A294528
a(n) is the smallest prime that begins a run of exactly n consecutive numbers having 2, 4, ..., 2*n divisors.
Original entry on oeis.org
2, 5, 61, 421, 1524085621
Offset: 1
a(3) = 61 because 61 (prime), 62 = 2*31, and 63 = 3^2*7 have 2, 4, and 6 divisors, respectively (and 64 does not have exactly 8 divisors, so 63 is the last number in the run), and there is no smaller number having this property.
a(5) = 1524085621 because the 5 consecutive integers 1524085621..1524085625 have 2, 4, 6, 8, and 10 divisors, respectively (and 1524085626 does not have exactly 12 divisors), and there is no smaller number having this property.
A341214
a(n) is the smallest prime p such that p, p - 1, p - 2, ..., p - n + 1 have 2, 4, 6, ..., 2*n divisors respectively.
Original entry on oeis.org
2, 7, 47, 1019, 55414379
Offset: 1
a(4) = 1019 because 1016, 1017, 1018 and 1019 have 8, 6, 4, and 2 divisors respectively and there is no smaller prime having this property (see A340872).
Cf.
A341213 (similar sequence for natural numbers).
A341212
Numbers m such that m, m - 1, m - 2, m - 3 and m - 4 have k, 2k, 3k, 4k and 5k divisors respectively.
Original entry on oeis.org
154379, 1075198, 4211518, 4700758, 4745227, 5954379, 6036043, 6330235, 6485998, 6524878, 6851227, 7846798, 8536027, 8556358, 11718598, 12100027, 12126838, 13584838, 14869379, 15320587, 16934998, 17074379, 18154379, 18904027, 19013129, 19774379, 19779995
Offset: 1
tau(154375) = 20, tau(154376) = 16, tau(154377) = 12, tau(154378) = 8, tau(154379) = 4.
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[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)]
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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 *)
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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
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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
Showing 1-3 of 3 results.
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