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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A007953 Digital sum (i.e., sum of digits) of n; also called digsum(n).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 8, 9, 10, 11, 12, 13, 14, 15
Offset: 0

Views

Author

R. Muller

Keywords

Comments

Do not confuse with the digital root of n, A010888 (first term that differs is a(19)).
Also the fixed point of the morphism 0 -> {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, 1 -> {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, 2 -> {2, 3, 4, 5, 6, 7, 8, 9, 10, 11}, etc. - Robert G. Wilson v, Jul 27 2006
For n < 100 equal to (floor(n/10) + n mod 10) = A076314(n). - Hieronymus Fischer, Jun 17 2007
It appears that a(n) is the position of 10*n in the ordered set of numbers obtained by inserting/placing one digit anywhere in the digits of n (except a zero before 1st digit). For instance, for n=2, the resulting set is (12, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 32, 42, 52, 62, 72, 82, 92) where 20 is at position 2, so a(2) = 2. - Michel Marcus, Aug 01 2022
Also the total number of beads required to represent n on a Russian abacus (schoty). - P. Christopher Staecker, Mar 31 2023
a(n) / a(2n) <= 5 with equality iff n is in A169964, while a(n) / a(3n) is unbounded, since if n = (10^k + 2)/3, then a(n) = 3*k+1, a(3n) = 3, so a(n) / a(3n) = k + 1/3 -> oo when k->oo (see Diophante link). - Bernard Schott, Apr 29 2023
Also the number of symbols needed to write number n in Egyptian numerals for n < 10^7. - Wojciech Graj, Jul 10 2025

Examples

			a(123) = 1 + 2 + 3 = 6, a(9875) = 9 + 8 + 7 + 5 = 29.

Links

Crossrefs

Cf. A003132, A055012, A055013, A055014, A055015, A010888, A007954, A031347, A055017, A076313, A076314, A054899, A138470, A138471, A138472, A000120, A004426, A004427, A054683, A054684, A069877, A179082-A179085, A108971, A169964, A179987, A179988, A180018, A180019, A217928, A216407, A037123, A074784, A231688, A231689, A225693, A254524 (ordinal transform).
Bisections: A004092, A004155.
For n + digsum(n) see A062028.

Programs

  • Haskell
    a007953 n | n < 10 = n
          | otherwise = a007953 n' + r where (n',r) = divMod n 10
-- Reinhard Zumkeller, Nov 04 2011, Mar 19 2011
  • Magma
    [ &+Intseq(n): n in [0..87] ];  // Bruno Berselli, May 26 2011
  • Maple
    A007953 := proc(n) add(d,d=convert(n,base,10)) ; end proc: # R. J. Mathar, Mar 17 2011
  • Mathematica
    Table[Sum[DigitCount[n][[i]] * i, {i, 9}], {n, 50}] (* Stefan Steinerberger, Mar 24 2006 *)
Table[Plus @@ IntegerDigits @ n, {n, 0, 87}] (* or *)
Nest[Flatten[# /. a_Integer -> Array[a + # &, 10, 0]] &, {0}, 2] (* Robert G. Wilson v, Jul 27 2006 *)
Total/@IntegerDigits[Range[0,90]] (* Harvey P. Dale, May 10 2016 *)
DigitSum[Range[0, 100]] (* Requires v. 14 *) (* Paolo Xausa, May 17 2024 *)
  • PARI
    a(n)=if(n<1, 0, if(n%10, a(n-1)+1, a(n/10))) \\ Recursive, very inefficient. A more efficient recursive variant: a(n)=if(n>9, n=divrem(n, 10); n[2]+a(n[1]), n)
  • PARI
    a(n, b=10)={my(s=(n=divrem(n, b))[2]); while(n[1]>=b, s+=(n=divrem(n[1], b))[2]); s+n[1]} \\ M. F. Hasler, Mar 22 2011
  • PARI
    a(n)=sum(i=1, #n=digits(n), n[i]) \\ Twice as fast. Not so nice but faster:
  • PARI
    a(n)=sum(i=1,#n=Vecsmall(Str(n)),n[i])-48*#n \\ M. F. Hasler, May 10 2015
/* Since PARI 2.7, one can also use: a(n)=vecsum(digits(n)), or better: A007953=sumdigits. [Edited and commented by M. F. Hasler, Nov 09 2018] */
  • PARI
    a(n) = sumdigits(n); \\ Altug Alkan, Apr 19 2018
  • Python
    def A007953(n):
    return sum(int(d) for d in str(n)) # Chai Wah Wu, Sep 03 2014
  • Python
    def a(n): return sum(map(int, str(n))) # Michael S. Branicky, May 22 2021
  • Scala
    (0 to 99).map(.toString.map(.toInt - 48).sum) // Alonso del Arte, Sep 15 2019
  • Smalltalk
    "Recursive version for general bases. Set base = 10 for this sequence."
digitalSum: base
| s |
base = 1 ifTrue: [^self].
(s := self // base) > 0
  ifTrue: [^(s digitalSum: base) + self - (s * base)]
  ifFalse: [^self]
"by Hieronymus Fischer, Mar 24 2014"
  • Swift
    A007953(n): String(n).compactMap{$0.wholeNumberValue}.reduce(0, +) // Egor Khmara, Jun 15 2021

Formula

a(A051885(n)) = n.
a(n) <= 9(log_10(n)+1). - Stefan Steinerberger, Mar 24 2006
From Benoit Cloitre, Dec 19 2002: (Start)
a(0) = 0, a(10n+i) = a(n) + i for 0 <= i <= 9.
a(n) = n - 9*(Sum_{k > 0} floor(n/10^k)) = n - 9*A054899(n). (End)
From Hieronymus Fischer, Jun 17 2007: (Start)
G.f. g(x) = Sum_{k > 0, (x^k - x^(k+10^k) - 9x^(10^k))/(1-x^(10^k))}/(1-x).
a(n) = n - 9*Sum_{10 <= k <= n} Sum_{j|k, j >= 10} floor(log_10(j)) - floor(log_10(j-1)). (End)
From Hieronymus Fischer, Jun 25 2007: (Start)
The g.f. can be expressed in terms of a Lambert series, in that g(x) = (x/(1-x) - 9*L[b(k)](x))/(1-x) where L[b(k)](x) = sum{k >= 0, b(k)*x^k/(1-x^k)} is a Lambert series with b(k) = 1, if k > 1 is a power of 10, else b(k) = 0.
G.f.: g(x) = (Sum_{k > 0} (1 - 9*c(k))*x^k)/(1-x), where c(k) = Sum_{j > 1, j|k} floor(log_10(j)) - floor(log_10(j-1)).
a(n) = n - 9*Sum_{0 < k <= floor(log_10(n))} a(floor(n/10^k))*10^(k-1). (End)
From Hieronymus Fischer, Oct 06 2007: (Start)
a(n) <= 9*(1 + floor(log_10(n))), equality holds for n = 10^m - 1, m > 0.
lim sup (a(n) - 9*log_10(n)) = 0 for n -> oo.
lim inf (a(n+1) - a(n) + 9*log_10(n)) = 1 for n -> oo. (End)
a(n) = A138530(n, 10) for n > 9. - Reinhard Zumkeller, Mar 26 2008
a(A058369(n)) = A004159(A058369(n)); a(A000290(n)) = A004159(n). - Reinhard Zumkeller, Apr 25 2009
a(n) mod 2 = A179081(n). - Reinhard Zumkeller, Jun 28 2010
a(n) <= 9*log_10(n+1). - Vladimir Shevelev, Jun 01 2011
a(n) = a(n-1) + a(n-10) - a(n-11), for n < 100. - Alexander R. Povolotsky, Oct 09 2011
a(n) = Sum_{k >= 0} A031298(n, k). - Philippe Deléham, Oct 21 2011
a(n) = a(n mod b^k) + a(floor(n/b^k)), for all k >= 0. - Hieronymus Fischer, Mar 24 2014
Sum_{n>=1} a(n)/(n*(n+1)) = 10*log(10)/9 (Shallit, 1984). - Amiram Eldar, Jun 03 2021

Extensions

More terms from Hieronymus Fischer, Jun 17 2007
Edited by Michel Marcus, Nov 11 2013

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