%I #15 May 31 2017 07:58:01
%S 1,3,9,14,26,36,63,74,103,118,149,169,210,233,280,302,357,392,464,489,
%T 553,591,673,713,796,844,941,987,1083,1134,1238,1292,1398,1463,1596,
%U 1652,1769,1840,1980,2046,2172,2250,2416,2492,2565,2715,2836,3051,3130,3298
%N Consider the triangle A136561: the nth diagonal (from the right) is the sequence of (signed) differences between pairs of consecutive terms in the (n1)th diagonal. The rightmost diagonal (A136562) is defined: A136562(1)=1; A136562(n) is the smallest integer > A136562(n1) such that any (signed) integer occurs at most once in the triangle A136561.
%C Requiring that the absolute values of the differences in the difference triangle only occur at most once each leads to the Zorach additive triangle. (See A035312.) The rightmost diagonal of the Zorach additive triangle is A035313.
%C It appears that a(n) is proportional to n^2.  _Andrey Zabolotskiy_, May 29 2017
%e The triangle begins:
%e 1,
%e 2,3,
%e 4,6,9,
%e 5,1,5,14,
%e 13,8,7,12,26,
%e 30,17,9,2,10,36.
%e Example:
%e Considering the rightmost value of the 4th row: Writing a 10 here instead, the first 4 rows of the triangle become:
%e 1
%e 2,3
%e 4,6,9
%e 9,5,1,10
%e But 1 already occurs earlier in the triangle. So 10 is not the rightmost element of row 4.
%e Checking 11,12,13,14; 14 is the smallest value that can be the rightmost element of row 4 and not have any elements of row 4 occur earlier in the triangle. So A136562(4) = 13.
%o (Python)
%o a, t = [1], [1]
%o for n in range(1, 100):
%o d = a[1]
%o while True:
%o d += 1
%o row = [d]
%o for j in range(n):
%o row.append(row[1]t[j1])
%o if row[1] in t:
%o break
%o else:
%o a.append(d)
%o t += reversed(row)
%o break
%o print(a)
%o # t contains the triangle
%o # [t[n*(n1)/2] for n in range(1, 100)] gives leftmost column
%o # _Andrey Zabolotskiy_, May 29 2017
%Y Cf. A035313, A136561, A136563.
%K nonn
%O 1,2
%A _Leroy Quet_, Jan 06 2008
%E More terms from _Andrey Zabolotskiy_, May 29 2017
