A060011 -id:A060011 - cp's OEIS frontend

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

N. J. A. Sloane

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)

			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

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).

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.

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

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

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 *)

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.

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

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

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

A052262 Partial sums of A014824.

Original entry on oeis.org

0, 1, 13, 136, 1370, 13715, 137171, 1371738, 13717416, 137174205, 1371742105, 13717421116, 137174211238, 1371742112471, 13717421124815, 137174211248270, 1371742112482836, 13717421124828513, 137174211248285301, 1371742112482853200, 13717421124828532210

Offset: 0

Barry E. Williams, Feb 03 2000

Comment from Peter Bala, Sep 18 2015: (Start)
This sequence may be viewed as a generalization of A014824 and appears to share similar properties. See also A262183.
It follows from the explicit formula for a(n) given below that the decimal expansion of 1/a(6*n+4) for n >= 1 begins with long strings of 0's interlaced successively with the digits of the numbers 3^6*(1458*n^2 + 2727*n + 1270)^k for k = 0,1,2,.... The strings of 0's gradually shorten in length so this pattern eventually breaks down. For example, for n = 3 we have 1/a(22) = 0.000...000729000...00016455717000...000371454899841000...00083848514541108930001... with 729 = 3^6, 16455717 = (3^6)*22573, 371454899841 = (3^6)*(22573^2) and 8384851454110893 = (3^6)*(22573^3). Similar results can be found for the decimal expansions of the reciprocal of a(6*n+r) for r = 0,1,2,3 and 5.
These results should be helpful in explaining the following empirical observations concerning the decimal expansions of certain roots of the sequence terms and of their reciprocals. The decimal expansion of a(6*n+4)^(1/6) begins with strings of repeated digits (giving the appearance of rationality) alternating with strings of apparently random digits. The strings of repeating digits gradually shorten in length until they disappear from the expansion and the pattern breaks down. Brown calls numbers with these properties schizophrenic numbers or mock-rational numbers. The powers (a(6*n+4)^1/6 )^k, k = 2,3,... may also be examples of schizophrenic numbers. The decimal expansions of the numbers a(6*n+4)^(2/3) are particularly interesting as they seem to begin with repeating strings of the digits 123456790. See example (1) below.
The decimal expansion of 1/a(3*n+1)^(1/3) for n >= 5 starts with 4 long strings of 0's interlaced with 3 blocks of digits. If we read these blocks of digits as ordinary integers and factorize them, we find the numbers are related in a surprising manner. See example (2) below. Next in the expansion we find blocks of apparently random digits interlaced with strings of repeated digits. As n increases the number of strings of repeating digits increases. The repeating digits appear to always be an initial subsequence of [5, 851, 975308641, ....] independent of the value of n. Cf. A014824, A060011 and A262183.
The decimal expansions of the numbers 1/a(6*n+4)^(1/6) also show the mock-irrational behavior described by Brown.
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 (10*a(6*n))^(1/3), (a(6*n+1))^(1/3), (100*a(6*n+2))^(1/3), (10*a(6*n+3))^(1/6), (a(6*n+4))^(1/6), (100*a(6*n+5))^(1/3), and their powers.
(End)

			From _Peter Bala_, Sep 18 2015: (Start)
(1) Repeating digits in the decimal expansion of a(64)^(2/3) = 1.23456790 123456790 123456790 123456790 123456790 123456790 123456775774... * 10^42.
(2) 1/a(121)^(1/3) = 9.000 ... 0001826553000 ... 000741399080402000 ... 000(351 ... 301)555 ... 555(733 ... 556)851851851 ... 851851851(945 ... 936)308641975308641975308641975 ... 308641975308641975308641975(30864202 ...).
The three blocks of digits [9, 1826553, 741399080402] at the start of the decimal expansion shown above factorize as [3^2, 3*608851, 3*608851^2] showing they are related. These 3 blocks of digits are interlaced with long strings of zeros. There then follows 4 blocks of apparently random digits (enclosed in parentheses above for clarity) interlaced with 3 blocks of repeating digits. The repeating digits are 5, 851 and 308641975.
(End)

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

Robert Israel, Table of n, a(n) for n = 0..990
K. S. Brown, Mock-rational numbers.
Wikipedia, Schizophrenic Number
Index entries for linear recurrences with constant coefficients, signature (13,-33,31,-10)

Cf. A014824, A002275, A011557, A060011, A262183.

A052262 := proc(n)
    100*(10^n-1)/729 - 10*n/81 - binomial(n+1, 2)/9 ;
end proc:
seq(A052262(n),n=0..10) ; # R. J. Mathar, Oct 02 2015

Table[(1/1458) (2 10^(n + 2) - 81 n^2 - 261 n - 200), {n, 0, 30}] (* Vincenzo Librandi, Sep 20 2015 *)

a(n) = (100*((10^n)-1)/729) - (10*n/81) - binomial(n+1, 2)/9;
vector(30, n, a(n-1)) \\ Altug Alkan, Oct 02 2015

a(n) = 100*(10^n-1)/729 - 10*n/81 - binomial(n+1, 2)/9.
a(n) = 10*a(n-1) + binomial(n+1,2).
G.f. : x/((1-10x)(1-x)^3); a(n) = Sum_{k=0..n-1} binomial(n+1, k+2)9^k. - Paul Barry, Aug 24 2004

A262183 a(0) = 0, a(n) = 10a(n-1) + n(n+1)*(n+2)/6.

Original entry on oeis.org

0, 1, 14, 150, 1520, 15235, 152406, 1524144, 15241560, 152415765, 1524157870, 15241578986, 152415790224, 1524157902695, 15241579027510, 152415790275780, 1524157902758616, 15241579027587129, 152415790275872430, 1524157902758725630, 15241579027587257840

Offset: 0

Peter Bala, Sep 14 2015

This sequence may be viewed as a generalization of A014824 and appears to share similar properties. See also A052262. We make the following empirical observations.
The decimal expansion of a(4*n+1)^(1/4) begins with strings of repeated digits (giving the appearance of rationality) alternating with strings of apparently random digits. The strings of repeated digits gradually shorten in length until they disappear and this pattern breaks down. See example (1) below. Brown calls numbers with these properties schizophrenic or mock-rational numbers. By increasing n we can prevent the disappearance of the repeating strings as long as we like. The repeating digits appear to always be an initial subsequence of [1, 3, 6, 7, 6, 3, 1, 9, 9, 3, 9, 9, 3, ....]. Cf. A060011.
It appears that the numbers (a(8*n+5))^(1/8) are also examples of Brown's schizophrenic numbers.
The decimal expansion of 1/(a(4*n+1))^(k/4) for k = 1,2,3,... has long strings of 0's (gradually shortening in length until they disappear) interlaced with blocks of digits. If we read these blocks of digits (for a fixed n and k) as ordinary integers and factorize them, we find the numbers are related in a surprising manner. See examples (2) through (5) below. Similar remarks apply to the decimal expansion of the numbers  1/(10*a(8*n))^(1/4), 1/(10*a(8*n+2))^(1/2), 1/(a(8*n+3))^(1/2), 1/(10*a(8*n+4))^(1/8), 1/(a(8*n+5))^(1/8), 1/(10*a(8*n+6))^(1/2), 1/(a(8*n+7))^(1/2) and their powers.
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(8*n+1))^(1/4), (a(8*n+3))^(1/2), (a(8*n+5))^(1/8), (a(8*n+7))^(1/2),
(10*a(8*n))^(1/4), (10*a(8*n+2))^(1/2), (10*a(8*n+4))^(1/8), (10*a(8*n+6))^(1/2) and their powers.

			(1) The decimal expansion of a(61)^(1/4) (with the blocks of 'random' digits enclosed in parentheses to aid readability) begins
1.111...111(026286308)333...333(2361974965884332291)666...666(4936365745813146737399105902)777...777(414516002742700195101894168058610026041)666...666(5834699239217156417791785081497321498627522786458)333..333(1... * 10^15.
The repeating digits are 1, 3, 6, 7, 6 and 3, an initial subsequence of A060011.
(2) The decimal expansion of 1/a(61)^(1/4) (with now the strings of 0's enclosed in parentheses) begins
9.(000..000)6870809025(000...000)131133379605615140625(000...000)300330802691003816294298046875(000...000)74515840736091563874877683318366943359375(000...000)193416219724333545001418899430083738351541748046875(000...000)5... * 10^(-16)
The long strings of 0's gradually shorten in length until they disappear and are interlaced with 5 strings of digits [6870809025, 131133379605615140625, 300330802691003816294298046875, 74515840736091563874877683318366943359375, 193416219724333545001418899430083738351541748046875]. Reading these strings as ordinary integers and factorizing we obtain [ (3^2)*(5^2)*30536929, (3^2)*(5^6)*(30536929)^2, (3^3)*(5^8)*(30536929)^3, (3^3)*(5^12)*13*(30536929)^4, (3^3)*(5^13)*13*17*(30536929)^3 ] showing how the numbers are related.
(3) The decimal expansion of 1/a(61)^(2/4) begins
8.1(000...000)12367456245(000...000)28324809994812870375(000...000)7207939264584091591063153125(000...000)192594788364052042015068473807471484375(000...000)52931280402387750233872466233174047490955859375(000...000)1... * 10^(-31).
The long strings of 0's gradually shorten in length and are interlaced with 5 strings of digits
[12367456245, 28324809994812870375, 7207939264584091591063153125, 192594788364052042015068473807471484375, 52931280402387750233872466233174047490955859375].
Reading these strings as ordinary integers and factorizing we obtain [ (3^4)*5*30536929, (3^5)*(5^3)*(30536929)^2, (3^4)*(5^5)*(30536929)^3, (3^4)*(5^8)*7*(30536929)^4, (3^6)*(5^8)*7*(30536929)^3 ].
(4) The decimal expansion of 1/a(61)^(3/4) begins
7.29(000...000)1669606593075(000...000)44611575741830270840625(000...000)124877547758919386815169127890625(000...000)35750407590077160299047085450511894287109375(000...000)1... * 10^(-46).
The long strings of 0's gradually shorten in length and are interlaced with 4 strings of digits [1669606593075, 44611575741830270840625, 124877547758919386815169127890625, 35750407590077160299047085450511894287109375]. Reading these strings as ordinary integers and factorizing we obtain [ (3^7)*(5^2)*30536929, (3^7)*(5^5)*7*(30536929)^2, (3^6)*(5^7)*7*11*(30536929)^3, (3^7)*(5^12)*7*11*(30536929)^4 ].
(5) The decimal expansion of 1/a(61) begins
6.561(000...000)200352791169(000...000)6118158958879580001(000...000)186829785738019654040356929(000...000)5705207902167118776034942675531041(000...000)174219528638716252198345946001761436313089(000...000)5320129376453944844526984070493622855630820563681(000...000)1... * 10^(-61).
The long strings of 0's gradually shorten in length and are interlaced with 6 strings of digits [200352791169, 6118158958879580001,  186829785738019654040356929, 5705207902167118776034942675531041, 174219528638716252198345946001761436313089, 5320129376453944844526984070493622855630820563681]. Reading these strings as ordinary integers and factorizing we obtain [ (3^8)*30536929, (3^8)*(30536929)^2, (3^8)*(30536929)^3, (3^8)*(30536929)^4, (3^8)*(30536929)^5, (3^8)*(30536929)^6 ].

Colin Barker, Table of n, a(n) for n = 0..1000
K. S. Brown, Mock-rational numbers.
Wikipedia, Schizophrenic Number
Index entries for linear recurrences with constant coefficients, signature (14,-46,64,-41,10).

Cf. A060011, A014824, A030512, A052262.

[0] cat [n eq 1 select 1  else 10*Self(n-1) + n*(n+1)*(n+2)/6: n in [1..30]]; // Vincenzo Librandi, Sep 20 2015

#A262183
seq((1/13122)*(2*10^(n+3)-243*n^3-1539*n^2-3096*n-2000), n = 0..22);

Table[(1/9^4) 10^(n + 3) - (243 n^3 + 1539 n^2 + 3096 n + 2000)/13122, {n, 0, 30}] (* Vincenzo Librandi, Sep 20 2015 *)
nxt[{n_,a_}]:={n+1,10a+((n+1)(n+2)(n+3))/6}; NestList[nxt,{0,0},20][[All,2]] (* or *) LinearRecurrence[ {14,-46,64,-41,10},{0,1,14,150,1520},30] (* Harvey P. Dale, Feb 29 2020 *)

concat(0, Vec(-x/((x-1)^4*(10*x-1)) + O(x^40))) \\ Colin Barker, Sep 20 2015

a(n) = (1/9^4)*10^(n+3) - (243*n^3 + 1539*n^2 + 3096*n + 2000)/13122.
O.g.f. x/((1 - 10*x)*(1 - x)^4).
a(n) = 14*a(n-1)-46*a(n-2)+64*a(n-3)-41*a(n-4)+10*a(n-5) for n>4. - Colin Barker, Sep 20 2015

cp's OEIS Frontend

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

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A052262 Partial sums of A014824.

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A262183 a(0) = 0, a(n) = 10*a(n-1) + n*(n+1)*(n+2)/6.

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A262183 a(0) = 0, a(n) = 10a(n-1) + n(n+1)*(n+2)/6.