

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




OFFSET

0,1


COMMENTS

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).
If we want to analyze more deeply,
A000798(p + 5) == a(5) mod p for all primes p.
A000798(7) == a(5) mod 2, that is, 9535241 == a(5) mod 2. So a(5) mod 2 == 1.
A000798(8) == a(5) mod 3, that is, 642779354 == a(5) mod 3. So a(5) mod 3 == 2.
A000798(10) == a(5) mod 5, that is, 8977053873043 == a(5) mod 5. So a(5) mod 5 == 3.
A000798(12) == a(5) mod 7, that is, 519355571065774021 == a(5) mod 7. So a(5) mod 7 == 5.
A000798(16) == a(5) mod 11, that is, 93411113411710039565210494095 == a(5) mod 11. So a(5) mod 11 == 5.
A000798(18) == a(5) mod 13, that is, 261492535743634374805066126901117203 == a(5) mod 13. So a(5) mod 13 == 2.
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.
(End)


LINKS



FORMULA



EXAMPLE

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


CROSSREFS



KEYWORD

nonn,hard,more


AUTHOR



EXTENSIONS



STATUS

approved



