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A189974 Numbers m such that d(m-1) = d(m+1) = 4, where d(k) is the number of divisors of k (A000005).

%I A189974 #26 Aug 12 2025 19:13:21
%S A189974 7,9,34,56,86,92,94,124,142,144,160,184,186,202,204,214,216,218,220,
%T A189974 236,248,266,300,302,304,320,322,328,340,342,392,394,412,414,416,446,
%U A189974 452,470,472,516,518,534,536,544,552,580,582,590,634,668,670,680,686
%N A189974 Numbers m such that d(m-1) = d(m+1) = 4, where d(k) is the number of divisors of k (A000005).
%C A189974 Numbers m such that m-1 and m+1 are both multiplicatively perfect numbers A007422.
%C A189974 Conjecture: all terms but the first two are even numbers. - _Harvey P. Dale_, Jul 21 2025
%C A189974 Proof of conjecture: if m is odd and > 10 then either m-1 or m+1 is divisible by 4 and > 8 as well. Let t be the number from {m-1, m+1} divisible by 4. Then t is a power of 2 that is > 8 and so has more than two divisors or it has an odd prime divisor such that it has more than 4 divisors. Both exclude the odd m > 8 from the sequence. - _David A. Corneth_, Aug 05 2025
%H A189974 Nathaniel Johnston, <a href="/A189974/b189974.txt">Table of n, a(n) for n = 1..10000</a>
%p A189974 with(numtheory): A189974 := proc(n) option remember: local k: if(n=1)then return 7:else k:=procname(n-1)+1: do if(tau(k-1)=4 and tau(k+1)=4)then return k: fi: k:=k+1: od: fi: end: seq(A189974(n),n=1..60); # _Nathaniel Johnston_, May 04 2011
%t A189974 Select[Range[2, 754], DivisorSigma[0, # - 1] == DivisorSigma[0, # + 1] == 4 &]
%t A189974 Flatten[Position[Partition[DivisorSigma[0,Range[700]],3,1],_?(#[[1]]==#[[3]]==4&),1,Heads->False]]+1 (* _Harvey P. Dale_, Jul 21 2025 *)
%Y A189974 Cf. A000005, A007422.
%K A189974 nonn
%O A189974 1,1
%A A189974 _Juri-Stepan Gerasimov_, May 03 2011