A384246 -id:A384246 - 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)).

A384248 The sum of the integers from 1 to n whose largest divisor that is an infinitary divisor of n is 1.

Original entry on oeis.org

1, 1, 3, 6, 10, 6, 21, 16, 36, 20, 55, 36, 78, 42, 60, 120, 136, 72, 171, 120, 126, 110, 253, 96, 300, 156, 243, 252, 406, 120, 465, 256, 330, 272, 420, 432, 666, 342, 468, 320, 820, 252, 903, 660, 720, 506, 1081, 720, 1176, 600, 816, 936, 1378, 486, 1100, 672

Offset: 1

Amiram Eldar, May 23 2025

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

Row sums of A384246.
Cf. A138302, A268335, A384247.
Analogous sequences: A023896, A200723, A333576.

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

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

a(n) = n * A384247(n) / 2, for n >= 2.
a(n) <= A333576(n), with equality if and only if n is in A138302.
a(n) >= A023896(n), with equality if and only if n is an exponentially odd number (A268335).
Sum_{k=1..n} a(k) ~ c * n^2 / 6, 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)).

A384245 Triangle read by rows: T(n, k) for 1 <= k <= n is the largest divisor of k that is an infinitary divisor of n.

Original entry on oeis.org

1, 1, 2, 1, 1, 3, 1, 1, 1, 4, 1, 1, 1, 1, 5, 1, 2, 3, 2, 1, 6, 1, 1, 1, 1, 1, 1, 7, 1, 2, 1, 4, 1, 2, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 2, 1, 2, 5, 2, 1, 2, 1, 10, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 3, 4, 1, 3, 1, 4, 3, 1, 1, 12, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13

Offset: 1

Amiram Eldar, May 23 2025

First differs from A384047 at n = 30.

			Triangle begins:
  1
  1, 2
  1, 1, 3
  1, 1, 1, 4
  1, 1, 1, 1, 5
  1, 2, 3, 2, 1, 6
  1, 1, 1, 1, 1, 1, 7
  1, 2, 1, 4, 1, 2, 1, 8
  1, 1, 1, 1, 1, 1, 1, 1, 9
  1, 2, 1, 2, 5, 2, 1, 2, 1, 10

Amiram Eldar, Table of n, a(n) for n = 1..10153 (first 142 rows flattened)

Cf. A050873, A064379, A077609, A384047, A384246 (positions of 1's).

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 *)
T[n_, k_] := Max[Intersection[infdivs[n], Divisors[k]]];
Table[T[n, k], {n, 1, 13}, {k, 1, n}] // 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
T(n, k) = vecmax(setintersect(infdivs(n), divisors(k)));

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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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A384248 The sum of the integers from 1 to n whose largest divisor that is an infinitary divisor of n is 1.

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Mathematica

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Formula

A384245 Triangle read by rows: T(n, k) for 1 <= k <= n is the largest divisor of k that is an infinitary divisor of n.

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Mathematica

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