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Sum of numbers less than or equal to n which are multiples of 3 or 5.
4

%I #31 Aug 10 2023 07:14:41

%S 0,0,3,3,8,14,14,14,23,33,33,45,45,45,60,60,60,78,78,98,119,119,119,

%T 143,168,168,195,195,195,225,225,225,258,258,293,329,329,329,368,408,

%U 408,450,450,450,495,495,495,543,543,593,644,644,644,698,753,753,810,810

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

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

%H Harvey P. Dale, <a href="/A126592/b126592.txt">Table of n, a(n) for n = 1..1000</a>

%H Project Euler, <a href="https://projecteuler.net/problem=1">Problem 1: Multiples of 3 and 5</a>

%H <a href="/index/Rec#order_31">Index entries for linear recurrences with constant coefficients</a>, 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).

%F 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).

%t 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}]

%t Accumulate[Table[If[Divisible[n, 3] || Divisible[n, 5], n, 0], {n, 60}]] (* _Harvey P. Dale_, Jun 09 2016 *)

%t Accumulate[Table[n Boole[GCD[n, 15] > 1], {n, 50}]] (* _Alonso del Arte_, Dec 23 2018 *)

%o (PARI) {b(n,x)=floor(n/x)*(1 + floor(n/x))};

%o for(n=1,30, print1((3*b(n,3) + 5*b(n,5) - 15*b(n,15))/2, ", ")) \\ _G. C. Greubel_, Mar 06 2018

%o (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

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

%Y Cf. A126073, A126590.

%K nonn,easy

%O 1,3

%A _Zak Seidov_, Mar 13 2007