A324922 a(n) = unique m such that m/A003963(m) = n, where A003963 is product of prime indices.

1, 2, 6, 4, 30, 12, 28, 8, 36, 60, 330, 24, 156, 56, 180, 16, 476, 72, 152, 120, 168, 660, 828, 48, 900, 312, 216, 112, 1740, 360, 10230, 32, 1980, 952, 840, 144, 888, 304, 936, 240, 6396, 336, 2408, 1320, 1080, 1656, 8460, 96, 784, 1800, 2856, 624, 848, 432

Offset

1,2

Comments

Every positive integer has a unique factorization into factors q(i) = prime(i)/i, i > 0 given by the rows of A324924. Then a(n) is the number obtained by encoding this factorization as a standard factorization into prime numbers (A112798).

Links

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Formula

a(n) = Product_t mg(t) where the product is over all (not necessarily distinct) terminal subtrees of the rooted tree with Matula-Goebel number n, and mg(t) is the Matula-Goebel number of t.

Completely multiplicative with a(prime(n)) = prime(n) * a(n). - Rémy Sigrist, Jul 18 2019

Mathematica

primeMS[n_]:=If[n==1, {}, Flatten[Cases[FactorInteger[n], {p_, k_}:>Table[PrimePi[p], {k}]]]];
difac[n_]:=If[n==1, {}, With[{m=Product[Prime[i]/i, {i, primeMS[n]}]}, Sort[Join[primeMS[n], difac[n/m]]]]];
Table[Times@@Prime/@difac[n], {n, 30}]

Prog

(PARI) a(n) = my (f=factor(n)); prod (i=1, #f~, (f[i, 1] * a(primepi(f[i, 1])))^f[i, 2]) \\ Rémy Sigrist, Jul 18 2019

Crossrefs

Sorting the sequence gives A324850.

Cf. A000081, A003963, A061775, A109129, A112798, A120383, A196050, A317713.

Cf. A324848, A324849, A324923, A324924, A324925, A324931, A324934.

Sequence in context: A253588 A228099 A227955 * A329886 A064538 A002790

Adjacent sequences: A324919 A324920 A324921 * A324923 A324924 A324925

Keyword

nonn,mult

Author

Gus Wiseman, Mar 20 2019

Extensions

Keyword mult added by Rémy Sigrist, Jul 18 2019

Status

approved