login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
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

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

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