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%I #29 Jul 07 2015 12:22:02
%S 11,123,695,991,1919,2839,3707,3841,7615,8047,8055,9347,10703,12847,
%T 16195,26743,27089,32127,42251,56419,59027,59179,59389,59815,62749,
%U 65113,74671,115289,119211,122847,126895,129495,168739,191051,219295,224281,232315,233729
%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 sigma(a + b) = sigma(n).
%C It appears that a or b is equal to 1.
%C The terms that have b=1 are 11, 695, 991, 2839, 3707, 9347, ...; see A232355. - _Michel Marcus_, Jun 12 2015
%C If b=1, the number n can be expressed as 2a+b=2a+1. We are looking for numbers that satisfy the relation sigma(a+1)=sigma(n), namely sigma(a+1)=sigma(2a+1). In A232355 we have the numbers such that sigma(n)=sigma((n+1)/2) that match sigma(2a+1)=sigma((2a+1+1)/2)=sigma(a+1). That's why the two "subsequences" are the same thing. - _Paolo P. Lava_ and _Michel Marcus_, Jun 16 2015
%H Paolo P. Lava, <a href="/A258843/b258843.txt">Table of n, a(n) for n = 1..110</a>
%e 11 in base 2 is 1011. If we take 1011 = concat(101,1) then 101 and 1 converted to base 10 are 5 and 1. Finally sigma(5 + 1) = sigma(6) = 12 = sigma(11).
%e 123 in base 2 is 1111011. If we take 1111011 = concat(1,111011) then 1 and 111011 converted to base 10 are 1 and 59. Finally sigma(1 + 59) = sigma(60) = 168 = sigma(123).
%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 sigma(a+b)=sigma(n) then print(n);
%p break; fi; fi; od; od; end: P(10^6);
%t f[n_] := Block[{d = IntegerDigits[n, 2], len, s}, len = Length@ d; s = FromDigits[#, 2] & /@ {Take[d, #], Take[d, -len + #]} & /@ Range[len - 1]; DeleteDuplicates[DivisorSigma[1, #1 + #2] == DivisorSigma[1, n] & @@@ s]]; Select[Range@ 250000, Length@ f@ # > 1 &] (* _Michael De Vlieger_, Jun 12 2015 *)
%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 (sigma(da+db) == sigma(n), return(1));););} \\ _Michel Marcus_, Jun 12 2015
%Y Cf. A232355, A258813, A258844.
%K nonn,base
%O 1,1
%A _Paolo P. Lava_, Jun 12 2015
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