A367167 Let Product_{i = 1..omega(n)} p_i^e_i be the prime factorization of n. Then a(n) = Sum_{i = 1..omega(n)} Product_{j = 1..n} p_i^(e_((i+j-1) mod omega(n) + 1)).

0, 2, 3, 4, 5, 12, 7, 8, 9, 20, 11, 30, 13, 28, 30, 16, 17, 30, 19, 70, 42, 44, 23, 78, 25, 52, 27, 126, 29, 90, 31, 32, 66, 68, 70, 72, 37, 76, 78, 290, 41, 126, 43, 286, 120, 92, 47, 210, 49, 70, 102, 390, 53, 78, 110, 742, 114, 116, 59, 300, 61, 124, 210, 64, 130

Offset

1,2

Comments

In other words, a(n) is constructed by incrementally shifting the elements of the ordered exponent multiset {e_1,e_2,...,e_k} to the right one place at a time, a total of k = A001221(n) times, and adding the results of these exponents applied to the distinct size ordered primes p_1,p_2,...,p_k which divide n. The first shift gives {e_k,e_1,e_2,...,e_(k-1)}, then {e_(k-1),e_k,...,e_(k-2)} ... and the last yields {e_1,e_2,...,e_k} (which is n itself). At each shift the last exponent of the previous shift becomes the first of the next. There are A001221(n) summands in the computation of a(n), each having the same rad (A007947) and omega (A001221) values as n.

a(n) >= n with equality for n > 1 in A000961.

Links

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^20.

Formula

a(n) = n for n > 1 in A000961.

a(n) = 2*n for n in A006881, and more generally if A007947(n) = n, a(n) = A001221(n)*n.

Let b(0) = n and let b(n+1) = A105119(b(n)) for n >= 0 and let omega(n) be the number of distinct prime factors of n. Then a(n) = Sum_{i = 1..omega(n)} b(i). - David A. Corneth, Nov 07 2023

Example

a(1) = 0, the empty sum.
a(6) = a(2^1*3^1) = 6 + 2^1*3^1 = 6 + 6 = 12.
a(12) = a(2^2*3^1) = 12 + 2^1*3^2 = 12 + 18 = 30.
a(2250) = a(2^1*3^2*5^3) = 2^3*3^1*5^2 + 2^2*3^3*5^1 + 2^1*3^2*5^3 = 3390.

Mathematica

{0}~Join~Array[Total@ Table[Times @@ Power @@@ Transpose@ {#1, RotateRight[#2, k]}, {k, PrimeNu[#]}] & @@ Transpose@ FactorInteger[#] &, 64, 2] (* Michael De Vlieger, Nov 09 2023 *)

Prog

(PARI)
a(n) = {
	my(f = factor(n), res = 0);
	for(i = 1, #f~,
		res+=prod(j = 1, #f~, f[j, 1]^f[(i+j-1)%#f~ + 1, 2])
	);
	res
} \\ David A. Corneth, Nov 07 2023

Crossrefs

Cf. A000961, A001221, A001222, A005117, A006881, A007947, A105119, A367140.

Sequence in context: A353694 A328260 A229347 * A135323 A052106 A064446

Adjacent sequences: A367164 A367165 A367166 * A367168 A367169 A367170

Keyword

nonn

Author

David James Sycamore and David A. Corneth Nov 07 2023

Extensions

More terms and revised name from David A. Corneth, Nov 07 2023

Status

approved