A343639 a(n) = (Sum of digits of 9*n) / 9.

0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2

Offset

0,12

Comments

Similar to, but different from A193582, A326307, ...

Consider g = A008585 = multiply by 3, and its left inverse h = A002264, h o g = id (but g o h = id only on (the range of) A008585). In the spirit of group theory, we can write ad(g) = (x -> h o x o g), then A343638 = ad(A008585)(A007953) and this A343639 = ad(A008585)(A343638) = ad(A008585)^2 (A007953).

Links

Table of n, a(n) for n=0..99.

Formula

a(n) = A007953(A008591(n))/9, by definition. - Felix Fröhlich, May 19 2021

a(n) = A343638(3*n)/3 = A002264(A343638(A008585(n))), i.e., A343639 = A002264 o A343638 o A008585 (just as A343638 = A002264 o A007953 o A008585).

Mathematica

a[n_] := Plus @@ IntegerDigits[9*n]/9; Array[a, 100, 0] (* Amiram Eldar, May 19 2021 *)

Prog

(PARI) A343639(n)=sumdigits(9*n)/9

Crossrefs

Cf. A007953 (digit sum), A008591 (9n), A343638 (similar for 3), A083824 (9*n reversed and divided by 9), A279777 (a(n)=3).

Sequence in context: A184340 A083911 A326307 * A095827 A193582 A091887

Adjacent sequences: A343636 A343637 A343638 * A343640 A343641 A343642

Keyword

nonn,base,easy

Author

M. F. Hasler, May 19 2021

Status

approved