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%I #96 Dec 16 2018 06:10:00
%S 0,1,1,2,1,2,4,3,5,2,5,5,9,7,11,13,1,7,9,9,12,9,11,14,10,2,13,13,16,
%T 12,16,12,16,19,18,15,2,21,18,20,19,25,19,20,25,26,19,24,26,3,20,25,
%U 25,31,28,36,30,33,33,37,38
%N Number of 1's in the first n digits of the binary expansion of sqrt(n).
%H Rainer Rosenthal, <a href="/A316997/b316997.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n^2) = A000120(n). - _Michel Marcus_, Dec 15 2018
%e For n = 7 we have sqrt(7) = 2.64575131... with binary expansion 10.1010010.... Of the first 7 digits there are a(7) = 3 digits equal to 1.
%p zaehle := proc(n) local e, p, c, i, z, m; Digits := n+5; e := evalf(sqrt(n)); p := [op(convert(e, binary))]; c := convert(p[1], base, 10); z := 0; m := min(n, nops(c)); for i to m do if c[-i] = 1 then z := z+1; fi; od; return z; end: seq(zaehle(n), n=0..60); # _Rainer Rosenthal_, Dec 14 2018
%p a := n -> StringTools:-CountCharacterOccurrences(convert(convert(evalf(sqrt(n), n+5), binary, n), string), "1"): seq(a(n),n=0..60); # _Peter Luschny_, Dec 15 2018
%t a[n_] := Count[RealDigits[Sqrt[n], 2, n][[1]], 1]; Array[a, 60, 0] (* _Amiram Eldar_, Dec 14 2018 *)
%o (PARI) a(n)=my(v=concat(binary(sqrt(n))));hammingweight(v[1..n]) \\ _Hugo Pfoertner_, Dec 16 2018
%Y Cf. A004539, A004547, A004555, A004609, A004569, A004585.
%Y Cf. A000120, A000290.
%K nonn,easy,base
%O 0,4
%A _Rainer Rosenthal_, Dec 14 2018
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