|
%I #72 Dec 24 2023 02:35:01
%S 1,1,1,3,1,3,1,3,4,3,1,6,1,3,4,7,1,6,1,7,4,3,1,10,6,3,4,7,1,11,1,7,4,
%T 3,6,16,1,3,4,12,1,12,1,7,9,3,1,16,8,8,4,7,1,12,6,14,4,3,1,21,1,3,11,
%U 15,6,12,1,7,4,15,1,24,1,3,9,7,8,12,1,20,13,3,1,23,6,3,4,15,1,26,8,7,4,3
%N a(n) = sum of positive divisors k of n with k <= sqrt(n).
%C Row sums of the table in A161906. - _Reinhard Zumkeller_, Mar 08 2013
%C Conjecture: a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 2. - _Omar E. Pol_, May 03 2020. This conjecture is true (the g.f. for these partitions agrees with the g.f. given below by _Michael Somos_). - _N. J. A. Sloane_, Dec 02 2020
%C Column 2 of A334466. - _Omar E. Pol_, Dec 03 2020
%H Harry J. Smith, <a href="/A066839/b066839.txt">Table of n, a(n) for n = 1..1000</a>
%H Douglas E. Iannucci, <a href="https://arxiv.org/abs/1910.11835">On sums of the small divisors of a natural number</a>, arXiv:1910.11835 [math.NT], 2019.
%F G.f.: Sum_{k>0} k*x^(k^2)/(1-x^k). - _Michael Somos_, Nov 19 2005
%F a(n) = Sum_{i=1..floor(sqrt(n))} (-(n mod i) + (n-1) mod i + 1). - _José de Jesús Camacho Medina_, Feb 21 2021
%F a(p^(2k+1)) = a(p^(2k)) = (p^(k+1)-1)/(p-1) = A000203(p^k) for k>=0 and p prime. - _Chai Wah Wu_, Dec 23 2023
%e a(9) = 4 = 1 + 3 because 1 and 3 are the positive divisors of 9 that are <= sqrt(9).
%e a(20) = 7: the divisors of 20 are 1, 2, 4, 5, 10 and 20. a(20) = 1 + 2 + 4 = 7.
%p with(numtheory):for n from 1 to 200 do c[n] := 0:d := divisors(n):for i from 1 to nops(d) do if d[i]<=n^.5+10^(-10) then c[n] := c[n]+d[i]:fi:od:od:seq(c[i],i=1..200);
%t f[n_] := Plus @@ Select[ Divisors@n, # <= Sqrt@n &]; Array[f, 94] (* _Robert G. Wilson v_, Mar 04 2010 *)
%o (PARI) a(n)=sumdiv(n,d, (d^2<=n)*d) /* _Michael Somos_, Nov 19 2005 */
%o (PARI) { for (n=1, 1000, d=divisors(n); s=sum(k=1, ceil(length(d)/2), d[k]); write("b066839.txt", n, " ", s) ) } \\ _Harry J. Smith_, Mar 31 2010
%o (Haskell)
%o a066839 = sum . a161906_row -- _Reinhard Zumkeller_, Mar 08 2013
%o (Sage) [sum(k for k in divisors(n) if k^2<=n) for n in (1..94)] # _Giuseppe Coppoletta_, Jan 21 2015
%o (Python)
%o from itertools import takewhile
%o from sympy import divisors
%o def A066839(n): return sum(takewhile(lambda x:x**2<=n,divisors(n))) # _Chai Wah Wu_, Dec 19 2023
%Y Cf. A000203, A070038, A038548, A072499, A334466.
%K nonn
%O 1,4
%A _Leroy Quet_, Jan 20 2002
%E More terms from Larry Reeves (larryr(AT)acm.org), Apr 12 2002
|
