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 1-4 of 4 results.

A175745 Numbers with 35 divisors.

Original entry on oeis.org

5184, 11664, 40000, 153664, 250000, 455625, 937024, 1265625, 1750329, 1827904, 1882384, 5345344, 8340544, 9529569, 10673289, 17909824, 20820969, 28344976, 37515625, 45265984, 59105344, 60886809, 73530625, 77228944, 95004009, 119946304, 143496441, 180848704
Offset: 1

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Author

Jaroslav Krizek, Aug 27 2010

Keywords

Comments

Numbers of the form p^34 and p^6*q^4 (A190464), where p and q are distinct primes.

Crossrefs

Programs

  • Mathematica
    Select[Range[9000000],DivisorSigma[0,#]==35&] (* Vladimir Joseph Stephan Orlovsky, May 06 2011 *)
  • PARI
    is(n)=numdiv(n)==35 \\ Charles R Greathouse IV, Jun 19 2016
    
  • Python
    from sympy import primepi, integer_nthroot, primerange
    def A175745(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x): return n+x-sum(primepi(integer_nthroot(x//p**6,4)[0]) for p in primerange(integer_nthroot(x,6)[0]+1))+primepi(integer_nthroot(x,10)[0])-primepi(integer_nthroot(x,34)[0])
        return bisection(f,n,n) # Chai Wah Wu, Feb 22 2025

Formula

A000005(a(n)) = 35.
Sum_{n>=1} 1/a(n) = P(4)*P(6) - P(10) + P(34) = 0.000320676..., where P is the prime zeta function. - Amiram Eldar, Jul 03 2022

Extensions

Extended by T. D. Noe, May 08 2011

A190465 Numbers with prime factorization p^5q^5.

Original entry on oeis.org

7776, 100000, 537824, 759375, 4084101, 5153632, 11881376, 39135393, 45435424, 52521875, 79235168, 90224199, 205962976, 345025251, 503284375, 601692057, 656356768, 916132832, 1160290625, 1564031349, 2219006624, 2706784157, 3707398432
Offset: 1

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Author

Keywords

Crossrefs

Programs

  • Mathematica
    f[n_]:=Sort[Last/@FactorInteger[n]]=={1,1}; Select[Range[10000],f]^5
  • PARI
    list(lim)=my(v=List(),t);forprime(p=2, lim^(1/10), t=p^5;forprime(q=p+1, (lim\t)^(1/5), listput(v,t*q^5))); vecsort(Vec(v)) \\ Charles R Greathouse IV, Jul 20 2011

Formula

Sum_{n>=1} 1/a(n) = (P(5)^2 - P(10))/2 = (A085965^2 - P(10))/2 = 0.000142..., where P is the prime zeta function. - Amiram Eldar, Jul 06 2020

A382292 Numbers k such that A382290(k) = 1.

Original entry on oeis.org

8, 24, 27, 32, 40, 54, 56, 64, 72, 88, 96, 104, 108, 120, 125, 135, 136, 152, 160, 168, 184, 189, 192, 200, 224, 232, 243, 248, 250, 264, 270, 280, 288, 296, 297, 312, 320, 328, 343, 344, 351, 352, 360, 375, 376, 378, 392, 408, 416, 424, 432, 440, 448, 456, 459, 472, 480, 486, 488, 500
Offset: 1

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Author

Amiram Eldar, Mar 21 2025

Keywords

Comments

First differs from A374590 and A375432 at n = 25: A374590(25) = A375432(25) = 216 is not a term of this sequence.
Numbers k such that A382291(k) = 2, i.e., numbers whose number of infinitary divisors is twice the number of their unitary divisors.
Numbers whose prime factorization has a single exponent that is a sum of two distinct powers of 2 (A018900) and all the other exponents, if they exist, are powers of 2. Equivalently, numbers of the form p^e * m, where p is a prime, e is a term in A018900, and m is a term in A138302 that is coprime to p.
If k is a term then k^2 is also a term. If m is a term in A138302 that is coprime to k then k * m is also a term. The primitive terms, i.e., the terms that cannot be generated from smaller terms using these rules, are the numbers of the form p^(2^i+1), where p is prime and i >= 1.
Analogous to A060687, which is the sequence of numbers k with prime excess A046660(k) = 2.
The asymptotic density of this sequence is A271727 * Sum_{p prime} (((1 - 1/p)/f(1/p)) * Sum_{k>=1} 1/p^A018900(k)) = 0.11919967112489084407..., where f(x) = 1 - x^3 + Sum_{k>=2} (x^(2^k)-x^(2^k+1)).

Crossrefs

Subsequences (numbers of the form): A030078 (p^3), A050997 (p^5), A030516 (p^6), A179665 (p^9), A030629 (p^10), A030631 (p^12), A065036 (p^3*q), A178740 (p^5*q), A189987 (p^6*q), A179692 (p^9*q), A143610 (p^2*q^3), A179646 (p^5*q^2), A189990 (p^2*q^6), A179702 (p^4*q^5), A179666 (p^4*q^3), A190464 (p^4*q^6), A163569 (p^3*q^2*r), A189975 (p*q*r^3), A190115 (p^2*q^3*r^4), A381315, A048109.

Programs

  • Mathematica
    f[p_, e_] := DigitCount[e, 2, 1] - 1; q[1] = False; q[n_] := Plus @@ f @@@ FactorInteger[n] == 1; Select[Range[500], q]
  • PARI
    isok(k) = vecsum(apply(x -> hammingweight(x) - 1, factor(k)[, 2])) == 1;

A355462 Powerful numbers divisible by exactly 2 distinct primes.

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, 968, 972, 1000, 1089, 1125, 1152, 1156, 1225, 1296, 1323, 1352, 1372, 1444, 1521, 1568, 1600, 1728, 1936, 1944, 2000, 2025, 2116, 2304, 2312, 2500
Offset: 1

Views

Author

Amiram Eldar, Jul 03 2022

Keywords

Comments

First differs from A286708 at n = 25.
Number of the form p^i * q^j, where p != q are primes and i,j > 1.
Numbers k such that A001221(k) = 2 and A051904(k) >= 2.
The possible values of the number of the divisors (A000005) of terms in this sequence is any composite number that is not 8 or twice a prime (A264828 \ {1, 8}).
675 = 3^3*5^2 and 676 = 2^2*13^2 are 2 consecutive integers in this sequence. There are no other such pairs below 10^22 (the lesser members of such pairs are terms of A060355).

Examples

			36 is a term since 36 = 2^2 * 3^2.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[2500], Length[(e = FactorInteger[#][[;; , 2]])] == 2 && Min[e] > 1 &]
  • PARI
    is(n) = {my(f=factor(n)); #f~ == 2 && vecmin(f[,2]) > 1};

Formula

Sum_{n>=1} 1/a(n) = ((Sum_{p prime} (1/(p*(p-1))))^2 - Sum_{p prime} (1/(p^2*(p-1)^2)))/2 = 0.1583860791... .
Showing 1-4 of 4 results.