cp's OEIS Frontend

A306363 For n > 1 having omega(n) = k and canonical prime factorization n = d_1*d_2*...*d_k, a(n) = Sum_{i=1..k} (d_i*a(n/d_i) + (n/d_i)); a(1)=0.

%I A306363 #21 Mar 10 2019 13:31:47
%S A306363 0,1,1,1,1,10,1,1,1,14,1,14,1,18,16,1,1,22,1,18,20,26,1,22,1,30,1,22,
%T A306363 1,155,1,1,28,38,24,26,1,42,32,26,1,205,1,30,28,50,1,38,1,54,40,34,1,
%U A306363 58,32,30,44,62,1,235,1,66,32,1,36,305,1,42,52,295,1,34,1,78
%N A306363 For n > 1 having omega(n) = k and canonical prime factorization n = d_1*d_2*...*d_k, a(n) = Sum_{i=1..k} (d_i*a(n/d_i) + (n/d_i)); a(1)=0.
%C A306363 Recursion similar to that of A006022. k=1 => a(n)=1; k=2 => a(n) = 2*(d_1 + d_2); claim: a(n)=A000522(k-1)*A066504(n); k = omega(n). Inductive proof on k (sketch): Let A=A000522 and B=A066504 = Sum_{i=1..k} (n/d_i). True for k=1,2 so assume true for arbitrary k. Then for n with omega(n)=k+1, a(n) = (Sum_{i=1..k+1} d_i*(n/d_i)) + B(n) = A(k-1)*k*B(n) + B(n) = ((A(k-1)*k) + 1)*B(n). But (A(k-1)k)+1) = A(k) by recursive formula for A000522, so a(n) = A(k)*B(n); hence true for k+1.
%H A306363 Antti Karttunen, <a href="/A306363/b306363.txt">Table of n, a(n) for n = 1..16384</a>
%H A306363 Antti Karttunen, <a href="/A306363/a306363.txt">Data supplement: n, a(n) computed for n = 1..65537</a>
%F A306363 a(n) = A000522(k-1)*A066504(n); k = omega(n) = A001221(n).
%e A306363 n=6 = 2*3, a(n) = 2*a(3) + 3*a(2) + 3 + 2 = 2*(2+3) = 10.
%e A306363 n=210=2*3*5*7; k=4, a(n)=A000522(3)*(2*3*5 + 2*3*7 + 2*5*7 + 3*5*7) = 16*247 = 3952.
%o A306363 (PARI) a(n) = if (n==1, 0, my(f=factor(n)); sum(k=1, #f~, my(dk=f[k,1]^f[k,2]); dk*a(n/dk) + (n/dk))); \\ _Michel Marcus_, Feb 19 2019
%Y A306363 Cf. A006022, A001221, A000522, A066504.
%K A306363 nonn
%O A306363 1,6
%A A306363 _David James Sycamore_, Feb 10 2019