A305996
Rectangular array, by antidiagonals; row n consists of the numbers R(n)/n, where R(n) is row n of the array at A305995.
Original entry on oeis.org
1, 10, 1, 65, 34, 1, 130, 260, 2, 1, 260, 884, 5, 10, 1, 340, 1300, 10, 26, 10, 2, 1105, 3380, 20, 260, 340, 20, 1, 1972, 8840, 50, 5140, 650, 52, 2, 1, 2210, 31300, 65, 8840, 1565, 100, 5, 260, 1, 4420, 82948, 68, 21320, 5525, 520, 10, 514, 2, 2
Offset: 1
Northwest corner:
1 10 65 130 260 340 1105
1 34 260 884 1300 3380 8840
1 2 5 10 20 50 65
1 10 26 260 5140 8840 21430
1 10 340 650 1565 5525 6260
2 30 52 100 520 10280 17680
1 2 5 10 20 25 50
-
z = 3000; r[n_] := Select[Divisors[n], GCD[#, n/#] == 1 &]; k[n_] := Length[r[n]];
t[n_] := Table[r[n][[k[n] + 1 - i]]/r[n][[k[1] + i - 1]], {i, 1, k[n]}];
s = Table[Plus @@ t[n], {n, 1, z}];
a[n_] := If[IntegerQ[s[[n]]], 1, 0];
u = Table[a[n], {n, 1, z}]; (*A229996*)
d = Denominator[s]; row[n_] := Flatten[Position[d, n]] (*A305995 array*)
rr[n_] := row[n]/n;
TableForm[Table[rr[n], {n, 1, 100}]] (* A305996 array *)
r1[n_, k_] := rr[n][[k]];
Flatten[Table[r1[n - k + 1, k], {n, 5}, {k, n, 1, -1}]] (* A305996 sequence *)
A306010
Let S(m) = d(k)/d(1) + ... + d(1)/d(k), where d(1)..d(k) are the unitary divisors of m; then a(n) is the number m when the sums S(m) are arranged in increasing order.
Original entry on oeis.org
1, 2, 3, 4, 5, 7, 8, 6, 9, 11, 10, 13, 12, 16, 17, 15, 14, 19, 20, 18, 23, 21, 25, 27, 24, 22, 29, 28, 31, 32, 26, 33, 37, 35, 36, 41, 40, 34, 43, 30, 39, 47, 44, 45, 38, 49, 53, 48, 52, 51, 46, 55, 56, 59, 42, 61, 50, 57, 64, 63, 67, 54, 65, 71, 68, 58, 73
Offset: 1
The first 8 pairs {m,S(m)} are {1, 1}, {2, 5/2}, {3, 10/3}, {4, 17/4}, {5, 26/5}, {6, 25/3}, {7, 50/7}, {8, 65/8}. When the numbers S(m) are arranged in increasing order, the pairs are {1, 1}, {2, 5/2}, {3, 10/3}, {4, 17/4}, {5, 26/5}, {7, 50/7}, {8, 65/8}, {6, 25/3}, so that the first 8 terms of (a(n)) are 1,2,3,4,5,7,8,6.
-
z = 100; r[n_] := Select[Divisors[n], GCD[#, n/#] == 1 &];
k[n_] := Length[r[n]];
t[n_] := Table[r[n][[k[n] + 1 - i]]/r[n][[k[1] + i - 1]], {i, 1, k[n]}];
s = Table[{n, Total[t[n]]}, {n, 1, z}]
v = SortBy[s, Last]
v1 = Table[v[[n]][[1]], {n, 1, z}] (* A306010 *)
w = Table[v[[n]][[2]], {n, 1, z}];
Numerator[w] (* A306011 *)
Denominator[w] (* A306012 *)
A306011
Let S(m) = d(k)/d(1) + ... + d(1)/d(k), where d(1)..d(k) are the unitary divisors of m; then a(n) is the numerator of S(m) when all the numbers S(m) are arranged in increasing order.
Original entry on oeis.org
1, 5, 10, 17, 26, 50, 65, 25, 82, 122, 13, 170, 85, 257, 290, 52, 125, 362, 221, 205, 530, 500, 626, 730, 325, 305, 842, 425, 962, 1025, 425, 1220, 1370, 260, 697, 1682, 169, 725, 1850, 130, 1700, 2210, 1037, 2132, 905, 2402, 2810, 1285, 1445, 2900, 1325
Offset: 1
The first 8 pairs {m,S(m)} are {1, 1}, {2, 5/2}, {3, 10/3}, {4, 17/4}, {5, 26/5}, {6, 25/3}, {7, 50/7}, {8, 65/8}. When the numbers S(m) are arranged in increasing order, the pairs are {1, 1}, {2, 5/2}, {3, 10/3}, {4, 17/4}, {5, 26/5}, {7, 50/7}, {8, 65/8}, {6, 25/3}, so that the first 8 numerators are 1,5,10,17,26,50,65,25.
-
z = 100; r[n_] := Select[Divisors[n], GCD[#, n/#] == 1 &];
k[n_] := Length[r[n]];
t[n_] := Table[r[n][[k[n] + 1 - i]]/r[n][[k[1] + i - 1]], {i, 1, k[n]}];
s = Table[{n, Total[t[n]]}, {n, 1, z}]
v = SortBy[s, Last]
v1 = Table[v[[n]][[1]], {n, 1, z}] (* A306010 *)
w = Table[v[[n]][[2]], {n, 1, z}];
Numerator[w] (* A306011 *)
Denominator[w] (* A306012 *)
A306012
Let S(m) = d(k)/d(1) + ... + d(1)/d(k), where d(1)..d(k) are the unitary divisors of m; then a(n) is the denominator of S(m) when all the numbers S(m) are arranged in increasing order.
Original entry on oeis.org
1, 2, 3, 4, 5, 7, 8, 3, 9, 11, 1, 13, 6, 16, 17, 3, 7, 19, 10, 9, 23, 21, 25, 27, 12, 11, 29, 14, 31, 32, 13, 33, 37, 7, 18, 41, 4, 17, 43, 3, 39, 47, 22, 45, 19, 49, 53, 24, 26, 51, 23, 55, 28, 59, 21, 61, 5, 57, 64, 63, 67, 27, 1, 71, 2, 29, 73, 3, 36, 69
Offset: 1
The first 8 pairs {m,S(m)} are {1, 1}, {2, 5/2}, {3, 10/3}, {4, 17/4}, {5, 26/5}, {6, 25/3}, {7, 50/7}, {8, 65/8}. When the numbers S(m) are arranged in increasing order, the pairs are {1, 1}, {2, 5/2}, {3, 10/3}, {4, 17/4}, {5, 26/5}, {7, 50/7}, {8, 65/8}, {6, 25/3}, so that the first 8 denominators are 1,2,3,4,5,7,8,3.
-
z = 100; r[n_] := Select[Divisors[n], GCD[#, n/#] == 1 &];
k[n_] := Length[r[n]];
t[n_] := Table[r[n][[k[n] + 1 - i]]/r[n][[k[1] + i - 1]], {i, 1, k[n]}];
s = Table[{n, Total[t[n]]}, {n, 1, z}]
v = SortBy[s, Last]
v1 = Table[v[[n]][[1]], {n, 1, z}] (* A306010 *)
w = Table[v[[n]][[2]], {n, 1, z}];
Numerator[w] (* A306011 *)
Denominator[w] (* A306012 *)
A306013
Let P(m) be the product of unitary divisors of m; then a(n) is the position of P(n) when all the numbers P(m) are arranged in increasing order.
Original entry on oeis.org
1, 2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17, 19, 23, 25, 27, 29, 31, 32, 36, 37, 41, 43, 47, 49, 53, 59, 61, 64, 67, 71, 73, 79, 81, 83, 89, 97, 100, 144, 196, 225, 324, 400, 441, 484, 576, 676, 784, 1089, 1156, 1225, 1296, 1444, 1521, 1600, 1936, 2025, 2116
Offset: 1
-
z = 100; r[n_] := Select[Divisors[n], GCD[#, n/#] == 1 &];
k[n_] := Length[r[n]];
Table[r[n], {n, 1, z}]
a[n_] := Apply[Times, r[n]]
u = Table[a[n], {n, 1, z}]
Sort[u]
Showing 1-5 of 5 results.
Comments