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

Author

David James Sycamore and David A. Corneth Nov 07 2023

Keywords

• nonn

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.

Examples

			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.
		

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.

Crossrefs

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

Programs

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 *)

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

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

Extensions

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