A379049 -id:A379049 - cp's OEIS frontend

A379260 Index of first appearance of n in sequence A379049.

0, 1, 3, 2, 9, 4, 6, 26, 24, 5, 18, 7, 78, 28, 12, 11, 54, 19, 216, 71, 15, 29, 162, 53, 21, 73, 231, 16, 486, 13, 51, 217, 84, 83, 36, 14, 33, 647, 57, 32, 4374, 31, 237, 649, 45, 22, 207, 236, 165, 1945, 693, 50, 2151, 212, 90, 46, 87, 160, 39366, 86, 63

Offset: 2

Lei Zhou, Dec 19 2024

a(n) is the smallest integer that makes A379049(a(n)) = n.

Conjecture: a(n) is defined for all integer n > 1.

			For n = 2, A379049(0) = 1 + 1 = 2.  Thus a(2) = 0;
For n = 3, A379049(1) = 2 + 1 = 3, since 1's balanced ternary representation is 1.  Thus a(3) = 1;
For n = 4, A379049(3) = 3 + 1 = 4, since 3's balanced ternary representation is 10.  Thus a(4) = 3;
...
For n = 60, A379049(39366) = 31 + 29 = 60, since 39366's balanced ternary representation is 1T000000000, where the 11's digit is 1 represents the 11's prime 31 in the term before the plus sign, and the 10's digit is T representing the 10's prime 29 in the term after the plus sign. And evaluation of A379049 found no number i smaller than 39366 can make A379049(i) = 60.  Thus a(60) = 39366.

Cf. A379049, A345128, A005812, A134022, A134023, A140267.

BTDigits[m_Integer, g_] :=  Module[{n = m, d, sign, t = g}, If[n != 0, If[n > 0, sign = 1, sign = -1; n = -n]; d = Ceiling[Log[3, n]]; If[3^d - n <= ((3^d - 1)/2), d++];    While[Length[t] < d, PrependTo[t, 0]]; t[[Length[t] + 1 - d]] = sign; t = BTDigits[sign*(n - 3^(d - 1)), t]]; t];
goal = 62; res = {}; ct = 1;
Do[AppendTo[res, 0], {i, 2, goal}]; i = -1; While[ct < goal, i++; BT = BTDigits[i, {0}]; BTl = Length[BT]; f = 1; b = 1;  Do[If[BT[[j]] == 1, f = f*Prime[BTl - j + 1]];   If[BT[[j]] == -1, b = b*Prime[BTl - j + 1]], {j, 1, BTl}]; d = f + b; If[(d <= goal) && (res[[d - 1]] == 0), res[[d - 1]] = i; ct++]];
res

A345128 Number of squarefree products s*t from all positive integer pairs (s,t), such that s + t = n, s <= t.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 3, 2, 2, 3, 4, 3, 3, 3, 2, 4, 4, 4, 4, 4, 4, 3, 5, 4, 5, 5, 4, 6, 4, 5, 7, 6, 5, 6, 8, 5, 9, 7, 6, 7, 8, 7, 8, 7, 5, 8, 10, 7, 6, 8, 7, 10, 9, 7, 11, 10, 8, 10, 8, 8, 11, 10, 9, 10, 11, 10, 13, 13, 9, 12, 10, 10, 14, 12, 12, 12, 13, 11, 12

Offset: 1

Wesley Ivan Hurt, Jun 08 2021

nonn

From Lei Zhou, Dec 19 2024: (Start)
a(n) is also the total number of appearance of n in A379049, by definition.
Conjecture: a(n) > 0 for all n > 1. (End)
A simple case of Zhou's conjecture: a(p) > 0 where p is prime. With some work this can be extended to a(n) > 0 for n with sum_{p | n} 1/p < 1/10 or so (the limit of the method is 6/Pi^2 - 1/2, so it can't prove the full conjecture). See my comment in A071068. - Charles R Greathouse IV, Dec 20 2024

			a(13) = 3; The partitions of 13 into two positive integer parts (s,t) where s <= t are (1,12), (2,11), (3,10), (4,9), (5,8), (6,7). The corresponding products are 1*12, 2*11, 3*10, 4*9, 5*8, and 6*7; 3 of which are squarefree.

Cf. A008683 (mu), A379049, A071068.

Table[Sum[MoebiusMu[k (n - k)]^2, {k, Floor[n/2]}], {n, 100}]

a(n)=my(s); forsquarefree(k=1,n\2, gcd(n,k[1])==1 && issquarefree(n-k[1]) && s++); s \\ Charles R Greathouse IV, Dec 20 2024

a(n) = Sum_{k=1..floor(n/2)} mu(k*(n-k))^2, where mu is the Möbius function (A008683).

a(n) <= A071068(n) and hence a(n) < 0.303967n for n > 3. - Charles R Greathouse IV, Dec 20 2024

cp's OEIS Frontend

A379260 Index of first appearance of n in sequence A379049.

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