A143270 a(n) = n*A002088(n).

1, 4, 12, 24, 50, 72, 126, 176, 252, 320, 462, 552, 754, 896, 1080, 1280, 1632, 1836, 2280, 2560, 2940, 3300, 3956, 4320, 5000, 5512, 6210, 6776, 7830, 8340, 9548, 10368, 11352, 12240, 13440, 14256, 15984, 17100, 18486, 19600, 21730, 22764, 25112

Offset: 1

Author

Gary W. Adamson, Aug 03 2008

Keywords

• nonn

Comments

Also number of reduced fractions with denominators <= n and values between 1/n and n (inclusive). - Reinhard Zumkeller, Jan 15 2009

Examples

			a(4) = 24 = n*A002088(n) = 4*6.
a(4) = 24 = sum of row 4 terms of triangle A143269: (4 + 4 + 8 + 8).
a(3) = #{1/3,1/2,2/3,1,4/3,3/2,5/3,2,7/3,5/2,8/3,3} = 12. - _Reinhard Zumkeller_, Jan 15 2009
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000010, A002088, A143269.

Programs

Mathematica

Module[{nn=50,ps},ps=Accumulate[EulerPhi[Range[nn]]];Times@@@Thread[{Range[nn],ps}]] (* Harvey P. Dale, Jun 04 2023 *)

PARI

list(nmax) = {my(s = vector(nmax, k, eulerphi(k))); for(k = 2, nmax, s[k] += s[k-1]); for(k = 1, nmax, s[k] = k*s[k]); s} \\ Amiram Eldar, Jun 01 2025

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A143270(n): # based on second formula in A018805
    if n == 0:
        return 0
    c, j = 0, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c += (j2-j)*(2*A143270(k1)//k1-1)
        j, k1 = j2, n//j2
    return n*(n*(n-1)-c+j)//2 # Chai Wah Wu, Mar 25 2021

Formula

a(n) = n*A002088(n), where A002088(n) = partial sums of phi(n).

Equals row sums of triangle A143269.

a(n) = Sum_{i=1..n} Sum_{j=1..i*n} 0^(gcd(i,j)-1). - Reinhard Zumkeller, Jan 15 2009

a(n) = (3/Pi^2) * n^3 + O(n^2*log(n)). - Amiram Eldar, Jun 01 2025

Extensions

More terms from Reinhard Zumkeller, Jan 15 2009