A014824 a(0) = 0; for n>0, a(n) = 10*a(n-1) + n.

0, 1, 12, 123, 1234, 12345, 123456, 1234567, 12345678, 123456789, 1234567900, 12345679011, 123456790122, 1234567901233, 12345679012344, 123456790123455, 1234567901234566, 12345679012345677, 123456790123456788, 1234567901234567899, 12345679012345679010, 123456790123456790121

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

The square roots of these numbers have some remarkable properties - see the link to Schizophrenic numbers.

Partial sums of A002275. - Jonathan Vos Post, Apr 25 2010

This sequence is the particular case of a(0) = 0, a(n) = r*a(n-1) + n, when r = 10. If now the first N terms are computed for (r > N) then the resulting set of numbers is readable as the smallest k-digits permutations (1 <= k <= N): Those built from the concatenation of the first k digits in base-r (see links). - R. J. Cano, Jan 09 2013

There is also an interesting structure to the decimal expansion of 1/sqrt(a(2*n+1)), which has long strings of 0's (that gradually shorten in length until they disappear) interspersed with strings of what, at first sight, appear to be random digits. However, if we factorize these blocks of 'random' digits we find they are related to each other. An example illustrating this is given below. - Peter Bala, Sep 13 2015

From Peter Bala, Sep 15 2015: (Start)

Extending the previous empirical observation, it appears that the decimal expansions of the numbers 1/ (a(4*n+1))^(1/2), 1/a((4*n+3))^(1/4), 1/(10*a(4*n))^(1/2), 1/(10*a(4*n+2))^(1/4), and their powers, have the same pattern of beginning with long strings of 0's (that gradually shorten in length until they disappear) interlaced with strings of digits which, when read as integers and factorized, turn out to be related to each other.

The following result, which is a consequence of Bottomley's explicit formula for a(n), should be helpful in explaining these observations: the decimal expansion of 1/a(2*n-1) for n >= 5 begins with long strings of 0's interlaced successively with the digits of the numbers 81*(18*n + 1)^k for k going from 0 up to approximately n/log_10(18*n). For example, for n = 7 we have 1/a(13) = 0.00000000000081000000000102870000000130644900000 165919023..., with 10287 = 81*127, 1306449 = 81*127^2 and 165919023 = 81*127^3. A similar result holds for the decimal expansion of the number 1/a(2*n).

It appears that a(4*n+3)^(1/4), a(4*n+3)^(3/4), (10*a(4*n))^(1/2), (10*a(4*n+2))^(1/4) and (10*a(4*n+2))^(3/4) are further examples of Brown's schizophrenic numbers.

A theorem of Kuzmin in the measure theory of continued fractions says that for a random real number alpha, the probability that some given partial quotient of alpha is equal to a positive integer k is given by 1/log(2)*( log(1 + 1/k) - log(1 + 1/(k+1)) ). Thus large partial quotients are the exception in continued fraction expansions. Empirically, we observe the presence of unexpectedly large partial quotients early in the continued fraction expansions of the numbers (a(4*n+1))^(1/2), (a(4*n+3))^(1/4), (10*a(4*n))^(1/2), (10*a(4*n+2))^(1/4) and their powers. An example is given below. (End)

From Ya-Ping Lu, Dec 21 2024: (Start)

To get a(n), concatenate the first n digits in the cyclic string '123456790' and subtract the number of occurrences of '9' from the concatenated number. For example, a(8) = 12345679 - 1 = 12345678.

There are 2 prime terms for n <= 20000: a(2497) and a(3301). (End)

Examples

			From _Peter Bala_, Sep 13 2015: (Start)
The decimal expansion of 1/sqrt(a(51)) begins 9.0...0211050...07423683750...02901423065625000... x 10^(-26). The long strings of 0's gradually shorten in length and are interspersed with eleven blocks of digits  [9, 21105, 742368375, 2901423065625, 1190671490555859375, 5025824361636282421875, 216068565680679841787109375, 940978603539360710982861328125, 4137365297437126626102768402099609375, 18326229731370116994398540261077880859375, 816525165681195562685426961332324981689453125]. Read as ordinary integers these numbers factorize as [3^2, (3^2)*5*7*67, (3^3)*(5^3)*(7^2)*(67^2), (3^2)*(5^5)*(7^3)*(67^3), (3^2)*(5^8)*(7^5)*(67^4), (3^4)*(5^8)*(7^6)*(67^5), (3^3)*(5^10)*(7^7)*11*(67^6), (3^3)*(5^11)*(7^7)*11*13*(67^7), (3^4)*(5^16)*(7^8)*11*13*(67^8), (3^2)*(5^17)*(7^9)*11*13*17*(67^9), (3^2)*(5^18)*(7^10)*11*13*17*19*(67^10)]. (End)
From _Peter Bala_, Sep 15 2015: (Start)
The continued fraction expansion of 1/sqrt(a(51)) begins [0; 11111111111111111111111111, 9, 47382136934375740345889, 2, 21, 3, 1, 7, 2, 1, 101028010521057015662, 5, 14, 9, 1, 1, 2, 2, 8, 5, 1, 1, 1, 1, 215411536292232442, 5, 1, 5, 1, 1, 2, 1, 1, 8, 1, 4, 3, 1, 4, 2, 1, 8, 1, 1, 3, 10, 459299650942926, 4, 1, 1, 4, 1, 20, 64, 5, 9, 2, 2, 1, 2, 1, 1, 1, 1, 30, 1, 11, 3, 979316952969, 1, 2, 93, 1, 5, 1, 1, 11, 1, 1, 1, 1, 5, 1, 29, 1, 29, 1, 1, 1, 2, 4, 1, 37, 1, 1, 2, 8, 2, 2088095848, 12, 1, 3, 1, 3, 2, 2, 3, 1, 5, 6, 1, 3, 1, 4, 2, 2, 1, 2, 2, 14, 4, 1, 2, 1, 50, 2, 6, 1, 11, 135, 4452229, 1, ...] and has several unexpectedly large partial quotients early on. (End)
For n=5, a(5) = 1*15 + 9*20 + 9^2*15 + 9^3*6 + 9^4*1 + 9^5*0 = 12345. - _Bruno Berselli_, Nov 13 2015
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See pp. 18-19.
• Kevin S. Brown, Schizophrenic numbers.
• R. J. Cano, Additional information (See appendix).
• Ralf Hinze, Concrete stream calculus: An extended study, J. Funct. Progr. 20 (5-6) (2010) pp. 463-535, doi, Section 5.1.
• László Tóth, On Schizophrenic Patterns in b-ary Expansions of Some Irrational Numbers, arXiv:2002.06584 [math.NT], 2020. See also Proc. Amer. Math. Soc. 148 (2020), pp. 461-469.
• Wikipedia, Schizophrenic Number.
• Index entries for linear recurrences with constant coefficients, signature (12,-21,10).

Crossrefs

Cf. A060011.

Cf. A002275. - Jonathan Vos Post, Apr 25 2010

Similar sequences in other bases are: (base-2) A000295, (base-3) A000340, (base-4) A014825, (base-5) A014827, (base-6) A014829. - R. J. Cano, Jan 11 2013

Differs from A007908, A035239, A057137, A060555, A138957 from n=10 on. - M. F. Hasler, Jan 17 2013

Cf. A030512.

Programs

Magma

[(10^n-1)*(10/81)-n/9: n in [0..20]]; // Vincenzo Librandi, Aug 23 2011

Maple

a:=n->sum((10^(n-j)-1^(n-j))/9,j=0..n): seq(a(n), n=0..17); # Zerinvary Lajos, Jan 15 2007
a:=n->sum(10^(n-j)*j,j=0..n): seq(a(n), n=0..16); # Zerinvary Lajos, Jun 05 2008

Mathematica

Table[Sum[10^i - 1, {i, n}]/9, {n, 18}] (* Robert G. Wilson v, Nov 20 2004 *)
CoefficientList[Series[x/(1 - 12*x + 21*x^2 - 10*x^3), {x, 0, 20}], x] (* Wesley Ivan Hurt, Sep 15 2015 *)

PARI

linrec01(p,u,base)={my(r=!p,A=1);for(j=2,u,A=A*base+r+p*j); A};
a(n)=(n!=0)*linrec01(1, n, 10); \\ R. J. Cano, Jan 09 2011; With (0, n, 10) it generates repunit numbers.

PARI

A014824(n)=(10^(n+1)\9-n)\9  \\ M. F. Hasler, Jan 17 2013

Python

def A014824(n): s = ''.join('123456790'[i%9] for i in range(n)); q, r = divmod(n, 9); return int(s) - q - r//8 # Ya-Ping Lu, Dec 21 2024

Formula

a(n) = (10^n-1)*(10/81) - n/9. - Henry Bottomley, Jul 04 2000

a(n)/10^n converges to 10/81 = 0.123456790123456790...

Let b(n) = if(n = 0, 1, if(n = 1, 10, 10*9^(n-2))). Then a(n) = Sum_{k=0..n} C(n, k)*b(k) (Binomial transform). - Paul Barry, Jan 29 2004

G.f.: x/(1-12*x+21*x^2-10*x^3). - Colin Barker, Jan 08 2012

a(n) = 12*a(n-1) - 21*a(n-2) + 10*a(n-3), n>2. - Wesley Ivan Hurt, Sep 15 2015

a(n) = Sum_{i=0..n} 9^i*binomial(n+1,n-1-i). - Bruno Berselli, Nov 13 2015

a(n) = Sum_{i=0..n} 10^(n-i)*i. - Ya-Ping Lu, Dec 21 2024

E.g.f.: exp(x)*(10*exp(9*x) - 9*x - 10)/81. - Elmo R. Oliveira, Mar 29 2025