A384246 Triangle in which the n-th row gives the numbers from 1 to n whose largest divisor that is an infinitary 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, 3, 5, 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

1,4

Links

Amiram Eldar, Table of n, a(n) for n = 1..10276 (first 175 rows flattened)

Example

Triangle begins:
  1
  1
  1, 2
  1, 2, 3
  1, 2, 3, 4
  1, 5
  1, 2, 3, 4, 5, 6
  1, 3, 5, 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

Mathematica

infdivs[n_] := If[n == 1, {1}, Sort@ Flatten@ Outer[Times, Sequence @@ (FactorInteger[n] /. {p_, m_Integer} :> p^Select[Range[0, m], BitOr[m, #] == m &])]]; (* Michael De Vlieger at A077609 *)
infGCD[n_, k_] := Max[Intersection[infdivs[n], Divisors[k]]];
row[n_] := Select[Range[n], infGCD[n, #] == 1 &]; Array[row, 16] // Flatten

Prog

(PARI) isidiv(d, f) = {if (d==1, return (1)); for (k=1, #f~, bne = binary(f[k, 2]); bde = binary(valuation(d, f[k, 1])); if (#bde < #bne, bde = concat(vector(#bne-#bde), bde)); for (j=1, #bne, if (! bne[j] && bde[j], return (0)); ); ); return (1); }
infdivs(n) = {my(f = factor(n), d = divisors(f), idiv = []); for (k=1, #d, if (isidiv(d[k], f), idiv = concat(idiv, d[k])); ); idiv; } \\ Michel Marcus at A077609
infgcd(n, k) = vecmax(setintersect(infdivs(n), divisors(k)));
row(n) = select(x -> infgcd(n, x) == 1, vector(n, i, i));

Crossrefs

Cf. A064379, A384046, A384245, A384247 (row lengths), A384248 (row sums).

Sequence in context: A064379 A384046 A278961 * A194973 A195113 A120418

Adjacent sequences: A384243 A384244 A384245 * A384247 A384248 A384249

Keyword

nonn,tabf,easy

Author

Amiram Eldar, May 23 2025

Status

approved