A136561
Triangle read by rows: 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.
Original entry on oeis.org
1, 2, 3, 4, 6, 9, -5, -1, 5, 14, 13, 8, 7, 12, 26, -30, -17, -9, -2, 10, 36, 75, 45, 28, 19, 17, 27, 63, -200, -125, -80, -52, -33, -16, 11, 74, 524, 324, 199, 119, 67, 34, 18, 29, 103, -1299, -775, -451, -252, -133, -66, -32, -14, 15, 118
Offset: 1
The triangle begins:
1,
2,3,
4,6,9,
-5,-1,5,14,
13,8,7,12,26,
-30,-17,-9,-2,10,36.
Example:
Considering the rightmost value of the 4th row: Writing a 10 here instead, the first 4 rows of the triangle become:
1
2,3
4,6,9
-9,-5,1,10
But 1 already occurs earlier in the triangle. So 10 is not the rightmost element of row 4.
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.
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.
Original entry on oeis.org
1, 3, 9, 14, 26, 36, 63, 74, 103, 118, 149, 169, 210, 233, 280, 302, 357, 392, 464, 489, 553, 591, 673, 713, 796, 844, 941, 987, 1083, 1134, 1238, 1292, 1398, 1463, 1596, 1652, 1769, 1840, 1980, 2046, 2172, 2250, 2416, 2492, 2565, 2715, 2836, 3051, 3130, 3298
Offset: 1
The triangle begins:
1,
2,3,
4,6,9,
-5,-1,5,14,
13,8,7,12,26,
-30,-17,-9,-2,10,36.
Example:
Considering the rightmost value of the 4th row: Writing a 10 here instead, the first 4 rows of the triangle become:
1
2,3
4,6,9
-9,-5,1,10
But 1 already occurs earlier in the triangle. So 10 is not the rightmost element of row 4.
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.
-
a, t = [1], [1]
for n in range(1, 100):
d = a[-1]
while True:
d += 1
row = [d]
for j in range(n):
row.append(row[-1]-t[-j-1])
if row[-1] in t:
break
else:
a.append(d)
t += reversed(row)
break
print(a)
# t contains the triangle
# [t[n*(n-1)/2] for n in range(1, 100)] gives leftmost column
# Andrey Zabolotskiy, May 29 2017
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