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.

A072984 Least k such that prime(n) appears in the factorization of A001008(k) (the numerator of the k-th harmonic number).

Original entry on oeis.org

2, 4, 6, 3, 12, 16, 18, 22, 13, 30, 17, 40, 13, 46, 22, 58, 10, 66, 70, 72, 78, 82, 88, 11, 100, 102, 106, 25, 112, 126, 130, 5, 138, 148, 150, 156, 162, 166, 71, 178, 180, 190, 192, 196, 38, 210, 222, 22, 228, 232, 238, 240, 250, 66, 262, 33, 58, 276, 280, 282
Offset: 2

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Author

Benoit Cloitre, Aug 21 2002

Keywords

Comments

a(n)<=n for n =2,5,14,18,25,29,33,46,49,...
For p = prime(n), Boyd defines J_p to be the set of numbers k such that p divides A001008(k). This sequence gives the smallest elements of J_p. The largest elements of J_p are given by A177734. The sizes of J_p are given by A092103.

Crossrefs

Cf. A092101 (harmonic primes), A092102 (non-harmonic primes), A092103 (size of Jp).

Programs

  • Mathematica
    A072984[n_] := Module[{p, k, sum},
       p = Prime[n]; k = 1; sum = 1/k;
       While[! Divisible[Numerator[sum], p],
        k++; sum += 1/k];
       Return[k]];
    Table[A072984[n], {n, 2, 61}] (* Robert Price, May 01 2019 *)
  • PARI
    a(n)=if(n<0,0,s=1; while(numerator(sum(k=1,s,1/k))%prime(n)>0,s++); s)

A092103 Number of values of k for which prime(n) divides A001008(k), the numerator of the k-th harmonic number.

Original entry on oeis.org

3, 3, 13, 638, 3, 3, 25, 3, 18, 26, 15, 3, 27, 24, 17, 23, 13, 3, 45, 3, 3, 43038, 7, 74, 44, 63, 3, 1273, 3, 3515, 7, 38, 3, 3, 7, 3, 74, 526, 288, 3, 19, 3, 3, 41, 11, 59, 3, 31, 65, 176, 3, 3, 3, 20, 3, 106, 55, 3, 3, 89, 3, 3, 3, 79, 3, 3, 3, 47, 3, 21, 253, 29, 7, 79, 41, 19, 701533, 13, 9, 703, 23, 3, 205, 105, 3, 3, 323, 3, 7, 3, 3, 3, 3, 3, 3, 13, 1763
Offset: 2

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Author

T. D. Noe, Feb 20 2004

Keywords

Comments

For p = prime(n), Boyd defines J_p to be the set of numbers k such that p divides A001008(k). This sequence gives the size of J_p. A072984 and A177734 give the smallest and largest elements of J_p, respectively.
A092101 gives primes prime(n) such that a(n) = 3 (i.e., a(A000720(A092101(m))) = 3 for all m). A092102 gives primes prime(n) such that a(n) > 3.
From Carlo Sanna, Apr 06 2016: (Start)
Eswarathasan and Levine conjectured that for any prime number p the set J_p is finite.
I proved that if J_p(x) is the number of integers in J_p that are less than x > 1, then J_p(x) < 129 p^(2/3) x^0.765 for any prime p. In particular, J_p has asymptotic density zero. (End)
Bing-Ling Wu and Yong-Gao Chen improved Sanna's (see previous comment) result showing that J_p(x) <= 3 x^(2/3 + 1/(25 log p)) for any prime p and any x > 1. - Carlo Sanna, Jan 12 2017

Examples

			a(2) = 3 because 3 divides A001008(k) for k = 2, 7, and 22.
a(4) = 13 because 7 divides A001008(k) for only the 13 values k = 6, 42, 48, 295, 299, 337, 341, 2096, 2390, 14675, 16731, 16735, and 102728. This is the 4th row in A229493.
		

Crossrefs

Cf. A092193 (number of generations for each prime).
Cf. A229493 (terms for each prime).

Extensions

a(8), a(15), and a(17) corrected by Max Alekseyev, Oct 23 2012
Terms a(23) onward from Carofiglio et al. (2025) added by Max Alekseyev, Apr 01 2025

A329061 Greatest k such that A002805(k) is not divisible by n, or a(n) = 0 if there's no such k.

Original entry on oeis.org

0, 1, 68, 3, 124, 68, 719102, 7, 206, 124, 11130347490407364042652446389727, 68, 2196, 719102, 124, 15, 4912, 206, 16128612858, 124, 719102, 11130347490407364042652446389727, 12166, 68, 624, 2196, 620, 719102, 20171036, 124, 27488495831, 31, 11130347490407364042652446389727
Offset: 1

Views

Author

Jinyuan Wang, Dec 07 2019

Keywords

Comments

There are two cases where a(n) = 0: (a) n divides A002805(k) for all k, which only happens for n = 1; (b) there are infinitely many k such that n does not divide A002805(k), which may happen for some primes p and their multiples.
For k > a(n) > 0, A002805(k) is always divisible by n.
For prime p and k >= p, A002805(k) = (the denominator of s + (Sum_{i=1..floor(k/p)} 1/i)/p) is not divisible by p if and only if p divides A001008(floor(k/p)) = (the numerator of Sum_{i=1..floor(k/p)} 1/i), because the denominator of s = Sum_{1 <= i <= k, i is not divisible by p} 1/i can never be divisible by p.
If k == -1 or 0 (mod p), then p divides A001008(k) iff p^2 divides A001008(floor(k/p)), otherwise p divides A001008(k) iff p divides the numerator of (Sum_{i=floor(k/p)*p+1..k} 1/i) + (Sum_{i=1..floor(k/p)} 1/i)/p, where p is an odd prime and k >= p. (Since Sum_{i=1..p-1} (p-1)!/i = (-1)^((p-1)/2)*((p-1)/2)!*(Sum_{i=1..(p-1)/2} ((p-1)/2)!/i) + ((p-1)/2)!*(Sum_{i=1..(p-1)/2} (-1)^((p-1)/2)*((p-1)/2)!/(-i)) == 0 (mod p), odd prime p divides the numerator of Sum_{1 <= i <= floor(k/p)*p, i is not divisible by p} 1/i.)

Examples

			For p = 3, 3 divides numerator(1+1/2), so 2*3, 2*3 + 1 and 2*3 + 2 are such k that A002805(k) can't be divisible by 3. Similarly, 7*3, 7*3 + 1 and 7*3 + 2 are such k. Mod(A001008(7), 3) > 0 and Mod(numerator(1/22 + (Sum_{i=1..7} 1/i)/3), 3) = 0, hence 3 divides A001008(22), which means 22*3, 22*3 + 1 and 22*3 + 2 are also such k. a(3) = 68 because A001008(k) can never be divisible by 3 for k = 66, 67 and 68.
		

Crossrefs

Formula

If n = Product_{j=1..i} p_j^e_j, p_1 < ... < p_i are primes and a(p_j^e_j) > 0, then a(n) = Max_{j=1..i} a(p_j^e_j).
a(p^e) = p^(e-1)*(a(p)+1) - 1 for prime p and a(p) > 0. Proof: A001008(k)/A002805(k) = (Sum_{1 <= i <= k, i is not divisible by p^e} 1/i) + (Sum_{i=1..floor(k/p^e)} 1/i)/p^e), hence A002805(k) is not divisible by p^e if and only if p divides A001008(floor(k/p^e)). From the comment, we know that (a(p)+1)/p - 1 is the greatest m such that p divides A001008(m). Therefore, a(p^e) = p^e*((a(p)+1)/p-1) + p^e - 1 = p^(e-1)*(a(p)+1) - 1.
a(prime(i)) = (A177734(i)+1)*prime(i) - 1, where prime(i) = A000040(i). - Jinyuan Wang, Feb 06 2020

Extensions

More terms from Jinyuan Wang, Feb 06 2020
Showing 1-3 of 3 results.