cp's OEIS Frontend

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.

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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.
		

Crossrefs

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

A007908 Triangle of the gods: to get a(n), concatenate the decimal numbers 1,2,3,...,n.

Original entry on oeis.org

1, 12, 123, 1234, 12345, 123456, 1234567, 12345678, 123456789, 12345678910, 1234567891011, 123456789101112, 12345678910111213, 1234567891011121314, 123456789101112131415, 12345678910111213141516, 1234567891011121314151617, 123456789101112131415161718
Offset: 1

Views

Author

R. Muller

Keywords

Comments

For the name "triangle of the gods" see Pickover link. - N. J. A. Sloane, Dec 15 2019
Number of digits: A058183(n) = A055642(a(n)); sums of digits: A037123(n) = A007953(a(n)). - Reinhard Zumkeller, Aug 10 2010
Charles Nicol and John Selfridge ask if there are infinitely many primes in this sequence - see the Guy reference. - Charles R Greathouse IV, Dec 14 2011
Stephan finds no primes in the first 839 terms. I checked that there are no primes in the first 5000 terms. Heuristically there are infinitely many, about 0.5 log log n through the n-th term. - Charles R Greathouse IV, Sep 19 2012 [Expanded search to 20000 without finding any primes. - Charles R Greathouse IV, Apr 17 2014] [Independent search extended to 64000 terms without finding any primes. - Dana Jacobsen, Apr 25 2014]
Elementary congruence arguments show that primes can occur only at indices congruent to 1, 7, 13, or 19 mod 30. - Roderick MacPhee, Oct 05 2015
A note on heuristics: I wrote a quick program to count primes in sequences which are like A007908 but start at k instead of 1. I ran this for k = 1 to 100 and counted the primes up to 1000 (1000 possibilities for k = 1, 999 for k = 2, etc. up to 901 for k = 100). I then compared this to the expected count which is 0 if the number N is divisible by 2, 3, or 5 and 15/(4 log N) otherwise. (If N < 43 I counted the number as 1 instead.) k = 1 has 1.788 expected primes but only 0 actual (of course). k = 2 has 2.268 expected but 4 actual (see A262571, A089987). In total the expectation is 111.07 and the actual count is 110, well within the expected error of +/- 10.5. - Charles R Greathouse IV, Sep 28 2015
Early bird numbers for n > 1: a(2) = A116700(1) = 12; a(3) = A116700(52) = 123; a(4) = A116700(725) = 1234; a(5) = A116700(8074) = 12345; a(6) = A116700(85846) = 123456. - Reinhard Zumkeller, Dec 13 2012
For n < 10^6, a(n)/A000217(n) is an integer for n = 1, 2, and 5. The integers are 1, 4, and 823 (a prime), respectively. - Derek Orr, Sep 04 2014; Max Alekseyev, Sep 30 2015
In order to be a prime, a(n) must end in a digit 1, 3, 7 or 9, so only 4 among 10 consecutive values can be prime. (But a(64000) already has A058183(64000) > 300000 digits.) Also, a(64001) and a(64011) and more generally a(64001+10k) is divisible by 3 unless k == 2 (mod 3), but for k = 2, 5, 8, ... 23 these are divisible by small primes < 999. a(64261) is the first serious candidate in this subsequence. - M. F. Hasler, Sep 30 2015
There are no primes in the first 10^5 terms. - Max Alekseyev, Oct 03 2015; Oct 11 2015
There are no primes in the first 200000 terms. - Serge Batalov, Oct 24 2015
There is a distributed project for continued search, using PRPNet/PFGW software; see the Mersenne Forum link below. - Serge Batalov, Oct 18 2015
It appears that the Mersenne Forum search reached n = 344869 without finding a prime, and was then abandoned. It would be nice if someone could recover the final version of that link from the Wayback machine - the Great Smarandache PRPrime search, http://99.121.249.54:1200 - so that we have a record of how far they searched. - N. J. A. Sloane, Apr 09 2018
The web page https://www.mersenneforum.org/showthread.php?t=20527&page=9 has a comment from Serge Balatov that seems to say that the search reached 10^6 without finding a prime. It would be nice to have this confirmed, and to get more details about how it was done. - N. J. A. Sloane, Dec 15 2019
The expected number of primes among the first million terms is about 0.6. - Ernst W. Mayer, Oct 09 2015
A few semiprimes exist among the early terms, but then become scarce: see A046461. For the base-2 analog of this sequence (A047778), there is a 15-decimal digit prime, but Hans Havermann has shown that the second prime would have more than 91000 digits. - N. J. A. Sloane, Oct 08 2015

References

  • R. K. Guy, Unsolved Problems in Number Theory, Section A3, page 15, of 3rd edition, Springer, 2010.

Crossrefs

See A057137 for another version.
Cf. A033307, A053064, A000422 (left concatenations)
If we concatenate 1 through n but leave out k, we get sequences A262571 (leave out 1) through A262582 (leave out 12), etc., and again we can ask for the smallest prime in each sequence. See A262300 for a summary of these results. Primes seem to exist if we search far enough. - N. J. A. Sloane, Sep 29 2015
Concatenation of first n numbers in other bases: 2: A047778, 3: A048435, 4: A048436, 5: A048437, 6: A048438, 7: A048439, 8: A048440, 9: A048441, 10: this sequence, 11: A048442, 12: A048443, 13: A048444, 14: A048445, 15: A048446, 16: A048447. - Dylan Hamilton, Aug 11 2010
Entries that give the primes in sequences of this type: A089987, A262298, A262300, A262552, A262555.
For semiprimes see A046461.
See also A007376 (the almost-natural numbers), A071620 (primes in that sequence).
See also A033307 (the Champernowne constant) and A176942 (the Champernowne primes). A262043 is a variant of the present sequence.
A002782 is an amusing cousin of this sequence.
Least prime factor: A075019.

Programs

  • Haskell
    a007908 = read . concatMap show . enumFromTo 1 :: Integer -> Integer
    -- Reinhard Zumkeller, Dec 13 2012
    
  • Magma
    [Seqint(Reverse(&cat[Reverse(Intseq(k)): k in [1..n]])): n in [1..17]];  // Bruno Berselli, May 27 2011
    
  • Maple
    A055642 := proc(n) max(1, ilog10(n)+1) ; end: A007908 := proc(n) if n = 1 then 1; else A007908(n-1)*10^A055642(n)+n ; fi ; end: seq(A007908(n),n=1..12) ; # R. J. Mathar, May 31 2008
    # second Maple program:
    a:= proc(n) a(n):= `if`(n=0, 0, parse(cat(a(n-1), n))) end:
    seq(a(n), n=1..22);  # Alois P. Heinz, Jan 12 2021
  • Mathematica
    Table[FromDigits[Flatten[IntegerDigits[Range[n]]]], {n, 20}] (* Alonso del Arte, Sep 19 2012 *)
    FoldList[#2 + #1 10^IntegerLength[#2] &, Range[20]] (* Eric W. Weisstein, Nov 06 2015 *)
    FromDigits /@ Flatten /@ IntegerDigits /@ Flatten /@ Rest[FoldList[List, {}, Range[20]]] (* Eric W. Weisstein, Nov 04 2015 *)
    FromDigits /@ Flatten /@ IntegerDigits /@ Rest[FoldList[Append, {}, Range[20]]] (* Eric W. Weisstein, Nov 04 2015 *)
  • Maxima
    a[1]:1$ a[n]:=a[n-1]*10^floor(log(10*n)/log(10))+n$ makelist(a[n],n,1,17);  /* Bruno Berselli, May 27 2011 */
    
  • PARI
    a(n)=my(s="");for(k=1,n,s=Str(s,k));eval(s) \\ Charles R Greathouse IV, Sep 19 2012
    
  • PARI
    A007908(n,a=0)={for(d=1,#Str(n),my(t=10^d);for(k=t\10,min(t-1,n),a=a*t+k));a} \\ M. F. Hasler, Sep 30 2015
    
  • Python
    def a(n): return int("".join(map(str, range(1, n+1))))
    print([a(n) for n in range(1, 18)]) # Michael S. Branicky, Jan 12 2021
    
  • Python
    from functools import reduce
    def A007908(n): return reduce(lambda i,j:i*10**len(str(j))+j,range(1,n+1)) # Chai Wah Wu, Feb 27 2023

Formula

a(n) = n + a(n-1)*10^A055642(n). - R. J. Mathar, May 31 2008
a(n) = floor(C*10^(A058183(n))) with C = A033307. - José de Jesús Camacho Medina, Aug 19 2015

Extensions

Name edited by N. J. A. Sloane, Dec 15 2019

A000788 Total number of 1's in binary expansions of 0, ..., n.

Original entry on oeis.org

0, 1, 2, 4, 5, 7, 9, 12, 13, 15, 17, 20, 22, 25, 28, 32, 33, 35, 37, 40, 42, 45, 48, 52, 54, 57, 60, 64, 67, 71, 75, 80, 81, 83, 85, 88, 90, 93, 96, 100, 102, 105, 108, 112, 115, 119, 123, 128, 130, 133, 136, 140, 143, 147, 151, 156, 159, 163, 167, 172, 176, 181, 186
Offset: 0

Views

Author

Keywords

Comments

Partial sums of A000120.
The graph of this sequence is a version of the Takagi curve: see Lagarias (2012), Section 9, especially Theorem 9.1. - N. J. A. Sloane, Mar 12 2016
a(n-1) is the largest possible number of ordered pairs (a,b) such that a/b is a prime in a subset of the positive integers with n elements. - Yifan Xie, Feb 21 2025

References

  • J.-P. Allouche & J. Shallit, Automatic sequences, Cambridge University Press, 2003, p. 94
  • R. Bellman and H. N. Shapiro, On a problem in additive number theory, Annals Math., 49 (1948), 333-340. See Eq. 1.9. [From N. J. A. Sloane, Mar 12 2009]
  • L. E. Bush, An asymptotic formula for the average sums of the digits of integers, Amer. Math. Monthly, 47 (1940), pp. 154-156. [From the bibliography of Stolarsky, 1977]
  • P. Cheo and S. Yien, A problem on the k-adic representation of positive integers (Chinese; English summary), Acta Math. Sinica, 5 (1955), pp. 433-438. [From the bibliography of Stolarsky, 1977]
  • M. P. Drazin and J. S. Griffith, On the decimal representation of integers, Proc. Cambridge Philos. Soc., (4), 48 (1952), pp. 555-565. [From the bibliography of Stolarsky, 1977]
  • E. N. Gilbert, Games of identification or convergence, SIAM Review, 4 (1962), 16-24.
  • Grabner, P. J.; Kirschenhofer, P.; Prodinger, H.; Tichy, R. F.; On the moments of the sum-of-digits function. Applications of Fibonacci numbers, Vol. 5 (St. Andrews, 1992), 263-271, Kluwer Acad. Publ., Dordrecht, 1993.
  • R. L. Graham, On primitive graphs and optimal vertex assignments, pp. 170-186 of Internat. Conf. Combin. Math. (New York, 1970), Annals of the NY Academy of Sciences, Vol. 175, 1970.
  • E. Grosswald, Properties of some arithmetic functions, J. Math. Anal. Appl., 28 (1969), pp.405-430.
  • Donald E. Knuth, The Art of Computer Programming, volume 3 Sorting and Searching, section 5.3.4, subsection Bitonic sorting, with C'(p) = a(p-1).
  • Hiu-Fai Law, Spanning tree congestion of the hypercube, Discrete Math., 309 (2009), 6644-6648 (see p(m) on page 6647).
  • Z. Li and E. M. Reingold, Solution of a divide-and-conquer maximin recurrence, SIAM J. Comput., 18 (1989), 1188-1200.
  • B. Lindström, On a combinatorial problem in number theory, Canad. Math. Bull., 8 (1965), 477-490.
  • Mauclaire, J.-L.; Murata, Leo; On q-additive functions. I. Proc. Japan Acad. Ser. A Math. Sci. 59 (1983), no. 6, 274-276.
  • Mauclaire, J.-L.; Murata, Leo; On q-additive functions. II. Proc. Japan Acad. Ser. A Math. Sci. 59 (1983), no. 9, 441-444.
  • M. D. McIlroy, The number of 1's in binary integers: bounds and extremal properties, SIAM J. Comput., 3 (1974), 255-261.
  • L. Mirsky, A theorem on representations of integers in the scale of r, Scripta Math., 15 (1949), pp. 11-12.
  • I. Shiokawa, On a problem in additive number theory, Math. J. Okayama Univ., 16 (1974), pp.167-176. [From the bibliography of Stolarsky, 1977]
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • K. B. Stolarsky, Power and exponential sums of digital sums related to binomial coefficient parity, SIAM J. Appl. Math., 32 (1977), 717-730.
  • Trollope, J. R. An explicit expression for binary digital sums. Math. Mag. 41 1968 21-25.

Crossrefs

For number of 0's in binary expansion of 0, ..., n see A059015.
The basic sequences concerning the binary expansion of n are A000120, A000788, A000069, A001969, A023416, A059015, A070939, A083652.

Programs

  • Haskell
    a000788_list = scanl1 (+) A000120_list
    -- Walt Rorie-Baety, Jun 30 2012
    
  • Haskell
    {a000788 0 = 0; a00788 n = a000788 n2 + a000788 (n-n2-1) + (n-n2) where n2 = n `div` 2}
    -- Walt Rorie-Baety, Jul 15 2012
    
  • Maple
    a:= proc(n) option remember; `if`(n=0, 0, a(n-1)+add(i, i=Bits[Split](n))) end:
    seq(a(n), n=0..62);  # Alois P. Heinz, Nov 11 2024
  • Mathematica
    a[n_] := Count[ Table[ IntegerDigits[k, 2], {k, 0, n}], 1, 2]; Table[a[n], {n, 0, 62}] (* Jean-François Alcover, Dec 16 2011 *)
    Table[Plus@@Flatten[IntegerDigits[Range[n], 2]], {n, 0, 62}] (* Alonso del Arte, Dec 16 2011 *)
    Accumulate[DigitCount[Range[0,70],2,1]] (* Harvey P. Dale, Jun 08 2013 *)
  • PARI
    A000788(n)={ n<3 && return(n); if( bittest(n,0) \\
    , n+1 == 1<A000788(n>>1)*2+n>>1+1 \\
    , n == 1<A000788(n>>=1)+A000788(n-1)+n )} \\ M. F. Hasler, Nov 22 2009
    
  • PARI
    a(n)=sum(k=1,n,hammingweight(k)) \\ Charles R Greathouse IV, Oct 04 2013
    
  • PARI
    a(n) = if (n==0, 0, m = logint(n, 2); r = n % 2^m; m*2^(m-1) + r + 1 + a(r)); \\ Michel Marcus, Mar 27 2018
    
  • PARI
    a(n)={n++; my(t, i, s); c=n; while(c!=0, i++; c\=2); for(j=1, i, d=(n\2^(i-j))%2; t+=(2^(i-j)*(s*d+d*(i-j)/2)); s+=d); t} \\ David A. Corneth, Nov 26 2024
    (C++) /* See David W. Wilson link. */
    
  • Python
    def A000788(n): return sum(i.bit_count() for i in range(1,n+1)) # Chai Wah Wu, Mar 01 2023
    
  • Python
    def A000788(n): return (n+1)*n.bit_count()+(sum((m:=1<>j)-(r if n<<1>=m*(r:=k<<1|1) else 0)) for j in range(1,n.bit_length()+1))>>1) # Chai Wah Wu, Nov 11 2024

Formula

McIlroy (1974) gives bounds and recurrences. - N. J. A. Sloane, Mar 24 2014
Stolarsky (1977) studies the asymptotics, and gives at least nine references to earlier work on the problem. I have added all the references that were not here already. - N. J. A. Sloane, Apr 06 2014
a(n) = Sum_{k=1..n} A000120(k). - Benoit Cloitre, Dec 19 2002
a(0) = 0, a(2n) = a(n)+a(n-1)+n, a(2n+1) = 2a(n)+n+1. - Ralf Stephan, Sep 13 2003
a(n) = n*log_2(n)/2 + O(n); a(2^n)=n*2^(n-1)+1. - Benoit Cloitre, Sep 25 2003 (The first result is due to Bellman and Shapiro, - N. J. A. Sloane, Mar 24 2014)
a(n) = n*log_2(n)/2+n*F(log_2(n)) where F is a nowhere differentiable continuous function of period 1 (see Allouche & Shallit). - Benoit Cloitre, Jun 08 2004
G.f.: (1/(1-x)^2) * Sum_{k>=0} x^2^k/(1+x^2^k). - Ralf Stephan, Apr 19 2003
a(2^n-1) = A001787(n) = n*2^(n-1). - M. F. Hasler, Nov 22 2009
a(4^n-2) = n(4^n-2).
For real n, let f(n) = [n]/2 if [n] even, n-[n+1]/2 otherwise. Then a(n) = Sum_{k>=0} 2^k*f((n+1)/2^k).
a(A000225(n)) = A173921(A000225(n)) = A001787(n); a(A000079(n)) = A005183(n). - Reinhard Zumkeller, Mar 04 2010
From Hieronymus Fischer, Jun 10 2012: (Start)
a(n) = (1/2)*Sum_{j=1..m+1} (floor(n/2^j + 1/2)*(2n + 2 - floor(n/2^j + 1/2))*2^j - floor(n/2^j)*(2n + 2 - (1 + floor(n/2^j)) * 2^j)), where m=floor(log_2(n)).
a(n) = (n+1)*A000120(n) - 2^(m-1) + 1/4 + (1/2)*Sum_{j=1..m+1} ((floor(n/2^j) + 1/2)^2 - floor(n/2^j + 1/2)^2)*2^j, where m=floor(log_2(n)).
a(2^m-1) = m*2^(m-1).
(This is the total number of '1' digits occurring in all the numbers with <= m bits.)
Generic formulas for the number of digits >= d in the base p representations of all integers from 0 to n, where 1<= d < p.
a(n) = (1/2)*Sum_{j=1..m+1} (floor(n/p^j + (p-d)/p)*(2n + 2 + ((p-2*d)/p - floor(n/p^j + (p-d)/p))*p^j) - floor(n/p^j)*(2n + 2 - (1+floor(n/p^j)) * p^j)), where m=floor(log_p(n)).
a(n) = (n+1)*F(n,p,d) + (1/2)*Sum_{j=1..m+1} ((((p-2*d)/p)*floor(n/p^j+(p-d)/p) + floor(n/p^j))*p^j - (floor(n/p^j+(p-d)/p)^2 - floor(n/p^j)^2)*p^j), where m=floor(log_p(n)) and F(n,p,d) = number of digits >= d in the base p representation of n.
a(p^m-1) = (p-d)*m*p^(m-1).
(This is the total number of digits >= d occurring in all the numbers with <= m digits in base p representation.)
G.f.: g(x) = (1/(1-x)^2)*Sum_{j>=0} (x^(d*p^j) - x^(p*p^j))/(1-x^(p*p^j)). (End)
a(n) = Sum_{k=1..n} A000120(A240857(n,k)). - Reinhard Zumkeller, Apr 14 2014
For n > 0, if n is written as 2^m + r with 0 <= r < 2^m, then a(n) = m*2^(m-1) + r + 1 + a(r). - Shreevatsa R, Mar 20 2018
a(n) = n*(n+1)/2 + Sum_{k=1..floor(n/2)} ((2k-1)((g(n,k)-1)*2^(g(n,k) + 1) + 2) - (n+1)*(g(n,k)+1)*g(n,k)/2), where g(n,k) = floor(log_2(n/(2k-1))). - Fabio Visonà, Mar 17 2020
From Jeffrey Shallit, Aug 07 2021: (Start)
A 2-regular sequence, satisfying the identities
a(4n+1) = -a(2n) + a(2n+1) + a(4n)
a(4n+2) = -2a(2n) + 2a(2n+1) + a(4n)
a(4n+3) = -4a(n) + 4a(2n+1)
a(8n) = 4a(n) - 8a(2n) + 5a(4n)
a(8n+4) = -9a(2n) + 5a(2n+1) + 4a(4n)
for n>=0. (End)
a(n) = Sum_{k=0..floor(log_2(n+1))} k * A360189(n,k). - Alois P. Heinz, Mar 06 2023

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Jan 15 2001

A034967 Sum of digits of numbers between 0 and (10^n)-1.

Original entry on oeis.org

0, 45, 900, 13500, 180000, 2250000, 27000000, 315000000, 3600000000, 40500000000, 450000000000, 4950000000000, 54000000000000, 585000000000000, 6300000000000000, 67500000000000000, 720000000000000000, 7650000000000000000, 81000000000000000000
Offset: 0

Views

Author

Keywords

References

  • Edward J. Barbeau, Murray S. Klamkin & William O. J. Moser, Five Hundred Mathematical Challenges, Problem 284 at pp. 142-143 (1995).

Crossrefs

Cf. A037123.

Programs

Formula

a(n) = 45*n*10^(n-1).
a(n) = 20*a(n-1) - 100*a(n-2), a(0)=0, a(1)=45. - Harvey P. Dale, Oct 09 2011
G.f.: (45*x)/(10*x-1)^2. - Harvey P. Dale, Oct 09 2011
a(n) = (9*n*10^n)/2. - Harvey P. Dale, Apr 23 2018

Extensions

More terms from James Sellers, Jan 19 2000

A061076 a(n) is the sum of the products of the digits of all the numbers from 1 to n.

Original entry on oeis.org

1, 3, 6, 10, 15, 21, 28, 36, 45, 45, 46, 48, 51, 55, 60, 66, 73, 81, 90, 90, 92, 96, 102, 110, 120, 132, 146, 162, 180, 180, 183, 189, 198, 210, 225, 243, 264, 288, 315, 315, 319, 327, 339, 355, 375, 399, 427, 459, 495, 495, 500, 510, 525, 545, 570, 600, 635
Offset: 1

Views

Author

Amarnath Murthy, Apr 14 2001

Keywords

Comments

What is the asymptotic behavior of this sequence? a(n) = a(n+1) for almost all n. A weak upper bound: a(n) << n^1.91. - Charles R Greathouse IV, Jan 13 2012
A check was done for k in {i^j | 1 <= i <= 10 AND 1 <= j <= 100}. For all these values, a(k) < k^1.733. Another check for k in {i^j | 101 <= i <= 110 AND 101 <= j <= 200} gave a(k) < k^1.65324. For k in {i | 10^6 <= i <= 10^7}, a(k) < k^1.6534. So I ask: is it true that a(n) < n^1.733 and a(n) -> n^(1.65323 + o(1)), or about n^(log(45)/log(10) + o(1))? - David A. Corneth, May 17 2016
For n = 10^(k-1), the closed-form formula from Mihai Teodor (see Formula section) gives a(n) = (45^k - 45)/44, so lim_{n->oo} log(a(n))/log_10(n) = log(45) = 3.80666248977.... - Jon E. Schoenfield, Apr 10 2022
For k >= 1, a(10^k-1) = a(10^k) = ... = a(10*R_k) where R = A002275; so there is a run of 10*R_{k-1} + 2 = A047855(k) consecutive terms equal to (45/44)*(45^k-1) when n runs from 10^k-1 up to 10*R_k, this is because those numbers have one or more 0's. Example: first runs with 2, 12, 112, 1112, ... consecutive terms equal to 45, 2070, 93195, 4193820, ... start at 9, 99, 999, 9999, ... and end at 10, 110, 1110, 11110, ... - Bernard Schott, Oct 18 2022

Examples

			a(9) = a(10) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 1*0 = 1+2+3+4+5+6+7+8+9 = 45.
		

References

  • Amarnath Murthy, Smarandache friendly numbers and a few more sequences, Smarandache Notions Journal, Vol. 12, No. 1-2-3, Spring 2001.

Crossrefs

Programs

  • Maple
    A007954:= n -> convert(convert(n,base,10),`*`):
    ListTools:-PartialSums(map(A007954,[$1..100])); # Robert Israel, May 17 2016
  • Mathematica
    Accumulate[Times@@IntegerDigits[#]& /@ Range[100]]
  • PARI
    pd(n) = my(d = digits(n)); prod(i=1, #d, d[i]);
    a(n) = sum(k=1, n, pd(k)); \\ Michel Marcus, Feb 01 2015
    
  • PARI
    a(n) = {n=digits(n); p=1; d=#n; for(i=1,#n,if(n[i]==0,d=i-1;break));
    (45/44) * (45^(#n-1)-1) + sum(i=1,d,p*=n[i]; p * (n[i]-1) * (45/44) * (45^(#n -i) - 45^(#n-i-1)) / 2)+p*(d==#n)} \\ David A. Corneth, May 17 2016
    
  • Python
    from math import prod
    def A061076(n): return sum(prod(int(d) for d in str(i)) for i in range(1,n+1)) # Chai Wah Wu, Mar 21 2022
  • Sage
    def A061076(n):
        p = 0
        i = 0
        while i < n + 1:
            p += prod(int(digit) for digit in str(i))
            i += 1
        return p # Daria Micovic, Apr 13 2016
    

Formula

a(n) = Sum_{k = 1..n} (product of the digits of k).
a(10^k-1) = (45/44)*(45^k-1). - Giovanni Resta, Oct 18 2012
From Robert Israel, May 17 2016: (Start)
Partial sums of A007954.
G.f.: (1-x)^(-1) * Sum_{n>=0} Product_{j=0..n} Sum_{k=1..9} k * x^(k*10^j).
G.f. satisfies A(x) = (x + 2*x^2 + ... + 9*x^9)*(1+(1-x^10)*A(x^10))/(1-x).
(End)
Let b(1), b(2), ..., b(k) be the digits of the base-10 expansion of n: n = b(1)*10^(k-1) + b(2)*10^(k-2) + ... + b(k). Then a(n) = b(1)*b(2)*...*b(k) + (45^k-45)/44 + (1/2)*Sum_{i=1..k} b(1)*b(2)*...*b(i)*(b(i)-1)*45^(k-i). - Mihai Teodor, Apr 09 2022

Extensions

Corrected and extended by Matthew Conroy, Apr 16 2001

A114136 Numbers k such that k divides the sum of digits of all numbers from 1 to k.

Original entry on oeis.org

1, 3, 5, 7, 9, 18, 21, 24, 38, 58, 78, 98, 298, 498, 501, 537, 698, 702, 707, 711, 716, 898, 1141, 1197, 1501, 1557, 1998, 2703, 2706, 2712, 2715, 3998, 4502, 4509, 4529, 4536, 5998, 7998, 8001, 8097, 9998, 29998, 36001, 36297, 49998, 54504, 54507, 54510
Offset: 1

Views

Author

David W. Wilson, Feb 02 2006

Keywords

Comments

a(n) | A037123(a(n)). Sequence is infinite, including 10^(2m)-2 for all m >= 1. For each m >= 1, there is at least one n with A037123(a(n))/a(n) = m.
Number of terms < 10^m for m >= 1: 5, 12, 22, 41, 58, 79, 105, 128, 176, 202, ..., . - Robert G. Wilson v, Jul 06 2018
Certain number forms appear repeatedly: 1(9e)8, 3(9e)8, 5(9e)8, 7(9e)8, 9(9e)8, 2(9o)8, 4(9o)8, 6(9o)8, 8(9o)8; where e represents an even number of times and o an odd number of times. - Robert G. Wilson v, Jul 06 2018

Crossrefs

Cf. A037123.

Programs

  • Mathematica
    lst = {}; k = 1; s = 0; While[k < 10001, If[ Mod[s += Plus @@ IntegerDigits@k, k++] == 0, AppendTo[lst, k -1]]]; lst (* Robert G. Wilson v, Jul 06 2018 *)
    nxt[{t_,n_}]:={t+Total[IntegerDigits[n+1]],n+1}; Select[NestList[nxt,{1,1},55000],Divisible[#[[1]],#[[2]]]&][[All,2]] (* Harvey P. Dale, Jan 18 2019 *)
  • PARI
    isok(n) = (sum(k=1, n, sumdigits(k)) % n) == 0; \\ Michel Marcus, Jul 16 2018

A071317 a(n) = a(n-1) + sum of digits of n^2.

Original entry on oeis.org

0, 1, 5, 14, 21, 28, 37, 50, 60, 69, 70, 74, 83, 99, 115, 124, 137, 156, 165, 175, 179, 188, 204, 220, 238, 251, 270, 288, 307, 320, 329, 345, 352, 370, 383, 393, 411, 430, 443, 452, 459, 475, 493, 515, 534, 543, 553, 566, 575, 582, 589, 598, 611, 630, 648, 658
Offset: 0

Views

Author

Labos Elemer, May 27 2002

Keywords

References

  • N. Agronomof, Question 4419, L'Intermédiaire des Math. 21 (1914) 147.

Crossrefs

Partial sums of A004159.

Programs

  • Haskell
    a071317 n = a071317_list !! n
    a071317_list = scanl1 (+) a004159_list
    -- Reinhard Zumkeller, Apr 12 2014
    
  • Mathematica
    s=0; Do[s=s+Apply[Plus, IntegerDigits[n^2]]; Print[s], {n, 1, 128}]
    nxt[{n_,a_}]:={n+1,a+Total[IntegerDigits[(n+1)^2]]}; NestList[nxt,{0,0},60][[All,2]] (* Harvey P. Dale, Mar 09 2017 *)
    FoldList[#1 + Total@ IntegerDigits[#2^2] &, 0, Range@ 55] (* Michael De Vlieger, Mar 25 2017 *)
    Accumulate[Plus @@@ IntegerDigits[Range[0, 50]^2]] (* Giovanni Resta, Mar 25 2017 *)
  • Python
    from itertools import count, islice, accumulate
    def A071317_gen(): # generator of terms
        return accumulate(map(lambda n:sum(map(int,str(n**2))),count(0)))
    A071317_list = list(islice(A071317_gen(),20)) # Chai Wah Wu, Mar 15 2023

Extensions

a(0) = 0 prepended by Reinhard Zumkeller, Apr 12 2014

A074784 a(n) = a(n-1) + square of the sum of digits of n.

Original entry on oeis.org

0, 1, 5, 14, 30, 55, 91, 140, 204, 285, 286, 290, 299, 315, 340, 376, 425, 489, 570, 670, 674, 683, 699, 724, 760, 809, 873, 954, 1054, 1175, 1184, 1200, 1225, 1261, 1310, 1374, 1455, 1555, 1676, 1820, 1836, 1861, 1897, 1946, 2010, 2091, 2191, 2312, 2456
Offset: 0

Views

Author

Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Sep 07 2002

Keywords

Comments

a(n) = Sum_{i=0..n} digsum(i)^2, where digsum(i) = A007953(i). - N. J. A. Sloane, Nov 13 2013

Crossrefs

Partial sums of A118881.

Programs

  • Magma
    [n eq 1 select n else Self(n-1)+(&+Intseq(n))^2: n in [1..48]];  // Bruno Berselli, Jul 12 2011
  • Maple
    See A037123.
  • Mathematica
    Accumulate @ Array[(Plus @@ IntegerDigits[#])^2 &, 50] (* Amiram Eldar, Jan 20 2022 *)

Formula

a(n) = Sum_{k=1..n} s(k)^2 = Sum_{k=1..n} A007953(k)^2, where s(k) denotes the sum of the digits of k in decimal representation.
Asymptotic expression: a(n-1) = Sum_{k=1..n-1} s(k)^2 = 20.25*n*log_10(n)^2 + O(n*log_10(n)).
In general: Sum_{k=1..n-1} s(k)^m = n*((9/2)*log_10(n))^m + O(n*log_10(n)^(m-1)).

Extensions

Offset changed to 0 and a(0) prepended by Amiram Eldar, Jan 20 2022

A177274 Periodic sequence: Repeat 1, 2, 3, 4, 5, 6, 7, 8, 9.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6
Offset: 0

Views

Author

Klaus Brockhaus, May 07 2010

Keywords

Comments

Interleaving of A131669 and A131669 without first five terms.
Continued fraction expansion of (684125+sqrt(635918528029))/1033802.
Decimal expansion of 13717421/111111111.
a(n) = A010888(n+1) = A010878(n)+1 = A117230(n+2)-1.
a(n) = A064806(n+1)-n-1.
Essentially first differences of A037123.

Crossrefs

Cf. A131669 (odd digits followed by positive even digits), A010888 (digital root of n), A010878 (n mod 9), A117230 (1 followed by (repeat 2, 3, 4, 5, 6, 7, 8, 9, 10), offset 1), A064806 (n + digital root of n), A037123, A177270 (decimal expansion of (684125+sqrt(635918528029))/1033802).

Programs

  • Magma
    &cat[ [1, 2, 3, 4, 5, 6, 7, 8, 9]: k in [1..12] ];
  • Mathematica
    PadRight[{},120,Range[9]] (* Paolo Xausa, Jan 08 2024 *)

Formula

a(n) = (n mod 9)+1.
a(n) = a(n-9) for n > 8; 1; a(n) = n+1 for n <= 8.
G.f.: (1+2*x+3*x^2+4*x^3+5*x^4+6*x^5+7*x^6+8*x^7+9*x^8)/(1-x^9). [corrected by Georg Fischer, May 11 2019]

A231688 a(n) = Sum_{i=0..n} digsum(i)^3, where digsum(i) = A007953(i).

Original entry on oeis.org

0, 1, 9, 36, 100, 225, 441, 784, 1296, 2025, 2026, 2034, 2061, 2125, 2250, 2466, 2809, 3321, 4050, 5050, 5058, 5085, 5149, 5274, 5490, 5833, 6345, 7074, 8074, 9405, 9432, 9496, 9621, 9837, 10180, 10692, 11421, 12421, 13752, 15480, 15544, 15669, 15885, 16228, 16740, 17469, 18469, 19800, 21528, 23725, 23850, 24066, 24409, 24921, 25650, 26650, 27981, 29709, 31906, 34650
Offset: 0

Views

Author

N. J. A. Sloane, Nov 13 2013

Keywords

References

  • Grabner, P. J.; Kirschenhofer, P.; Prodinger, H.; Tichy, R. F.; On the moments of the sum-of-digits function. Applications of Fibonacci numbers, Vol. 5 (St. Andrews, 1992), 263-271, Kluwer Acad. Publ., Dordrecht, 1993.

Crossrefs

Partial sums of A118880.

Programs

  • Maple
    See A037123.
  • Mathematica
    Accumulate[Table[Total[IntegerDigits[n]]^3,{n,0,60}]] (* Harvey P. Dale, Aug 06 2021 *)
  • PARI
    a(n) = sum(i=0, n, sumdigits(i)^3); \\ Michel Marcus, Jan 07 2017
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