A053829 Sum of digits of (n written in base 8).
0, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 8, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 5, 6, 7, 8, 9, 10, 4, 5, 6, 7, 8, 9, 10, 11, 5, 6, 7, 8, 9, 10, 11, 12, 6, 7, 8, 9, 10, 11, 12, 13, 7, 8, 9, 10, 11, 12, 13, 14, 1, 2, 3, 4, 5, 6, 7, 8, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 5, 6, 7, 8, 9, 10, 4, 5, 6, 7, 8, 9, 10
Offset: 0
Examples
a(20)=2+4=6 because 20 is written as 24 base 8. From _Omar E. Pol_, Feb 21 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, 1,2,3,4,5,6,7,8,2,3,4,5,6,7,8,9,3,4,5,6,7,8,9,10,4,5,6,7,8,9,10,11,5,6,7,8,9,10,11,12,6,7,8,9,10,11,12,13,7,8,9,10,11,12,13,14, 1,2,3,4,5,6,7,8,2,3,4,5,6,7,8,9,3,4,5,6,7,8,9,10,4,5,6,7,8,9,10... where the rows converge to A173528. (End)
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
- 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.
- Robert Walker, Self Similar Sloth Canon Number Sequences
- Eric Weisstein's World of Mathematics, Digit Sum.
- Eric Weisstein's World of Mathematics, Octal.
Crossrefs
Programs
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Haskell
a053829 n = q 0 $ divMod n 8 where q r (0, d) = r + d q r (m, d) = q (r + d) $ divMod m 8 -- Reinhard Zumkeller, May 15 2011
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Mathematica
Table[Plus @@ IntegerDigits[n, 8], {n, 0, 95}] (* or *) Nest[ Flatten[ #1 /. a_Integer -> Table[a + i, {i, 0, 7}]] &, {0}, 4] (* Robert G. Wilson v, Jul 27 2006 *) -
PARI
a(n)=if(n<1,0,if(n%8,a(n-1)+1,a(n/8)))
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PARI
a(n) = sumdigits(n, 8); \\ Michel Marcus, Jul 10 2022
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Python
def A053829(n): return sum(int(d) for d in oct(n)[2:]) # Chai Wah Wu, Jul 09 2022
Formula
From Benoit Cloitre, Dec 19 2002: (Start)
a(0) = 0, a(8n+i) = a(n)+i for 0 <= i <= 7.
a(n) = n-7*(Sum_{k>0} floor(n/8^k)) = n-7*A054897(n). (End)
a(n) = A138530(n,8) for n > 7. - Reinhard Zumkeller, Mar 26 2008
a(n) = Sum_k>=0 {A031045(n,k)}. - Philippe Deléham, Oct 21 2011
a(0) = 0; a(n) = a(n - 8^floor(log_8(n))) + 1. - Ilya Gutkovskiy, Aug 24 2019
Sum_{n>=1} a(n)/(n*(n+1)) = 8*log(8)/7 (Shallit, 1984). - Amiram Eldar, Jun 03 2021
Comments