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 A298369 a(n) = a(n-1) + a(n-2) + 2*a(floor(n/2)) + 3*a(floor(n/3)) + ... +  n*a(floor(n/n)), where a(0) = 1, a(1) = 1, a(2) = 1. 2
 1, 1, 1, 7, 17, 38, 87, 164, 318, 576, 1040, 1773, 3134, 5241, 8877, 14728, 24579, 40298, 66585, 108610, 178004, 289717, 472312, 766643, 1247081, 2021980, 3281557, 5316888, 8619474, 13957420, 22611507, 36603571, 59270152, 95931095, 155290091, 251310597 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS 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. LINKS Clark Kimberling, Table of n, a(n) for n = 0..1000 MATHEMATICA a[0] = 1; a[1] = 1; a[2] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] + Sum[k*a[Floor[n/k]], {k, 2, n}]; Table[a[n], {n, 0, 30}]  (* A298369 *) PROG (Python) from functools import lru_cache @lru_cache(maxsize=None) def A298369(n):     if n <= 2:         return 1     c, j = A298369(n-1)+A298369(n-2), 2     k1 = n//j     while k1 > 1:         j2 = n//k1 + 1         c += (j2*(j2-1)-j*(j-1))*A298369(k1)//2         j, k1 = j2, n//j2     return c+(n*(n+1)-j*(j-1))//2 # Chai Wah Wu, Mar 31 2021 CROSSREFS Cf. A001622, A000045, A298338. Sequence in context: A140121 A102770 A253663 * A213789 A058273 A058274 Adjacent sequences:  A298366 A298367 A298368 * A298370 A298371 A298372 KEYWORD nonn,easy AUTHOR Clark Kimberling, Feb 10 2018 STATUS approved

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Last modified September 16 12:21 EDT 2021. Contains 347472 sequences. (Running on oeis4.)