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

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

Offset: 1

Author

R. J. Mathar, Jan 18 2024

Keywords

• nonn
• easy

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

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

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

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000

Crossrefs

Subsequence of A369274 and hence of A369273 and A026309.

Cf. A096444.

Programs

Mathematica

Select[Range[0,2350], Sin[#]James C. McMahon, Jan 20 2024 *)
Flatten[Position[Partition[Sin[Range[2500]],5,1],?(Min[Differences[#]]>0&)]]//Quiet (* _Harvey P. Dale, Dec 10 2024 *)

PARI

first(n) = {
	my(res = List(), streak = 1, s = sin(1));
	for(i = 2, oo,
		c = sin(i);
		if(c > s,
			streak++;
			if(streak >= 5,
				listput(res, i-4);
				if(#res >= n,
					return(res)));
		,
			streak = 1);
s = c); res} \\ David A. Corneth, Dec 11 2024

Formula

a(n) ~ k*n where k = 2*Pi/(Pi - 3) = 44.375... by the Equidistribution Theorem. - Charles R Greathouse IV, Dec 10 2024