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a(n) = floor(5^n/4^(n-1)).
1

%I #17 Jun 07 2023 21:15:17

%S 5,6,7,9,12,15,19,23,29,37,46,58,72,90,113,142,177,222,277,346,433,

%T 542,677,847,1058,1323,1654,2067,2584,3231,4038,5048,6310,7888,9860,

%U 12325,15407,19259,24074,30092,37615,47019,58774,73468,91835,114794,143492,179366,224207,280259

%N a(n) = floor(5^n/4^(n-1)).

%C a(n) is the curvature (rounded down) of circles inscribed in minor segment where chord length equal to sagitta length starting from a unit circle, the next iterations are nested down at scale factor 4/5. The curvature of circles inscribed in major segment would be A065565: floor((5/4)^n). See illustrations.

%H G. C. Greubel, <a href="/A241203/b241203.txt">Table of n, a(n) for n = 1..1000</a>

%H Kival Ngaokrajang, <a href="/A241203/a241203.pdf">Illustrations of initial terms</a>

%F a(n) = floor(5^n/4^(n-1)), n >= 1.

%t Floor[4*(5/4)^Range[60]] (* _G. C. Greubel_, Jun 07 2023 *)

%o (PARI) for(n=1,100,print1(floor(5^n/4^(n-1)),", "))

%o (Magma) [Floor(4*(5/4)^n): n in [1..60]]; // _G. C. Greubel_, Jun 07 2023

%o (SageMath) [(5^n//4^(n-1)) for n in range(1,61)] # _G. C. Greubel_, Jun 07 2023

%Y Cf. A065565.

%K nonn,easy

%O 1,1

%A _Kival Ngaokrajang_, Aug 08 2014