1,2

a(n) = digital root of (1+floor(n/9))*(1+((n-1) mod 9)).

Sequence is periodic, period length is 81. The 9 by 9 array of the first 81 terms is symmetric.

To determine the digital root of a product a*b by means of this sequence (or rather by means of the 9 by 9 array) reduce both a and b modulo 9 - if a result is zero replace it by 9 - to obtain c and d, then choose the d-th element of the c-th row, or alternatively the c-th element of the d-th row, i.e. A(c,d) or A(d,c); commutativity of multiplication is reflected in the symmetry of the array.

Array begins:

1 2 3 4 5 6 7 8 9

2 4 6 8 1 3 5 7 9

3 6 9 3 6 9 3 6 9

4 8 3 7 2 6 1 5 9

5 1 6 2 7 3 8 4 9

6 3 9 6 3 9 6 3 9

7 5 3 1 8 6 4 2 9

8 7 6 5 4 3 2 1 9

9 9 9 9 9 9 9 9 9

Find the digital root of 197*799

(a) customarily: digital root of 197*799 = digital root of (digital root of 197)*(digital root of 799) = digital root of 8*7 = digital root of 56 = 2.

(b) using the array: 197 mod 9 = 8, 799 mod 9 = 7; A(8,7) = A(7,8) = 2.

(PARI) {digitalroot(n) = if(n<1, 0, (n-1)%9+1)} {a(n) = digitalroot((1+floor(n/9))*(1+((n-1)%9)))} {for(n=1, 105, print1(a(n), ", "))} /* Klaus Brockhaus, Mar 28 2007 */

Cf. A010888 (digital root of n).

Sequence in context: A068505 A080719 A049105 this_sequence A043269 A122481 A110805

Adjacent sequences: A125956 A125957 A125958 this_sequence A125960 A125961 A125962

nonn,tabf

Simon Alexander (simonalexander2005(AT)hotmail.com), Feb 07 2007

Edited by Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Mar 28 2007