A126592 Sum of numbers less than or equal to n which are multiples of 3 or 5.

0, 0, 3, 3, 8, 14, 14, 14, 23, 33, 33, 45, 45, 45, 60, 60, 60, 78, 78, 98, 119, 119, 119, 143, 168, 168, 195, 195, 195, 225, 225, 225, 258, 258, 293, 329, 329, 329, 368, 408, 408, 450, 450, 450, 495, 495, 495, 543, 543, 593, 644, 644, 644, 698, 753, 753, 810, 810

Offset: 1

Author

Zak Seidov, Mar 13 2007

Keywords

• nonn
• easy

Comments

Sum of numbers m <= n such that (m mod 3) * (m mod 5) = 0.

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000
• Project Euler, Problem 1: Multiples of 3 and 5
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,0,0,0,2,-2,0,0,0,0,0,0,0,0,0,0,0,0,0,-1,1).

Crossrefs

Cf. A126073, A126590.

Programs

Magma

[(3*Floor(n/3)*(1 + Floor(n/3)) + 5*Floor(n/5)*(1 + Floor(n/5)) - 15*Floor(n/15)*(1 + Floor(n/15)))/2: n in [1..30]]; // G. C. Greubel, Mar 06 2018

Mathematica

an[n_, d_] := d * Floor[n/d]; sn[n_, d_] := (an[n, d] * (an[n, d] + d))/(2 * d); Table[sn[n, 3] + sn[n, 5] - sn[n, 15], {n, 1000}]
Accumulate[Table[If[Divisible[n, 3] || Divisible[n, 5], n, 0], {n, 60}]] (* Harvey P. Dale, Jun 09 2016 *)
Accumulate[Table[n Boole[GCD[n, 15] > 1], {n, 50}]] (* Alonso del Arte, Dec 23 2018 *)

PARI

{b(n,x)=floor(n/x)*(1 + floor(n/x))};
for(n=1,30, print1((3*b(n,3) + 5*b(n,5) - 15*b(n,15))/2, ", ")) \\ G. C. Greubel, Mar 06 2018

Scala

(for (n <- 2 to 50) yield if ((n % 3) * (n % 5) == 0) { n } else { 0 }).scanLeft(0)( + ) // Alonso del Arte, Dec 23 2018

Formula

an(n, d) = d * floor(n/d), sn(n, d) = (an(n, d) * (an(n, d) + d))/(2*d), a(n) = sn(n, 3) + sn(n, 5) - sn(n, 15).