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A366472
Irregular triangle read by rows: T(n,k) (n >= 1, k >= 1) = number of increasing geometric progressions in {1,2,3,...,n} of length k with rational ratio.
2
1, 2, 1, 3, 3, 4, 6, 1, 5, 10, 1, 6, 15, 1, 7, 21, 1, 8, 28, 2, 1, 9, 36, 4, 1, 10, 45, 4, 1, 11, 55, 4, 1, 12, 66, 5, 1, 13, 78, 5, 1, 14, 91, 5, 1, 15, 105, 5, 1, 16, 120, 8, 2, 1, 17, 136, 8, 2, 1, 18, 153, 10, 2, 1, 19, 171, 10, 2, 1, 20, 190, 11, 2, 1, 21, 210, 11, 2, 1, 22, 231, 11, 2, 1, 23, 253, 11, 2, 1, 24, 276, 12, 3, 1
OFFSET
1,2
FORMULA
T(n,k) = Sum_{p=2..floor(n^(1/(k-1)))} phi(p)*floor(n/p^(k-1)) where phi is the Euler phi-function A000010 and k runs from 1 to 1+floor(log_2(n))}.
EXAMPLE
Triangle begins:
[1],
[2, 1],
[3, 3],
[4, 6, 1],
[5, 10, 1],
[6, 15, 1],
[7, 21, 1],
[8, 28, 2, 1],
[9, 36, 4, 1],
[10, 45, 4, 1],
[11, 55, 4, 1],
[12, 66, 5, 1],
[13, 78, 5, 1],
[14, 91, 5, 1],
[15, 105, 5, 1],
[16, 120, 8, 2, 1],
...
MAPLE
with(numtheory);
A366472 := proc(n) local v, u2, u1, k, i, p;
v := Array(1..100, 0);
v[1] := n;
u1 := 1+floor(log(n)/log(2));
for k from 2 to u1 do
u2 := floor(n^(1/(k-1)));
v[k] := add(phi(p)*floor(n/p^(k-1)), p=2..u2);
od;
[seq(v[i], i=1..u1)];
end;
for n from 1 to 36 do lprint(A366472(n)); od:
CROSSREFS
Row sums give A366471.
First three columns are A000027, A000217, A132345.
Cf. A000010.
Sequence in context: A350844 A083041 A318611 * A130067 A282906 A032303
KEYWORD
nonn,tabf
AUTHOR
STATUS
approved