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Let s(x) be the Maclaurin series for cos(x); then a(n) is the index k for which the (k+1)-st partial sum of s(2*n*Pi) is greatest among all partial sums.
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%I #5 Sep 30 2024 12:40:57

%S 2,6,8,12,14,18,22,24,28,30,34,36,40,44,46,50,52,56,58,62,66,68,72,74,

%T 78,80,84,88,90,94,96,100,102,106,110,112,116,118,122,124,128,132,134,

%U 138,140,144,146,150,154,156,160,162,166,168,172,176,178,182

%N Let s(x) be the Maclaurin series for cos(x); then a(n) is the index k for which the (k+1)-st partial sum of s(2*n*Pi) is greatest among all partial sums.

%F |a(n)-A376457(n)| = 1 for n>=1.

%e For n = 2 the partial sums (of which the 1st is for k=0) are approximately 1, -18.7, 46.2, -39.2, 20.9, -5.4,..., where the greatest, 46.2..., is the 3rd, so that a(2) = 2.

%t z = 200; r = Pi;

%t f[n_, m_] := f[n, m] = N[Sum[(-1)^k (2 n r)^(2 k)/(2 k)!, {k, 0, m}], 10]

%t t[n_] := Table[f[n, m], {m, 1, z}]

%t g[n_] := Select[Range[z], f[n, #] == Max[t[n]] &]

%t h[n_] := Select[Range[z], f[n, #] == Min[t[n]] &]

%t Flatten[Table[g[n], {n, 1, 60}]] (* this sequence *)

%t Flatten[Table[h[n], {n, 1, 60}]] (* A376457 *)

%Y Cf. A376457.

%K nonn

%O 1,1

%A _Clark Kimberling_, Sep 26 2024