|
Search: id:A110661
|
|
|
| A110661 |
|
Triangle read by rows: T(n,k) = total number of divisors of k,k+1,...,n (1<=k<=n). |
|
+0
2
|
|
| 1, 3, 2, 5, 4, 2, 8, 7, 5, 3, 10, 9, 7, 5, 2, 14, 13, 11, 9, 6, 4, 16, 15, 13, 11, 8, 6, 2, 20, 19, 17, 15, 12, 10, 6, 4, 23, 22, 20, 18, 15, 13, 9, 7, 3, 27, 26, 24, 22, 19, 17, 13, 11, 7, 4, 29, 28, 26, 24, 21, 19, 15, 13, 9, 6, 2, 35, 34, 32, 30, 27, 25, 21, 19, 15, 12, 8, 6, 37, 36, 34
(list; table; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
T(n,n)=tau(n)=A000005(n) =number of divisors of n. T(n,1)=sum_{j=1..n} tau(j) = A006218(n).
Equals A000012 * (A000005 * 0^(n-k)) * A000012, 1<=k<=n. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jul 26 2008
Row sums = A143127: (1, 5, 11, 23, 33, 57,...) - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jul 26 2008
|
|
FORMULA
|
T(n, k)=sum(tau(j), j=k..n), where tau(j) is the number of divisors of j.
|
|
EXAMPLE
|
T(4,2)=7 because 2 has 2 divisors, 3 has 2 divisors, and 4 has 3 divisors.
Triangle begins:
1;
3,2;
5,4,2;
8,7,5,3;
10,9,7,5,2;
|
|
MAPLE
|
with(numtheory): T:=(n, k)->add(tau(j), j=k..n): for n from 1 to 13 do seq(T(n, k), k=1..n) od; # yields sequence in triangular form
|
|
CROSSREFS
|
Cf. A000005, A006218.
Sequence in context: A131025 A070151 A130912 this_sequence A128076 A076243 A140061
Adjacent sequences: A110658 A110659 A110660 this_sequence A110662 A110663 A110664
Cf. A143127.
|
|
KEYWORD
|
nonn,tabl
|
|
AUTHOR
|
Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 02 2005
|
|
|
Search completed in 0.002 seconds
|
