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%I #15 Jun 29 2015 17:21:48
%S 9,36,49,100,225,784,961,1296,2601,3969,7225,8281,14400,16129,18496,
%T 21609,29241,34969,42025,65025,116964,123201,133225,246016,261121,
%U 278784,465124,508369,672400,700569,828100,1046529,1368900,1590121,1782225,4064256,4190209,4326400
%N Numbers n with the property that it is possible to write the base 2 expansion of n as concat(a_2,b_2), with a_2>0 and b_2>0 such that, converting a_2 and b_2 to base 10 as a and b, we have (a+b)^2 = n.
%C Obviously all terms are squares.
%C The terms that have b=1 are: 9, 49, 225, 961, 3969, 16129, 65025, ...; see A060867 ((2^n-1)^2). - _Michel Marcus_, Jun 13 2015
%e 9 in base 2 is 1001. If we take 1001 = concat(10,01) then 10 and 01 converted to base 10 are 2 and 1. Finally (2 + 1)^2 = 3^2 = 9;
%e 36 in base 2 is 100100. If we take 100100 = concat(10,0100) then 10 and 0100 converted to base 10 are 2 and 4. Finally (2 + 4)^2 = 6^2 = 36.
%p with(numtheory): P:=proc(q) local a,b,c,k,n;
%p for n from 1 to q do c:=convert(n,binary,decimal);
%p for k from 1 to ilog10(c) do
%p a:=convert(trunc(c/10^k),decimal,binary);
%p b:=convert((c mod 10^k),decimal,binary);
%p if a*b>0 then if (a+b)^2=n then print(n); break;
%p fi; fi; od; od; end: P(10^6);
%o (PARI) isok(n) = {b = binary(n); if (#b > 1, for (k=1, #b-1, vba = Vecrev(vector(k, i, b[i])); vbb = Vecrev(vector(#b-k, i, b[k+i])); da = sum(i=1, #vba, vba[i]*2^(i-1)); db = sum(i=1, #vbb, vbb[i]*2^(i-1)); if ((da+ db)^2 == n, return(1));););} \\ _Michel Marcus_, Jun 13 2015
%Y Cf. A258813, A258843.
%K nonn,base
%O 1,1
%A _Paolo P. Lava_, Jun 12 2015
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