|
Search: id:A110662
|
|
|
| A110662 |
|
Triangle read by rows: T(n,k) = sum of the sums of divisors of k,k+1,...,n (1<=k<=n). |
|
+0
2
|
|
| 1, 4, 3, 8, 7, 4, 15, 14, 11, 7, 21, 20, 17, 13, 6, 33, 32, 29, 25, 18, 12, 41, 40, 37, 33, 26, 20, 8, 56, 55, 52, 48, 41, 35, 23, 15, 69, 68, 65, 61, 54, 48, 36, 28, 13, 87, 86, 83, 79, 72, 66, 54, 46, 31, 18, 99, 98, 95, 91, 84, 78, 66, 58, 43, 30, 12, 127, 126, 123, 119, 112
(list; table; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
T(n,n)=sigma(n)=A000203(n) =sum of divisors of n. T(n,1)=sum_{j=1..n} sigma(j) = A024916(n).
Equals A000012 * (A000203 * 0^(n-k)) * A000012, 1<=k<=n. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jul 26 2008
Row sums = A143128: (1, 7, 19, 47, 77,...) - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jul 26 2008
|
|
FORMULA
|
T(n, k)=sum(sigma(j), j=k..n), where sigma(j) is the sum of the divisors of j.
|
|
EXAMPLE
|
T(4,2)=14 because the divisors of 2 are {1,2}, the divisors of 3 are {1,3} and the divisors of 4 are {1,2,4}; sum of all these divisors is 14.
Triangle begins:
1;
4,3;
8,7,4;
15,14,11,7;
21,20,17,13,6
|
|
MAPLE
|
with(numtheory): T:=(n, k)->add(sigma(j), j=k..n): for n from 1 to 12 do seq(T(n, k), k=1..n) od; # yields sequence in triangular form
|
|
CROSSREFS
|
Cf. A000203, A024916.
Cf. A143128.
Sequence in context: A114562 A011451 A117956 this_sequence A132021 A089368 A116583
Adjacent sequences: A110659 A110660 A110661 this_sequence A110663 A110664 A110665
|
|
KEYWORD
|
nonn,tabl
|
|
AUTHOR
|
Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 02 2005
|
|
|
Search completed in 0.002 seconds
|
