A384248 -id:A384248 - cp's OEIS frontend

A384247 The number of integers from 1 to n whose largest divisor that is an infinitary divisor of n is 1.

1, 1, 2, 3, 4, 2, 6, 4, 8, 4, 10, 6, 12, 6, 8, 15, 16, 8, 18, 12, 12, 10, 22, 8, 24, 12, 18, 18, 28, 8, 30, 16, 20, 16, 24, 24, 36, 18, 24, 16, 40, 12, 42, 30, 32, 22, 46, 30, 48, 24, 32, 36, 52, 18, 40, 24, 36, 28, 58, 24, 60, 30, 48, 48, 48, 20, 66, 48, 44, 24

Offset: 1

Amiram Eldar, May 23 2025

Analogous to A047994, as A064380 is analogous to A116550.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Row lengths of A384246.

Cf. A000010, A001146, A006519, A047994, A064380, A091732, A116550, A138302, A268335, A384248.

f[p_, e_] := p^e*(1 - 1/p^(2^(IntegerExponent[e, 2]))); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

a(n) = {my(f = factor(n)); n * prod(i = 1, #f~, (1 - 1/f[i,1]^(1 << valuation(f[i,2], 2))));}

Multiplicative with a(p^e) = p^e * (1 - 1/p^A006519(e)).
a(n) >= A091732(n), with equality if and only if n is in A138302.
a(n) <= A047994(n), with equality if and only if n is in A138302.
a(n) >= A000010(n), with equality if and only if n is an exponentially odd number (A268335).
a(n) is odd if and only if n = 1 or 2^(2^k) for k >= 0 (A001146). a(2^(2^k)) = 2^(2^k)-1.
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = Product_{p prime} f(1/p) = 0.66718130416373472394..., and f(x) = 1 - (1-x)*Sum_{k>=1} x^(2^k)/(1-x^(2^k)).

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.

Original entry on oeis.org

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

Amiram Eldar, May 23 2025

			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

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

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

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

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));

cp's OEIS Frontend

A384247 The number of integers from 1 to n whose largest divisor that is an infinitary divisor of n is 1.

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