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Indices m for which A177961(m) = 4.
3

%I #33 Nov 27 2024 07:26:10

%S 2,13,17,28,32,43,47,58,62,73,77,88,92,103,107,118,122,133,137,148,

%T 152,163,167,178,182,193,197,208,212,223,227,238,242,253,257,268,272,

%U 283,287,298,302,313,317,328,332,343,347,358,362,373,377,388,392,403,407,418,422

%N Indices m for which A177961(m) = 4.

%C Note that 4 is the smallest value of A177961.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).

%F a(n+2) = a(n)+15.

%F a(n) == (-1)^n (mod 3).

%F a(n) = 15*(n/2-1/4)+7*(-1)^n/4. - _R. J. Mathar_, Oct 25 2010

%F k such that k == 2 or -2 (mod 15). - _Robert Israel_, Jul 31 2015

%F Sum_{n>=1} (-1)^(n+1)/a(n) = tan(3*Pi/10)*Pi/15 = sqrt(1+2/sqrt(5))*Pi/15. - _Amiram Eldar_, Feb 28 2023

%F From _Amiram Eldar_, Nov 25 2024: (Start)

%F Product_{n>=1} (1 - (-1)^n/a(n)) = cos(3*Pi/10)*sec(11*Pi/30).

%F Product_{n>=1} (1 + (-1)^n/a(n)) = sec(Pi/15)/2. (End)

%p seq(seq(15*i+j, j=[2,13]),i=0..100); # _Robert Israel_, Jul 31 2015

%t Table[15 (n/2 - 1/4) + 7 (-1)^n/4, {n, 60}] (* _Vincenzo Librandi_, Aug 01 2015 *)

%t LinearRecurrence[{1,1,-1},{2,13,17},80] (* _Harvey P. Dale_, Nov 01 2023 *)

%o (Magma) [15*(n/2-1/4)+7*(-1)^n/4: n in [1..60]]; // _Vincenzo Librandi_, Aug 01 2015

%Y Cf. A177961.

%K nonn,easy

%O 1,1

%A _Vladimir Shevelev_, May 16 2010

%E More terms from _R. J. Mathar_, Oct 25 2010