|
|
A306568
|
|
Regular triangle read by rows where T(n,k) is obtained by partitioning the set of the first n squares into k groups, then summing up each group and taking the minimum width, where the width is the difference between its greatest and its smallest term.
|
|
0
|
|
|
3, 4, 8, 2, 11, 15, 3, 11, 20, 24, 1, 10, 22, 31, 35, 0, 4, 23, 35, 44, 48, 0, 4, 19, 38, 50, 59, 63, 1, 3, 15, 36, 55, 67, 76, 80, 1, 2, 6, 34, 55, 74, 86, 95, 99, 0, 2, 4, 27, 55, 76, 95, 107, 116, 120, 0, 1, 3, 19, 50, 78, 99, 118, 130, 139, 143, 1, 0, 3, 8, 44, 75, 103, 124, 143, 155, 164, 168
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
2,1
|
|
COMMENTS
|
Collatz proves that for a sufficiently large n (depending on k), the k-th column is periodic with period 2*k^2, but remarks that column 2 and 3 appear to have periods 4 and 9, so he asks if the period is actually k^2.
|
|
LINKS
|
|
|
FORMULA
|
T(n,n) = n^2 - 1.
|
|
EXAMPLE
|
For n=3 and k=2, the partitions are (1,4; 9) or (1; 4,9) or (1,9; 4);
the sums are (5,9) or (1,13) or (4,10);
the widths are 4, 12, 6; so a(3,2) = 4, the minimum.
The triangle begins:
3;
4, 8;
2, 11, 15;
3, 11, 20, 24;
1, 10, 22, 31, 35;
0, 4, 23, 35, 44, 48;
0, 4, 19, 38, 50, 59, 63;
1, 3, 15, 36, 55, 67, 76, 80;
...
|
|
PROG
|
(PARI) vj(j, b, n, p) = {vd = select(x->(x==j), b, 1); sum(k=1, #vd, vd[k]^2); }
a(n, p) = {my(res = sum(k=1, n, k^2), b, vone); for (k=0, p^n-1, b = digits(k, p); if (#b < n, b = concat(vector(n-#b), b)); if (#Set(b) == p, vone = vector(p, j, vj(j-1, b, n, p)); res = min(res, vecmax(vone) - vecmin(vone)); ); ); res; }
tabl(nn) = for (n=2, nn, for (p=2, n, print1(a(n, p), ", ")); print); \\ Michel Marcus, Feb 24 2019
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|