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A369275 Nonnegative numbers k satisfying sin(k) < sin(k+1) < sin(k+2) < sin(k+3) < sin(k+4).

%I A369275 #41 Dec 27 2024 01:35:21
%S A369275 42,86,130,199,243,287,331,375,419,463,507,532,576,620,664,708,752,
%T A369275 796,840,909,953,997,1041,1085,1129,1173,1217,1242,1286,1330,1374,
%U A369275 1418,1462,1506,1550,1619,1663,1707,1751,1795,1839,1883,1927,1952,1996,2040,2084,2128,2172,2216,2260,2329
%N A369275 Nonnegative numbers k satisfying sin(k) < sin(k+1) < sin(k+2) < sin(k+3) < sin(k+4).
%C A369275 This is a subsequence of terms in A369274: The smaller of two consecutive terms there.
%C A369275 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.
%C A369275 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
%C A369275 Both patterns break. The first at n = 7035 and the second at n = 7019. In general this sequence is not periodic mod m for m > 1 as the (period of sin(x))/(1) is irrational. - _David A. Corneth_, Dec 10 2024
%C A369275 Mathar’s conjecture is true: these are just the numbers which are between Pi*3/2 - 1/2 and Pi*5/2 - 7/2 mod 2*Pi, and reducing the upper bound by 1 leaves it empty. - _Charles R Greathouse IV_, Dec 10 2024
%H A369275 David A. Corneth, <a href="/A369275/b369275.txt">Table of n, a(n) for n = 1..10000</a>
%F A369275 a(n) ~ k*n where k = 2*Pi/(Pi - 3) = 44.375... by the Equidistribution Theorem. - _Charles R Greathouse IV_, Dec 10 2024
%t A369275 Select[Range[0,2350], Sin[#]<Sin[#+1]<Sin[#+2]<Sin[#+3]<Sin[#+4]&] (* _James C. McMahon_, Jan 20 2024 *)
%t A369275 Flatten[Position[Partition[Sin[Range[2500]],5,1],_?(Min[Differences[#]]>0&)]]//Quiet (* _Harvey P. Dale_, Dec 10 2024 *)
%o A369275 (PARI) first(n) = {
%o A369275 	my(res = List(), streak = 1, s = sin(1));
%o A369275 	for(i = 2, oo,
%o A369275 		c = sin(i);
%o A369275 		if(c > s,
%o A369275 			streak++;
%o A369275 			if(streak >= 5,
%o A369275 				listput(res, i-4);
%o A369275 				if(#res >= n,
%o A369275 					return(res)));
%o A369275 		,
%o A369275 			streak = 1);
%o A369275 s = c); res} \\ _David A. Corneth_, Dec 11 2024
%Y A369275 Subsequence of A369274 and hence of A369273 and A026309.
%Y A369275 Cf. A096444.
%K A369275 nonn,easy
%O A369275 1,1
%A A369275 _R. J. Mathar_, Jan 18 2024