A094798 -id:A094798 - cp's OEIS frontend

A014778 Numbers k equal to the number of 1's in the decimal digits of all numbers <= k.

0, 1, 199981, 199982, 199983, 199984, 199985, 199986, 199987, 199988, 199989, 199990, 200000, 200001, 1599981, 1599982, 1599983, 1599984, 1599985, 1599986, 1599987, 1599988, 1599989, 1599990, 2600000, 2600001, 13199998, 35000000

Offset: 1

Yves Babe, Maurice Protat, Olivier Gérard

The full list of 84 terms is given in the b-file.
It can be proved that this sequence is finite. (The main idea of the proof is that the number of 1's used in positive integers <= k is greater than or equal to A(k) = (1/10)*(number of digits in positive integers from 1 to k) = (1/10) Sum_{i=1..k} (1+floor(log_10 i)). By considering the area below a logarithmic function and the corresponding integral, it can be shown that A(k)/k goes to infinity.) - Joseph L. Pe, Nov 05 2002
Fixed points of A094798. Sequence consists of six runs of ten consecutive numbers, ten pairs of consecutive numbers and four isolated numbers. - David Wasserman, Jun 29 2007

			a(5)=199983 because the number of 1's in the decimal digits of the numbers from 0 to 199983 is 199983 and this is the 5th such number.

Maurice Protat, "Des Olympiades à l'Agrégation", Editions Ellipses, Paris 1997, p. 183.

Graeme McRae, May 26 2007, Table of n, a(n) for n = 1..84 (complete sequence)
Tanya Khovanova and Gregory Marton, Archive Labeling Sequences, arXiv:2305.10357 [math.HO], 2023.
Ed Pegg Jr. and Eric W. Weisstein, Mathematica's Google Aptitude, MathWorld Headline news, Oct 13 2004.

Cf. A101639, A101640, A101641, A130427, A130428, A130429, A130430, A130431; cf. A130432 for the number of numbers in these sequences.
Cf. A094798.
Cf. A165617 for the sequence generalized to an arbitrary base. - Martin J. Erickson (erickson(AT)truman.edu), Oct 08 2010

Join[{0},With[{nn=35*10^6},Position[Thread[{Accumulate[ DigitCount[ Range[nn],10,1]], Range[nn]}],{x_,x_}]]]//Flatten (* Harvey P. Dale, Oct 14 2017 *)

from itertools import count, islice
def agen(s=0): # generator of terms
    yield from (k for k in count(0) if (s:=s+str(k).count('1'))==k)
print(list(islice(agen(),26))) # Michael S. Branicky, Oct 02 2023

Corrected and extended by Deepan Majmudar (deepan.majmudar(AT)hp.com), Nov 19 2004
41 further terms from Ryan Propper, Dec 07 2004, who observed that there are no more terms <= 10^9
The final (84th) term 1111111110 was sent by Lambrecht Kok (L.P.Kok(AT)rug.nl), Jan 13 2005. He says: "H. van Haeringen and I showed that this list of 84 terms is complete on Dec 15 2004".
Independently shown to be complete by Ryan Propper and Vaughan Pratt, Jan 08 2005
Edited by M. F. Hasler, Feb 12 2013

A053541 a(n) = n*10^(n-1).

Original entry on oeis.org

1, 20, 300, 4000, 50000, 600000, 7000000, 80000000, 900000000, 10000000000, 110000000000, 1200000000000, 13000000000000, 140000000000000, 1500000000000000, 16000000000000000, 170000000000000000

Offset: 1

Barry E. Williams, Jan 15 2000

This sequence gives the number of 1's (or any other nonzero digit) required to write all integers from 0 up to 10^n-1. - Jason D. W. Taff (jtaff(AT)jburroughs.org), Dec 05 2004 (improved by Bernard Schott, Nov 17 2022)

The corresponding number of 0's required to write all these integers from 0 up to 10^n-1 is A033714(n). - Bernard Schott, Nov 17 2022

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Vincenzo Librandi, Table of n, a(n) for n = 1..100
Frank Ellermann, Illustration of binomial transforms.
Index entries for linear recurrences with constant coefficients, signature (20,-100).

Cf. A001787, A033714, A038303, A053464, A053469, A081045, A094798.

List([1..20], n-> n*10^(n-1)) # G. C. Greubel, May 16 2019

[n*10^(n-1): n in [1..30]]; // Vincenzo Librandi, Jun 06 2011

seq(n*10^(n-1), n = 1 .. 40); # Bernard Schott, Nov 17 2022

f[n_]:=n*10^(n-1);f[Range[40]] (* Vladimir Joseph Stephan Orlovsky, Feb 09 2011*)
LinearRecurrence[{20,-100},{1,20},20] (* Harvey P. Dale, Aug 08 2023 *)

a(n)=n*10^(n-1) \\ Charles R Greathouse IV, Dec 05 2011

[n*10^(n-1) for n in (1..20)] # G. C. Greubel, May 16 2019

a(n) = 20*a(n-1) - 100*a(n-2), with a(0)=0, a(1)=1, a(2)=20.
From Jason D. W. Taff (jtaff(AT)jburroughs.org), Dec 05 2004: (Start)
a(n) = 10*a(n-1) + 10*(n-1).
a(n) = Sum_{k=1..n} k*binomial(n,k)*9^(n-k).
a(n) = A094798(10^n - 1). (End)
From G. C. Greubel, May 16 2019: (Start)
G.f.: x/(1-10*x)^2.
E.g.f.: x*exp(10*x). (End)
From Amiram Eldar, Oct 28 2020: (Start)
Sum_{n>=1} 1/a(n) = 10*log(10/9).
Sum_{n>=1} (-1)^(n+1)/a(n) = 10*log(11/10). (End)
a(n) = Sum_{k=1..n} A081045(k-1). - Bernard Schott, Nov 17 2022

Offset changed from 0 to 1 by Vincenzo Librandi, Jun 06 2011

A094799 First term of a run of 10 consecutive numbers such that for each m in the 10-tuple exactly m 1's are used in writing out all numbers 1 through m.

Original entry on oeis.org

199981, 1599981, 35199981, 500199981, 501599981, 535199981

Offset: 1

Lekraj Beedassy, Jun 11 2004

The sequence is complete. - David Wasserman, Jun 29 2007

M. Protat, Des Olympiades a l'Agregation, Nombre de "1", Problem 89, pp. 182-183, Ellipses, Paris 1997.

Cf. A014778.

Cf. A094798.

A331375 a(n) is the number of times the digit 1 appears in the concatenation of integers from 0 to n, minus the number of times the next most frequent digit appears.

Original entry on oeis.org

-1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 9, 9, 7, 6, 5, 4, 3, 2, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 0

Scott R. Shannon, Jan 14 2020

Other than a(0) = 0 the digit 1 is the most frequently seen digit in the concatenation of the integers from 0 to n. See A094798 for the exact number of times. This sequence is the difference between that number and the number of times the next most frequent digit appears. For almost all numbers the next most frequent digit is 2. That only changes to the digit 0 once per order of magnitude, after reaching the number consisting of two or more 1's followed by 0. The digit 0 keeps this record for the next number, a repunit, after which the number of appearances of 2 again either equals or surpasses the number of appearances of 0.

When concatenating the integers from 0 to 10^k, with k >= 2, this sequence reaches its maximum value of 10^(k-1) at n = 10^k/5-1.

			a(0) = -1 as after '0' the digit 0 has appeared once while 1 has not appeared, so a(0) = 0 - 1 = -1.
a(10) = 0 as after '012345678910' the digits 0 and 1 have both appeared two times, so a(10) = 2 - 2 = 0.
a(11) = 2 as after '01234567891011' the digit 1 has appeared four times and the digit 0 two times, so a(11) = 4 - 2 = 2.

Scott R. Shannon, Table of n, a(n) for n = 1..10000

Cf. A094798, A007376.

Deleted a conjectured but incorrect g.f. and recurrence. - N. J. A. Sloane, Jan 17 2020

A094800 First term of a run of exactly two consecutive numbers such that for each m in the run, exactly m 1's are used in writing out all numbers 1 through m.

Original entry on oeis.org

0, 200000, 2600000, 35000000, 35200000, 500000000, 500200000, 502600000, 535000000, 535200000

Offset: 1

Lekraj Beedassy, Jun 11 2004

Numbers n such that n and n+1 are members of A014778, but n-1 and n+2 are not. - David Wasserman

M. Protat, Des Olympiades a l'Agregation, Nombre de "1", Problem 89, pp. 182-183, Ellipses, Paris 1997.

Cf. A014778.

Cf. A094798, A094799, A094801.

Corrected by David Wasserman, Jun 29 2007. There are no further terms.

A094801 Numbers k such that k is a term of A014778, but k-1 and k+1 are not.

Original entry on oeis.org

13199998, 117463825, 513199998, 1111111110

Offset: 1

Lekraj Beedassy, Jun 11 2004

M. Protat, Des Olympiades a l'Agregation, Nombre de "1", Problem 89, pp. 182-183, Ellipses, Paris 1997.

Cf. A014778.

Cf. A094798, A094799, A094800.

Corrected by David Wasserman, Jun 29 2007

A365097 Smallest k > 1 such that the total number of digits "1" required to write the numbers 1..k in base n is equal to k.

Original entry on oeis.org

2, 4, 25, 181, 421, 3930, 8177, 102772, 199981, 3179142, 5971945, 143610511, 210826981, 4754446846, 8589934561, 222195898593, 396718580701, 13494919482970, 20479999999961, 764527028941797, 1168636602822613, 41826814261329722, 73040694872113105, 2855533828630999398

Offset: 2

Andrew Pope, Aug 21 2023

a(10) = A014778(3), being the smallest term > 1 there.

An upper bound is a(n) <= A226238(n) = u, since the digits of u show there are u 1's in numbers 1..u (in base n). - Kevin Ryde, Sep 28 2023

			For n=2, the first k=2 positive integers are 1 = 1_2 and 2 = 10_2, which have a total of two 1's, so a(2) = 2.
For n=3, the first k=4 positive integers, which are 1_3, 2_3, 10_3, and 11_3, have a total of four 1's, which is equal to k, so a(3) = 4.
For n=4, a total of 25 1's occur in the first k=25 positive integers (they occur in 1_4, 10_4, 11_4, 12_4, 13_4, 21_4, 31_4, 100_4, 101_4, 102_4, 103_4, 110_4, 111_4, 112_4, 113_4, 120_4, and 121_4 = 25), so a(4) = 25.

Jon E. Schoenfield, Table of n, a(n) for n = 2..200

Cf. A014778, A094798, A226238.

a[n_] := Module[{k = 1, sum = 1}, While[sum == 1 || sum != k, k++; sum += Count[IntegerDigits[k, n], 1]]; k]; Array[a, 6, 2] (* Amiram Eldar, Aug 29 2023 *)

from itertools import count
from sympy.ntheory.factor_ import digits
def A365097(n):
    c, a, q, m = 1, 1, 0, 1
    for k in count(2):
        m += 1
        if m == n:
            m = 0
            q += 1
            a = digits(q,n).count(1)
        elif m==1:
            a += 1
        elif m==2:
            a -= 1
        c += a
        if c == k:
            return k # Chai Wah Wu, Sep 28 2023

For even n > 2, a(n) = 2*n^(n/2) - 2*n + 1. - Jon E. Schoenfield, Sep 30 2023

a(11)-a(15) from Amiram Eldar, Aug 29 2023
a(16)-a(19) from Chai Wah Wu, Sep 29 2023
a(20)-a(25) from Jon E. Schoenfield, Sep 30 2023

A094797 Number of times 1 is used in writing out all numbers 1 through 10^n.

Original entry on oeis.org

1, 2, 21, 301, 4001, 50001, 600001, 7000001, 80000001, 900000001, 10000000001, 110000000001, 1200000000001, 13000000000001, 140000000000001, 1500000000000001, 16000000000000001, 170000000000000001, 1800000000000000001, 19000000000000000001

Offset: 0

Lekraj Beedassy, Jun 11 2004

Index entries for linear recurrences with constant coefficients, signature (21,-120,100).

Cf. A072290, A078427.

Cf. A011557, A094798.

Table[ n*10^(n - 1) + 1, {n, 0, 17}] (* Robert G. Wilson v, Jun 15 2004 *)
LinearRecurrence[{21,-120,100},{1,2,21},20] (* Harvey P. Dale, Sep 07 2022 *)

Vec(-(99*x^2-19*x+1)/((x-1)*(10*x-1)^2) + O(x^100)) \\ Colin Barker, May 23 2014

a(n) = n*10^(n-1) + 1.
a(n) = 21*a(n-1)-120*a(n-2)+100*a(n-3). - Colin Barker, May 23 2014
G.f.: -(99*x^2-19*x+1) / ((x-1)*(10*x-1)^2). - Colin Barker, May 23 2014
a(n) = A094798(A011557(n)). - Michel Marcus, Oct 03 2023

More terms from Robert G. Wilson v, Jun 15 2004

Further terms from Colin Barker, May 23 2014

cp's OEIS Frontend

A014778 Numbers k equal to the number of 1's in the decimal digits of all numbers <= k.

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A053541 a(n) = n*10^(n-1).

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A094799 First term of a run of 10 consecutive numbers such that for each m in the 10-tuple exactly m 1's are used in writing out all numbers 1 through m.

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A331375 a(n) is the number of times the digit 1 appears in the concatenation of integers from 0 to n, minus the number of times the next most frequent digit appears.

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A094800 First term of a run of exactly two consecutive numbers such that for each m in the run, exactly m 1's are used in writing out all numbers 1 through m.

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A094801 Numbers k such that k is a term of A014778, but k-1 and k+1 are not.

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A365097 Smallest k > 1 such that the total number of digits "1" required to write the numbers 1..k in base n is equal to k.

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A094797 Number of times 1 is used in writing out all numbers 1 through 10^n.

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