A176409 Multiples of 3 or 7.

0, 3, 6, 7, 9, 12, 14, 15, 18, 21, 24, 27, 28, 30, 33, 35, 36, 39, 42, 45, 48, 49, 51, 54, 56, 57, 60, 63, 66, 69, 70, 72, 75, 77, 78, 81, 84, 87, 90, 91, 93, 96, 98, 99, 102, 105, 108, 111, 112, 114, 117, 119, 120, 123, 126, 129, 132, 133, 135, 138, 140, 141, 144, 147, 150, 153

Offset

1,2

Comments

Therefore, this sequence also includes multiples of 21.

First differences: 3, 3, 1, 2, 3, 2, 1, 3, 3, 3, 3, 1, 2, 3, 2, 1, 3, 3, 3, 3, 1, 2, 3, 2, 1, 3, 3, 3, 3, 1, 2, 3, 2, 1, 3, 3, 3 have period {3, 3, 1, 2, 3, 2, 1, 3, 3}.

In general, sequences of numbers divisible by primes p and q will be of the form a(n+p+q-1) = a(n)+p*q. - Gary Detlefs, Oct 07 2013

Links

Daniel Starodubtsev, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,1,-1).

Formula

a(n+9) = a(n) + 21.

a(n) = 21*floor((n-1)/9) + 2*((n-1) mod 9) + s(((n-1) mod 9)-1) + 1 - floor(((n-2) mod 9)/8), where s(n) = floor(n*sqrt(2)) - 2*floor(n/sqrt(2)). - Gary Detlefs, Oct 07 2013

a(n) = 7n/3 + O(1). - Charles R Greathouse IV, Feb 13 2017

From Colin Barker, Mar 21 2020: (Start)

G.f.: x^2*(3 + 3*x + x^2 + 2*x^3 + 3*x^4 + 2*x^5 + x^6 + 3*x^7 + 3*x^8) / ((1 - x)^2*(1 + x + x^2)*(1 + x^3 + x^6)).

a(n) = a(n-1) + a(n-9) - a(n-10) for n>10.

(End)

Mathematica

Union@Flatten@{Table[3n, {n, 70}], Table[7n, {n, 30}]}
Select[ Range@ 153, Mod[#, 3] == 0 || Mod[#, 7] == 0 &]

Prog

(PARI) concat(0, Vec(x^2*(3 + 3*x + x^2 + 2*x^3 + 3*x^4 + 2*x^5 + x^6 + 3*x^7 + 3*x^8) / ((1 - x)^2*(1 + x + x^2)*(1 + x^3 + x^6)) + O(x^70))) \\ Colin Barker, Mar 21 2020

Crossrefs

Cf. A047229, A065520, A083035.

Sequence in context: A208327 A131392 A131397 * A284473 A120684 A324927

Adjacent sequences: A176406 A176407 A176408 * A176410 A176411 A176412

Keyword

nonn,easy

Author

Zak Seidov, Dec 07 2010

Extensions

0 inserted by Daniel Starodubtsev, Mar 21 2020

Status

approved