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%I #20 Apr 20 2016 00:33:09
%S 0,0,1,1,2,0,3,3,4,0,1,5,5,0,6,0,0,7,7,3,8,0,0,1,9,9,0,0,10,0,5,0,11,
%T 11,0,0,12,0,0,3,13,13,7,0,1,14,0,0,0,0,15,15,0,0,0,16,0,9,5,0,17,17,
%U 0,0,0,18,0,0,0,3,19,19,11,0,0,1,20,0,0,7,0,0
%N Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the odd numbers interleaved with k-1 zeros, except the first column which lists 0 together with the nonnegative integers, and the first element of column k is in row k*(k+1)/2.
%C Alternating row sums give the Chowla's function, i.e., sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A048050(n).
%C Row n has length A003056(n) hence column k starts in row A000217(k).
%C Column 1 gives 0 together with A001477.
%C Column 2 is A193356.
%C The number of positive terms in row n is A001227(n), if n >= 3. - _Omar E. Pol_, Apr 18 2016
%F T(n,k) = A196020(n,k), if k >= 2. - _Omar E. Pol_, Apr 18 2016
%e Triangle begins (row n = 1..24):
%e 0;
%e 0;
%e 1, 1;
%e 2, 0;
%e 3, 3;
%e 4, 0, 1;
%e 5, 5, 0;
%e 6, 0, 0;
%e 7, 7, 3;
%e 8, 0, 0, 1;
%e 9, 9, 0, 0;
%e 10, 0, 5, 0;
%e 11, 11, 0, 0;
%e 12, 0, 0, 3;
%e 13, 13, 7, 0, 1;
%e 14, 0, 0, 0, 0;
%e 15, 15, 0, 0, 0;
%e 16, 0, 9, 5, 0;
%e 17, 17, 0, 0, 0;
%e 18, 0, 0, 0, 3;
%e 19, 19, 11, 0, 0, 1;
%e 20, 0, 0, 7, 0, 0;
%e 21, 21, 0, 0, 0, 0;
%e 22, 0, 13, 0, 0, 0;
%e ...
%e For n = 15 the divisors of 15 are 1, 3, 5, 15 therefore the sum of divisors of 15 except 1 and 15 is 3 + 5 = 8. On the other hand the 15th row of triangle is 13, 13, 7, 0, 1, hence the alternating row sum is 13 - 13 + 7 - 0 + 1 = 8, equalling the sum of divisors of 15 except 1 and 15.
%e If n is even then the alternating sum of the n-th row of triangle is simpler than the sum of divisors of n, except 1 and n. Example: the sum of divisors of 24 except 1 and 24 is 2 + 3 + 4 + 6 + 8 + 12 = 35, and the alternating sum of the 24th row of triangle is 22 - 0 + 13 - 0 + 0 - 0 = 35.
%Y Cf. A000203, A000217, A001065, A001227, A001477, A193356, A196020, A211343, A212119, A228813, A231345, A231347, A235791, A235794, A236104, A236106, A236112.
%K nonn,tabf
%O 1,5
%A _Omar E. Pol_, Mar 15 2014