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A053824
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Sum of digits of (n written in base 5).
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37
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0, 1, 2, 3, 4, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 4, 5, 6, 7, 8, 5, 6, 7, 8, 9, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 4, 5, 6, 7, 8, 5, 6, 7, 8, 9, 6, 7, 8, 9, 10, 3, 4, 5, 6, 7, 4, 5, 6, 7, 8, 5, 6, 7, 8, 9, 6, 7, 8, 9, 10, 7, 8, 9, 10, 11, 4, 5, 6
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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}, 1->{1,2,3,4,5}, 2->{2,3,4,5,6}, 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(5n+i) = a(n) + i for 0 <= i <= 4;
a(n) = n - 4*Sum_{k>=1} floor(n/5^k) = n - 4*A027868(n). (End)
a(0) = 0; a(n) = a(n - 5^floor(log_5(n))) + 1. - Ilya Gutkovskiy, Aug 23 2019
Sum_{n>=1} a(n)/(n*(n+1)) = 5*log(5)/4 (Shallit, 1984). - Amiram Eldar, Jun 03 2021
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EXAMPLE
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a(20) = 4 + 0 = 4 because 20 is written as 40 in base 5.
It appears that this can be written as a triangle:
0,
1,2,3,4,
1,2,3,4,5,2,3,4,5,6,3,4,5,6,7,4,5,6,7,8,
1,2,3,4,5,2,3,4,5,6,3,4,5,6,7,4,5,6,7,8,5,6,7,8,9,2,3,4,5,6,3,4,5,6,7,4,5,...
See the conjecture in the entry A000120. (End)
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MATHEMATICA
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Table[Plus @@ IntegerDigits[n, 5], {n, 0, 100}] (* or *)
Nest[Flatten[ #1 /. a_Integer -> Table[a + i, {i, 0, 4}]] &, {0}, 4] (* Robert G. Wilson v, Jul 27 2006 *)
f[n_] := n - 4 Sum[Floor[n/5^k], {k, n}]; Array[f, 103, 0]
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PROG
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(PARI) a(n)=if(n<1, 0, if(n%5, a(n-1)+1, a(n/5)))
(Haskell)
a053824 0 = 0
a053824 x = a053824 x' + d where (x', d) = divMod x 5
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CROSSREFS
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Sum of digits of n written in bases 2-16: A000120, A053735, A053737, this sequence, A053827, A053828, A053829, A053830, A007953, A053831, A053832, A053833, A053834, A053835, A053836.
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KEYWORD
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AUTHOR
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STATUS
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approved
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