A006590 a(n) = Sum_{k=1..n} ceiling(n/k).

1, 3, 6, 9, 13, 16, 21, 24, 29, 33, 38, 41, 48, 51, 56, 61, 67, 70, 77, 80, 87, 92, 97, 100, 109, 113, 118, 123, 130, 133, 142, 145, 152, 157, 162, 167, 177, 180, 185, 190, 199, 202, 211, 214, 221, 228, 233, 236, 247, 251, 258, 263, 270, 273, 282, 287, 296, 301

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• nice
• easy

Comments

The following sequences all have the same parity: A004737, A006590, A027052, A071028, A071797, A078358, A078446. - Jeremy Gardiner, Mar 16 2003

Given the fact that ceiling(x) <= x+1, we can, using well known results for the harmonic series, easily derive that n*log(n) <= a(n) <= n*(1+log(n)) + n = n(log(n)+2). - Stefan Steinerberger, Apr 08 2006

References

• Marc LeBrun, personal communication.
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• T. D. Noe, Table of n, a(n) for n = 1..1000
• M. Le Brun, Email to N. J. A. Sloane, Jul 1991

Crossrefs

Cf. A000005, A006218.

Programs

Haskell

a006590 n = sum $ map f [1..n] where
   f x = y + 1 - 0 ^ r where (y, r) = divMod n x
-- Reinhard Zumkeller, Feb 18 2013

Magma

[&+[Ceiling(n/j): j in [1..n]] : n in [1..60]]; // G. C. Greubel, Nov 07 2019

Maple

seq(add(ceil(n/j), j = 1..n), n = 1..60); # G. C. Greubel, Nov 07 2019

Mathematica

Table[Sum[Ceiling[n/i], {i, 1, n}], {n, 1, 60}] (* Stefan Steinerberger, Apr 08 2006 *)
nxt[{n_,a_}]:={n+1,a+DivisorSigma[0,n]+1}; Transpose[NestList[nxt,{1,1},60]][[2]] (* Harvey P. Dale, Aug 23 2013 *)

PARI

first(n)=my(v=vector(n,i,i),s); for(i=1,n-1,v[i+1]+=s+=numdiv(i)); v \\ Charles R Greathouse IV, Feb 07 2017

PARI

a(n) = n + sum(k=1, n-1, (n-1)\k); \\ Michel Marcus, Oct 10 2021

Python

from math import isqrt
def A006590(n): return (lambda m: n+2*sum((n-1)//k for k in range(1, m+1))-m*m)(isqrt(n-1)) # Chai Wah Wu, Oct 09 2021

Sage

[sum(ceil(n/j) for j in (1..n)) for n in (1..60)] # G. C. Greubel, Nov 07 2019

Formula

a(n) = n+Sum_{k=1..n-1} tau(k). - Vladeta Jovovic, Oct 17 2002

a(n) = 1 + a(n-1) + tau(n-1), a(n) = A006218(n-1) + n. - T. D. Noe, Jan 05 2007

a(n) = a(n-1) + A000005(n) + 1 for n >= 2. a(n) = A161886(n) - A000005(n) + 1 = A161886(n-1) + 2 = A006218(n) + A049820(n) for n >= 1. - Jaroslav Krizek, Nov 14 2009

Extensions

More terms from Stefan Steinerberger, Apr 08 2006