A384046 Triangle in which the n-th row gives the numbers from 1 to n whose largest divisor that is a unitary divisor of n is 1.
1, 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 5, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 8, 1, 3, 7, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 5, 7, 10, 11, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 3, 5, 9, 11, 13, 1, 2, 4, 7, 8, 11, 13, 14, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15
Offset: 1
Examples
Triangle begins: 1, 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 5, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 8, 1, 3, 7, 9
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10204 (first 170 rows flattened)
- Eckford Cohen, Arithmetical functions associated with the unitary divisors of an integer, Mathematische Zeitschrift, Vol. 74 (1960), pp. 66-80.
Crossrefs
Programs
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Mathematica
udiv[n_] := Select[Divisors[n], CoprimeQ[#, n/#] &]; uGCD[n_, k_] := Max[Intersection[udiv[n], Divisors[k]]]; row[n_] := Select[Range[n], uGCD[n, #] == 1 &]; Array[row, 10] // Flatten
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PARI
udiv(n) = select(x -> gcd(x, n/x) == 1, divisors(n)); ugcd(n, k) = vecmax(setintersect(udiv(n), divisors(k))); row(n) = select(x -> ugcd(n, x) == 1, vector(n, i, i));
Formula
T(n, 1) = 1.