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.

A061299 Least number whose number of divisors is A007304(n) (the n-th number that is the product of 3 distinct primes).

Original entry on oeis.org

720, 2880, 46080, 25920, 184320, 2949120, 129600, 414720, 11796480, 1658880, 188743680, 3732480, 2073600, 26542080, 12079595520, 14929920, 48318382080, 106168320, 8294400, 3092376453120, 1698693120, 18662400, 238878720
Offset: 1

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Author

Labos Elemer, Jun 05 2001

Keywords

Comments

All terms are divisible by a(1) = 720, the first entry.
All terms [=a(j)], not only arguments [=j] have 3 distinct prime factors at exponents determined by the p,q,r factors of their arguments: a(pqr) = RPQ.

Examples

			For n = 5: A007304(5) = 78 = 2*3*13, A005179(78) = 184320 = (2^12)*(3^2)*(5^1) = a(5).
		

Crossrefs

Formula

a(n) = A005179(A007304(n)).
Min{x; A000005(x) = pqr} p, q, r are distinct primes. If k = pqr and p > q > r then A005179(k) = 2^(p-1)*3^(q-1)*5^(r-1).
From Reinhard Zumkeller, Jul 15 2004: (Start)
A000005(a(n)) = A007304(n).
A000005(m) != A007304(n) for m < a(n).
a(n) = A005179(A007304(n)).
a(p*m*q) = 2^(q-1) * 3^(m-1) * 5^(p-1) for primes p < m < q.
a(A000040(i)*A000040(j)*A000040(k)) = 2^(A084127(k)-1) * 3^(A084127(j)-1) * 5^(A084127(i)-1) for i < j < k. (End)

Extensions

Edited by N. J. A. Sloane, Apr 20 2007

A061234 Smallest number with prime(n)^2 divisors where prime(n) is the n-th prime.

Original entry on oeis.org

6, 36, 1296, 46656, 60466176, 2176782336, 2821109907456, 101559956668416, 131621703842267136, 6140942214464815497216, 221073919720733357899776, 10314424798490535546171949056, 13367494538843734067838845976576
Offset: 1

Views

Author

Labos Elemer, Jun 01 2001

Keywords

Examples

			1296 = 2*2*2*2*3*3*3*3 is the smallest number with 25 divisors.
		

Crossrefs

Formula

a(n) = Min_{x : d(x) = A000005(x) = p(n)^2} = 6^(p(n)-1) because x = 2^(pp-1) > 2^(p-1)3^(p-1) holds if p > 1.
a(n) = A005179(A001248(n)). - Amiram Eldar, Jun 21 2024

A061218 Least number whose number of divisors is n-th term from A014613 (numbers of form p*q*r*s, products of exactly 4 primes, counted with multiplicity).

Original entry on oeis.org

120, 360, 1260, 1680, 6300, 6720, 5040, 44100, 20160, 107520, 25200, 45360, 430080, 100800, 322560, 176400, 6881280, 181440, 226800, 27525120, 1290240, 440401920, 705600, 1632960, 1612800, 20643840, 907200, 2903040, 1587600, 82575360, 28185722880, 6451200, 112742891520
Offset: 1

Views

Author

Labos Elemer, Jun 06 2001

Keywords

Examples

			p*q*r*s = 210 is the 27th term in A014613; the smallest number with 210 divisors is 907200 = 2*2*2*2*2*2*3*3*3*3*5*5*7.
		

Crossrefs

Programs

  • Python
    from math import prod, isqrt
    from sympy import primepi, primerange, integer_nthroot, isprime, divisors, prime
    def A061218(n):
        def f(x): return int(n+x-sum(primepi(x//(k*m*r))-c for a,k in enumerate(primerange(integer_nthroot(x,4)[0]+1)) for b,m in enumerate(primerange(k,integer_nthroot(x//k,3)[0]+1),a) for c,r in enumerate(primerange(m,isqrt(x//(k*m))+1),b)))
        def mult_factors(n):
            if isprime(n):
                return [(n,)]
            c = []
            for d in divisors(n,generator=True):
                if 1Chai Wah Wu, Aug 17 2024

Formula

a(n) = A005179(A014613(n)).

Extensions

Corrected and extended by Michel Marcus, Sep 05 2017

A061236 Smallest number with prime(n)^3 divisors where prime(n) is n-th prime.

Original entry on oeis.org

24, 900, 810000, 729000000, 590490000000000, 531441000000000000, 430467210000000000000000, 387420489000000000000000000, 313810596090000000000000000000000, 228767924549610000000000000000000000000000, 205891132094649000000000000000000000000000000
Offset: 1

Views

Author

Labos Elemer, Jun 01 2001

Keywords

Examples

			If p = 2, then d(128) = d(24) = d(30) = 8 and a(1) = 24 < 30 is the smallest.
If p = 5, then 2^124 > (2^24)*(3^4) > 30^4 = 810000 = a(3).
		

Crossrefs

Formula

For p = 2, 24 is the solution. If a prime p > 2, the suitable powers of 30 are the least solutions: a(n) = Min{x | d(x) = A000005(x) = p(n)^3} = 30^(prime(n)-1). d(2^(ppp-1)) = d(2^(pp-1)*3^(p-1)) = d(30^(p-1)) = p^3 and 2^(ppp-1) > 2^(pp-1)*3^(p-1) > 30^(p-1) holds if p > 2.
a(n) = A005179(A030078(n)) = A005179(prime(n)^3). - Amiram Eldar, Jan 23 2025

Extensions

a(10)-a(11) from Amiram Eldar, Jan 23 2025
Showing 1-4 of 4 results.