A090526 -id:A090526 - cp's OEIS frontend

A090525 Least number k such that floor((n^n)/k) is prime, or 0 if no such number exists.

2, 2, 11, 9, 11, 20, 10, 42, 13, 16, 57, 7, 35, 41, 53, 35, 171, 18, 141, 45, 19, 55, 212, 236, 94, 265, 13, 76, 26, 9, 13, 21, 160, 21, 21, 24, 378, 100, 66, 52, 75, 54, 214, 6, 678, 193, 137, 123, 138, 59, 605, 87, 165, 109, 417, 403, 100, 57, 778, 719, 79, 12, 83, 450

Offset: 2

Amarnath Murthy, Dec 07 2003

nonn

Conjecture: No term is zero.

The conjecture is true: If p is a prime factor of n, k = (n^n)/p gives an upper bound. - James Rayman, Mar 01 2023

Robert Israel, Table of n, a(n) for n = 2..577

Cf. A090526, A090527, A090528.

f:= proc(n) local t,k;
  t:= n^n;
  for k from 2 do if isprime(floor(t/k)) then return k fi od
end proc:
map(f, [$2..100]); # Robert Israel, Mar 02 2023

lnk[n_]:=Module[{k=1,nn=n^n},While[!PrimeQ[Floor[nn/k]],k++];k]; Array[lnk,70,2] (* Harvey P. Dale, Nov 07 2022 *)

a(n)=for(i=1,1000,if(isprime(floor((n^n)/i)),print1(i,",");break()))

A090527 Smallest prime p such that floor((n^n)/p) is prime, or 0 if no such number exists.

Original entry on oeis.org

2, 2, 11, 29, 11, 137, 79, 149, 13, 17, 181, 7, 71, 41, 53, 541, 197, 61, 149, 149, 19, 541, 1663, 829, 229, 599, 13, 563, 113, 137, 13, 1129, 421, 1759, 683, 389, 919, 877, 233, 1933, 2137, 97, 331, 881, 1753, 193, 137, 521, 1063, 59

Offset: 2

Amarnath Murthy, Dec 07 2003

nonn

Conjecture: No term is zero.

As long as p(j+1)/p(j) < 2 for all j, then for any integer n >= 4, there exists at least one p such that p and floor(n/p) are both prime. (I do not know a proof for the premise above; however, it seems quite weak compared to other conjectures and theorems about primes. It may be that it follows from the results in sequence A038458.) In fact, there exists a prime p such that either floor(n/p) = 2 or floor(n/p) = 3. Outline of proof: (1) If p is a prime number, then for all n with 2p <= n < 3p, floor(n/p) = 2, which is prime. (2) In addition, for all n with 3p <= n < 4p, floor(n/p) = 3, which is prime. So for any n >= 4, consider the largest prime, p, with 2p <= n. (3) floor(n/p) can't be less than 2, since 2 <= n/p. (4) If floor(n/p) = 2, then p and floor(n/p) are both prime, so we are done. (5) Similarly, if floor(n/p) = 3, we are done. The only remaining case is that 4p <= n. Let p_1 be the next prime after p. (6) p_n must not meet 2(p_1) <= n, since p is the largest that does. Therefore 2(p_1) > n. (7) 4p <= n < 2(p_1) (8) (p_1 / p) > 2 (9) As long as p(j+1)/p(j) < 2 for all j, the case of 4p <= n is not possible. - Weston Markham (WMarkham(AT)paradigmgenetics.com), Jun 15 2004

Cf. A090525, A090526, A090528.

Mathematica
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<Ryan Propper, Aug 02 2005 *)
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a(7)-a(14) from Weston Markham (WMarkham(AT)paradigmgenetics.com), Jun 15 2004

a(15)-a(51) from Ryan Propper, Aug 02 2005

A090528 Primes arising in A090527.

Original entry on oeis.org

2, 13, 23, 107, 4241, 6011, 212369, 2600137, 769230769, 16783039447, 49260223471, 43267872370321, 156507138388141, 10680338789777057, 348051774975651917, 1529094753948866477, 199727959773078870941, 32433109109185468674163

Offset: 1

Amarnath Murthy, Dec 07 2003

nonn

Conjecture: No term is zero.

Cf. A090525, A090526, A090527.

Do[p = n^n; i = 1; While[ !ProvablePrimeQ[Floor[p/Prime[i]]], i++ ]; Print[Floor[p/Prime[i]]], {n, 2, 50}] (* Ryan Propper, Aug 02 2005 *)

More terms from Ryan Propper, Aug 02 2005

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A090525 Least number k such that floor((n^n)/k) is prime, or 0 if no such number exists.

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A090527 Smallest prime p such that floor((n^n)/p) is prime, or 0 if no such number exists.

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A090528 Primes arising in A090527.

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