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A300059 Numbers k such that k and k+1 are terms of A279767.

%I A300059 #41 May 19 2024 02:12:46
%S A300059 17,54,247,339,347,721,849,948,1060,1147,1203,1204,1240,1276,1313,
%T A300059 1314,1666,1671,2147,2715,2947,2948,3257,3580,3649,3650,4043,4264,
%U A300059 4626,4841,5091,5092,5450,5740,5764,6243,6354,6355,6564,6639,6850,6962,7361,7403,7756,7771,7849,7939
%N A300059 Numbers k such that k and k+1 are terms of A279767.
%C A300059 For certain values of n, a(n) - a(n - 1) = 1. The smallest such term is a(12). The next two terms are a(16) and a(32).
%C A300059 a(12)-a(11) = 1 tells us there are three consecutive terms of A279767 that are consecutive numbers and these are A279767(149) = 1203, A279767(150) = 1204, and A279767(151) = 1205.
%C A300059 a(16) and a(32) tells us other such triples exist in A279767.
%C A300059 The first triple of consecutive terms in this sequence begins with a(197). This tells us there are four consecutive terms of A279767 that are consecutive numbers and these are A279767(4605) = 41058, A279767(4606) = 41059, A279767(4607) = 41060, and A279767(4068) = 41061.
%C A300059 The first run of four consecutive terms of this sequence that are also consecutive numbers begins at a(1557).
%C A300059 Conjecture: If a run of consecutive terms that are also consecutive numbers of length m exists in this sequence, then there also exists a run of length m + 1.
%C A300059 The smallest term where a(n)-a(n-1) = a(n-2)-a(n-3) = 1 is a(53). The next such term is a(162).
%H A300059 Amiram Eldar, <a href="/A300059/b300059.txt">Table of n, a(n) for n = 1..10000</a>
%H A300059 Wikipedia, <a href="http://en.wikipedia.org/wiki/Prime_signature">Prime signature</a>.
%H A300059 <a href="/index/Pri#prime_signature">Index to sequences related to prime signature</a>.
%e A300059 17 is in the list because 17 and 19 have the same prime signature {1}, and 18 and 20 have the same prime signature {1,2}. Note, order doesn't matter.
%t A300059 With[{s = Array[Sort[FactorInteger[#][[All, -1]]] &, 10^4]}, Function[t, t[[Flatten@ Position[Differences@ t, 1] ]] ]@ Select[Range[Length@ s - 2], s[[#]] == s[[# + 2]] &]] (* _Michael De Vlieger_, Feb 26 2018 *)
%o A300059 (PARI) isok1(n) = vecsort(factor(n)[,2]) == vecsort(factor(n+2)[,2]);
%o A300059 isok(n) = isok1(n) && isok1(n+1); \\ _Michel Marcus_, Feb 25 2018
%Y A300059 Cf. A279767.
%K A300059 nonn
%O A300059 1,1
%A A300059 _Torlach Rush_, Feb 23 2018
%E A300059 More terms from _Michel Marcus_, Feb 25 2018