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 A265042 a(n) = the unique number k such that T(p + n) == k mod p for all primes p, where T(n) = A000798(n) = number of topologies on n points. 1

%I

%S 2,7,51,634,12623

%N a(n) = the unique number k such that T(p + n) == k mod p for all primes p, where T(n) = A000798(n) = number of topologies on n points.

%C From _Altug Alkan_, Dec 20 2015: (Start)

%C From the inequality in the formula section, since A000798(6) = 209527, we have 209527 < a(5) < 419054. The same inequality shows that a(17) has 36 digits (A000798 is currently known only for n <= 18).

%C If we want to analyze more deeply,

%C A000798(p + 5) == a(5) mod p for all primes p.

%C A000798(7) == a(5) mod 2, that is, 9535241 == a(5) mod 2. So a(5) mod 2 == 1.

%C A000798(8) == a(5) mod 3, that is, 642779354 == a(5) mod 3. So a(5) mod 3 == 2.

%C A000798(10) == a(5) mod 5, that is, 8977053873043 == a(5) mod 5. So a(5) mod 5 == 3.

%C A000798(12) == a(5) mod 7, that is, 519355571065774021 == a(5) mod 7. So a(5) mod 7 == 5.

%C A000798(16) == a(5) mod 11, that is, 93411113411710039565210494095 == a(5) mod 11. So a(5) mod 11 == 5.

%C A000798(18) == a(5) mod 13, that is, 261492535743634374805066126901117203 == a(5) mod 13. So a(5) mod 13 == 2.

%C In conclusion, a(5) is a number of the form 2*3*5*7*11*13*n - 2767, that is, 30030*n - 2767. Moreover we know that 209527 < a(5) < 419054. So a(5) is the one of these numbers: 237473, 267503, 297533, 327563, 357593, 387623, 417653. If we take into consideration the first four inequalities, which are 4 < 7 < 8, 29 < 51 < 58, 355 < 634 < 710, 6942 < 12623 < 13884, then 387623 seems a strong candidate for a(5) because of relevant proportions in inequalities.

%C (End)

%H M. Y. Kizmaz, <a href="http://arxiv.org/abs/1503.08359">On The Number Of Topologies On A Finite Set</a>, arXiv preprint arXiv:1503.08359 [math.NT], 2015.

%F A000798(n+1) < a(n) < 2*A000798(n+1), for n > 0. - _Altug Alkan_, Dec 17 2015

%e From _Altug Alkan_, Dec 17 2015: (Start)

%e A000798(p^k) == k+1 mod p for all primes p. If k=1, A000798(p^1) == 2 mod p, that is, A000798(p) == 2 mod p. So a(0) = 2.

%e a(1) = 7 because A000798(p + 1) == 7 mod p for all primes p.

%e (End)

%Y Cf. A000798, A001035.

%K nonn,hard,more

%O 0,1

%A _N. J. A. Sloane_, Dec 16 2015

%E a(0) = 2 prepended by _Altug Alkan_, Dec 17 2015

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Last modified October 23 23:51 EDT 2019. Contains 328379 sequences. (Running on oeis4.)