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Cf. property described by Gary Detlefs in A113801: more generally, these a(n) are of the form (2*h*n + (h-4)*(-1)^n-h)/4 (h,n natural numbers). Therefore a(n)^2 - 1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 8). Also a(n)^2 - 1 == 0 (mod 16). - Bruno Berselli, Nov 17 2010
S(a(n+1)/2, 0) = (1/2)*(S(a(n+1), sqrt(2)) - S(a(n+1) - 2, sqrt(2))) = T(a(n+1), sqrt(2)/2) = cos(a(n+1)*Pi/4) = sqrt(2)/2 = A010503, identically for n >= 0, where S is the Chebyshev polynomial (A049310) here extended to fractional n, evaluated at x = 0. (for For T see A053120.) - Wolfdieter Lang, Jun 04 2023
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S(a(n+1)/2, 0) = (1/2)*(S(a(n+1), sqrt(2)) - S(a(n+1) - 2, sqrt(2))) = T(a(n+1), sqrt(2)/2) = cos(a(n+1)*Pi/4) = sqrt(2)/2 = A010503, identically for n >= 0, where S(n, x) is the Chebyshev polynomial (A049310) here extended to fractional n, evaluated at x = 0. (for T see A053120) - Wolfdieter Lang, Jun 03 04 2023
S(a(n+1)/2, 0) = cos(a(n+1)*Pi/4) = sqrt(2)/2 = A010503, identically for n >= 0, where S (n, x) is the Chebyshev polynomial (A049310) extended to fractional n, evaluated at x = 0. - Wolfdieter Lang, Jun 03 2023
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S(a(n+1)/2, 0) = sqrt(2)/2 = A010503, identically for n >= 0, where S is the Chebyshev polynomial (A049310) extended to fractional n, evaluated at x = 0. - Wolfdieter Lang, Jun 03 2023
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