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Numbers m such that numbers m, m + 1 and m + 2 have k, 2k and 3k divisors respectively.
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%I #16 Sep 08 2022 08:46:26

%S 61,73,277,421,458,493,583,1234,1393,1418,1658,1909,1954,2066,2138,

%T 2234,2329,2386,2533,2594,2773,2797,2846,3013,3073,3265,3394,3841,

%U 4322,4333,4538,4586,4633,4717,4754,4766,5029,5223,5245,5342,5378,5554,5893,5906,6169

%N Numbers m such that numbers m, m + 1 and m + 2 have k, 2k and 3k divisors respectively.

%C Numbers m such that tau(m) = tau(m + 1) / 2 = tau(m + 2) / 3, where tau(k) = the number of divisors of k (A000005).

%C Corresponding values of tau(a(n)): 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2, 4, ...

%C Triplets of [tau(a(n)), tau(a(n) + 1), tau(a(n) + 2)] = [tau(a(n)), 2*tau(a(n)), 3*tau(a(n)]: [2, 4, 6], [2, 4, 6], [2, 4, 6], [2, 4, 6], [4, 8, 12], [4, 8, 12], [4, 8, 12], [4, 8, 12], [4, 8, 12], ...

%H David A. Corneth, <a href="/A339778/b339778.txt">Table of n, a(n) for n = 1..10000</a>

%e tau(61) = 2, tau(62) = 4, tau(63) = 6.

%t Select[Range[6000], Equal @@ (DivisorSigma[0, # + {0, 1, 2}]/{1, 2, 3}) &] (* _Amiram Eldar_, Dec 16 2020 *)

%o (Magma) [m: m in [1..10^5] | #Divisors(m) eq #Divisors(m + 1) / 2 and #Divisors(m) eq #Divisors(m + 2) / 3]

%o (PARI) isok(m) = my(nb = numdiv(m)); (numdiv(m+1) == 2*nb) && (numdiv(m+2) == 3*nb); \\ _Michel Marcus_, Dec 18 2020

%Y Cf. A000005, A100363.

%Y Subsequence of A063446.

%K nonn

%O 1,1

%A _Jaroslav Krizek_, Dec 16 2020