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: 5, 7, 13, 37, 61, 73, 157, 193, 277, 313, 397, 421, ...
n = 3: 23, 29, 41, 53, 101, 103, 109, 113, 127, 137, 151, ...
n = 4: 167, 263, 269, 311, 383, 389, 439, 461, 509, 569, ...
n = 5: 839, 1319, 1511, 1559, 1847, 1889, 2039, 2309, 2687, ...
Conjecture: a(n) is also the smallest number m such that tau(m+1) = tau(m)^n.
LINKS
David A. Corneth, Table of n, a(n) for n = 1..31
EXAMPLE
a(4) = 167 because 167 is the smallest prime p such that tau(p+1) = 16 = 2^4.
MATHEMATICA
Do[p = 1; While[DivisorSigma[0, Prime[p] + 1] != 2^n, p++]; Print[n, " ", Prime[p]], {n, 1, 9}] (* Vaclav Kotesovec, Apr 03 2021 *)
PROG
(Magma) Ax:=func<n|exists(r){m:m in[1..10^6] | IsPrime(m) and #Divisors(m + 1) eq 2 ^ n} select r else 0>; [Ax(n): n in [1..7]]
(PARI) a(n) = my(t=2^n); forprime(p=2, oo, if(numdiv(p+1)==t, return(p))); \\ Jinyuan Wang, Apr 02 2021
(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 True
sizeok=False
cnt=True
while cnt:
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)
ok=solve(v*f, k-1, i, j)
stack.pop()
sizeok|=ok
cnt&=ok
else:
cnt=False
return sizeok
def a343020(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 02 2021
EXTENSIONS
a(11) from Jinyuan Wang, Apr 02 2021
More terms from David A. Corneth, Apr 09 2021
STATUS
approved