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A110661 Triangle read by rows: T(n,k) = total number of divisors of k,k+1,...,n (1<=k<=n). 1
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; refs; listen; history; internal format)
OFFSET

1,2

COMMENTS

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.

Cf. A143127.

Sequence in context: A130912 A178844 A143956 * A143124 A205400 A205850

Adjacent sequences:  A110658 A110659 A110660 * A110662 A110663 A110664

KEYWORD

nonn,tabl

AUTHOR

Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 02 2005

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Last modified February 14 03:03 EST 2012. Contains 205567 sequences.