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A079044
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Numbers k such that Sum_{j=0..k} sin(j/Pi) < 0.
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0
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19, 39, 59, 78, 98, 118, 138, 157, 177, 197, 217, 236, 256, 276, 296, 315, 335, 355, 375, 394, 414, 434, 454, 473, 493, 513, 532, 552, 572, 592, 611, 631, 651, 671, 690, 710, 730, 750, 769, 789, 809, 829, 848, 868, 888, 908, 927, 947, 967, 986
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OFFSET
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1,1
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LINKS
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FORMULA
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a(n) is asymptotic to c*n where c=19.7...
Conjecture: a(n) = floor( 2 * Pi^2 * n ), checked for n <= 10^4. - Vincenzo Librandi, Sep 03 2015
Numbers k such that sin(k/(2*Pi)) * sin((k+1)/(2*Pi)) < 0.
Numbers k such that cos((2*k+1)/(2*Pi)) > cos(1/(2*Pi)).
Numbers k such that k+1 > 2*Pi^2*(floor(k/(2*Pi^2))+1).
Numbers k such that k mod (2*Pi^2) > 2*Pi^2 - 1.
(End)
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MATHEMATICA
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Select[Range[1, 1000], Sum[Sin[k/Pi], {k, 0, #}] < 0&] (* Vaclav Kotesovec, Feb 15 2019 *)
Select[Range[1, 1000], Cos[(2*# + 1)/(2*Pi)] > Cos[1/(2*Pi)]&] (* Vaclav Kotesovec, Feb 15 2019 *)
Select[Range[1, 1000], Mod[(2*# + 1)/(2*Pi), 2*Pi] < 1/(2*Pi) || Mod[(2*# + 1)/(2*Pi), 2*Pi] > 2*Pi - 1/(2*Pi) &] (* Vaclav Kotesovec, Feb 15 2019 *)
Select[Range[1, 1000], # + 1 > 2*Pi^2*(Floor[#/(2*Pi^2)] + 1) &] (* Vaclav Kotesovec, Feb 15 2019 *)
Select[Range[1, 1000], Mod[#, 2*Pi^2] > 2*Pi^2 - 1 &] (* Vaclav Kotesovec, Feb 15 2019 *)
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PROG
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(PARI) isok(n) = sum(k=0, n, sin(k/Pi)) < 0; \\ Michel Marcus, Nov 30 2013
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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