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 1-3 of 3 results.

A333072 Least k such that Sum_{i=1..n} k^i / i is a positive integer.

Original entry on oeis.org

1, 2, 6, 6, 30, 10, 70, 70, 210, 168, 1848, 1848, 18018, 8580, 2574, 2574, 102102, 102102, 831402, 2771340, 3233230, 587860, 43266496, 117630786, 162249360, 145088370, 145088370, 2897310, 672175920, 672175920, 18232771830, 18232771830, 44279588730, 8886561060
Offset: 1

Views

Author

Jinyuan Wang, Mar 10 2020

Keywords

Comments

Note that the denominator of (Sum_{i=1..n} k^i/i) - k^p/p can never be divisible by p, where n/2 < p <= n. Therefore, for the expression to be an integer, such p must divide k. Thus, a(n) = k is divisible by A055773(n).

Crossrefs

Programs

  • PARI
    a(n) = {my(m = prod(i=primepi(n/2)+1, primepi(n), prime(i)), k = m); while (denominator(sum(i=2, n, k^i/i)) != 1, k += m); k; }
    
  • Python
    from sympy import primorial, lcm
    def A333072(n):
        f = 1
        for i in range(1,n+1):
            f = lcm(f,i)
        f, glist = int(f), []
        for i in range(1,n+1):
            glist.append(f//i)
        m = 1 if n < 2 else primorial(n,nth=False)//primorial(n//2,nth=False)
        k = m
        while True:
            p,ki = 0, k
            for i in range(1,n+1):
                p = (p+ki*glist[i-1]) % f
                ki = (k*ki) % f
            if p == 0:
                return k
            k += m # Chai Wah Wu, Apr 04 2020

Formula

a(n) <= A034386(n).

A330030 Least k such that Sum_{i=0..n} k^n / i! is a positive integer.

Original entry on oeis.org

1, 1, 2, 3, 6, 30, 30, 42, 210, 42, 210, 2310, 2310, 30030, 30030, 30030, 30030, 39270, 510510, 1939938, 9699690, 9699690, 9699690, 17160990, 223092870, 903210, 223092870, 223092870, 223092870, 6469693230, 6469693230, 200560490130, 200560490130, 10555815270, 200560490130
Offset: 0

Views

Author

Jinyuan Wang, Mar 07 2020

Keywords

Comments

Least k > 0 such that k^n/A061355(n) is an integer.

Examples

			For n = 7, the denominator of Sum_{i=0..7} 1/i! is 252 = 2^2*3^2*7, so a(7) = 2*3*7 = 42.
		

Crossrefs

Programs

  • PARI
    a(n) = factorback(factorint(denominator(sum(i=2, n, 1/i!)))[, 1]);

Formula

a(n) = A007947(A061355(n)).

A382213 Largest squarefree number dividing the numerator of harmonic number H(n).

Original entry on oeis.org

1, 3, 11, 5, 137, 7, 33, 761, 7129, 671, 83711, 6617, 1145993, 1171733, 1195757, 143327, 42142223, 751279, 275295799, 55835135, 18858053, 830139, 444316699, 1347822955, 34052522467, 34395742267, 312536252003, 10876020307, 214582477009, 300151059037, 290774257297357
Offset: 1

Views

Author

Ali Sada, Mar 22 2025

Keywords

Examples

			A001008(4) = 25, and the largest squarefree divisor of 25 is 5, so a(4) = 5.
		

Crossrefs

Programs

  • Maple
    a:= n-> mul(i[1], i=ifactors(numer(harmonic(n)))[2]):
    seq(a(n), n=1..35);  # Alois P. Heinz, Apr 23 2025
  • PARI
    a(n) = factorback(factor(numerator(sum(k=1, n, 1/k)))[, 1]); \\ Michel Marcus, Apr 24 2025

Formula

a(n) = A007947(A001008(n)).
Showing 1-3 of 3 results.