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 1-10 of 29 results. Next

A007913 Squarefree part of n: a(n) is the smallest positive number m such that n/m is a square.

Original entry on oeis.org

1, 2, 3, 1, 5, 6, 7, 2, 1, 10, 11, 3, 13, 14, 15, 1, 17, 2, 19, 5, 21, 22, 23, 6, 1, 26, 3, 7, 29, 30, 31, 2, 33, 34, 35, 1, 37, 38, 39, 10, 41, 42, 43, 11, 5, 46, 47, 3, 1, 2, 51, 13, 53, 6, 55, 14, 57, 58, 59, 15, 61, 62, 7, 1, 65, 66, 67, 17, 69, 70, 71, 2, 73, 74, 3, 19, 77
Offset: 1

Views

Author

R. Muller, Mar 15 1996

Keywords

Comments

Also called core(n). [Not to be confused with the squarefree kernel of n, A007947.]
Sequence read mod 4 gives A065882. - Philippe Deléham, Mar 28 2004
This is an arithmetic function and is undefined if n <= 0.
A note on square roots of numbers: we can write sqrt(n) = b*sqrt(c) where c is squarefree. Then b = A000188(n) is the "inner square root" of n, c = A007913(n), lcm(A007947(b),c) = A007947(n) = "squarefree kernel" of n and bc = A019554(n) = "outer square root" of n. [Corrected by M. F. Hasler, Mar 01 2018]
If n > 1, the quantity f(n) = log(n/core(n))/log(n) satisfies 0 <= f(n) <= 1; f(n) = 0 when n is squarefree and f(n) = 1 when n is a perfect square. One can define n as being "epsilon-almost squarefree" if f(n) < epsilon. - Kurt Foster (drsardonicus(AT)earthlink.net), Jun 28 2008
a(n) is the smallest natural number m such that product of geometric mean of the divisors of n and geometric mean of the divisors of m are integers. Geometric mean of the divisors of number n is real number b(n) = Sqrt(n). a(n) = 1 for infinitely many n. a(n) = 1 for numbers from A000290: a(A000290(n)) = 1. For n = 8; b(8) = sqrt(8), a(n) = 2 because b(2) = sqrt(2); sqrt(8) * sqrt(2) = 4 (integer). - Jaroslav Krizek, Apr 26 2010
Dirichlet convolution of A010052 with the sequence of absolute values of A055615. - R. J. Mathar, Feb 11 2011
Booker, Hiary, & Keating outline a method for bounding (on the GRH) a(n) for large n using L-functions. - Charles R Greathouse IV, Feb 01 2013
According to the formula a(n) = n/A000188(n)^2, the scatterplot exhibits the straight lines y=x, y=x/4, y=x/9, ..., i.e., y=x/k^2 for all k=1,2,3,... - M. F. Hasler, May 08 2014
The Dirichlet inverse of this sequence is A008836(n) * A063659(n). - Álvar Ibeas, Mar 19 2015
a(n) = 1 if n is a square, a(n) = n if n is a product of distinct primes. - Zak Seidov, Jan 30 2016
All solutions of the Diophantine equation n*x=y^2 or, equivalently, G(n,x)=y, with G being the geometric mean, are of the form x=k^2*a(n), y=k*sqrt(n*a(n)), where k is a positive integer. - Stanislav Sykora, Feb 03 2016
If f is a multiplicative function then Sum_{d divides n} f(a(d)) is also multiplicative. For example, A010052(n) = Sum_{d divides n} mu(a(d)) and A046951(n) = Sum_{d divides n} mu(a(d)^2). - Peter Bala, Jan 24 2024

Crossrefs

See A000188, A007947, A008833, A019554, A117811 for related information, specific to n.
See A027746, A027748, A124010 for factorization data for n.
Analogous sequences: A050985, A053165, A055231.
Cf. A002734, A005117 (range of values), A059897, A069891 (partial sums), A090699, A350389.
Related to A006519 via A225546.

Programs

  • Haskell
    a007913 n = product $
                zipWith (^) (a027748_row n) (map (`mod` 2) $ a124010_row n)
    -- Reinhard Zumkeller, Jul 06 2012
    
  • Magma
    [ Squarefree(n) : n in [1..256] ]; // N. J. A. Sloane, Dec 23 2006
    
  • Maple
    A007913 := proc(n) local f,a,d; f := ifactors(n)[2] ; a := 1 ; for d in f do if type(op(2,d),'odd') then a := a*op(1,d) ; end if; end do: a; end proc: # R. J. Mathar, Mar 18 2011
    # second Maple program:
    a:= n-> mul(i[1]^irem(i[2], 2), i=ifactors(n)[2]):
    seq(a(n), n=1..100);  # Alois P. Heinz, Jul 20 2015
    seq(n / expand(numtheory:-nthpow(n, 2)), n=1..77);  # Peter Luschny, Jul 12 2022
  • Mathematica
    data = Table[Sqrt[n], {n, 1, 100}]; sp = data /. Sqrt[] -> 1; sfp = data/sp /. Sqrt[x] -> x (* Artur Jasinski, Nov 03 2008 *)
    Table[Times@@Power@@@({#[[1]],Mod[ #[[2]],2]}&/@FactorInteger[n]),{n,100}] (* Zak Seidov, Apr 08 2009 *)
    Table[{p, e} = Transpose[FactorInteger[n]]; Times @@ (p^Mod[e, 2]), {n, 100}] (* T. D. Noe, May 20 2013 *)
    Sqrt[#] /. (c_:1)*a_^(b_:0) -> (c*a^b)^2& /@ Range@100 (* Bill Gosper, Jul 18 2015 *)
  • PARI
    a(n)=core(n)
    
  • Python
    from sympy import factorint, prod
    def A007913(n):
        return prod(p for p, e in factorint(n).items() if e % 2)
    # Chai Wah Wu, Feb 03 2015
    
  • Sage
    [squarefree_part(n) for n in (1..77)] # Peter Luschny, Feb 04 2015

Formula

Multiplicative with a(p^k) = p^(k mod 2). - David W. Wilson, Aug 01 2001
a(n) modulo 2 = A035263(n); a(A036554(n)) is even; a(A003159(n)) is odd. - Philippe Deléham, Mar 28 2004
Dirichlet g.f.: zeta(2s)*zeta(s-1)/zeta(2s-2). - R. J. Mathar, Feb 11 2011
a(n) = n/( Sum_{k=1..n} floor(k^2/n)-floor((k^2 -1)/n) )^2. - Anthony Browne, Jun 06 2016
a(n) = rad(n)/a(n/rad(n)), where rad = A007947. This recurrence relation together with a(1) = 1 generate the sequence. - Velin Yanev, Sep 19 2017
From Peter Munn, Nov 18 2019: (Start)
a(k*m) = A059897(a(k), a(m)).
a(n) = n / A008833(n).
(End)
a(A225546(n)) = A225546(A006519(n)). - Peter Munn, Jan 04 2020
From Amiram Eldar, Mar 14 2021: (Start)
Theorems proven by Copil and Panaitopol (2007):
Lim sup_{n->oo} a(n+1)-a(n) = oo.
Lim inf_{n->oo} a(n+1)-a(n) = -oo.
Sum_{k=1..n} 1/a(k) ~ c*sqrt(n) + O(log(n)), where c = zeta(3/2)/zeta(3) (A090699). (End)
a(n) = A019554(n)^2/n. - Jianing Song, May 08 2022
Sum_{k=1..n} a(k) ~ c * n^2, where c = Pi^2/30 = 0.328986... . - Amiram Eldar, Oct 25 2022
a(n) = A007947(A350389(n)). - Amiram Eldar, Jan 20 2024

Extensions

More terms from Michael Somos, Nov 24 2001
Definition reformulated by Daniel Forgues, Mar 24 2009

A362974 Decimal expansion of Product_{p prime} (1 + 1/p^(4/3) + 1/p^(5/3)).

Original entry on oeis.org

4, 6, 5, 9, 2, 6, 6, 1, 2, 2, 5, 0, 0, 6, 5, 6, 9, 4, 1, 2, 7, 7, 4, 3, 1, 1, 0, 8, 9, 1, 3, 6, 2, 5, 8, 6, 2, 1, 3, 0, 5, 4, 3, 3, 6, 7, 2, 8, 3, 2, 5, 6, 5, 3, 8, 4, 7, 5, 7, 6, 9, 2, 4, 0, 1, 5, 3, 0, 3, 4, 1, 8, 0, 8, 6, 5, 7, 3, 5, 2, 3, 8, 7, 2, 1, 8, 0, 7, 7, 5, 8, 9, 0, 2, 6, 8, 4, 6, 2, 3, 4, 9, 0, 9, 7
Offset: 1

Views

Author

Amiram Eldar, May 11 2023

Keywords

Comments

The coefficient c_0 of the leading term in the asymptotic formula for the number of cubefull numbers (A036966) not exceeding x, N(x) = c_0 * x^(1/3) + c_1 * x^(1/4) + c_2 * x^(1/5) + o(x^(1/8)) (Bateman and Grosswald, 1958; Finch, 2003).

Examples

			4.65926612250065694127743110891362586213054336728325...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, section 2.6.1, pp. 113-115.

Crossrefs

Cf. A036966, A090699 (analogous constant for powerful numbers), A244000, A337736, A362973, A362975 (c_1), A362976 (c_2).
Cf. A051904.

Programs

  • Mathematica
    $MaxExtraPrecision = 1000; m = 1000; c = LinearRecurrence[{0, 0, 0, -1, -1}, {0, 0, 0, 4, 5}, m]; RealDigits[(1 + 1/2^(4/3) + 1/2^(5/3)) * (1 + 1/3^(4/3) + 1/3^(5/3)) * Exp[NSum[Indexed[c, n]*(PrimeZetaP[n/3] - 1/2^(n/3) - 1/3^(n/3))/n, {n, 4, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 120][[1]]
  • PARI
    prodeulerrat(1 + 1/p^4 + 1/p^5, 1/3)

Formula

Equals 1 + lim_{m->oo} (1/m) Sum_{k=1..m} A337736(k).

A338325 Biquadratefree powerful numbers: numbers whose exponents in their prime factorization are either 2 or 3.

Original entry on oeis.org

1, 4, 8, 9, 25, 27, 36, 49, 72, 100, 108, 121, 125, 169, 196, 200, 216, 225, 289, 343, 361, 392, 441, 484, 500, 529, 675, 676, 841, 900, 961, 968, 1000, 1089, 1125, 1156, 1225, 1323, 1331, 1352, 1369, 1372, 1444, 1521, 1681, 1764, 1800, 1849, 2116, 2197, 2209
Offset: 1

Views

Author

Amiram Eldar, Oct 22 2020

Keywords

Comments

Equivalently, numbers k such that if a prime p divides k then p^2 divides k but p^4 does not divide k.
Each term has a unique representation as a^2 * b^3, where a and b are coprime squarefree numbers.
Dehkordi (1998) refers to these numbers as "2-full and 4-free numbers".

Examples

			4 = 2^2 is a term since the exponent of its only prime factor is 2.
72 = 2^3 * 3^2 is a terms since the exponents of the primes in its prime factorization are 2 and 3.
		

Crossrefs

Intersection of A001694 and A046100.
Subsequences: A062503, A062838.

Programs

  • Mathematica
    Select[Range[2500], # == 1 || AllTrue[FactorInteger[#][[;; , 2]], MemberQ[{2, 3}, #1] &] &]

Formula

The number of terms not exceeding x is asymptotically (zeta(3/2)/zeta(3)) * J_2(1/2) * x^(1/2) + (zeta(2/3)/zeta(2)) * J_2(1/3) * x^(1/3), where J_2(s) = Product_{p prime} (1 - p^(-4*s) - p^(-5*s) - p^(-6*s) + p^(-7*s) + p^(-8*s)) (Dehkordi, 1998).
Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + 1/p^2 + 1/p^3) = 1.748932... (A330595).

A119241 Number of powerful numbers (A001694) between consecutive squares n^2 and (n+1)^2.

Original entry on oeis.org

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

Views

Author

T. D. Noe, May 09 2006

Keywords

Comments

Is there an upper bound on the number of powerful numbers between consecutive squares? Pettigrew conjectures that there is no bound. See A119242.
This sequence is unbounded. For each k >= 0 the sequence of solutions to a(x) = k has a positive asymptotic density (Shiu, 1980). - Amiram Eldar, Jul 10 2020

Examples

			a(5) = 2 because the two powerful numbers 27 and 32 are between 25 and 36.
		

References

  • József Sándor, Dragoslav S. Mitrinovic and Borislav Crstici, Handbook of Number Theory I, Springer Science & Business Media, 2005, chapter VI, p. 226.

Crossrefs

Programs

  • Mathematica
    Powerful[n_] := (n==1) || Min[Transpose[FactorInteger[n]][[2]]]>1; Table[Length[Select[Range[k^2+1, k^2+2k], Powerful[ # ]&]], {k,130}]
  • Python
    from math import isqrt
    from sympy import integer_nthroot, factorint
    def A119241(n):
        def f(x): return int(sum(isqrt(x//k**3) for k in range(1, integer_nthroot(x, 3)[0]+1) if all(d<=1 for d in factorint(k).values())))
        return f((n+1)**2-1)-f(n**2) # Chai Wah Wu, Sep 10 2024

Formula

Asymptotic mean: lim_{n->oo} (1/n) Sum_{k=1..n} a(k) = zeta(3/2)/zeta(3) - 1 = A090699 - 1 = 1.173254... - Amiram Eldar, Oct 24 2020

A118896 Number of powerful numbers <= 10^n.

Original entry on oeis.org

1, 4, 14, 54, 185, 619, 2027, 6553, 21044, 67231, 214122, 680330, 2158391, 6840384, 21663503, 68575557, 217004842, 686552743, 2171766332, 6869227848, 21725636644, 68709456167, 217293374285, 687174291753, 2173105517385, 6872112993377, 21731852479862, 68722847672629, 217322225558934, 687236449779456, 2173239433013146
Offset: 0

Views

Author

Eric W. Weisstein, May 05 2006

Keywords

Comments

These numbers agree with the asymptotic formula c*sqrt(x), with c=2.1732...(A090699). - T. D. Noe, May 09 2006
Bateman & Grosswald proved that the number of powerful numbers up to x is zeta(3/2)/zeta(3) * x^1/2 + zeta(2/3)/zeta(2) * x^1/3 + o(x^1/6). This approximates the series very closely: up to a(24), all absolute errors are less than 75. - Charles R Greathouse IV, Sep 23 2008

Crossrefs

Programs

  • Maple
    f:= m -> nops({seq(seq(a^2*b^3, b=1..floor((m/a^2)^(1/3))),a=1..floor(sqrt(m)))}):
    seq(f(10^n),n=0..10); # Robert Israel, Aug 12 2014
  • Mathematica
    f[n_] := Block[{max = 10^n}, Length@ Union@ Flatten@ Table[ a^2*b^3, {b, max^(1/3)}, {a, Sqrt[ max/b^3]}]]; Array[f, 13, 0] (* Robert G. Wilson v, Aug 11 2014 *)
    powerfulNumberPi[n_] := Sum[ If[ SquareFreeQ@ i, Floor[ Sqrt[ n/i^3]], 0], {i, n^(1/3)}]; Array[ powerfulNumberPi[10^#] &, 27, 0] (* Robert G. Wilson v, Aug 12 2014 *)
  • PARI
    a(n)=n=10^n;sum(k=1, floor((n+.5)^(1/3)), if(issquarefree(k), sqrtint(n\k^3))) \\ Charles R Greathouse IV, Sep 23 2008
    
  • Python
    from math import isqrt
    from sympy import integer_nthroot, factorint
    def A118896(n):
        m = 10**n
        return sum(isqrt(m//x**3) for x in range(1,integer_nthroot(m,3)[0]+1) if max(factorint(x).values(),default=0)<=1) # Chai Wah Wu, May 13 2023
    
  • Python
    # faster program
    def A118896(n):
        def squarefreepi(n): return int(sum(mobius(k)*(n//k**2) for k in range(1, isqrt(n)+1)))
        m, c, l = 10**n, 0, 0
        j = isqrt(m)
        while j>1:
            k2 = integer_nthroot(m//j**2,3)[0]+1
            w = squarefreepi(k2-1)
            c += j*(w-l)
            l, j = w, isqrt(m//k2**3)
        c += squarefreepi(integer_nthroot(m,3)[0])-l
        return c # Chai Wah Wu, Sep 09 2024

Formula

Pi(x) = Sum_{i=1..x^(1/3)} floor(sqrt(x/i^3)) only if i is squarefree. - Robert G. Wilson v, Aug 12 2014

Extensions

More terms from T. D. Noe, May 09 2006
a(13)-a(24) from Charles R Greathouse IV, Sep 23 2008
a(25)-a(29) from Charles R Greathouse IV, May 30 2011
a(30) from Charles R Greathouse IV, May 31 2011

A180114 a(n) = sigma(A001694(n)), sum of divisors of the powerful number A001694(n).

Original entry on oeis.org

1, 7, 15, 13, 31, 31, 40, 63, 91, 57, 127, 195, 121, 217, 280, 133, 156, 255, 403, 183, 399, 465, 600, 403, 364, 511, 819, 307, 847, 400, 381, 855, 961, 1240, 741, 931, 1092, 1023, 553, 1651, 781, 1815, 1240, 1281, 1093, 1767, 1953, 871, 2520, 2821, 993, 1995
Offset: 1

Views

Author

Walter Kehowski, Aug 10 2010

Keywords

Examples

			Sigma(2^2) = 7, sigma(2^3) = 15, sigma(3^2) = 13.
		

Crossrefs

Programs

  • Maple
    emin := proc(n::posint) local L; if n>1 then L:=ifactors(n)[2]; L:=map(z->z[2],L); min(L) else 0 fi end: L:=[]: for w to 1 do for n from 1 to 144 do sn:=sigma(n); if emin(n)>1 then L:=[op(L),sn]; print(n,ifactor(n),sn,ifactor(sn)) fi; od; od;
  • Mathematica
    pwfQ[n_] := n == 1 || Min[Last /@ FactorInteger[n]] > 1; DivisorSigma[1, Select[ Range@ 1000, pwfQ]] (* Giovanni Resta, Feb 06 2018 *)
  • PARI
    lista(kmax) = for(k = 1, kmax, if(k==1 || vecmin(factor(k)[, 2]) > 1, print1(sigma(k), ", "))); \\ Amiram Eldar, May 12 2023
    
  • Python
    from itertools import count, islice
    from math import prod
    from sympy import factorint
    def A180114_gen(): # generator of terms
        for n in count(1):
            f = factorint(n)
            if all(e>1 for e in f.values()):
                yield prod((p**(e+1)-1)//(p-1) for p,e in f.items())
    A180114_list = list(islice(A180114_gen(),20)) # Chai Wah Wu, May 21 2023
    
  • Python
    from math import isqrt
    from sympy import mobius, integer_nthroot, divisor_sigma
    def A180114(n):
        def squarefreepi(n):
            return int(sum(mobius(k)*(n//k**2) for k in range(1, isqrt(n)+1)))
        def bisection(f, kmin=0, kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x):
            c, l = n+x, 0
            j = isqrt(x)
            while j>1:
                k2 = integer_nthroot(x//j**2, 3)[0]+1
                w = squarefreepi(k2-1)
                c -= j*(w-l)
                l, j = w, isqrt(x//k2**3)
            c -= squarefreepi(integer_nthroot(x, 3)[0])-l
            return c
        return divisor_sigma(bisection(f, n, n)) # Chai Wah Wu, Sep 10 2024

Formula

From Amiram Eldar, May 12 2023: (Start)
Sum_{A001694(k) < x} a(k) = c * x^(3/2) + O(x^(23/18 + eps)), where c = A362984 * A090699 / 3 = 1.5572721108... (Jakimczuk and Lalín, 2022). [corrected Sep 21 2024]
Sum_{k=1..n} a(k) ~ c * n^3, where c = A362984 / (3 * A090699^2) = 0.151716514097... . (End)

Extensions

a(1)=1 prepended by and more terms from Giovanni Resta, Feb 06 2018

A368710 The maximal exponent in the prime factorization of the powerful numbers.

Original entry on oeis.org

0, 2, 3, 2, 4, 2, 3, 5, 2, 2, 6, 3, 4, 2, 3, 2, 3, 7, 4, 2, 2, 3, 3, 2, 5, 8, 5, 2, 4, 3, 2, 3, 4, 4, 2, 2, 3, 9, 2, 6, 4, 4, 3, 2, 6, 4, 5, 2, 5, 2, 2, 3, 5, 3, 10, 2, 3, 7, 2, 2, 4, 3, 3, 3, 2, 3, 2, 2, 5, 6, 2, 6, 2, 3, 2, 4, 5, 4, 4, 11, 2, 7, 3, 2, 8, 3, 4
Offset: 1

Views

Author

Amiram Eldar, Jan 04 2024

Keywords

Crossrefs

Programs

  • Mathematica
    s[n_] := If[n == 1, 0, Max @@ Last /@ FactorInteger[n]]; s /@ Select[Range[3000], # == 1 || Min[FactorInteger[#][[;;, 2]]] > 1 &]
    (* or *)
    f[n_] := Module[{e = FactorInteger[n][[;; , 2]]}, If[n == 1, 0, If[Min[e] > 1, Max[e], Nothing]]]; Array[f, 3000]
  • PARI
    lista(kmax) = {my(e); for(k = 1, kmax, e = factor(k)[,2]; if(k == 1, print1(0, ", "), if(vecmin(e) > 1, print1(vecmax(e), ", "))));}

Formula

a(n) = A051903(A001694(n)).
a(n) >= 2 for n >= 2.
Sum_{a(n)<=x} = D_{2,1} * sqrt(x) + O(sqrt(x)), where D_{2,1} = (6/Pi^2) * (2 + Sum_{k>=1} (A051903(k)+2)/(sqrt(k) * A048250(k))) (Jakimczuk, 2018; Theorem 2.1 and Remark 2.3).
Asymptotic mean (consequence of the formula above): Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = D_{2,1} * zeta(3)/zeta(3/2) = D_{2,1} / A090699.
The sum in the formula for D_{2,1} converges slowly: for k up to 10^8, 10^9 and 10^10 the sums are 14.845..., 14.908... and 14.938..., respectively. Thus, a lower bound for the value of this mean, calculated by summing over k=1..10^10, is 4.738... .

A328013 Decimal expansion of the growth constant for the partial sums of powerful part of n (A057521).

Original entry on oeis.org

3, 5, 1, 9, 5, 5, 5, 0, 5, 8, 4, 1, 7, 1, 0, 6, 6, 4, 7, 1, 9, 7, 5, 2, 9, 4, 0, 3, 6, 9, 8, 5, 7, 8, 1, 7, 1, 8, 6, 0, 3, 9, 8, 0, 8, 2, 2, 5, 4, 0, 7, 8, 1, 4, 7, 1, 1, 4, 6, 4, 0, 3, 1, 4, 5, 4, 1, 7, 8, 3, 9, 8, 4, 7, 9, 7, 3, 5, 4, 0, 8, 9, 7, 7, 1, 3, 5, 8, 0, 3, 7, 5, 3, 6, 4, 6, 1, 6, 2, 0, 1, 1, 4, 5, 5
Offset: 1

Views

Author

Amiram Eldar, Oct 01 2019

Keywords

Examples

			3.51955505841710664719752940369857817186039808225407...
		

References

  • D. Suryanarayana and P. Subrahmanyam, The maximal k-full divisor of an integer, Indian J. Pure Appl. Math., Vol. 12, No. 2 (1981), pp. 175-190.

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = 500; m = 500; c = LinearRecurrence[{0, 0, -2, 0, 1}, {0, 0, 6, 0, -5}, m]; RealDigits[(1 + 2/2^(3/2) - 1/2^(5/2))*(1 + 2/3^(3/2) - 1/3^(5/2))* Exp[NSum[Indexed[c, n]*(PrimeZetaP[n/2] - 1/2^(n/2) - 1/3^(n/2))/n, {n, 3, m}, NSumTerms -> m, WorkingPrecision -> m]], 10, 100][[1]]
  • PARI
    prodeulerrat(1 + 2/p^3 - 1/p^5, 1/2) \\ Amiram Eldar, Jun 29 2023

Formula

The constant d1 in the paper by Cloutier et al. such that Sum_{k=1..x} A057521(k) = (d1/3)*x^(3/2) + O(x^(4/3)).
Sum_{k=1..x} 1/A055231(k) = d1*x^(1/2) + O(x^(1/3)).
Equals Product_{primes p} (1 + 2/p^(3/2) - 1/p^(5/2)).
Equals (zeta(3/2)/zeta(3)) * Product_{p prime} (1 + (sqrt(p)-1)/(p*(p-sqrt(p)+1))). - Amiram Eldar, Dec 26 2024

Extensions

More terms from Vaclav Kotesovec, May 29 2020

A362975 Decimal expansion of zeta(3/4) * Product_{p prime} (1 + 1/p^(5/4) - 1/p^2 - 1/p^(9/4)) (negated).

Original entry on oeis.org

5, 8, 7, 2, 6, 1, 8, 8, 2, 0, 8, 1, 3, 8, 4, 2, 3, 9, 1, 0, 7, 4, 1, 3, 8, 1, 4, 2, 6, 6, 7, 8, 3, 5, 6, 1, 1, 4, 8, 6, 2, 6, 4, 3, 1, 1, 0, 8, 2, 9, 3, 5, 3, 5, 1, 7, 0, 7, 9, 8, 0, 4, 6, 6, 9, 0, 3, 9, 8, 2, 0, 5, 3, 5, 0, 1, 1, 2, 5, 3, 5, 6, 8, 6, 3, 3, 7, 5, 7, 9, 1, 7, 5, 1, 3, 0, 1, 2, 1, 3, 1, 6, 8, 4, 3
Offset: 1

Views

Author

Amiram Eldar, May 11 2023

Keywords

Comments

The coefficient c_1 of the second term in the asymptotic formula for the number of cubefull numbers (A036966) not exceeding x, N(x) = c_0 * x^(1/3) + c_1 * x^(1/4) + c_2 * x^(1/5) + o(x^(1/8)) (Bateman and Grosswald, 1958; Finch, 2003).

Examples

			-5.87261882081384239107413814266783561148626431108293...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, section 2.6.1, pp. 113-115.

Crossrefs

Programs

  • PARI
    zeta(3/4) * prodeulerrat(1 + 1/p^5 - 1/p^8 - 1/p^9 ,1/4)

A362976 Decimal expansion of zeta(3/5) * zeta(4/5) * Product_{p prime} (1 - 1/p^(8/5) - 1/p^(9/5) - 1/p^2 + 1/p^(13/5) + 1/p^(14/5)).

Original entry on oeis.org

1, 6, 8, 2, 4, 4, 1, 5, 1, 0, 2, 3, 5, 9, 3, 2, 9, 3, 8, 9, 5, 6, 0, 0, 2, 0, 3, 4, 3, 1, 7, 7, 1, 2, 4, 5, 3, 3, 7, 2, 3, 3, 6, 2, 1, 3, 5, 7, 9, 9, 4, 9, 4, 3, 8, 5, 1, 5, 8, 3, 5, 4, 3, 9, 7, 4, 9, 6, 9, 8, 9, 7, 7, 6, 7, 6, 0, 1, 0, 6, 4, 7, 8, 5, 6, 2, 7, 7, 7, 7, 5, 4, 1, 9, 7, 6, 4, 3, 9, 5, 5, 6, 7, 5, 2
Offset: 1

Views

Author

Amiram Eldar, May 11 2023

Keywords

Comments

The coefficient c_2 of the third term in the asymptotic formula for the number of cubefull numbers (A036966) not exceeding x, N(x) = c_0 * x^(1/3) + c_1 * x^(1/4) + c_2 * x^(1/5) + o(x^(1/8)) (Bateman and Grosswald, 1958; Finch, 2003).

Examples

			1.68244151023593293895600203431771245337233621357994...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, section 2.6.1, pp. 113-115.

Crossrefs

Programs

  • PARI
    zeta(3/5) * zeta(4/5) * prodeulerrat(1 - 1/p^8 - 1/p^9 - 1/p^10 + 1/p^13 + 1/p^14, 1/5)
Showing 1-10 of 29 results. Next