OFFSET
1,1
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
FORMULA
a(n) ~ k*n where k = 2*Pi/(Pi - 3) = 44.375... by the Equidistribution Theorem. - Charles R Greathouse IV, Dec 10 2024
MATHEMATICA
Select[Range[0, 2350], Sin[#]<Sin[#+1]<Sin[#+2]<Sin[#+3]<Sin[#+4]&] (* James C. McMahon, Jan 20 2024 *)
Flatten[Position[Partition[Sin[Range[2500]], 5, 1], _?(Min[Differences[#]]>0&)]]//Quiet (* Harvey P. Dale, Dec 10 2024 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
R. J. Mathar, Jan 18 2024
STATUS
approved
