A014824 -id:A014824 - cp's OEIS frontend

A277830 Number of digits '0' in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

1, 1, 2, 23, 344, 4665, 58986, 713307, 8367628, 96021949, 1083676272, 12071330614, 133058985146, 1454046641578, 15775034317010, 170096022182442, 1824417011947874, 19478738020713306, 207133059219478738, 2194787382318244170, 23182441724417009624, 244170096256515775267

Offset: 0

M. F. Hasler, Nov 01 2016

The first 10 terms are given by a simple explicit formula and linear recurrence, which does not hold for n > 9. Note that A007908 (concat(1..n)) differs from A014824 (a(n) = a(n-1)*10 + n) for n > 9. - M. F. Hasler, Nov 07 2020

M. F. Hasler, Table of n, a(n) for n = 0..100

Cf. A277831 - A277838, A277849, A277635, A272525, A083449, A014824.

print1(c=1);N=0;for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==0,digits(k))))) \\ For purpose of illustration.

apply( A277830(n)={A061217(A014824(n)+!n)+1}, [0..22]) \\ Thanks to Kevin Ryde's formula. - M. F. Hasler, Nov 07 2020

a(n) = A083449(n) + 1 for n <= 9.

a(n) = 1 + A061217(A014824(n)), taking A061217(0)=0. - Kevin Ryde, Nov 07 2020

Incorrect data, b-file, links, formulas and programs deleted by M. F. Hasler, following observations by Kevin Ryde, Nov 07 2020

A277849 Number of digits '9' in the set of all numbers from 0 to A014824(n) = sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Original entry on oeis.org

0, 0, 1, 22, 343, 4664, 58985, 713306, 8367627, 96021949, 1083676281, 12071330713, 133058986145, 1454046651577, 15775034417009, 170096023182441, 1824417021947873, 19478738120713305, 207133060219478737, 2194787392318244180, 23182441824417009723

Offset: 0

M. F. Hasler, Nov 01 2016

			For n = 2 there is only one digit '9' in the sequence 0, 1, 2, ..., 12.
For n = 3 there are 11 + 10 = 21 more digits '9' in { 19, 29, ..., 89, 90, ..., 99, 109, 119 }, where 99 accounts for two '9's.

Lars Blomberg, Table of n, a(n) for n = 0..997

Cf. A277830 - A277838, A277635, A272525, A083449, A014824.

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==9,digits(k)))))

A014824(n)=(10^n-1)*(10/81)-n/9;
A102684(n)=my(pow,f,g,h);sum(j=1,#Str(n),pow=10^j;f=floor(n/pow);g=floor(n/pow+1/10);h=(4/5+g)*pow;g*(2*n+2-h)-f*(2*n+2-(1+f)*pow))/2;
A277849(n)=A102684(A014824(n));
vector(50,n,A277849(n-1)) \\ Lars Blomberg, Nov 11 2020

a(n) = A083449(n) = A277830(n) - 1 for 0 < n < 9.
a(n) = A277838(n) for n < 8, and a(8) = A277838(8) - 1.
More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

Replaced incorrect b-file by Lars Blomberg, Nov 11 2020

A277838 Number of '8' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Original entry on oeis.org

0, 0, 1, 22, 343, 4664, 58985, 713306, 8367628, 96021959, 1083676380, 12071331701, 133058996022, 1454046750343, 15775035404664, 170096033058985, 1824417120713306, 19478739108367627, 207133070096021958, 2194787491083676380, 23182442812071331701

Offset: 0

M. F. Hasler, Nov 01 2016

			For n=2 there is only one digit '8' in the sequence 0, 1, 2, ..., 12.
For n=3 there are 11 + 10 = 21 more digits '8' in { 18, 28, ..., 78, 80, ..., 89, 98, 108, 118 }, where 88 accounts for two '8's.

David A. Corneth, Table of n, a(n) for n = 0..998

Cf. A277830 - A277837, A277849, A277635, A272525, A083449, A014824.

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==8,digits(k)))))

A277838(n,m=8)=if(n>m,A277838(n,m+1)+(m+2)*10^(n-m-1),A277830(n)-(m>n)) \\ M. F. Hasler, Nov 02 2016

a(n) = A277849(n) = A083449(n) = A277830(n) - 1 for n < 8, a(8) = A277849(8) + 1 = A277837(8) - 9.

More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

Removed incorrect b-file. - David A. Corneth, Dec 31 2020

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

A060011 Schizophrenic sequence: these are the repeating digits in the decimal expansion of sqrt(f(2n+1)), where f(m) = A014824(m).

Original entry on oeis.org

1, 5, 6, 2, 4, 9, 6, 3, 9, 2, 1, 3, 7, 5, 9, 9, 9, 9, 6, 3, 9, 3, 6, 9, 9, 9, 9, 2, 1, 3, 4, 8, 9, 3, 6, 9, 7, 8, 6, 2, 4, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 6, 3, 9, 3, 6, 9, 9, 9, 9, 3, 6, 9, 6, 3, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 2, 1, 3, 4, 8, 9

Offset: 0

Jason Earls, Mar 15 2001

The repeating strings that form the sequence 1, 5, 6, 2, 4, 9, 6, 3, 9, ... become progressively smaller and the irregular strings increase, until eventually the repeating strings disappear. With larger odd values of n however, the demise of the repeating digits slows down.
From Peter Bala, Sep 27 2015: (Start)
Conjecture: same as the repeating digits in the decimal expansion of 1/9*sqrt(1 - 1/10^n).
As n increases, the decimal expansion of 1/9*sqrt(1 - 1/10^n) begins with long strings of repeating digits of 1's, 5's, 6's, 2's,..., which appear to be taken from an initial subsequence of the present sequence, interlaced with the digit strings [0, 41, 597, 178819, 140624, 77213541, 487630208, 1878662109374, 87877739800347, 1191830105251736, 02212270100911458, ...]. An example is given below. Empirical observations: for a fixed value of n, the lengths of the repeating strings gradually shorten until they eventually disappear; as n increases, the number of repeating strings of digits increases. (End)
Conjecture: same as the digital root of the trisection of the Catalan numbers: a(n) = A130856(3*n). - Christian Krause, Nov 26 2022

			From _Peter Bala_, Sep 27 2015: (Start)
Decimal expansion of 1/9*sqrt(1 - 1/10^20) with repeating strings of digits shown in parentheses for clarity:
0.(111...111)0(555...555)41(666...666)597(222...222)178819(444...444)140624(999...999)77213541(666...666)487630208(333...333)1878662109374(999...999)87877739800347(222222)1191830105251736(1111)02212270100911458(333)2....
Repeating digits 1, 5, 6, 2, 4, 9, 6, 3, 9, 2, 1, 3. (End)

J. Earls, Mathematical Bliss, Pleroma Publications, 2009, pages 29-36. ASIN: B002ACVZ6O [From Jason Earls, Nov 22 2009]
C. A. Pickover, Wonders of Numbers, Oxford University Press, NY, 2001. p. 210-211.

K. S. Brown, Mock-rational numbers.
Sean A. Irvine, Java program (github)
C. A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Zentralblatt review

Cf. A014824.

sqrt(f(n)) where f(n) = 10 * f(n-1) + n, for odd integers n. 1, 5, 6, 2, 4, 9, 6, 3, 9, 2, ... are the repeating digits that alternate with random looking strings.

Corrected by Martin Renner, Apr 15 2007

More terms from Jinyuan Wang, Oct 11 2020

A277837 Number of '7' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Original entry on oeis.org

0, 0, 1, 22, 343, 4664, 58985, 713307, 8367637, 96022049, 1083677281, 12071340713, 133059086145, 1454047651577, 15775044417009, 170096123182441, 1824418021947873, 19478748120713314, 207133160219478837, 2194788392318245180, 23182451824417019723

Offset: 0

M. F. Hasler, Nov 01 2016

This sequence is very similar (actually equal, for 1 <= n <= 9) to A277635, which was the original motivation for considering the family A277830 - A277838 and A277849. The main difference is that A277635 is based on A007908 (where 123456789 is followed by 12345678910) while this family is based on A014824, starts as the latter at offset 0, and therefore has a strongly different growth for n > 9.

			For n=2 there is only one digit '7' in the sequence 0, 1, 2, ..., 12.
For n=3 there are 11 + 10 = 21 more digits '7' in { 17, 27, ..., 67, 70, ..., 79, 87, 97, 107, 117 }, where 77 accounts for two '7's.

David A. Corneth, Table of n, a(n) for n = 0..998

Cf. A277830 - A277838, A277849, A277635, A272525, A083449, A014824.

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==7,digits(k)))))

A277837(n,m=7)=if(n>16,error("n>16 not yet implemented"), n>m,A277837(n,m+1)+(m+2)*10^(n-m-1),(9*n-11)*(10^n+1)\729+2-(m>n)) \\ Edited by M. F. Hasler, Dec 29 2020

a(n) = A277849(n) = A083449(n) = A277830(n) - 1 for n < 7,
a(n) = A277836(n) - 8*10^(n-7) [for n >= 7] = A277838(n) + 9*10^(n-8) [for n >= 8].
More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

Removed incorrect b-file. - David A. Corneth, Dec 31 2020

A068995 Integer parts of the square roots of the schizophrenic numbers (A014824).

Original entry on oeis.org

1, 3, 11, 35, 111, 351, 1111, 3513, 11111, 35136, 111111, 351364, 1111111, 3513641, 11111111, 35136418, 111111111, 351364184, 1111111111, 3513641844, 11111111111, 35136418446, 111111111111, 351364184463, 1111111111111

Offset: 1

Joseph L. Pe, Mar 14 2002

a(n) appears to result from (alternately) intermeshing two subsequences, one of the form 11, 111, 1111, ..., the other of the form 35, 351, 3513, .... In both subsequences, the current term is an initial segment of the next term. If the first k (k an even number) terms are deleted from a(n), a(n) can be reconstructed from the resulting sequence by deleting appropriate digits from the end of terms. In this sense, a(n) is self-similar.

			123 is the third schizophrenic number; its square root has integer part 11.

Cf. A014824.

h[n_ /; n == 0] := 0; h[n_ /; n > 0] := 10*h[n - 1] + n; t = Table[Floor[Sqrt[h[i]]], {i, 1, 40}]

From Christopher Hohl, Jun 27 2019: (Start)
a(2n-1) = A014824(n) - A014824(n-1), for n>=1;
a(2n-2) = floor(a(2n-1) / sqrt(10)), for n>=2. (End)

A277831 Number of '1' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Original entry on oeis.org

0, 1, 5, 57, 689, 8121, 93553, 1058985, 11824417, 130589849, 1429355281, 15528120716, 167626886179, 1799725651922, 19231824420465, 204663923217008, 2170096022293551, 22935528124170094, 241700960254046637, 2540466392663923180, 26639231827873799724

Offset: 0

M. F. Hasler, Nov 01 2016

			For n=2 are counted the same '1' as for n=1, plus the 4 additional digits '1' in 10, 11 and 12.

David A. Corneth, Table of n, a(n) for n = 0..998
M. F. Hasler, "Digits d in 0 through 123...n", OEIS Wiki, Nov. 2016.

Cf. A277830 - A277838, A277849, A277635, A272525, A083449, A014824.

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==1,digits(k)))))

A277831(n)=if(n<2,n, n<11, A277832(n)+3*10^(n-2), error("n > 10 not yet implemented")) \\ M. F. Hasler, Nov 02 2016, edited Dec 28 2020

a(n) = A277832(n) + 3*10^(n-2), for 2 <= n <= 10.

More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

Removed incorrect b-file. - David A. Corneth, Dec 31 2020

A277832 Number of '2' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Original entry on oeis.org

0, 0, 2, 27, 389, 5121, 63553, 758985, 8824417, 100589849, 1129355281, 12528120713, 137626886149, 1499725651622, 16231824417465, 174663923187008, 1870096021993551, 19935528121170094, 211700960224046637, 2240466392363923180, 23639231824873799723

Offset: 0

M. F. Hasler, Nov 01 2016

			For n=2 are counted the two '2's in { 2, 12 }.

David A. Corneth, Table of n, a(n) for n = 0..998

Cf. A102685, A277830 - A277838, A277849, A277635, A272525, A083449, A014824.

Array[Total@ DigitCount[Range[Sum[10^i - 1, {i, #}]/9], 10, 2] &, 7] (* Michael De Vlieger, Dec 31 2020 *)

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==2,digits(k)))))

A277832(n)=if(n<3,(n==2)*2, n<13,A277833(n)+4*10^(n-3), error("n > 12 not yet implemented")) \\ M. F. Hasler, Nov 02 2016, edited Dec 28 2020

a(n) = {if(n == 0, return(0)); n = (10^(n+1)\9-n)\9; f(n, 2) }
f(n, {c = 2}) = { my(d = digits(n), res = 0); for(i = 1, #d - 1, res += d[i] * (#d - i)*10^(#d - i - 1); if(d[i]==c, res+=(n % (10^(#d - i)) + 1); ); if(d[i] > c, res+=(10^(#d - i)) ); ); if(d[#d] >= c, res++); res } \\ David A. Corneth, Dec 31 2020

a(n) = A277831(n) - 3*10^(n-2) [for n >= 2] = A277833(n) + 4*10^(n-3) [n >= 3].

More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

Removed incorrect b-file. - David A. Corneth, Dec 31 2020

A277833 Number of '3' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Original entry on oeis.org

0, 0, 1, 23, 349, 4721, 59553, 718985, 8424417, 96589849, 1089355281, 12128120713, 133626886145, 1459725651582, 15831824417065, 170663923183008, 1830096021953551, 19535528120770094, 207700960220046637, 2200466392323923180, 23239231824473799723

Offset: 0

M. F. Hasler, Nov 01 2016

			For n=2 there is only one digit '3' in the sequence 0, 1, 2, *3*, 4, ..., 12.
For n=3 there are 12 + 10 = 22 more digits '3' in { 13, 23, 30, ..., 39, 43, 53, ..., 123 }, where 33 accounts for two '3's.

David A. Corneth, Table of n, a(n) for n = 0..998

Cf. A277830 - A277838, A277849, A277635, A272525, A083449, A014824.

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==3,digits(k)))))

A277833(n,m=3)=if(n>12, error("not yet implemented"), n>m, A277833(n,m+1)+(m+2)*10^(n-m-1), (9*n-11)*(10^n+1)\729+2-(m>n)) \\ M. F. Hasler, Nov 02 2016, edited Dec 29 2020

a(n) = A277832(n) - 4*10^(n-3) [for n >= 3] = A277834(n) + 5*10^(n-4) [for n >= 4].

More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

Removed incorrect b-file. - David A. Corneth, Dec 31 2020

cp's OEIS Frontend

A277830 Number of digits '0' in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

PARI

Formula

Extensions

A277849 Number of digits '9' in the set of all numbers from 0 to A014824(n) = sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

Extensions

A277838 Number of '8' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

Extensions

A052262 Partial sums of A014824.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A060011 Schizophrenic sequence: these are the repeating digits in the decimal expansion of sqrt(f(2n+1)), where f(m) = A014824(m).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Formula

Extensions

A277837 Number of '7' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

Extensions

A068995 Integer parts of the square roots of the schizophrenic numbers (A014824).

Views

Author

Keywords

Comments