This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A325973 #44 Feb 22 2024 02:18:16
%S A325973 1,3,4,4,6,12,8,6,7,18,12,16,14,24,24,10,18,21,20,24,32,36,24,24,16,
%T A325973 42,16,32,30,72,32,18,48,54,48,31,38,60,56,36,42,96,44,48,42,72,48,40,
%U A325973 29,48,72,56,54,48,72,48,80,90,60,96,62,96,56,34,84,144,68,72,96,144,72,51,74,114,64,80,96,168,80,60,43,126
%N A325973 Arithmetic mean of {sum of unitary divisors} and {sum of squarefree divisors}: a(n) = (1/2) * (A034448(n) + A048250(n)).
%C A325973 This is not multiplicative: a(4) = 4, a(9) = 7, but a(36) = 31, not 28. However, the function acts multiplicatively on certain subsequences of natural numbers, like for example when restricted to A048107, where this sequence coincides with A326043.
%H A325973 Antti Karttunen, <a href="/A325973/b325973.txt">Table of n, a(n) for n = 1..20000</a>
%F A325973 a(n) = (1/2) * (A034448(n) + A048250(n)).
%F A325973 a(n) = A000203(n) - A325974(n).
%F A325973 a(n) = n + A325977(n).
%F A325973 a(A048107(n)) = A326043(A048107(n)).
%F A325973 For n >= 1, a(2^n) = A052548(n-1) = 2^(n-1) + 2.
%F A325973 For n >= 1, a(3^n) = A289521(n) = (3^n + 5)/2.
%F A325973 Sum_{k=1..n} a(k) ~ c * n^2, where c = (zeta(2)/zeta(3) + 1)/4 = 0.5921081944... . - _Amiram Eldar_, Feb 22 2024
%e A325973 For n = 36, its divisors are 1, 2, 3, 4, 6, 9, 12, 18, 36. Of these, unitary divisors are 1, 4, 9 and 36, so A034448(36) = 1+4+9+36 = 50, while the squarefree divisors are 1, 2, 3 and 6, so A048250(36) = 1+2+3+6 = 12, thus a(36) = (50+12)/2 = 31.
%e A325973 For n = 495, its divisors are 1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 165, 495. Of these, unitary are 1, 5, 9, 11, 45, 55, 99, 495, whose sum is A034448(495) = 720, while the squarefree divisors are 1, 3, 5, 11, 15, 33, 55, 165, and their sum is A048250(495) = 288. Thus a(495) = (720+288)/2 = 504. Also for 495, whose prime factorization is 3^2 * 5^1 * 11^1 this can be computed faster as the average of ((3^2)+1)*(5+1)*(11+1) and (3+1)*(5+1)*(11+1), thus (1/2)*(3+(3^2)+2)*(5+1)*(11+1) = 504.
%t A325973 Array[(1/2) If[# == 1, 2, Times @@ (1 + Power @@@ #) + Times @@ (1 + #[[;; , 1]]) &@ FactorInteger[#]] &, 90] (* _Michael De Vlieger_, Jun 06 2019, after _Giovanni Resta_ at A034448 and _Amiram Eldar_ at A048250. *)
%o A325973 (PARI)
%o A325973 A034448(n) = { my(f=factorint(n)); prod(k=1, #f~, 1+(f[k, 1]^f[k, 2])); }; \\ After code in A034448
%o A325973 A048250(n) = factorback(apply(p -> p+1,factor(n)[,1]));
%o A325973 A325973(n) = ((A034448(n)+A048250(n))/2);
%o A325973 (PARI) A325973(n) = (1/2)*sumdiv(n, d, d*(issquarefree(d) + (1==gcd(d, n/d))));
%Y A325973 Cf. A000203, A034448, A048107, A048250, A052548, A289521, A306633, A325974, A325975, A325977, A325978, A325981, A326043.
%K A325973 nonn
%O A325973 1,2
%A A325973 _Antti Karttunen_, Jun 02 2019