This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A265042 #43 Jun 15 2024 11:41:49 %S A265042 2,7,51,634,12623 %N 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. %C A265042 From _Altug Alkan_, Dec 20 2015: (Start) %C A265042 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 A265042 If we want to analyze more deeply, %C A265042 A000798(p + 5) == a(5) mod p for all primes p. %C A265042 A000798(7) == a(5) mod 2, that is, 9535241 == a(5) mod 2. So a(5) mod 2 == 1. %C A265042 A000798(8) == a(5) mod 3, that is, 642779354 == a(5) mod 3. So a(5) mod 3 == 2. %C A265042 A000798(10) == a(5) mod 5, that is, 8977053873043 == a(5) mod 5. So a(5) mod 5 == 3. %C A265042 A000798(12) == a(5) mod 7, that is, 519355571065774021 == a(5) mod 7. So a(5) mod 7 == 5. %C A265042 A000798(16) == a(5) mod 11, that is, 93411113411710039565210494095 == a(5) mod 11. So a(5) mod 11 == 5. %C A265042 A000798(18) == a(5) mod 13, that is, 261492535743634374805066126901117203 == a(5) mod 13. So a(5) mod 13 == 2. %C A265042 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 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 A265042 (End) %H A265042 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 A265042 A000798(n+1) < a(n) < 2*A000798(n+1), for n > 0. - _Altug Alkan_, Dec 17 2015 %e A265042 From _Altug Alkan_, Dec 17 2015: (Start) %e A265042 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 A265042 a(1) = 7 because A000798(p + 1) == 7 mod p for all primes p. %e A265042 (End) %Y A265042 Cf. A000798, A001035. %K A265042 nonn,hard,more %O A265042 0,1 %A A265042 _N. J. A. Sloane_, Dec 16 2015 %E A265042 a(0) = 2 prepended by _Altug Alkan_, Dec 17 2015