A236112 - OEIS

A236112

Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists k+1 copies of the squares in nondecreasing order, and the first element of column k is in row k(k+1)/2.

%I #35 Nov 05 2024 05:40:24

%S 0,0,1,0,1,0,4,0,4,1,0,9,1,0,9,1,0,16,4,0,16,4,1,0,25,4,1,0,25,9,1,0,

%T 36,9,1,0,36,9,4,0,49,16,4,1,0,49,16,4,1,0,64,16,4,1,0,64,25,9,1,0,81,

%U 25,9,1,0,81,25,9,4,0,100,36,9,4,1,0,100,36,16,4,1,0,121,36,16,4,1,0,121,49,16,4,1,0

%N Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists k+1 copies of the squares in nondecreasing order, and the first element of column k is in row k(k+1)/2.

%C Gives an identity for the sum of remainders of n mod k, for k = 1,2,3,...,n. Alternating sum of row n equals A004125(n), i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A004125(n).

%C Row n has length A003056(n) hence the first element of column k is in row A000217(k).

%e Triangle begins:

%e 0;

%e 1, 0;

%e 4, 0;

%e 4, 1, 0;

%e 9, 1, 0;

%e 16, 4, 0;

%e 16, 4, 1, 0;

%e 25, 4, 1, 0;

%e 25, 9, 1, 0;

%e 36, 9, 1, 0;

%e 36, 9, 4, 0;

%e 49, 16, 4, 1, 0;

%e 64, 16, 4, 1, 0;

%e 64, 25, 9, 1, 0;

%e 81, 25, 9, 1, 0;

%e 81, 25, 9, 4, 0;

%e 100, 36, 9, 4, 1, 0;

%e 100, 36, 16, 4, 1, 0;

%e 121, 36, 16, 4, 1, 0;

%e 121, 49, 16, 4, 1, 0;

%e ...

%e For n = 24 the 24th row of triangle is 121, 49, 16, 4, 1, 0 therefore the alternating row sum is 121 - 49 + 16 - 4 + 1 - 0 = 85 equaling A004125(24).

%Y Cf. A000203, A000217, A000290, A003056, A004125, A120444, A196020, A211343, A228813, A231345, A231347, A235791, A235794, A236104, A236106, A237048, A237591, A237593, A261699.

%K nonn,tabf

%O 1,7

%A _Omar E. Pol_, Jan 23 2014