cp's OEIS Frontend

A298407 a(n) = 2*a(n-1) - a(n-3) + a(floor(n/2)) + a(floor(n/3)) + ... + a(floor(n/n)), where a(0) = 1, a(1) = 2, a(2) = 3.

%I A298407 #10 Mar 31 2021 19:19:37
%S A298407 1,2,3,9,23,52,113,223,431,794,1442,2532,4433,7589,12924,21730,36411,
%T A298407 60440,100125,164816,270863,443390,724846,1181713,1925113,3130488,
%U A298407 5087530,8258585,13400782,21728136,35221342,57065559,92441545,149701409,242400952,392424193
%N A298407 a(n) = 2*a(n-1) - a(n-3) + a(floor(n/2)) + a(floor(n/3)) + ... + a(floor(n/n)), where a(0) = 1, a(1) = 2, a(2) = 3.
%C A298407 a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio (A001622), so that (a(n)) has the growth rate of the Fibonacci numbers (A000045). See A298338 for a guide to related sequences.
%H A298407 Clark Kimberling, <a href="/A298407/b298407.txt">Table of n, a(n) for n = 0..1000</a>
%t A298407 a[0] = 1; a[1] = 2; a[2] = 3;
%t A298407 a[n_] := a[n] = 2*a[n - 1] - a[n - 3] + Sum[a[Floor[n/k]], {k, 2, n}];
%t A298407 Table[a[n], {n, 0, 90}]  (* A298407  *)
%o A298407 (Python)
%o A298407 from functools import lru_cache
%o A298407 @lru_cache(maxsize=None)
%o A298407 def A298407(n):
%o A298407     if n <= 2:
%o A298407         return n+1
%o A298407     c, j = 2*A298407(n-1)-A298407(n-3), 2
%o A298407     k1 = n//j
%o A298407     while k1 > 1:
%o A298407         j2 = n//k1 + 1
%o A298407         c += (j2-j)*A298407(k1)
%o A298407         j, k1 = j2, n//j2
%o A298407     return c+2*(n-j+1) # _Chai Wah Wu_, Mar 31 2021
%Y A298407 Cf. A001622, A000045, A298338, A298406.
%K A298407 nonn,easy
%O A298407 0,2
%A A298407 _Clark Kimberling_, Feb 10 2018