%I #49 Mar 18 2023 16:27:17
%S 1,11,52174,260515,37362253,42781604,2685575996367
%N Numbers k for which sec(k) > k.
%C This sequence includes abs(m) for many terms m from A088306, including 1, 11, 52174, 260515, 37362253, 42781604, 2685575996367, 65398140378926, 214112296674652, 12055686754159438, 18190586279576483, 1538352035865186794, 1428599129020608582548671, 103177264599407569664999125, 9322105473781932574489648896, .... - _Jon E. Schoenfield_, Feb 12 2021
%C From _Wolfe Padawer_, Jan 05 2023: (Start)
%C For any given value in this sequence, it is extremely unlikely that it or its negation is not also in A088306. Take the following facts:
%C [1] |sec(x)| > |tan(x)| for any finite value of sec(x) and tan(x).
%C [2] |sec(x)| - |tan(x)| approaches 0, and |sec(x)| and |tan(x)| approach infinity, as x approaches (0.5 + n)*Pi where n is any integer.
%C [3] Any integer k where |sec(k)| > k or |tan(k)| > k must be close to some value of (0.5 + n)*Pi, increasingly so with larger k.
%C [4] sec(2685575996367) - |tan(2685575996367)| is approximately 8.437*10^-14.
%C Therefore, for any integer k > 2685575996367 where sec(k) > k, it must be that sec(k) - |tan(k)| < 8.437*10^-14. In order for sec(k) > k but |tan(k)| < k, it must be that k + 8.437*10^-14 > sec(k) > k, a very small interval that only gets smaller as k increases.
%C It is thus extremely likely, but not yet explicitly proven, that a(8) = 65398140378926, a(9) = 214112296674652, and a(10) = 12055686754159438. Assuming it exists, the smallest k for which sec(k) > k but |tan(k)| < k is probably very large, and it is unknown whether it is currently computable. (End)
%e sec(1) = 1.8508... so 1 is a term.
%t Select[Range[10^6], Sec[#] > # &] (* _Amiram Eldar_, Aug 21 2020 *)
%o (Python)
%o import math
%o i = 1
%o while True:
%o if 1 / math.cos(i) > i:
%o print(i)
%o i += 1
%o (PARI) isok(m) = 1/cos(m) > m; \\ _Michel Marcus_, Aug 27 2020
%Y Subsequence of A337371.
%Y Cf. A337249, A092328, A249836.
%K nonn,more
%O 1,2
%A _Joseph C. Y. Wong_, Aug 21 2020
%E a(7) from _Wolfe Padawer_, Jan 05 2023