

A195609


Numbers n such that sum_{i=1..n} A(i) = A(n)*A(n+1)/4, where A(n) = A000069(n).


0



3, 4, 5, 9, 15, 16, 17, 23, 27, 28, 29, 33, 39, 43, 44, 45, 51, 52, 53, 57, 63, 64, 65, 71, 75, 76, 77, 83, 84, 85, 89, 95, 99, 100, 101, 105, 111, 112, 113, 119, 123, 124, 125, 129, 135, 139, 140, 141, 147, 148, 149, 153, 159, 163, 164, 165, 169, 175, 176
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

Conjectures: 1) there are only 3 different first differences 1,4,6; 2) the sequence contains either isolated series of terms, e.g., {9},{23},{33},{39},..., or series of 3 consecutive integers, e.g., {3,4,5}, {15,16,17}, etc.; 3)the first terms m of every series satisfy the condition A(m+1)A(m1)=5, where A(n)=A000069(n).


LINKS

Table of n, a(n) for n=1..59.


MATHEMATICA

a = Select[Range[1000], OddQ[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 *)


CROSSREFS

Cf. A000069, A174531, A001969, A195608.
Sequence in context: A136259 A099560 A050161 * A117125 A000692 A195971
Adjacent sequences: A195606 A195607 A195608 * A195610 A195611 A195612


KEYWORD

nonn


AUTHOR

Vladimir Shevelev, Sep 21 2011


STATUS

approved



