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Nonnegative numbers k satisfying sin(k) < sin(k+1) < sin(k+2) < sin(k+3).
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5, 11, 17, 24, 30, 36, 42, 43, 49, 55, 61, 68, 74, 80, 86, 87, 93, 99, 105, 112, 118, 124, 130, 131, 137, 143, 149, 156, 162, 168, 174, 181, 187, 193, 199, 200, 206, 212, 218, 225, 231, 237, 243, 244, 250, 256, 262, 269, 275, 281, 287, 288, 294, 300, 306, 313, 319, 325, 331, 332, 338, 344, 350
COMMENTS
Subsequence of terms in A026309: The smaller of three consecutive terms there. Also a subsequence of terms in A369273: The smaller of two consecutive terms there.
MATHEMATICA
Select[Range[500], Less @@ Sin[Range[#, #+3]] &] (* Paolo Xausa, Mar 19 2024 *)
Nonnegative numbers k satisfying sin(k) < sin(k+1) < sin(k+2) < sin(k+3) < sin(k+4).
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42, 86, 130, 199, 243, 287, 331, 375, 419, 463, 507, 532, 576, 620, 664, 708, 752, 796, 840, 909, 953, 997, 1041, 1085, 1129, 1173, 1217, 1242, 1286, 1330, 1374, 1418, 1462, 1506, 1550, 1619, 1663, 1707, 1751, 1795, 1839, 1883, 1927, 1952, 1996, 2040, 2084, 2128, 2172, 2216, 2260, 2329
COMMENTS
This is a subsequence of terms in A369274: The smaller of two consecutive terms there. Conjecture: there is no followup sequence with sin(k) < sin(k+1) < ... < sin(k+5), i.e., there are no two consecutive integers in this sequence.
It appears that, starting at n=4, a(n) mod 2 has a period of 16 with a periodic part consisting of 8 ones followed by 8 zeros. Similarly, starting at n=4, a(n) mod 4 has a period of 32 with a periodic part consisting of 8 threes followed by 8 zeros followed by 8 ones followed by 8 twos. Tested to 5688 terms. - Gary Detlefs, Jan 20 2024
MATHEMATICA
Select[Range[0, 2350], Sin[#]<Sin[#+1]<Sin[#+2]<Sin[#+3]<Sin[#+4]&] (* James C. McMahon, Jan 20 2024 *)
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