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A343087
a(n) is the smallest prime p such that tau(p-1) = 2^n.
2
3, 7, 31, 211, 1321, 7561, 120121, 1580041, 24864841, 328648321, 7558911361, 162023621761, 5022732274561, 93163582512001, 4083134943888001, 151075992923856001, 5236072827921936001, 188391763176048432001, 8854412869274276304001, 469283882071536644112001, 29844457947060064452144001, 1917963226026370264485744001
OFFSET
1,1
COMMENTS
tau(m) = the number of divisors of m (A000005).
Sequences of primes p such that tau(p-1) = 2^n for 2 <= n <= 5:
n = 2: 7, 11, 23, 47, 59, 83, 107, 167, 179, ... (A005385(k) for k >= 2).
n = 3: 31, 41, 43, 67, 71, 79, 89, 103, 131, 137, 139, 191, ...
n = 4: 211, 271, 281, 313, 331, 379, 409, 457, 463, 521, 547, ...
n = 5: 1321, 2281, 2311, 2377, 2689, 2731, 2857, 2971, 3001, ...
Conjecture: a(n) is also the smallest number m such that tau(m-1) = tau(m)^n.
LINKS
EXAMPLE
For n = 4; a(4) = 211 because 211 is the smallest prime p such that tau(p - 1) = 2^4; tau(210) = 16.
PROG
(Magma) Ax:=func<n|exists(r){m: m in[1..10^8] | IsPrime(m) and #Divisors(m - 1) eq 2 ^ n} select r else 0>; [Ax(n): n in [1..9]]
(Python)
from sympy import isprime, nextprime
primes=[2]
def solve(v, k, i, j):
global record, stack, primes
if k==0:
if isprime(v+1):
record=v
return
while True:
if i>=len(primes):
primes.append(nextprime(primes[-1]))
if j<len(stack) and stack[j]<primes[i]:
f=stack[j] ; j+=1
else:
f=primes[i] ; i+=1
if record==None or v * f**k < record:
stack.append(f**2)
solve(v*f, k-1, i, j)
stack.pop()
else:
return
def a343087(n):
global record, stack
record, stack = None, []
solve(1, n, 0, 0)
return record+1
# Bert Dobbelaere, Apr 11 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Jaroslav Krizek, Apr 04 2021 (following a suggestion of Vaclav Kotesovec)
EXTENSIONS
a(11) from Vaclav Kotesovec, Apr 05 2021
More terms from Bert Dobbelaere, Apr 11 2021
STATUS
approved