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"Upper positive integers": n is in the sequence iff in the representation n=A000695(k)+2*A000695(l) satisfies inequality A000695(k) < A000695(l)
4

%I #14 Dec 16 2018 08:41:33

%S 2,8,9,10,11,14,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,50,56,

%T 57,58,59,62,128,129,130,131,132,133,134,135,136,137,138,139,140,141,

%U 142,143,144,145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160

%N "Upper positive integers": n is in the sequence iff in the representation n=A000695(k)+2*A000695(l) satisfies inequality A000695(k) < A000695(l)

%C In the mapping: every integer m corresponds to a unique pair (k,l) with m=A000695(k)+2*A000695(l) (k=A059905(m), l=A059906(m)), the numbers a(n) are mapped into the lattice points lying upper the diagonal l=k. If the binary expansion of N is Sum b_j*2^j, then N is in the sequence iff Sum b_(2j)*2^j<Sum b_(2j+1)*2^j. Therefore "in average" satisfies the condition of A139370. This explains, somewhat, why many terms of the sequence are in A139370 as well.

%H Vincenzo Librandi, <a href="/A152754/b152754.txt">Table of n, a(n) for n = 1..8128</a>

%t fh[n_,h_] := If[h==1, Mod[n,2], If[Mod[n,4]>=2,1,0]]; half[n_, h_ ] := Module[{t=1, s=0, m=n}, While[m>0, s += fh[m,h]*t; m=Quotient[m,4]; t *= 2]; s]; mb[n_] := FromDigits[Riffle[IntegerDigits[n, 2], 0], 2]; aQ[n_] := mb[half[n,1]] < mb[half[n, 2]]; Select[Range[160], aQ] (* _Amiram Eldar_, Dec 16 2018 from the PARI code *)

%o (PARI) a000695(n) = fromdigits(binary(n), 4);

%o half1(n) = { my(t=1, s=0); while(n>0, s += (n%2)*t; n \= 4; t *= 2); (s); }; \\ A059905

%o half2(n) = { my(t=1, s=0); while(n>0, s += ((n%4)>=2)*t; n \= 4; t *= 2); (s); }; \\ A059906

%o isok(n) = a000695(half1(n)) < a000695(half2(n)); \\ _Michel Marcus_, Dec 15 2018

%Y Cf. A000695, A139370.

%K nonn

%O 1,1

%A _Vladimir Shevelev_, Dec 12 2008

%E Missing 9 and more terms from _Michel Marcus_, Dec 15 2018