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A136562 Consider the triangle A136561: the n-th diagonal (from the right) is the sequence of (signed) differences between pairs of consecutive terms in the (n-1)th diagonal. The rightmost diagonal (A136562) is defined: A136562(1)=1; A136562(n) is the smallest integer > A136562(n-1) such that any (signed) integer occurs at most once in the triangle A136561.

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