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

%I A014824 #130 Mar 29 2025 17:24:40
%S A014824 0,1,12,123,1234,12345,123456,1234567,12345678,123456789,1234567900,
%T A014824 12345679011,123456790122,1234567901233,12345679012344,
%U A014824 123456790123455,1234567901234566,12345679012345677,123456790123456788,1234567901234567899,12345679012345679010,123456790123456790121
%N A014824 a(0) = 0; for n>0, a(n) = 10*a(n-1) + n.
%C A014824 The square roots of these numbers have some remarkable properties - see the link to Schizophrenic numbers.
%C A014824 Partial sums of A002275. - _Jonathan Vos Post_, Apr 25 2010
%C A014824 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
%C A014824 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
%C A014824 From _Peter Bala_, Sep 15 2015: (Start)
%C A014824 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.
%C A014824 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).
%C A014824 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.
%C A014824 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)
%C A014824 From _Ya-Ping Lu_, Dec 21 2024: (Start)
%C A014824 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.
%C A014824 There are 2 prime terms for n <= 20000: a(2497) and a(3301). (End)
%H A014824 Vincenzo Librandi, <a href="/A014824/b014824.txt">Table of n, a(n) for n = 0..1000</a>
%H A014824 Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, <a href="https://arxiv.org/abs/2501.06675">Chip-Firing on Infinite k-ary Trees</a>, arXiv:2501.06675 [math.CO], 2025. See pp. 18-19.
%H A014824 Kevin S. Brown, <a href="http://www.mathpages.com/home/kmath404.htm">Schizophrenic numbers</a>.
%H A014824 R. J. Cano, <a href="http://oeis.org/w/images/c/cd/Canotheorem.pdf">Additional information (See appendix)</a>.
%H A014824 Ralf Hinze, <a href="https://www.cs.ox.ac.uk/people/ralf.hinze/publications/CSC.pdf">Concrete stream calculus: An extended study</a>, J. Funct. Progr. 20 (5-6) (2010) pp. 463-535, <a href="https://doi.org/10.1017/S0956796810000213">doi</a>, Section 5.1.
%H A014824 László Tóth, <a href="https://arxiv.org/abs/2002.06584">On Schizophrenic Patterns in b-ary Expansions of Some Irrational Numbers</a>, arXiv:2002.06584 [math.NT], 2020. See also <a href="https://doi.org/10.1090/proc/14863">Proc. Amer. Math. Soc.</a> 148 (2020), pp. 461-469.
%H A014824 Wikipedia, <a href="http://wikipedia.org/wiki/Schizophrenic_number">Schizophrenic Number</a>.
%H A014824 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (12,-21,10).
%F A014824 a(n) = (10^n-1)*(10/81) - n/9. - _Henry Bottomley_, Jul 04 2000
%F A014824 a(n)/10^n converges to 10/81 = 0.123456790123456790...
%F A014824 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
%F A014824 G.f.: x/(1-12*x+21*x^2-10*x^3). - _Colin Barker_, Jan 08 2012
%F A014824 a(n) = 12*a(n-1) - 21*a(n-2) + 10*a(n-3), n>2. - _Wesley Ivan Hurt_, Sep 15 2015
%F A014824 a(n) = Sum_{i=0..n} 9^i*binomial(n+1,n-1-i). - _Bruno Berselli_, Nov 13 2015
%F A014824 a(n) = Sum_{i=0..n} 10^(n-i)*i. - _Ya-Ping Lu_, Dec 21 2024
%F A014824 E.g.f.: exp(x)*(10*exp(9*x) - 9*x - 10)/81. - _Elmo R. Oliveira_, Mar 29 2025
%e A014824 From _Peter Bala_, Sep 13 2015: (Start)
%e A014824 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)
%e A014824 From _Peter Bala_, Sep 15 2015: (Start)
%e A014824 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)
%e A014824 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
%p A014824 a:=n->sum((10^(n-j)-1^(n-j))/9,j=0..n): seq(a(n), n=0..17); # _Zerinvary Lajos_, Jan 15 2007
%p A014824 a:=n->sum(10^(n-j)*j,j=0..n): seq(a(n), n=0..16); # _Zerinvary Lajos_, Jun 05 2008
%t A014824 Table[Sum[10^i - 1, {i, n}]/9, {n, 18}] (* _Robert G. Wilson v_, Nov 20 2004 *)
%t A014824 CoefficientList[Series[x/(1 - 12*x + 21*x^2 - 10*x^3), {x, 0, 20}], x] (* _Wesley Ivan Hurt_, Sep 15 2015 *)
%o A014824 (Magma) [(10^n-1)*(10/81)-n/9: n in [0..20]]; // _Vincenzo Librandi_, Aug 23 2011
%o A014824 (PARI)
%o A014824 linrec01(p,u,base)={my(r=!p,A=1);for(j=2,u,A=A*base+r+p*j); A};
%o A014824 a(n)=(n!=0)*linrec01(1, n, 10); \\ _R. J. Cano_, Jan 09 2011; With (0, n, 10) it generates repunit numbers.
%o A014824 (PARI) A014824(n)=(10^(n+1)\9-n)\9  \\ _M. F. Hasler_, Jan 17 2013
%o A014824 (Python)
%o A014824 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
%Y A014824 Cf. A060011.
%Y A014824 Cf. A002275. - _Jonathan Vos Post_, Apr 25 2010
%Y A014824 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
%Y A014824 Differs from A007908, A035239, A057137, A060555, A138957 from n=10 on. - _M. F. Hasler_, Jan 17 2013
%Y A014824 Cf. A030512.
%K A014824 nonn,easy
%O A014824 0,3
%A A014824 _N. J. A. Sloane_