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A337248
Numbers k for which sec(k) > k.
1
1, 11, 52174, 260515, 37362253, 42781604, 2685575996367
OFFSET
1,2
COMMENTS
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
From Wolfe Padawer, Jan 05 2023: (Start)
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:
[1] |sec(x)| > |tan(x)| for any finite value of sec(x) and tan(x).
[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.
[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.
[4] sec(2685575996367) - |tan(2685575996367)| is approximately 8.437*10^-14.
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.
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)
EXAMPLE
sec(1) = 1.8508... so 1 is a term.
MATHEMATICA
Select[Range[10^6], Sec[#] > # &] (* Amiram Eldar, Aug 21 2020 *)
PROG
(Python)
import math
i = 1
while True:
if 1 / math.cos(i) > i:
print(i)
i += 1
(PARI) isok(m) = 1/cos(m) > m; \\ Michel Marcus, Aug 27 2020
CROSSREFS
Subsequence of A337371.
Sequence in context: A198626 A343119 A295176 * A102367 A055322 A348644
KEYWORD
nonn,more
AUTHOR
Joseph C. Y. Wong, Aug 21 2020
EXTENSIONS
a(7) from Wolfe Padawer, Jan 05 2023
STATUS
approved