A195608 Numbers n such that Sum_{i=1..n} A(i) = A(n)*A(n+1)/4, where A(n) = A001969(n).

1, 7, 11, 12, 13, 19, 20, 21, 25, 31, 35, 36, 37, 41, 47, 48, 49, 55, 59, 60, 61, 67, 68, 69, 73, 79, 80, 81, 87, 91, 92, 93, 97, 103, 107, 108, 109, 115, 116, 117, 121, 127, 131, 132, 133, 137, 143, 144, 145, 151, 155, 156, 157, 161, 167, 171, 172, 173, 179

Offset: 1

Author

Vladimir Shevelev, Sep 21 2011

Keywords

• nonn
• base

Comments

Conjectures: 1) there are only 3 different first differences 1,4,6; 2) the sequence contains either isolated series of terms, e.g., {1},{7},{25},{31},..., or series of 3 consecutive integers, e.g., {11,12,13}, {19,20,21}, etc.; 3)the first terms m of every series satisfy the condition A(m+1)-A(m-1)=5, where A(n)=A001969(n).

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001969, A195467.

Programs

Mathematica

a = Select[Range[0, 1000], EvenQ[DigitCount[#, 2][[1]]] &]; t = {}; s = 0; Do[s = s + a[[n]]; If[s == a[[n]] a[[n + 1]]/4, AppendTo[t, n]], {n, Length[a] - 1}]; t (* T. D. Noe, Sep 23 2011 *)