A078309 Numbers that are congruent to {1, 4, 7} mod 10.

1, 4, 7, 11, 14, 17, 21, 24, 27, 31, 34, 37, 41, 44, 47, 51, 54, 57, 61, 64, 67, 71, 74, 77, 81, 84, 87, 91, 94, 97, 101, 104, 107, 111, 114, 117, 121, 124, 127, 131, 134, 137, 141, 144, 147, 151, 154, 157, 161, 164, 167, 171, 174, 177, 181, 184, 187, 191, 194, 197

Offset: 1

Author

Jochen Ziegenbalg, Dec 23 2002

Keywords

• nonn
• easy

Comments

The formula 1 + 3*mod(n-1, 3) + floor(2/7 + (n-1)/n) * 10^floor((n-1)/3) produces a sequence with the same first 6 terms.

Numbers with last digit 1, 4, or 7. - Wesley Ivan Hurt, Jun 14 2016

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Crossrefs

The sequence begins with the same first 6 terms as the 'Straight' sequence (A028373).

Programs

Magma

[3*n + Floor((n-1)/3) - 2: n in [1..60]]; // Vincenzo Librandi, Apr 23 2014

Maple

A078309:=n->(30*n-24-3*cos(2*n*Pi/3)+sqrt(3)*sin(2*n*Pi/3))/9: seq(A078309(n), n=1..100); # Wesley Ivan Hurt, Jun 14 2016

Mathematica

Table[1 + 3*Mod[n - 1, 3] + 10*Floor[(n - 1)/3], {n, 55}]
Select[Range[200], MemberQ[{1,4,7}, Mod[#,10]]&] (* or *) LinearRecurrence[ {1,0,1,-1}, {1,4,7,11}, 60] (* Harvey P. Dale, Apr 21 2014 *)
CoefficientList[Series[(-1 - 3 x (1 + x + x^2))/(-1 + x + x^3 - x^4), {x, 0, 100}], x] (* Vincenzo Librandi, Apr 23 2014 *)

PARI

a(n)=1+n--%3*3+n\3*10 \\ Charles R Greathouse IV, Sep 25 2012

Formula

a(n) = 1 + 3*mod(n-1, 3) + 10*floor((n-1)/3).

From Arkadiusz Wesolowski, Sep 21 2012: (Start)

a(n) = a(n-3) + 10 for n>4.

a(n) = a(n-1) + a(n-3) - a(n-4) for n>4.

a(n) = 3*n + floor((n-1)/3) - 2.

G.f.: (-x - 3*(x^2 + x^3 + x^4))/(-1 + x + x^3 - x^4). (End)

From Wesley Ivan Hurt, Jun 14 2016: (Start)

a(n) = (30*n-24-3*cos(2*n*Pi/3)+sqrt(3)*sin(2*n*Pi/3))/9.

a(3k) = 10k-3, a(3k-1) = 10k-6, a(3k-2) = 10k-9. (End)

Extensions

Edited by Robert G. Wilson v, Dec 24 2002