%I #36 Apr 13 2021 23:19:05

%S 1,2,3,5,7,9,10,12,15,17,21,27,31,33,38,42,45,51,52,56,63,65,73,85,93,

%T 99,107,119,127,129,142,150,153,165,170,178,189,195,204,212,219,231,

%U 232,240,255,257,273,297,313,325,341,365,381

%N Positive integers whose base-2 run lengths form a palindrome.

%C This sequence exactly contains those positive integers in A006995 (positive binary palindromes) together with the terms of A035928 (those positive integers n where reversing the order of the binary digits produces the binary complement of n). - _Leroy Quet_, Sep 14 2009

%C Also the indices of the compositions that are palindromic. For the definition of the index of a composition see A298644. For example, 93 is in the sequence since its binary form is 1011101 and the composition [1,1,3,1,1] is palindromic. On the other hand, 132 is not in the sequence since its binary form is 10000100 and the composition [1,4,1,2] is not palindromic. The command c(n) from the Maple program yields the composition having index n. - _Emeric Deutsch_, Jan 28 2018

%H Rémy Sigrist, <a href="/A044918/b044918.txt">Table of n, a(n) for n = 1..10000</a>

%H Rémy Sigrist, <a href="/A044918/a044918.png">Logarithmic scatterplot of (n, a(n+1)-a(n)) for n = 1..9999</a>

%p Runs:=proc(L) local j,r,i,k:j:=1: r[j]:=L[1]: for i from 2 to nops(L) do if L[i]=L[i-1] then r[j]:=r[j], L[i] else j:=j+1: r[j]:=L[i] end if end do: [seq([r[k]],k=1..j)] end proc: RunLengths:=proc(L) map(nops,Runs(L)) end proc: c:=proc(n) ListTools:-Reverse(convert(n,base,2)): RunLengths(`%`) end proc: A:={}: for n from 1 to 500 do crev(n):=[seq(c(n)[1+ nops(c(n))-j],j=1..nops(c(n)))] od: for n from 1 to 500 do if c(n)=crev(n) then A:=A union {n} else fi od: A; # most of the Maple program is due to W. Edwin Clark. # _Emeric Deutsch_, Jan 28 2018

%t Position[Array[Length /@ Split@ IntegerDigits[#, 2] &, 400], _? PalindromeQ, 1] // Flatten (* _Michael De Vlieger_, Jan 28 2018 *)

%o (PARI) ispal(v) = {for(i=1, #v\2, if (v[i] != v[#v-i+1], return(0));); return(1);}

%o isok(n) = {b = binary(n); lastb = b[1]; vrun = vector(1); vrun[1] = 1; for (i=2, #b, if (b[i] != lastb, vrun = concat(vrun, 1); lastb = b[i];, vrun[#vrun]++;)); return (ispal(vrun));} \\ _Michel Marcus_, Jul 10 2013

%Y Cf. A006995, A035928. - _Leroy Quet_, Sep 14 2009

%Y Cf. A298644, A101211. - _Emeric Deutsch_, Jan 28 2018

%K nonn,base

%O 1,2

%A _Clark Kimberling_