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A053830
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Sum of digits of (n written in base 9).
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26
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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
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OFFSET
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0,3
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COMMENTS
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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
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LINKS
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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.
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FORMULA
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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(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
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EXAMPLE
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a(20) = 2+2 = 4 because 20 is written as 22 base 9.
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)
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MATHEMATICA
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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 *)
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PROG
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(PARI) a(n)=if(n<1, 0, if(n%9, a(n-1)+1, a(n/9)))
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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STATUS
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approved
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