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Values of A076336(n) such that A076336(n) = A076336(n+1) - 14.
3

%I #30 Sep 11 2022 12:04:28

%S 7523267,18708077,29892887,41077697,52262507,63447317,74632127,

%T 85816937,97001747,108186557,119371367,130556177,141740987,152925797,

%U 164110607,175295417,186480227,197665037,208849847,220034657,231219467,242404277,253589087,264773897,275958707,287143517,298328327,309513137

%N Values of A076336(n) such that A076336(n) = A076336(n+1) - 14.

%C See A270971 for the motivation behind this sequence.

%C Riesel showed that there are infinitely many integers such that k*(2^m) - 1 is not prime for any integer m. He showed that the number 509203 has this property, as does 509203 plus any positive integer multiple of 11184810.

%C In this sequence, the lesser of (provable) Sierpiński pairs appears with the linear formula a(n) = 7523267 + 11184810*(n-1).

%C Since 7523267 is a term of A244561, for every integer k > 0, 7523267*2^k+1 has a divisor in the set {3, 5, 7, 13, 17, 241}. Because 11184810 = 2*3*5*7*13*17*241, a(n)*2^k+1 = 7523267*2^k+1 + 11184810*(n-1)*2^k+1 always has a divisor in the set {3, 5, 7, 13, 17, 241}. Since a(n) is always odd because of its definition, a(n) is a Sierpiński number. Additionally, 7523267 + 14 = 7523281 is also a term of A244561. So a(n) + 14 is a Sierpiński number too, with the same proof.

%C In conclusion, if the minimum difference between consecutive (provable) Sierpiński numbers is 14 (see comment section of A270971 for the reason behind this claim), a(n) and a(n) + 14 must be consecutive and a(n) = 7523267 + 11184810*(n-1) is the formula for this sequence.

%e 7523267 is a term because 7523267 and 7523267 + 14 = 7523281 are consecutive (provable) Sierpiński numbers.

%Y Cf. A076336, A270971.

%K nonn

%O 1,1

%A _Altug Alkan_, Mar 28 2016