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 18 results. Next

A051904 Minimal exponent in prime factorization of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Labos Elemer, Dec 16 1999

Keywords

Comments

The asymptotic mean of this sequence is 1 (Niven, 1969). - Amiram Eldar, Jul 10 2020
Let k = A007947(n), then for n > 1 k^a(n) is the greatest power of k which divides n; see example. - David James Sycamore, Sep 07 2023

Examples

			For n = 72 = 2^3*3^2, a(72) = min(exponents) = min(3,2) = 2.
For n = 72, using alternative definition: rad(72) = 6; and 6^2 = 36 divides 72 but no higher power of 6 divides 72, so a(72) = 2.
For n = 432, rad(432) = 6 and 6^3 = 216 divides 432 but no higher power of 6 divides 432, therefore a(432) = 3. - _David James Sycamore_, Sep 08 2023
		

Crossrefs

Programs

  • Haskell
    a051904 1 = 0
    a051904 n = minimum $ a124010_row n  -- Reinhard Zumkeller, Jul 15 2012
    
  • Maple
    a := proc (n) if n = 1 then 0 else min(seq(op(2, op(j, op(2, ifactors(n)))), j = 1 .. nops(op(2, ifactors(n))))) end if end proc: seq(a(n), n = 1 .. 100); # Emeric Deutsch, May 20 2015
  • Mathematica
    Table[If[n == 1, 0, Min @@ Last /@ FactorInteger[n]], {n, 100}] (* Ray Chandler, Jan 24 2006 *)
  • PARI
    a(n)=vecmin(factor(n)[,2]) \\ Charles R Greathouse IV, Nov 19 2012
    
  • Python
    from sympy import factorint
    def a(n):
        f = factorint(n)
        l = [f[p] for p in f]
        return 0 if n == 1 else min(l)
    print([a(n) for n in range(1, 51)]) # Indranil Ghosh, Jul 13 2017
  • Scheme
    (define (A051904 n) (cond ((= 1 n) 0) ((= 1 (A001221 n)) (A001222 n)) (else (min (A067029 n) (A051904 (A028234 n)))))) ;; Antti Karttunen, Jul 12 2017
    

Formula

a(n) = min_{k=1..A001221(n)} A124010(n,k). - Reinhard Zumkeller, Aug 27 2011
a(1) = 0, for n > 1, if A001221(n) = 1 (when n is in A000961), a(n) = A001222(n), otherwise a(n) = min(A067029(n), a(A028234(n))). - Antti Karttunen, Jul 12 2017
Sum_{k=1..n} a(k) ~ n + zeta(3/2)*n^(1/2)/zeta(3) + (zeta(2/3)/zeta(2) + c0)*n^(1/3), where c0 = A362974 = Product_{p prime} (1 + 1/p^(4/3) + 1/p^(5/3)) [Cao Hui-Zhong, 1991]. - Vaclav Kotesovec, Mar 24 2025

A337736 The number of cubefull numbers (A036966) between the consecutive cubes n^3 and (n+1)^3.

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Sep 17 2020

Keywords

Comments

For each k >= 0 the sequence of solutions to a(x) = k has a positive asymptotic density (Shiu, 1991).

Examples

			a(2) = 1 since there is one cubefull number, 16 = 2^4, between 2^3 = 8 and 3^3 = 27.
		

Crossrefs

Programs

  • Mathematica
    cubQ[n_] := Min[FactorInteger[n][[;; , 2]]] > 2; a[n_] := Count[Range[n^3 + 1, (n + 1)^3 - 1], _?cubQ]; Array[a, 100]
  • Python
    from math import gcd
    from sympy import integer_nthroot, factorint
    def A337736(n):
        def f(x):
            c = 0
            for w in range(1,integer_nthroot(x,5)[0]+1):
                if all(d<=1 for d in factorint(w).values()):
                    for y in range(1,integer_nthroot(z:=x//w**5,4)[0]+1):
                        if gcd(w,y)==1 and all(d<=1 for d in factorint(y).values()):
                            c += integer_nthroot(z//y**4,3)[0]
            return c
        return f((n+1)**3-1)-f(n**3) # Chai Wah Wu, Sep 10 2024

Formula

Asymptotic mean: lim_{m->oo} (1/m) Sum_{k=1..m} a(k) = A362974 - 1 = 3.659266... . - Amiram Eldar, May 11 2023

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)

A362986 a(n) = A000203(A036966(n)), the sum of divisors of the n-th cubefull number A036966(n).

Original entry on oeis.org

1, 15, 31, 40, 63, 127, 121, 156, 255, 600, 364, 511, 400, 1240, 1023, 781, 1815, 1093, 2520, 2340, 2047, 3751, 1464, 5080, 5460, 4836, 4095, 3280, 2380, 2801, 7623, 6000, 3906, 6240, 10200, 11284, 9828, 8191, 5220, 11715, 15367, 12400, 16395, 9841, 7240, 20440
Offset: 1

Views

Author

Amiram Eldar, May 12 2023

Keywords

Crossrefs

Programs

  • Mathematica
    DivisorSigma[1, Select[Range[10^4], # == 1 || Min[FactorInteger[#][[;; , 2]]] > 2 &]]
  • PARI
    lista(kmax) = for(k = 1, kmax, if(k==1 || vecmin(factor(k)[, 2]) > 2, print1(sigma(k), ", ")));
    
  • Python
    from itertools import count, islice
    from math import prod
    from sympy import factorint
    def A362986_gen(): # generator of terms
        for n in count(1):
            f = factorint(n)
            if all(e>2 for e in f.values()):
                yield prod((p**(e+1)-1)//(p-1) for p,e in f.items())
    A362986_list = list(islice(A362986_gen(),20)) # Chai Wah Wu, May 21 2023

Formula

Sum_{A036966(k) < x} a(k) = c * x^(4/3) + O(x^(113/96 + eps)), where c = A362985 * A362974 / 4 = 2.8912833599... (Jakimczuk and Lalín, 2022). [corrected Sep 21 2024]
Sum_{k=1..n} a(k) ~ c * n^4, where c = A362985 / (4 * A362974^3) = 0.006135085083... .

A376363 The number of distinct prime factors of the cubefull numbers.

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Sep 21 2024

Keywords

Crossrefs

Programs

  • Mathematica
    f[k_] := Module[{e = If[k == 1, {}, FactorInteger[k][[;; , 2]]]}, If[AllTrue[e, # > 2 &], Length[e], Nothing]]; Array[f, 60000]
  • PARI
    lista(kmax) = {my(e, is); for(k = 1, kmax, e = factor(k)[, 2]; is = 1; for(i = 1, #e, if(e[i] < 3, is = 0; break)); if(is, print1(#e, ", ")));}

Formula

a(n) = A001221(A036966(n)).
Sum_{A036966(k) <= x} a(k) = c * x^(1/3) * (log(log(x)) + B - log(3) + L(3, 4) - L(3, 6)) + O(x^(1/3)/log(x)), where c = A362974, B is Mertens's constant (A077761), L(h, r) = Sum_{p prime} 1/(p^(r/h - 1) * (p - p^(1 - 1/h) + 1)), L(3, 4) = 1.65235055631578303808..., and L(3, 6) = 0.67060646664392140547... (Das et al., 2024).

A362973 The number of cubefull numbers (A036966) not exceeding 10^n.

Original entry on oeis.org

1, 2, 7, 20, 51, 129, 307, 713, 1645, 3721, 8348, 18589, 41136, 90619, 198767, 434572, 947753, 2062437, 4480253, 9718457, 21055958, 45575049, 98566055, 213028539, 460160083, 993533517, 2144335391, 4626664451, 9980028172, 21523027285, 46408635232, 100053270534
Offset: 0

Views

Author

Amiram Eldar, May 11 2023

Keywords

Comments

The number of cubefull numbers not exceeding x is N(x) = c_0 * x^(1/3) + c_1 * x^(1/4) + c_2 * x^(1/5) + o(x^(1/8)), where c_0 (A362974), c_1 (A362975) and c_2 (A362976) are constants (Bateman and Grosswald, 1958; Finch, 2003).
The digits of a(3k) converge to A362974 as k -> oo. - Chai Wah Wu, May 13 2023

Examples

			There are 2 cubefull numbers not exceeding 10, 1 and 8, therefore a(1) = 2.
		

References

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

Crossrefs

Similar sequences: A070428, A118896.

Programs

  • Mathematica
    a[n_] := Module[{max = 10^n}, CountDistinct@ Flatten@ Table[i^5 * j^4 * k^3, {i, Surd[max, 5]}, {j, Surd[max/i^5, 4]}, {k, CubeRoot[max/(i^5*j^4)]}]]; Array[a, 15, 0]
  • Python
    from math import gcd
    from sympy import factorint, integer_nthroot
    def A362973(n):
        m, c = 10**n, 0
        for x in range(1,integer_nthroot(m,5)[0]+1):
            if all(d<=1 for d in factorint(x).values()):
                for y in range(1,integer_nthroot(z:=m//x**5,4)[0]+1):
                    if gcd(x,y)==1 and all(d<=1 for d in factorint(y).values()):
                        c += integer_nthroot(z//y**4,3)[0]
        return c # Chai Wah Wu, May 11-13 2023

Extensions

a(23)-a(31) from Chai Wah Wu, May 11 2023

A371186 Indices of the cubes in the sequence of cubefull numbers.

Original entry on oeis.org

1, 2, 4, 6, 8, 10, 13, 15, 18, 20, 23, 24, 29, 32, 34, 38, 39, 43, 45, 48, 50, 54, 57, 58, 61, 67, 69, 73, 75, 77, 81, 85, 88, 90, 94, 96, 99, 102, 105, 107, 110, 113, 117, 124, 126, 128, 130, 135, 137, 139, 143, 147, 149, 153, 158, 160, 163, 167, 169, 172, 176
Offset: 1

Views

Author

Amiram Eldar, Mar 14 2024

Keywords

Comments

Equivalently, the number of cubefull numbers that do not exceed n^3.
The asymptotic density of this sequence is 1 / A362974 = 0.214626074... .
If k is a term of A371187 then a(k) and a(k+1) are consecutive integers, i.e., a(k+1) = a(k) + 1.

Examples

			The first 4 cubefull numbers are 1, 8, 16, and 27. The 1st, 2nd, and 4th, 1, 8, and 27, are the first 3 cubes. Therefore, the first 3 terms of this sequence are 1, 2, and 4.
		

Crossrefs

Similar sequences: A361936, A371185.

Programs

  • Mathematica
    cubQ[n_] := n == 1 || AllTrue[FactorInteger[n], Last[#] >= 3 &]; Position[Select[Range[10^6], cubQ], _?(IntegerQ[Surd[#1, 3]] &)] // Flatten
    (* or *)
    seq[max_] := Module[{cubs = Union[Flatten[Table[i^5*j^4*k^3, {i, 1, Surd[max, 5]}, {j, 1, Surd[max/i^5, 4]}, {k, Surd[max/(i^5*j^4), 3]}]]], s = {}}, Do[If[IntegerQ[Surd[cubs[[k]], 3]], AppendTo[s, k]], {k, 1, Length[cubs]}]; s]; seq[10^6]
  • PARI
    iscub(n) = n == 1 || vecmin(factor(n)[, 2]) >= 3;
    lista(kmax) = {my(f, c = 0); for(k = 1, kmax, if(iscub(k), c++; if(ispower(k, 3), print1(c, ", "))));}

Formula

A036966(a(n)) = A000578(n) = n^3.
a(n+1) - a(n) = A337736(n) + 1.
a(n) ~ c * n, where c = A362974.

A376364 The number of unitary divisors that are cubes of primes applied to the cubefull numbers.

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Sep 21 2024

Keywords

Crossrefs

Programs

  • Mathematica
    f[k_] := Module[{e = If[k == 1, {}, FactorInteger[k][[;; , 2]]]}, If[AllTrue[e, # > 2 &], Count[e, 3], Nothing]]; Array[f, 60000]
  • PARI
    lista(kmax) = {my(e, is); for(k = 1, kmax, e = factor(k)[, 2]; is = 1; for(i = 1, #e, if(e[i] < 3, is = 0; break)); if(is, print1(#select(x -> x == 3, e), ", ")));}

Formula

a(n) = A295883(A036966(n)).
Sum_{A036966(k) <= x} a(k) = c * sqrt(x) * (log(log(x)) + B - log(3) - L(3, 6)) + O(x^(1/3)/log(x)), where c = A362974, B is Mertens's constant (A077761), L(h, r) = Sum_{p prime} 1/(p^(r/h - 1) * (p - p^(1 - 1/h) + 1)), and L(3, 6) = 0.67060646664392140547... (Das et al., 2025).

A370787 Cubefull numbers with an even number of prime factors (counted with multiplicity).

Original entry on oeis.org

1, 16, 64, 81, 216, 256, 625, 729, 864, 1000, 1024, 1296, 1944, 2401, 2744, 3375, 3456, 4000, 4096, 5184, 6561, 7776, 9261, 10000, 10648, 10976, 11664, 13824, 14641, 15625, 16000, 16384, 17496, 17576, 20736, 25000, 28561, 30375, 31104, 35937, 38416, 39304, 40000
Offset: 1

Views

Author

Amiram Eldar, Mar 02 2024

Keywords

Comments

Jakimczuk (2024) proved:
The number of terms that do not exceed x is N(x) = c * x^(1/3) / 2 + o(x^(1/3)) where c = A362974.
The relative asymptotic density of this sequence within the cubefull numbers is 1/2.
In general, the relative asymptotic density of the s-full numbers (numbers whose exponents in their prime factorization are all >= s) with an even number of prime factors (counted with multiplicity) within the s-full numbers is 1/2 when s is odd.

Crossrefs

Intersection of A036966 and A028260.
Complement of A370788 within A036966.
Subsequence of A370785.
Cf. A362974.

Programs

  • Mathematica
    q[n_] := Module[{e = FactorInteger[n][[;; , 2]]}, AllTrue[e, # > 2 &] && OddQ[Total[e]]]; Select[Range[30000], q]
  • PARI
    is(n) = {my(e = factor(n)[, 2]); n > 1 && vecmin(e) > 2 && vecsum(e)%2;}
Showing 1-10 of 18 results. Next