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 A298370 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) = 2, a(2) = 3. 2
 1, 2, 3, 15, 38, 83, 190, 356, 695, 1254, 2267, 3861, 6829, 11417, 19340, 32076, 53545, 87784, 145048, 236589, 387765, 631106, 1028866, 1670013, 2716595, 4404599, 7148426, 11582096, 18776334, 30404300, 49256015, 79735758, 129111774, 208972513, 338277831 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 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] = 2; a[2] = 3; 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}]  (* A298370 *) PROG (Python) from functools import lru_cache @lru_cache(maxsize=None) def A298370(n):     if n <= 2:         return n+1     c, j = A298370(n-1)+A298370(n-2), 2     k1 = n//j     while k1 > 1:         j2 = n//k1 + 1         c += (j2*(j2-1)-j*(j-1))*A298370(k1)//2         j, k1 = j2, n//j2     return c+2*(n*(n+1)-j*(j-1))//2 # Chai Wah Wu, Mar 31 2021 CROSSREFS Cf. A001622, A000045, A298338. Sequence in context: A296296 A143880 A037388 * A221860 A216339 A048076 Adjacent sequences:  A298367 A298368 A298369 * A298371 A298372 A298373 KEYWORD nonn,easy AUTHOR Clark Kimberling, Feb 10 2018 STATUS approved

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Last modified May 24 17:23 EDT 2022. Contains 354042 sequences. (Running on oeis4.)