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Digital root of 2*n^2.
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%I #15 Jan 21 2021 10:04:12

%S 2,8,9,5,5,9,8,2,9,2,8,9,5,5,9,8,2,9,2,8,9,5,5,9,8,2,9,2,8,9,5,5,9,8,

%T 2,9,2,8,9,5,5,9,8,2,9,2,8,9,5,5,9,8,2,9,2,8,9,5,5,9,8,2,9,2,8,9,5,5,

%U 9,8,2,9,2,8,9,5,5,9,8,2,9,2,8,9,5,5,9,8,2,9

%N Digital root of 2*n^2.

%C Period 9: repeat [2, 8, 9, 5, 5, 9, 8, 2, 9].

%C This sequence is also right if digital root of factor in "2*n^2" is 2 (for example, "11*n^2", "20*n^2", "29*n^2", etc.)

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,0,0,1).

%F a(n) = A010888(A001105(n)).

%e For n=5, 2*5^2 = 50 and digital root is 5, so a(5) = 5.

%o (PARI) dr(n) = if(n, (n-1)%9+1); \\ A010888

%o a(n) = dr(2*n^2); \\ _Michel Marcus_, Jan 21 2021

%Y Cf. A001105, A010888, A056992.

%K nonn,base,easy

%O 1,1

%A _Radaev Nikita_, Jan 10 2021