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A343020
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a(n) is the smallest prime p such that tau(p+1) = 2^n.
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4
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2, 5, 23, 167, 839, 7559, 128519, 1081079, 20540519, 397837439, 8031343319, 188972783999, 3212537327999, 125568306863999, 2888071057871999, 190487121512687999, 4381203794791823999, 215961289494494543999, 13283916764437951631999, 540119185025730854543999, 26465840066260811872655999, 1356699703068812438127791999
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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a(4) = 167 because 167 is the smallest prime p such that tau(p+1) = 16 = 2^4.
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MATHEMATICA
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Do[p = 1; While[DivisorSigma[0, Prime[p] + 1] != 2^n, p++]; Print[n, " ", Prime[p]], {n, 1, 9}] (* Vaclav Kotesovec, Apr 03 2021 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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