A053824 Sum of digits of (n written in base 5).
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
Offset: 0
Examples
a(20) = 4 + 0 = 4 because 20 is written as 40 in base 5. From _Omar E. Pol_, Feb 21 2010: (Start) 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)
Links
- Tanar Ulric, Table of n, a(n) for n = 0..10000 (terms 0..3125=5^5 from Reinhard Zumkeller).
- 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.
Crossrefs
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.
Cf. A173525. - Omar E. Pol, Feb 21 2010
Cf. A173670 (last nonzero decimal digit of (10^n)!). - Washington Bomfim, Jan 01 2011
Programs
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Haskell
a053824 0 = 0 a053824 x = a053824 x' + d where (x', d) = divMod x 5 -- Reinhard Zumkeller, Jan 31 2014
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Magma
[&+Intseq(n, 5):n in [0..100]]; // Marius A. Burtea, Aug 24 2019
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Mathematica
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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PARI
a(n)=if(n<1,0,if(n%5,a(n-1)+1,a(n/5)))
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PARI
a(n) = sumdigits(n, 5); \\ Michel Marcus, Aug 24 2019
Formula
From Benoit Cloitre, Dec 19 2002: (Start)
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(n) = A138530(n,5) for n > 4. - Reinhard Zumkeller, Mar 26 2008
If i >= 2, a(2^i) mod 4 = 0. - Washington Bomfim, Jan 01 2011
a(n) = Sum_{k>=0} A031235(n,k). - Philippe Deléham, Oct 21 2011
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
Comments