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a(n) = (Sum of digits of 9*n) / 9.
2

%I #16 Jun 04 2021 07:13:57

%S 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,

%T 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,

%U 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

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

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

%C 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).

%F a(n) = A007953(A008591(n))/9, by definition. - _Felix Fröhlich_, May 19 2021

%F 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).

%t a[n_] := Plus @@ IntegerDigits[9*n]/9; Array[a, 100, 0] (* _Amiram Eldar_, May 19 2021 *)

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

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

%K nonn,base,easy

%O 0,12

%A _M. F. Hasler_, May 19 2021