views

Author

John M. Campbell, Dec 26 2011

Keywords

Comments

Sums of coefficients from (3n+1)th moments of binomial(m,k)*binomial(2m,k): see Maple code below.

Examples

			The evaluation of sum(k=0..n, k^7*binomial(n,k)*binomial(2*n,k)) involves the polynomial 32*n^7+96*n^6-336*n^5-360*n^4+1020*n^3-42*n^2-455*n+63, the sum of the coefficients of which is 18 = a(2).
		

Crossrefs

Programs

  • Maple
    with(PolynomialTools); polyn := proc (q) options operator, arrow; 3^q*Pi*GAMMA(2*n)*(sum(k^q*binomial(n, k)*binomial(2*n, k), k = 0 .. n))/(27^n*sqrt(3)*GAMMA(n-floor((1/3)*q+1/3)+2/3)*GAMMA(n-floor((1/3)*q)+1/3)) end proc; coefl := proc (q) options operator, arrow; CoefficientList(expand(polyn(q)), n) end proc; coe := proc (j, h) options operator, arrow; coefl(j)[h] end proc; seq(sum(coe(3*r+1, k), k = 1 .. 5*r+1), r = 1 .. 8) ;
  • PARI
    print1(a=3);for(n=2,10,print1(", ",a*=27*n*(n-3)+60)) \\ Charles R Greathouse IV, Dec 26 2011

Formula

a(n) = (1/18)*27^n*Gamma(n-1/3)*Gamma(n-2/3)*sqrt(3)/Pi.

A202948 a(n+1) = 3*A136016*a(n).

Original entry on oeis.org

-3, -72, -7560, -1814400, -778377600, -523069747200, -506854585036800, -669048052248576000, -1154107890128793600000, -2520571632041285222400000, -6797981691615346244812800000
Offset: 1

Views

Author

John M. Campbell, Dec 26 2011

Keywords

Comments

Sums of coefficients from (3n+2)th moments of binomial(m,k)*binomial(2m,k): see Maple code below.

Examples

			The evaluation of sum(k^8 binomial(n,k) binomial(2n,k), k=0..n) involves the polynomial 64n^10 + 192n^9 - 1344n^8 - 1056n^7 + 8256n^6 - 3696n^5 - 9940n^4 + 7551n^3 + 348n^2 - 507n + 60, the sum of the coefficients of which is -72=a(2).
		

Crossrefs

Programs

  • Maple
    with(PolynomialTools); polyn := proc (q) options operator, arrow; 3^q*Pi*GAMMA(2*n)*(sum(k^q*binomial(n, k)*binomial(2*n, k), k = 0 .. n))/(27^n*sqrt(3)*GAMMA(n-floor((1/3)*q+1/3)+2/3)*GAMMA(n-floor((1/3)*q)+1/3)) end proc; coefl := proc (q) options operator, arrow; CoefficientList(expand(polyn(q)), n) end proc; coe := proc (j, h) options operator, arrow; coefl(j)[h] end proc; seq(sum(coe(3*r+2, k), k = 1 .. 5*r+3), r = 1 .. 8) ;
  • PARI
    print1(a=-3);for(n=2,20,print1(", ",a*=27*n*(n-2)+24)) \\ Charles R Greathouse IV, Dec 27 2011

Formula

a(n)=-(1/6)*27^n*GAMMA(n-1/3)*GAMMA(n+1/3)*sqrt(3)/Pi.

A001525 a(n) = (3n)!/(3!n!).

Original entry on oeis.org

1, 60, 10080, 3326400, 1816214400, 1482030950400, 1689515283456000, 2564684200286208000, 5001134190558105600000, 12182762888199545241600000, 36255902355281846639001600000, 129433571408356192501235712000000, 545950804200446419970212233216000000
Offset: 1

Views

Author

Keywords

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Equals A064350/3!, row sums of A157704/3! and row sums of A157705/3!.

Programs

  • Mathematica
    Table[(3 n)!/(3! n!), {n, 20}] (* Harvey P. Dale, May 25 2011 *)

Extensions

More terms from Harvey P. Dale, May 25 2011

A066387 Triangle T(n,m) (1<=m<=n) giving number of maps f:N -> N such that f^m(X)=X+n for all natural numbers X.

Original entry on oeis.org

1, 1, 2, 1, 0, 6, 1, 12, 0, 24, 1, 0, 0, 0, 120, 1, 120, 360, 0, 0, 720, 1, 0, 0, 0, 0, 0, 5040, 1, 1680, 0, 20160, 0, 0, 0, 40320, 1, 0, 60480, 0, 0, 0, 0, 0, 362880, 1, 30240, 0, 0, 1814400, 0, 0, 0, 0, 3628800, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 39916800
Offset: 1

Views

Author

Floor van Lamoen, Dec 23 2001

Keywords

Examples

			Triangle T(n,m) begins:
  1;
  1,    2;
  1,    0,   6;
  1,   12,   0,    24;
  1,    0,   0,     0, 120;
  1,  120, 360,     0,   0, 720;
  1,    0,   0,     0,   0,   0, 5040;
  1, 1680,   0, 20160,   0,   0,    0, 40320;
  ...
		

Crossrefs

Row sums give A057625.
Main diagonal gives A000142.
m-section of column m=2-4 (for n>0) gives: A001813, A064350, A166338.

Programs

  • Mathematica
    t[n_, m_] /; Divisible[n, m] := n!/(n/m)!; t[, ] = 0; Flatten[Table[t[n, m], {n, 1, 11}, {m, 1, n}]] (* Jean-François Alcover, Nov 29 2011 *)

Formula

T(n,m) = n!/(n/m)! if m|n, T(n,m) = 0 otherwise.

A066991 Square array read by descending antidiagonals of number of ways of dividing n*k labeled items into k unlabeled orders with n items in each order.

Original entry on oeis.org

1, 1, 2, 1, 12, 6, 1, 120, 360, 24, 1, 1680, 60480, 20160, 120, 1, 30240, 19958400, 79833600, 1814400, 720, 1, 665280, 10897286400, 871782912000, 217945728000, 239500800, 5040, 1, 17297280, 8892185702400, 20274183401472000
Offset: 1

Views

Author

Henry Bottomley, Feb 01 2002

Keywords

Comments

T(p,k) = (pk)!/k! is divisible by p^k but not p^(k+1) for p prime; e.g., T(3,4) = 3^4*11*10*8*7*5*4*2*1 = 19958400.

Examples

			The array begins:
  n\k|   1        2             3                   4  ...
  --------------------------------------------------------
   1 |   1,       1,            1,                  1, ...
   2 |   2,      12,          120,               1680, ...
   3 |   6,     360,        60480,           19958400, ...
   4 |  24,   20160,     79833600,       871782912000, ...
   5 | 120, 1814400, 217945728000, 101370917007360000, ...
  ...
		

Crossrefs

Rows include A000012, A001813, A064350.
Columns include A000142, A002674, A065961.

Programs

  • Mathematica
    Table[((n-k+1)*k)!/k!, {n, 10}, {k, n, 1, -1}] (* Paolo Xausa, Feb 19 2024 *)

Formula

T(n,k) = (n*k)!/k!.

Extensions

Edited by Paolo Xausa, Feb 19 2024
Showing 1-9 of 9 results.