OFFSET
1,1
COMMENTS
Numbers m such that tau(m) = tau(m + 1)/2 = tau(m + 2)/3 = tau(m + 3)/4 = tau(m + 4)/5, where tau(k) = the number of divisors of k (A000005).
Quintuples of [tau(a(n)), tau(a(n) + 1), tau(a(n) + 2), tau(a(n) + 3), tau(a(n) + 4)] = [tau(a(n)), 2*tau(a(n)), 3*tau(a(n)), 4*tau(a(n)), 5*tau(a(n))]: [4, 8, 12, 16, 20], [4, 8, 12, 16, 20], [4, 8, 12, 16, 20], [8, 16, 24, 32, 40], [4, 8, 12, 16, 20], [4, 8, 12, 16, 20], ...
Corresponding values of numbers k: 4, 4, 4, 8, 4, 4, 4, 4, 4, 8, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 8, 4, 4, 4, 4, 4, 4, 4, ...
1524085621 is the smallest prime term (see A294528).
EXAMPLE
tau(211082) = 4, tau(211083) = 8, tau(211084) = 12, tau(211085) = 16, tau(211086) = 20.
MATHEMATICA
Select[Range[5*10^6], Equal @@ (DivisorSigma[0, # + {0, 1, 2, 3, 4}]/{1, 2, 3, 4, 5}) &] (* Amiram Eldar, Dec 30 2020 *)
PROG
(Magma) [m: m in [1..10^6] | #Divisors(m) eq #Divisors(m + 1)/2 and #Divisors(m) eq #Divisors(m + 2)/3 and #Divisors(m) eq #Divisors(m + 3)/4 and #Divisors(m) eq #Divisors(m + 4)/5]
(PARI) isok(m) = my(k = numdiv(m)); (numdiv(m+1) == 2*k) && (numdiv(m+2) == 3*k) && (numdiv(m+3) == 4*k) && (numdiv(m+4) == 5*k); \\ Michel Marcus, Jan 16 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Jaroslav Krizek, Dec 29 2020
STATUS
approved