|
|
A246636
|
|
Numbers k such that C(k+2,2) divides 2^(k+1) - 1.
|
|
2
|
|
|
0, 1, 5, 17, 41, 125, 161, 377, 485, 881, 1457, 2645, 3077, 3941, 5417, 9197, 11825, 14405, 16757, 18521, 24965, 26405, 37337, 39365, 42461, 71441, 77657, 95921, 99077, 113777, 117305, 143261, 174761, 175445, 184841, 265481, 304037, 308825, 318401, 351917
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,3
|
|
COMMENTS
|
These are the numbers k such that mean of the numbers in the first k rows of Pascal' s triangle is an integer. All such k except 1 are congruent to -1 mod 6.
|
|
LINKS
|
|
|
EXAMPLE
|
The sum of the numbers in Pascal's triangle, from row 0 through row 17, is 2^18 - 1 = 262143; the number of such numbers is C(19,2) = 171, and 262143/171 = 1533; thus 17 is in this sequence, and 1533 is in A246637.
|
|
MAPLE
|
select(k -> 2 &^(k+1) - 1 mod ((k+1)*(k+2)/2) = 0, [$0..10^6]); # Robert Israel, Nov 30 2023
|
|
MATHEMATICA
|
z = 1000;
t = Select[Range[0, z], IntegerQ[(2^(# + 1) - 1)/Binomial[# + 2, 2]] &]
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|