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
Examples
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
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10276 (first 175 rows flattened)
Programs
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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 -
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));