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A177964
Indices m for which A177961(m) = 4.
3
2, 13, 17, 28, 32, 43, 47, 58, 62, 73, 77, 88, 92, 103, 107, 118, 122, 133, 137, 148, 152, 163, 167, 178, 182, 193, 197, 208, 212, 223, 227, 238, 242, 253, 257, 268, 272, 283, 287, 298, 302, 313, 317, 328, 332, 343, 347, 358, 362, 373, 377, 388, 392, 403, 407, 418, 422
OFFSET
1,1
COMMENTS
Note that 4 is the smallest value of A177961.
FORMULA
a(n+2) = a(n)+15.
a(n) == (-1)^n (mod 3).
a(n) = 15*(n/2-1/4)+7*(-1)^n/4. - R. J. Mathar, Oct 25 2010
k such that k == 2 or -2 (mod 15). - Robert Israel, Jul 31 2015
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
From Amiram Eldar, Nov 25 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = cos(3*Pi/10)*sec(11*Pi/30).
Product_{n>=1} (1 + (-1)^n/a(n)) = sec(Pi/15)/2. (End)
MAPLE
seq(seq(15*i+j, j=[2, 13]), i=0..100); # Robert Israel, Jul 31 2015
MATHEMATICA
Table[15 (n/2 - 1/4) + 7 (-1)^n/4, {n, 60}] (* Vincenzo Librandi, Aug 01 2015 *)
LinearRecurrence[{1, 1, -1}, {2, 13, 17}, 80] (* Harvey P. Dale, Nov 01 2023 *)
PROG
(Magma) [15*(n/2-1/4)+7*(-1)^n/4: n in [1..60]]; // Vincenzo Librandi, Aug 01 2015
CROSSREFS
Cf. A177961.
Sequence in context: A352572 A018459 A037384 * A174050 A122487 A109181
KEYWORD
nonn,easy,changed
AUTHOR
Vladimir Shevelev, May 16 2010
EXTENSIONS
More terms from R. J. Mathar, Oct 25 2010
STATUS
approved