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Trajectory of 2 under repeated application of the map n -> n + square excess of n.
+10
3
2, 3, 5, 6, 8, 12, 15, 21, 26, 27, 29, 33, 41, 46, 56, 63, 77, 90, 99, 117, 134, 147, 150, 156, 168, 192, 215, 234, 243, 261, 266, 276, 296, 303, 317, 345, 366, 371, 381, 401, 402, 404, 408, 416, 432, 464, 487, 490, 496, 508, 532, 535, 541, 553, 577, 578, 580, 584, 592, 608
MAPLE
a[0]:= 2:
for n from 1 to 100 do a[n]:= f(a[n-1]) od:
PROG
(PARI) lista(nn) = {print1(n=2, ", "); for (k=2, nn, m = 2*n - sqrtint(n)^2; print1(m, ", "); n = m; ); } \\ Michel Marcus, Oct 23 2015
Trajectory of 7 under repeated application of the map n --> n + square excess of n.
+10
2
7, 10, 11, 13, 17, 18, 20, 24, 32, 39, 42, 48, 60, 71, 78, 92, 103, 106, 112, 124, 127, 133, 145, 146, 148, 152, 160, 176, 183, 197, 198, 200, 204, 212, 228, 231, 237, 249, 273, 290, 291, 293, 297, 305, 321, 353, 382, 403, 406, 412, 424, 448, 455, 469, 497, 510, 536, 543
REFERENCES
H. Brocard, Note 2837, L'Intermédiaire des Mathématiciens, 11 (1904), p. 239.
PROG
(PARI) lista(nn) = {print1(n=7, ", "); for (k=2, nn, m = 2*n - sqrtint(n)^2; print1(m, ", "); n = m; ); } \\ Michel Marcus, Oct 24 2015
Lower triangular matrix describing the shape of a half hyperbola in the Dirichlet divisor problem.
+10
0
1, 1, 1, 1, 1, 1, 0, 2, 1, 1, 0, 2, 1, 1, 1, 0, 1, 2, 1, 1, 1, 0, 1, 2, 1, 1, 1, 1, 0, 1, 1, 2, 1, 1, 1, 1, 0, 0, 2, 2, 1, 1, 1, 1, 1, 0, 0, 2, 1, 2, 1, 1, 1, 1, 1, 0, 0, 2, 1, 2, 1, 1, 1, 1, 1, 1, 0, 0, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1
COMMENTS
The sum of terms of row n is n. Length of row n is n.
A006218(n) = (n^2 - ((2*Sum_{kk=1..n} Sum_{k=1..kk} T(n,k)) - n)) + 2*n - round(1 + (1/2)*(-3 + sqrt(n) + sqrt(1 + n))).
A006218(n) = -((n^2 - ((2*Sum_{kk=1..n} Sum_{k=1..kk} T(n,n-k+1)) - n)) - 2*n + round(1 + (1/2)*(-3 + sqrt(n) + sqrt(1 + n)))).
(End)
A006218(n) = (n^2 - (2*(Sum_{k=1..n} T(n, k)*(n - k + 1)) - n)) + 2*n - round(1 + (1/2)*(-3 + sqrt(n) + sqrt(1 + n))).
A006218(n) = -((n^2 - (2*(Sum_{k=1..n} T(n, n - k + 1)*(n - k + 1)) - n)) - 2*n + round(1 + (1/2)*(-3 + sqrt(n) + sqrt(1 + n)))).
(End)
It appears that:
The number of 0's in row n is equal to the number of 2's in row n and their number is given by A000196(n) - 1.
The number of 1's in column k is given by A152948(k+2).
The number of 2's in column k is given by A000096(k-1).
The row index of the last nonzero entry in column k is given by A005563(k).
(End)
The smallest k such that T(n,k)=2 is given by A079643(n) = floor(n/floor(sqrt(n))).
<=> A006218(n) >= 2*n - (floor(sqrt(n)))^2 + floor(n/floor(sqrt(n)))*2*floor(sqrt(n)-1).
The average of k:s such that T(n,k)=2, for n>3 is given by:
b(n) = Sum_{k=1..n} (k*floor(abs(T(n, k)-1/2)))/floor(sqrt(n)-1).
This gives A006218(n) = 2*n - (floor(sqrt(n)))^2 + b(n)*2*floor(sqrt(n)-1) = 2*n - (floor(sqrt(n)))^2 + (Sum_{k=1..n} (k*floor(abs(T(n, k)-1/2))))*2, for n>3.
The largest k such that T(n,k)=2 is given by A004526(n) = floor(n/2).
<=> A006218(n) <= 2*n - (floor(sqrt(n)))^2 + floor(n/2)*2*floor(sqrt(n)-1).
The lower bound starts: 1, 3, 5, 8, 10, 14, 16, 20, 21, 23, ...
Sequence A006218 starts: 1, 3, 5, 8, 10, 14, 16, 20, 23, 27, ...
The upper bound starts: 1, 3, 5, 8, 10, 14, 16, 20, 25, 31, ...
(End)
EXAMPLE
1;
1, 1;
1, 1, 1;
0, 2, 1, 1;
0, 2, 1, 1, 1;
0, 1, 2, 1, 1, 1;
0, 1, 2, 1, 1, 1, 1;
0, 1, 1, 2, 1, 1, 1, 1;
0, 0, 2, 2, 1, 1, 1, 1, 1;
0, 0, 2, 1, 2, 1, 1, 1, 1, 1;
0, 0, 2, 1, 2, 1, 1, 1, 1, 1, 1;
0, 0, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1;
MATHEMATICA
nn = 12;
T = Table[
Sum[Table[
If[And[If[n*k <= r, If[n >= k, 1, 0], 0] == 1,
If[(n + 1)*(k + 1) <= r, If[n >= k, 1, 0], 0] == 0], 1, 0], {n,
1, r}], {k, 1, r}], {r, 1, nn}];
Flatten[T]
A006218a = Table[(n^2 - (2*Sum[Sum[T[[n, k]], {k, 1, kk}], {kk, 1, n}] -
n)) + 2*n - Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])], {n,
1, nn}];
A006218b = -Table[(n^2 - (2*
Sum[Sum[T[[n, n - k + 1]], {k, 1, kk}], {kk, 1, n}] - n)) -
2*n + Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])], {n, 1, nn}];
(A006218b - A006218a);
(* (End) *)
nn = 12;
T = Table[
Sum[Table[
If[And[If[n*k <= r, If[n >= k, 1, 0], 0] == 1,
If[(n + 1)*(k + 1) <= r, If[n >= k, 1, 0], 0] == 0], 1, 0], {n,
1, r}], {k, 1, r}], {r, 1, nn}];
Flatten[T]
A006218a = Table[(n^2 - (2*Sum[T[[n, k]]*(n - k + 1), {k, 1, n}] - n)) +
2*n - Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])], {n, 1, nn}];
A006218b = Table[-((n^2 - (2*Sum[T[[n, n - k + 1]]*(n - k + 1), {k, 1, n}] -
n)) - 2*n +
Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])]), {n, 1, nn}];
(A006218b - A006218a);
(* (End) *)
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