A000026 -id:A000026 - cp's OEIS frontend

A191278 Count of Mosaic numbers that equal n.

1, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 6, 1, 3, 3, 1, 1, 6, 1, 6, 3, 3, 1, 10, 1, 3, 1, 6, 1, 16, 1, 1, 3, 3, 3, 20, 1, 3, 3, 10, 1, 16, 1, 6, 6, 3, 1, 15, 1, 6, 3, 6, 1, 10, 3, 10, 3, 3, 1, 50, 1, 3, 6, 1, 3, 16, 1, 6, 3, 16, 1, 50, 1, 3, 6, 6, 3, 16, 1, 15, 1, 3, 1, 50, 3, 3, 3, 10, 1, 50

Offset: 1

R. J. Mathar, May 29 2011

The number of solutions x to A000026(x)=n.

R. J. Mathar, Table of n, a(n) for n = 1..1000

A191278 := proc(n)
    local f, beta, a, j ;
    f := ifactors(n)[2] ;
    beta := A073093(n) ;
    a := 1/beta ;
    for j in ifactors(n)[2] do
        a := a*binomial(beta, op(2, j) ) ;
    end do:
    a ;
end proc:

Let n=product_j p_j^e(j) be the prime factorization of n and beta=A073093(n). Then a(n)*beta = product_j binomial(beta,e(j)). [Gordon-Robertson in A000026, Theorem 1]

A333267 If n = Product (p_j^k_j) then a(n) = Product (a(pi(p_j)) * k_j), where pi = A000720.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 3, 2, 1, 1, 2, 1, 2, 1, 4, 2, 2, 3, 2, 2, 1, 2, 3, 2, 1, 3, 4, 1, 1, 1, 5, 1, 2, 2, 4, 2, 3, 1, 3, 1, 2, 2, 2, 2, 2, 1, 4, 4, 2, 2, 2, 4, 3, 1, 6, 3, 1, 2, 2, 2, 1, 4, 6, 1, 1, 3, 4, 2, 2, 2, 6, 2, 2, 2, 6, 2, 1, 1, 4, 4, 1, 2, 4, 2, 2, 1, 3, 3, 2, 2, 4, 1, 1, 3, 5, 2, 4, 2, 4

Offset: 1

Ilya Gutkovskiy, Mar 13 2020

			a(36) = a(2^2 * 3^2) = a(prime(1)^2 * prime(2)^2) = a(1) * 2 * a(2) * 2 = 4.

Cf. A000026, A000720, A002110, A003963, A005361, A054725, A109129, A276625 (positions of 1's), A282446, A304117, A318046, A328880.

a:= proc(n) option remember;
      mul(a(numtheory[pi](i[1]))*i[2], i=ifactors(n)[2])
    end:
seq(a(n), n=1..120);  # Alois P. Heinz, Mar 13 2020

a[1] = 1; a[n_] := a[n] = Times @@ (a[PrimePi[#[[1]]]] #[[2]] & /@ FactorInteger[n]); Table[a[n], {n, 1, 100}]

a(n) = A005361(n) * Product_{p|n, p prime} a(pi(p)).
a(n) = a(prime(n)).
a(p^k) = k * a(p), where p is prime.
a(A002110(n)) = Product_{k=1..n} a(k).

A341042 Multiplicative projection of odd part of n.

Original entry on oeis.org

1, 1, 3, 1, 5, 3, 7, 1, 6, 5, 11, 3, 13, 7, 15, 1, 17, 6, 19, 5, 21, 11, 23, 3, 10, 13, 9, 7, 29, 15, 31, 1, 33, 17, 35, 6, 37, 19, 39, 5, 41, 21, 43, 11, 30, 23, 47, 3, 14, 10, 51, 13, 53, 9, 55, 7, 57, 29, 59, 15, 61, 31, 42, 1, 65, 33, 67, 17, 69, 35, 71, 6, 73, 37, 30

Offset: 1

Ilya Gutkovskiy, Feb 03 2021

			a(54) = a(2 * 3^3) = 3 * 3 = 9.

Cf. A000026, A000079 (positions of 1's), A000265, A056911 (fixed points), A204455.

a:= n-> mul(`if`(i[1]=2, 1, i[1]*i[2]), i=ifactors(n)[2]):
seq(a(n), n=1..75);  # Alois P. Heinz, Feb 03 2021

a[n_] := Times @@ (#[[1]] #[[2]] & /@ FactorInteger[n/2^IntegerExponent[n, 2]]); Table[a[n], {n, 75}]

a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,1]==2, 1, f[i,1] * f[i,2]));} \\ Amiram Eldar, Nov 12 2022

a(n) = A000026(A000265(n)).
a(n) = A000026(n) if n odd, a(n) = a(n/2) if n even.
From Amiram Eldar, Nov 12 2022: (Start)
Multiplicative with a(2^e) = 1 and a(p^e) = e*p for p > 2.
Sum_{k=1..n} a(k) ~ c * n^2, where c = (6*zeta(2)^2/17) * Product_{p prime} (1 - 3/p^2 + 2/p^3 + 1/p^4 - 1/p^5) = 0.2947570019... . (End)

A385644 Swap multiplication and exponentiation in the canonical prime factorization of n.

Original entry on oeis.org

2, 3, 4, 5, 8, 7, 6, 6, 32, 11, 64, 13, 128, 243, 8, 17, 64, 19, 1024, 2187, 2048, 23, 216, 10, 8192, 9, 16384, 29, 14134776518227074636666380005943348126619871175004951664972849610340958208, 31, 10, 177147, 131072, 78125, 4096, 37, 524288, 1594323, 7776, 41

Offset: 2

Jens Ahlström, Jul 06 2025

nonn

In the canonical prime factorization of n larger than one, swap multiplication and exponentiation and calculate the result.

			a(6) = a(2 * 3) = 2^3 = 8,
a(24) = a(2^3 * 3) = (2 * 3)^3 = 216,
a(30) = a(2 * 3 * 5) = 2^3^5 = 2^243.

Jens Ahlström, Table of n, a(n) for n = 2..65

Cf. A001414, A000026, A005361, A008474.

f[{p_,e_}]:=p*e;a[n_]:=Module[{pp=f/@FactorInteger[n]},r=pp[[-1]];Do[r=pp[[Length[pp]-i]]^r,{i,1,Length[pp]-1}];r];Array[a,40,2] (* James C. McMahon, Jul 11 2025 *)
A385644[n_] := Power @@ Times @@@ FactorInteger[n];
Array[A385644, 40, 2] (* Paolo Xausa, Jul 14 2025 *)

from sympy import factorint
from functools import reduce
def rpow(a, b):
    return b**a
def a(n):
    return reduce(rpow, [p*e for p, e in reversed(factorint(n).items())])
print([a(n) for n in range(2, 42)])

A039785 Numbers n such that phi(n) is equal to the multiplicative projection of n.

Original entry on oeis.org

9, 16, 50, 54, 100, 108, 144, 294, 450, 588, 900

Offset: 1

Olivier Gérard

Next term if it exists is greater than 1000000. - C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 29 2004
From Lambert Herrgesell (zero815(AT)googlemail.com), Oct 18 2006: (Start)
There is a simple proof that there are no other terms. Let m(n) be the multiplicative projection of n.
The key point is that if a prime in the factorization of n has a "big" power, either phi(n) > m(n) or omega(n) is "big", so a solution would be "more difficult".
It is easy to show that 3^2 and 2^4 are the only solutions of the form p^k.
Consider solutions of the form 2^x * 3^y: phi(2^x * 3^y) = 2^x * 3^(y-1) = 2*x*3*y.
Since p^k is done, assume x >= 1, y >= 2 and simplify to (*) 2^(x-1) * 3^(y-2) = xy.
A quick search yields the only solutions 2^1 * 3^3, 2^2 * 3^3 and 2^4 * 3^2.
Now try adding a third prime p and assume phi(p) of the form 2^m * 3^k:
phi(2^x * 3^y * p^z) = 2^(x+m) * 3^(y-1+k) * p^(z-1). We find 2^(x+m) * 3^(y-1+k) * p^(z-1) = 2*x*3*y*p*z and, dividing by 2*3*p, 2^(x-1+m) * 3^(y-2+k) * p^(z-2) = xyz.
So z >= 2, but by the above remark we may assume z=2. We get 2^(x-1+m) * 3^(y-2+k) * p^0 = x*y*2 and (**) 2^(x-2+m) * 3^(y-2+k) = xy.
(**) is independent of p and reduces to solving the 2^x * 3^y case in (*), but additional powers of 2 and 3 are needed.
Using the bounds for phi(p^z), the only possible values for p are 5 or 7 and the only solutions are 2^1 * 5^2, 2^2 * 5^2, 2^1 * 3^2 * 5^2, 2^2 * 3^2 * 5^2 and 2^1 * 3^1 * 7^2, 2^2 * 3^1 * 7^2.
Now adding a further prime q, the difficulty from (*) and (**) is that both q has to be small and phi(q) has to produce a lot of factors for the next (**) like expression.
If we also spare no effort and try this for all the remaining numbers of step (**), we see that there are no more results.
More generally: If n is a member, omega(n) > 2 and n = Product(p_i^e_i)q^e, p_i, q primes, then Product(p_i^e_i) is also a member up to some powers e^i of some p_i, which have to be greater. Also, for omega(n) >= 2, if 2^2 is a factor of n, then n/2 is also in a(n).
All in all this implies that there are no other terms. (End)

			phi(50)=20, 50 = 2^1 * 5^2, 2*1*5*2 = 20.

Cf. A000010, A000026.

for(n=1, 1000000, f=factor(n); l=#f[,1]; if( eulerphi(n)==prod(i=1, l, f[i,1]) * prod(i=1,l,f[i,2]), print1(n,","))) \\ C. Ronaldo

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A191278 Count of Mosaic numbers that equal n.

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A333267 If n = Product (p_j^k_j) then a(n) = Product (a(pi(p_j)) * k_j), where pi = A000720.

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A341042 Multiplicative projection of odd part of n.

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A385644 Swap multiplication and exponentiation in the canonical prime factorization of n.

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A039785 Numbers n such that phi(n) is equal to the multiplicative projection of n.

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