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).
Corresponding values of numbers k: 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, ...
First prime term 55414379 (= A341214(5)) of this sequence is the smallest prime p such that p, p - 1, p - 2, p - 3 and p - 4 have 2, 4, 6, 8 and 10 divisors respectively.
LINKS
Robert G. Wilson v, Table of n, a(n) for n = 1..2590
EXAMPLE
tau(154375) = 20, tau(154376) = 16, tau(154377) = 12, tau(154378) = 8, tau(154379) = 4.
MATHEMATICA
seq[max_, n_] := Module[{d = DivisorSigma[0, Range[n]], s = {}}, Do[If[Length @ Union[d/Range[n, 1, -1]] == 1, AppendTo[s, k - 1]]; d = Join[Rest@d, {DivisorSigma[0, k]}], {k, n + 1, max}]; s]; seq[5*10^6, 5] (* Amiram Eldar, Feb 08 2021 *)
PROG
(Magma) [m: m in [5..10^6] | #Divisors(m - 1) eq 2*#Divisors(m) and #Divisors(m - 2) eq 3*#Divisors(m) and #Divisors(m - 3) eq 4*#Divisors(m) and #Divisors(m - 4) eq 5*#Divisors(m)]
(Python)
def tau(n): # A000005
d, t = 1, 0
while d*d < n:
if n%d == 0:
t = t+2
d = d+1
if d*d == n:
t = t+1
return t
n, a = 1, 2
while n <= 27:
nn, t1 = 1, tau(a)
while nn < 5 and tau(a-nn) == (nn+1)*t1:
nn = nn+1
if nn == 5:
print(n, a)
n = n+1
a = a+1 # A.H.M. Smeets, Feb 07 2021
(PARI) isok(m) = if (m>5, my(nb=numdiv(m)); (numdiv(m-1) == 2*nb) && (numdiv(m-2) == 3*nb) && (numdiv(m-3) == 4*nb) && (numdiv(m-4) == 5*nb)); \\ Michel Marcus, Apr 01 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Jaroslav Krizek, Feb 07 2021
STATUS
approved