This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
For n = 6, a(6) = 5 because A337226(5) = 3 = A337226(5 + 6).
For n = 6, a(6) = 3 because A337226(5) = A337226(5 + 6) = 3.
a(1)=1; no pair of terms exists yet.
a(2)=1 creates the pair [1, 1], which encloses 0 elements. This means that no consecutive equal values can occur again, since this would create another set of 0 elements.
a(3)=2 because if a(3)=1, this would create a second pair enclosing 0 elements.
a(4)=1 creates two new sets: [1, 2, 1], enclosing 1 element {2}, and [1, 1, 2, 1], enclosing 2 elements {1, 2}.
a(5) cannot be 1 as this would again create a pair enclosing 0 elements [1,1]. 2 would create the pair [2, 1, 2] which encloses 1 element {1}, which has been impossible since a(4). So a(5)=3, which has not occurred before.
See Links section.
from itertools import islice
def agen(): # generator of terms
e, a = set(), []
while True:
an, allnew = 0, False
while not allnew:
allnew, an, ndset = True, an+1, set()
for i in range(len(a)):
if an == a[i]:
nd = len(set(a[i+1:]))
if nd in e or nd in ndset: allnew = False; break
ndset.add(nd)
yield an; a.append(an); e |= ndset
print(list(islice(agen(), 72))) # Michael S. Branicky, Oct 25 2023
Initial locations and the (by definition) distinct terms that they reach:
n| 1 2 3 4 5 6 7 8 9
a(n)| 1 1 2 2 3 4 2 5 6
=>1=>2====>3
====>4
=======>5
====>6
When we evaluate a(i+a(i)) with each index i, we get a distinct value. When i=1, for example, a(1+a(1))=a(1+1)=a(2)=1; no other i gives 1 as the solution to a(i+a(i)). When i=4, a(4+a(4))=a(4+2)=a(6)=4, and 4 is likewise a solution unique to i=4.
function a = A367467( max_n ) a = [1 1:2*max_n]; for n = 3:max_n a(n) = 1; while consistency(a, n) == false a(n) = a(n)+1; end end a = a(1:max_n); end function ok = consistency(a, n) v = a([1:n] + a(1:n)); ok = (n == length(unique(v))); end % Thomas Scheuerle, Nov 21 2023
Triangle begins: 0; 0, 0; 1, 2, 1; 0, 3, 4, 0; 3, 5, 2, 6, 3; 2, 7, 8, 5, 1, 9; ...
T(n)={my(v=vector(n), S=Set(), L=List());
for(n=1, #v, v[n]=vector(n); for(k=1, n, my(i=1);
while(i<=#L, my(P=Set([[n-p[1], k-p[2]] | p<-L[i]])); if(!#setintersect(P,S), S = setunion(S,P); break); i++);
if(i>#L, listput(L, []));
L[i] = concat(L[i], [[n,k]]);
v[n][k] = i-1 )); v
}
concat(T(12)) \\ Andrew Howroyd, Sep 24 2020
See Links section.
a(13) = 3: a(13) cannot be 1 as i = 4,13 would have the same standard deviation as i = 1,4,8,13 (namely 4.5). We cannot have a(13) = 2 because i = 3,6 would have the same standard deviation as i = 10,13 (namely 1.5). With a(13) = 3, we find that no two subsets of i = 7,9,12,13 have the same standard deviation, so a(13) = 3.
The spiral begins:
10--13---8--12---9--11---7
| |
11 2---6---5---4---3 10
| | | |
14 7 0---1---0 2 8
| | | | | |
15 4 2 0---0 3 9
| | | | |
12 6 1---3---4---1 7
| | |
13 5---7---8---9---5---6
|
10--14--11--15--16--17--12
See Links section.
a(7)=5: We cannot have a(7)=1 because then, for example, the unordered pair (1,2) would have the same absolute distance twice at distinct indices:
1, 1, 2, 3, 4, 2, 1
1--2 2--1
a(7) could not equal 2 because again the pair (1,2) would have the same absolute distance twice at different indices (i=6-1=5 and i=7-2=5):
1, 1, 2, 3, 4, 2, 2
1--------------2
1--------------2
a(7) cannot be 3 because of the following two equal unordered pairs, which would have the same distance:
1, 1, 2, 3, 4, 2, 3
2--3 2--3
a(7) cannot be 4, or we would have two equal unordered pairs both with distance 1:
1, 1, 2, 3, 4, 2, 4
4--2
2--4
a(7) can be 5 without restriction since 5 is a first occurrence and every unordered pair with 5 has a distinct distance.
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