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A053830
Sum of digits of (n written in base 9).
27
0, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 9, 2, 3, 4, 5, 6, 7, 8, 9, 10, 3, 4, 5, 6, 7, 8, 9, 10, 11, 4, 5, 6, 7, 8, 9, 10, 11, 12, 5, 6, 7, 8, 9, 10, 11, 12, 13, 6, 7, 8, 9, 10, 11, 12, 13, 14, 7, 8, 9, 10, 11, 12, 13, 14, 15, 8, 9, 10, 11, 12, 13, 14, 15, 16, 1, 2, 3, 4, 5, 6, 7, 8, 9
OFFSET
0,3
COMMENTS
Also the fixed point of the morphism 0->{0,1,2,3,4,5,6,7,8}, 1->{1,2,3,4,5,6,7,8,9}, 2->{2,3,4,5,6,7,8,9,10}, etc. - Robert G. Wilson v, Jul 27 2006
LINKS
Jeffrey O. Shallit, Problem 6450, Advanced Problems, The American Mathematical Monthly, Vol. 91, No. 1 (1984), pp. 59-60; Two series, solution to Problem 6450, ibid., Vol. 92, No. 7 (1985), pp. 513-514.
Eric Weisstein's World of Mathematics, Digit Sum.
FORMULA
From Benoit Cloitre, Dec 19 2002: (Start)
a(0) = 0, a(9n+i) = a(n) + i for 0 <= i <= 8;
a(n) = n - 8*Sum_{k>=1} floor(n/9^k) = n - 8*A054898(n). (End)
a(n) = A138530(n,9) for n > 8. - Reinhard Zumkeller, Mar 26 2008
a(n) = Sum_{k>=0} A031087(n,k). - Philippe Deléham, Oct 21 2011
a(0) = 0; a(n) = a(n - 9^floor(log_9(n))) + 1. - Ilya Gutkovskiy, Aug 24 2019
Sum_{n>=1} a(n)/(n*(n+1)) = 9*log(9)/8 (Shallit, 1984). - Amiram Eldar, Jun 03 2021
EXAMPLE
a(20) = 2+2 = 4 because 20 is written as 22 base 9.
From Omar E. Pol, Feb 23 2010: (Start)
It appears that this can be written as a triangle (see the conjecture in the entry A000120):
0;
1,2,3,4,5,6,7,8;
1,2,3,4,5,6,7,8,9,2,3,4,5,6,7,8,9,10,3,4,5,6,7,8,9,10,11,4,5,6,7,8,9,10,11,...
where the rows converge to A173529. (End)
MATHEMATICA
Table[Plus @@ IntegerDigits[n, 9], {n, 0, 100}] (* or *)
Nest[ Flatten[ #1 /. a_Integer -> Table[a + i, {i, 0, 8}]] &, {0}, 3] (* Robert G. Wilson v, Jul 27 2006 *)
PROG
(PARI) a(n)=if(n<1, 0, if(n%9, a(n-1)+1, a(n/9)))
(Magma) [&+Intseq(n, 9):n in [0..100]]; // Marius A. Burtea, Aug 24 2019
KEYWORD
base,nonn
AUTHOR
Henry Bottomley, Mar 28 2000
STATUS
approved