cp's OEIS Frontend

A110661 Triangle read by rows: T(n,k) = total number of divisors of k, k+1, ..., n (1 <= k <= n).

%I A110661 #16 Feb 07 2022 03:43:53
%S A110661 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,
%T A110661 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,
%U A110661 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
%N A110661 Triangle read by rows: T(n,k) = total number of divisors of k, k+1, ..., n (1 <= k <= n).
%C A110661 Equals A000012 * (A000005 * 0^(n-k)) * A000012, 1 <= k <= n. - _Gary W. Adamson_, Jul 26 2008
%C A110661 Row sums = A143127. - _Gary W. Adamson_, Jul 26 2008
%H A110661 G. C. Greubel, <a href="/A110661/b110661.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%F A110661 T(n, k) = Sum_{j=k..n} tau(j), where tau(j) is the number of divisors of j, and 1 <= k <= n.
%F A110661 T(n,n) = tau(n) = A000005(n) = number of divisors of n.
%F A110661 T(n,1) = Sum_{j=1..n} tau(j) = A006218(n).
%e A110661 T(4,2)=7 because 2 has 2 divisors, 3 has 2 divisors and 4 has 3 divisors.
%e A110661 Triangle begins:
%e A110661    1;
%e A110661    3, 2;
%e A110661    5, 4, 2;
%e A110661    8, 7, 5, 3;
%e A110661   10, 9, 7, 5, 2;
%e A110661   ...
%p A110661 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
%t A110661 T[n_, n_] := DivisorSigma[0, n]; T[n_, k_] := Sum[DivisorSigma[0, j], {j, k, n}]; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* _G. C. Greubel_, Sep 03 2017 *)
%Y A110661 Cf. A000005, A006218.
%Y A110661 Cf. A143127.
%K A110661 nonn,tabl
%O A110661 1,2
%A A110661 _Emeric Deutsch_, Aug 02 2005