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A300059
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Numbers k such that k and k+1 are terms of A279767.
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0
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17, 54, 247, 339, 347, 721, 849, 948, 1060, 1147, 1203, 1204, 1240, 1276, 1313, 1314, 1666, 1671, 2147, 2715, 2947, 2948, 3257, 3580, 3649, 3650, 4043, 4264, 4626, 4841, 5091, 5092, 5450, 5740, 5764, 6243, 6354, 6355, 6564, 6639, 6850, 6962, 7361, 7403, 7756, 7771, 7849, 7939
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OFFSET
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1,1
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COMMENTS
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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).
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.
a(16) and a(32) tells us other such triples exist in A279767.
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.
The first run of four consecutive terms of this sequence that are also consecutive numbers begins at a(1557).
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.
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).
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(PARI) isok1(n) = vecsort(factor(n)[, 2]) == vecsort(factor(n+2)[, 2]);
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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