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 461 results. Next

A286708 Powerful numbers (A001694) that are not prime powers (A000961).

Original entry on oeis.org

36, 72, 100, 108, 144, 196, 200, 216, 225, 288, 324, 392, 400, 432, 441, 484, 500, 576, 648, 675, 676, 784, 800, 864, 900, 968, 972, 1000, 1089, 1125, 1152, 1156, 1225, 1296, 1323, 1352, 1372, 1444, 1521, 1568, 1600, 1728, 1764, 1800, 1936, 1944, 2000, 2025, 2116, 2304, 2312, 2500, 2592, 2601, 2700, 2704, 2744
Offset: 1

Views

Author

Ilya Gutkovskiy, May 13 2017

Keywords

Comments

If a prime p divides a(n) then p^2 must also divide a(n) and number of distinct primes dividing a(n) > 1.
Intersection of A001694 and A024619.

Examples

			-------------------------------
| n | a(n) | prime            |
|   |      | factorization    |
|------------------------------
| 1 | 36   | {{2, 2}, {3, 2}} |
| 2 | 72   | {{2, 3}, {3, 2}} |
| 3 | 100  | {{2, 2}, {5, 2}} |
| 4 | 108  | {{2, 2}, {3, 3}} |
| 5 | 144  | {{2, 4}, {3, 2}} |
| 6 | 196  | {{2, 2}, {7, 2}} |
| 7 | 200  | {{2, 3}, {5, 2}} |
| 8 | 216  | {{2, 3}, {3, 3}} |
| 9 | 225  | {{3, 2}, {5, 2}} |
-------------------------------
a(n) = p_1^e_1*p_2^e_2*... : {{p_1, e_1}, {p_2, e_2}, ...}.
		

Crossrefs

Programs

  • Maple
    N:= 10000:
    S:= {1}: P:= {1}:
    p:= 1:
    do
      p:= nextprime(p);
      if p^2 > N then break fi;
      S:= map(s -> (s, seq(s*p^k, k = 2 .. floor(log[p](N/s)))), S);
      P:= P union {seq(p^k, k=2..floor(log[p](N)))}:
    od:
    sort(convert(S minus P, list)); # Robert Israel, May 14 2017
  • Mathematica
    Select[Range@2750, Min@FactorInteger[#][[All, 2]] > 1 && ! PrimePowerQ[#] &]
    (* Second program *)
    nn = 2^25; Select[Rest@ Union@ Flatten@ Table[a^2*b^3, {b, nn^(1/3)}, {a, Sqrt[nn/b^3]}], ! PrimePowerQ[#] &] (* Michael De Vlieger, Jun 22 2022 *)
  • Python
    from sympy import primefactors, factorint
    print([n for n in range(4,2745) if len(primefactors(n)) > 1 and min(list(factorint(n).values())) > 1]) # Karl-Heinz Hofmann, Feb 07 2023
    
  • Python
    from math import isqrt
    from sympy import integer_nthroot, primepi, mobius
    def A286708(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+1+sum(primepi(integer_nthroot(x, k)[0]) for k in range(2, x.bit_length()))
        return bisection(f,n,n) # Chai Wah Wu, Sep 10 2024

Formula

Sum_{n>=1} 1/a(n) = zeta(2)*zeta(3)/zeta(6) - Sum_{p prime} 1/(p*(p-1)) - 1 = A082695 - A136141 - 1 = 0.17043976777096407719... - Amiram Eldar, Feb 12 2021

A076445 The smaller of a pair of powerful numbers (A001694) that differ by 2.

Original entry on oeis.org

25, 70225, 130576327, 189750625, 512706121225, 13837575261123, 99612037019889, 1385331749802025, 3743165875258953025, 10114032809617941274225, 8905398244301708746029223, 27328112908421802064005625, 73840550964522899559001927225
Offset: 1

Views

Author

Jud McCranie, Oct 15 2002

Keywords

Comments

Erdos conjectured that there aren't three consecutive powerful numbers and no examples are known. There are an infinite number of powerful numbers differing by 1 (cf. A060355). A requirement for three consecutive powerful numbers is a pair that differ by 2 (necessarily odd). These pairs are much more rare.
Sentance gives a method for constructing families of these numbers from the solutions of Pell equations x^2-my^2=1 for certain m whose square root has a particularly simple form as a continued fraction. Sentance's result can be generalized to any m such that A002350(m) is even. These m, which generate all consecutive odd powerful numbers, are in A118894. - T. D. Noe, May 04 2006

Examples

			25=5^2 and 27=3^3 are powerful numbers differing by 2, so 25 is in the sequence.
		

References

  • R. K. Guy, Unsolved Problems in Number Theory, B16

Crossrefs

Extensions

a(8)-a(10) from Geoffrey Reynolds (geoff(AT)hisplace.co.nz), Feb 15 2005
More terms from T. D. Noe, May 04 2006

A076871 Sum of two powerful numbers (definition (1), A001694).

Original entry on oeis.org

2, 5, 8, 9, 10, 12, 13, 16, 17, 18, 20, 24, 25, 26, 28, 29, 31, 32, 33, 34, 35, 36, 37, 40, 41, 43, 44, 45, 48, 50, 52, 53, 54, 57, 58, 59, 61, 63, 64, 65, 68, 72, 73, 74, 76, 80, 81, 82, 85, 88, 89, 90, 91, 96, 97, 98, 99, 100, 101, 104, 106, 108, 109, 112, 113, 116, 117
Offset: 1

Views

Author

N. J. A. Sloane, Nov 25 2002

Keywords

Comments

Complement of A085253. - Reinhard Zumkeller, Jun 23 2003

References

  • Aleksandar Ivić, The Riemann Zeta-Function, Wiley, NY, 1985, see p. 439.

Crossrefs

Different from A070049.

Programs

  • Mathematica
    With[{m = 120}, pow = Select[Range[m], # == 1 || Min[FactorInteger[#][[;; , 2]]] > 1 &]; Select[Union[Plus @@@ Tuples[pow, {2}]], # <= m &]] (* Amiram Eldar, Jan 30 2023 *)

Formula

A085252(a(n)) > 0. - Reinhard Zumkeller, Jun 23 2003
Blomer shows that there are x/log^k x sums of two powerful numbers up to x, where k = 0.20629947... is A261883. - Charles R Greathouse IV, Sep 04 2015

Extensions

More terms from Vladeta Jovovic, Nov 25 2002

A320966 Powerful numbers A001694 divisible by a cube > 1.

Original entry on oeis.org

8, 16, 27, 32, 64, 72, 81, 108, 125, 128, 144, 200, 216, 243, 256, 288, 324, 343, 392, 400, 432, 500, 512, 576, 625, 648, 675, 729, 784, 800, 864, 968, 972, 1000, 1024, 1125, 1152, 1296, 1323, 1331, 1352, 1372, 1568, 1600, 1728, 1800, 1936, 1944, 2000, 2025, 2048, 2187, 2197, 2304, 2312, 2401, 2500
Offset: 1

Views

Author

Hugo Pfoertner, Oct 25 2018

Keywords

Comments

Powerful numbers that are not squares of squarefree numbers. - Amiram Eldar, Jun 25 2022

Crossrefs

Intersection of A001694 and A046099.

Programs

  • Mathematica
    Select[Range[2500], (m = MinMax[FactorInteger[#][[;; , 2]]])[[1]] > 1 && m[[2]] > 2 &] (* Amiram Eldar, Jun 25 2022 *)
  • PARI
    isA001694(n)=n=factor(n)[, 2]; for(i=1, #n, if(n[i]==1, return(0))); 1 \\ from Charles R Greathouse IV
    isA046099(n)=n=factor(n)[, 2]; for(i=1, #n, if(n[i]>2, return(1)));0
    for (k=1,2500,if(isA001694(k)&&isA046099(k),print1(k,", ")))
    
  • Python
    from math import isqrt
    from sympy import mobius, integer_nthroot
    def A320966(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+squarefreepi(isqrt(x))-squarefreepi(integer_nthroot(x,3)[0]), 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)
            return c+l
        return bisection(f,n,n) # Chai Wah Wu, Sep 15 2024

Formula

Sum_{n>=1} 1/a(n) = zeta(2)*zeta(3)/zeta(6) - 15/Pi^2 = 0.4237786821... . - Amiram Eldar, Jun 25 2022

A085253 Numbers having no representation as sum of two powerful numbers (A001694).

Original entry on oeis.org

1, 3, 4, 6, 7, 11, 14, 15, 19, 21, 22, 23, 27, 30, 38, 39, 42, 46, 47, 49, 51, 55, 56, 60, 62, 66, 67, 69, 70, 71, 75, 77, 78, 79, 83, 84, 86, 87, 92, 93, 94, 95, 102, 103, 105, 107, 110, 111, 114, 115, 118, 119, 120, 123, 131, 138, 139, 142, 143, 147, 151, 154
Offset: 1

Views

Author

Reinhard Zumkeller, Jun 23 2003

Keywords

Comments

Complement of A076871.

Crossrefs

Different from A075434.

Programs

  • Mathematica
    With[{m = 160}, pow = Select[Range[m], # == 1 || Min[FactorInteger[#][[;; , 2]]] > 1 &]; Complement[Range[m], Select[Union[Plus @@@ Tuples[pow, {2}]], # <= m &]]] (* Amiram Eldar, Jan 30 2023 *)

Formula

A085252(a(n)) = 0.

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

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

A258599 a(n) is the index m such that A001694(m) = prime(n)^2.

Original entry on oeis.org

2, 4, 6, 10, 16, 20, 28, 31, 39, 48, 51, 65, 71, 75, 84, 94, 107, 110, 120, 129, 133, 145, 152, 163, 180, 187, 191, 199, 202, 212, 238, 246, 258, 261, 282, 286, 297, 309, 319, 330, 342, 344, 366, 372, 377, 382, 407, 431, 440, 443, 450, 463, 468, 487, 498
Offset: 1

Views

Author

Reinhard Zumkeller, Jun 06 2015

Keywords

Examples

			.   n |  p |  a(n) | A001694(a(n)) = A001248(n) = p^2
. ----+----+-------+---------------------------------
.   1 |  2 |     2 |             4
.   2 |  3 |     4 |             9
.   3 |  5 |     6 |            25
.   4 |  7 |    10 |            49
.   5 | 11 |    16 |           121
.   6 | 13 |    20 |           169
.   7 | 17 |    28 |           289
.   8 | 19 |    31 |           361
.   9 | 23 |    39 |           529
.  10 | 29 |    48 |           841
.  11 | 31 |    51 |           961
.  12 | 37 |    65 |          1369
.  13 | 41 |    71 |          1681
.  14 | 43 |    75 |          1849
.  15 | 47 |    84 |          2209
.  16 | 53 |    94 |          2809
.  17 | 59 |   107 |          3481
.  18 | 61 |   110 |          3721
.  19 | 67 |   120 |          4489
.  20 | 71 |   129 |          5041
.  21 | 73 |   133 |          5329
.  22 | 79 |   145 |          6241
.  23 | 83 |   152 |          6889
.  24 | 89 |   163 |          7921
.  25 | 97 |   180 |          9409  .
		

Crossrefs

Programs

  • Haskell
    import Data.List (elemIndex); import Data.Maybe (fromJust)
    a258599 = (+ 1) . fromJust . (`elemIndex` a258567_list) . a000040
    
  • Mathematica
    With[{m = 60}, c = Select[Range[Prime[m]^2], Min[FactorInteger[#][[;; , 2]]] > 1 &]; 1 + Flatten[FirstPosition[c, #] & /@ (Prime[Range[m]]^2)]] (* Amiram Eldar, Feb 07 2023 *)
  • Python
    from math import isqrt
    from sympy import prime, integer_nthroot, factorint
    def A258599(n):
        m = prime(n)**2
        return int(sum(isqrt(m//k**3) for k in range(1, integer_nthroot(m, 3)[0]+1) if all(d<=1 for d in factorint(k).values()))) # Chai Wah Wu, Sep 10 2024

Formula

A001694(a(n)) = A001248(n) = prime(n)^2.
A001694(m) mod prime(n) > 0 for m < a(n).
Also smallest number m such that A258567(m) = prime(n):
A258567(a(n)) = A000040(n) and A258567(m) != A000040(n) for m < a(n).

A323333 The Euler phi function values of the powerful numbers, A000010(A001694(n)).

Original entry on oeis.org

1, 2, 4, 6, 8, 20, 18, 16, 12, 42, 32, 24, 54, 40, 36, 110, 100, 64, 48, 156, 84, 80, 72, 120, 162, 128, 96, 272, 108, 294, 342, 168, 160, 144, 252, 220, 200, 256, 506, 192, 500, 216, 360, 312, 486, 336, 320, 812, 288, 240, 930, 440, 324, 400, 512, 660, 600
Offset: 1

Views

Author

Amiram Eldar, Jan 11 2019

Keywords

Comments

The sum of the reciprocals of all the terms of this sequence is Murata's constant Product_{p prime}(1 + 1/(p-1)^2) (A065485).
Sequence is injective: no value occurs more than once. - Amiram Eldar and Antti Karttunen, Sep 30 2019

Crossrefs

Cf. A000010, A001694, A002618 (a subsequence), A065485, A082695, A112526, A323332.

Programs

  • Mathematica
    EulerPhi /@ Join[{1}, Select[Range@ 1200, Min@ FactorInteger[#][[All, 2]] > 1 &]] (* after Harvey P. Dale at A001694 *)
  • PARI
    lista(nn) = apply(x->eulerphi(x), select(x->ispowerful(x), vector(nn, k, k))); \\ Michel Marcus, Jan 11 2019

A360720 a(n) is the sum of unitary divisors of n that are powerful (A001694).

Original entry on oeis.org

1, 1, 1, 5, 1, 1, 1, 9, 10, 1, 1, 5, 1, 1, 1, 17, 1, 10, 1, 5, 1, 1, 1, 9, 26, 1, 28, 5, 1, 1, 1, 33, 1, 1, 1, 50, 1, 1, 1, 9, 1, 1, 1, 5, 10, 1, 1, 17, 50, 26, 1, 5, 1, 28, 1, 9, 1, 1, 1, 5, 1, 1, 10, 65, 1, 1, 1, 5, 1, 1, 1, 90, 1, 1, 26, 5, 1, 1, 1, 17, 82
Offset: 1

Views

Author

Amiram Eldar, Feb 18 2023

Keywords

Comments

The number of these divisors is given by A323308.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e == 1, 1, p^e + 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i, 2] == 1, 1, f[i, 1]^f[i, 2] + 1));}
    
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, (1 - p^3*X^4 - p^2*X^3 + p^3*X^3) / ((1 - X) * (1 - p^2*X^2)))[n], ", ")) \\ Vaclav Kotesovec, Feb 18 2023

Formula

Multiplicative with a(p) = 1 and a(p^e) = p^e + 1 for e > 1.
a(n) <= A034448(n), with equality if and only if n is powerful (A001694).
a(n) <= A183097(n), with equality if and only if n is cubefree (A004709).
Dirichlet g.f.: zeta(s)*zeta(s-1)*Product_{p prime} (1 - p^(1-s) + p^(2-2*s) - p^(2-3*s)).
From Vaclav Kotesovec, Feb 18 2023: (Start)
Dirichlet g.f.: zeta(s) * zeta(2*s-2) * Product_{primes p} (1 - p^(3-4*s) - p^(2-3*s) + p^(3-3*s)).
Sum_{k=1..n} a(k) ~ c * zeta(3/2) * n^(3/2) / 3, where c = Product_{primes p} (1 + 1/p^(3/2) - 1/p^(5/2) - 1/p^3) = 1.48039182258752809541724060173644... (End)
a(n) = A034448(A057521(n)) (the sum of unitary divisors of the powerful part of n). - Amiram Eldar, Dec 12 2023
a(n) = A034448(n)/A092261(n). - Amiram Eldar, Jun 19 2025
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