close
login
A376181
Array read by antidiagonals: Start from 1 and thereafter add gnomons of terms for each t >= 2 with width a(t-1).
0
1, 2, 2, 3, 2, 3, 3, 3, 3, 3, 4, 3, 3, 3, 4, 4, 4, 3, 3, 4, 4, 5, 4, 4, 3, 4, 4, 5, 5, 5, 4, 4, 4, 4, 5, 5, 5, 5, 5, 4, 4, 4, 5, 5, 5, 6, 5, 5, 5, 4, 4, 5, 5, 5, 6, 6, 6, 5, 5, 5, 4, 5, 5, 5, 6, 6, 7, 6, 6, 5, 5, 5, 5, 5, 5, 6, 6, 7, 7, 7, 6, 6, 5, 5, 5, 5, 5, 6, 6, 7, 7, 7, 7, 7, 6, 6, 5, 5, 5, 5
OFFSET
1,2
COMMENTS
A gnomon is an L-shaped group of terms that are equal to each other and surrounds the top-left corner.
The sequence seems to oscillate a lot and it seems that log(a(n))/log(n) doesn't converge. However, log(max{a(i); i < n})/log(n) appears to converge to around ~0.387.
The formula a(d(d+1)/2) = n, where d = 1 + (a(1) + ... + a(n-1)), suggests that the sequence might be growing like k*n^c on average, where c = (sqrt(3)-1)/2.
The construction is similar to the Golomb sequence (A001462) in that sequence terms themselves determine repetitions, but here those repetitions are gnomon widths.
FORMULA
a(d(d+1)/2) = n, where d = 1 + (a(1) + ... + a(n-1)), for n > 1.
a(n) ~ k*n^c, on average, where c = (sqrt(3)-1)/2 = 0.366... (heuristic).
max{a(i); i < n} ~ K*n^e with e ~ 0.387... (empirical).
EXAMPLE
Array begins:
k=1 2 3 4 5 6
n=1: 1 2 3 3 4 4
n=2: 2 2 3 3 4 4
n=3: 3 3 3 3 4 4
n=4: 3 3 3 3 4 4
n=5: 4 4 4 4 4 4
n=6: 4 4 4 4 4 4
The first three terms by antidiagonals are 1,2,2 and they are the widths of the gnomons comprising terms 2,3,4 respectively.
CROSSREFS
Sequence in context: A202472 A235613 A343901 * A322418 A019569 A003434
KEYWORD
nonn,tabl
AUTHOR
Bryle Morga, Sep 14 2024
STATUS
approved