%I #17 Feb 08 2022 12:50:27
%S 0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,
%T 400,441,484,529,576,625,676,729,784,841,900,961,24,89,156,225,296,
%U 369,444,521,600,681,764,849,936,25,116,209,304,401,500,601,704,809,916,25
%N a(n) = n^2 mod 1000.
%C a(n) = A000290(n) for n < 32, but a(32) = 24;
%C A008959(n) = a(n) mod 10; A002015(n) = a(n) mod 100;
%C periodic with period 500: a(n+500)=a(n) and a(250*n+k)=a(250*n-k) for k <= 250*n;
%C a(n) = (n mod 1000)^2 mod 1000;
%C a(m*n) = a(m)*a(n) mod 1000;
%C A122986 gives the range of this sequence;
%C a(n) = n for n = 0, 1, and 376.
%H R. Zumkeller, <a href="/A174452/b174452.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Fi#final">Index entries for sequences related to final digits of numbers</a>
%F a(n) = ((n mod 100)^2 + 200 * (floor(n/100) mod 10) * (n mod 10)) mod 1000.
%e Some calculations for n=982451653, to be realized by hand:
%e a(n) = (53^2 + 200*6*3) mod 1000 = 6409 mod 1000 = 409;
%e a(n) = (653^2) mod 1000 = 426409 mod 1000 = 409;
%e a(n) = a(n mod 500) = a(153) = 409;
%e a(n) = 965211250482432409 mod 1000 = 409.
%p seq(n^2 mod 1000, n=0..55); # _Nathaniel Johnston_, Jun 22 2011
%t PowerMod[Range[0,60],2,1000] (* _Harvey P. Dale_, Feb 08 2022 *)
%o (Haskell)
%o a174452 = (`mod` 1000) . (^ 2) -- _Reinhard Zumkeller_, Jul 06 2011
%o (PARI) a(n)=n^2%1000 \\ _Charles R Greathouse IV_, Apr 06 2016
%Y Cf. A053879, A070430, A070431, A070432, A070433, A070434, A070435, A070438, A070442, A070452, A159852.
%K nonn,easy,look
%O 0,3
%A _Reinhard Zumkeller_, Mar 21 2010