A174438 Numbers that are congruent to {0, 2, 5, 8} mod 9.

0, 2, 5, 8, 9, 11, 14, 17, 18, 20, 23, 26, 27, 29, 32, 35, 36, 38, 41, 44, 45, 47, 50, 53, 54, 56, 59, 62, 63, 65, 68, 71, 72, 74, 77, 80, 81, 83, 86, 89, 90, 92, 95, 98, 99, 101, 104, 107, 108, 110, 113, 116, 117, 119, 122, 125, 126, 128, 131, 134, 135, 137

Offset: 1

Author

Gary Detlefs, Mar 19 2010

Keywords

• nonn
• easy

Comments

Also the set of nonnegative integers expressible as (x + 2y)(2x + y) for integer x and y, where integers of the form 3k + 2 are given by x = 2k + 1, y = -k, and integers of the form 9k are given by x = 2k - 1, y = 2 - k. - Drake Thomas, Nov 10 2022

From Klaus Purath, Feb 21 2023: (Start)

The sum of digits of any term belongs to the sequence. Also the products of an odd number of terms as well as products of one term each of this sequence and one term of A056991 are members. The products of an even number of terms belong to A056991.

Nonnegative integers of the forms 2*x^2 + (2*m+1)*x*y + ((m^2+m-2)/2)*y^2, for integers m. This includes the formula given by Drake Thomas above.

The union of this sequence and A056991 is closed under multiplication. (End)

Links

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

Programs

Magma

[n : n in [0..150] | n mod 9 in [0, 2, 5, 8]]; // Wesley Ivan Hurt, Jun 07 2016

Maple

seq(3*(n-floor(n/4))-(3-I^n-(-I)^n-(-1)^n)/4, n=0..100);

Mathematica

Table[(18n-15+I^(2n)+(3-I)*I^(-n)+(3+I)*I^n)/8, {n, 80}] (* Wesley Ivan Hurt, Jun 07 2016 *)
Select[Range[0,150],MemberQ[{0,2,5,8},Mod[#,9]]&] (* Harvey P. Dale, Jan 02 2019 *)
LinearRecurrence[{1,0,0,1,-1},{0,2,5,8,9},70] (* Harvey P. Dale, Jan 15 2022 *)

Python

def A174438(n): return (0,2,5,8)[n&3]+9*(n>>2) # Chai Wah Wu, Jan 30 2023

Formula

a(n) = 3*(n-floor(n/4)) - (3 - i^n - (-i)^n - (-1)^n)/4 where i=sqrt(-1), offset=0.

From Wesley Ivan Hurt, Jun 07 2016: (Start)

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

a(n) = a(n-1) + a(n-4) - a(n-5) for n > 5.

a(n) = (18*n - 15 + i^(2*n) + (3-i)*i^(-n) + (3+i)*i^n)/8 where i=sqrt(-1). (End)

Extensions

a(23) corrected by Chai Wah Wu, Jun 10 2016