A014824 -id:A014824 - cp's OEIS frontend

A277835 Number of '5' 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, ...).

0, 0, 1, 22, 343, 4665, 58993, 713385, 8368417, 96029849, 1083755281, 12072120713, 133066886145, 1454125651577, 15775824417009, 170103923182448, 1824496021947951, 19479528120714094, 207140960219486637, 2194866392318323180, 23183231824417799723

Offset: 0

M. F. Hasler, Nov 01 2016

			For n = 2 there is only one digit '5' in the sequence 0, 1, 2, ..., 12.
For n = 3 there are 11 + 10 = 21 more digits '5' in { 15, 25, ..., 45, 50, ..., 59, 65, ..., 115 }, where 55 accounts for two '5'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==5,digits(k)))))

A277835(n,m=5)=if(n>m,A277835(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 < 5,
a(5) = A277836(5) + 1, a(n) = A277836(n) + 7*10^(n-6) for n >= 6.
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

A277836 Number of '6' 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, 58986, 713315, 8367717, 96022849, 1083685281, 12071420713, 133059886145, 1454055651577, 15775124417009, 170096923182441, 1824426021947881, 19478828120713394, 207133960219479637, 2194796392318253180, 23182531824417099723

Offset: 0

M. F. Hasler, Nov 01 2016

			For n=2 there is only one digit '6' in the sequence 0, 1, 2, ..., 12.
For n=3 there are 11 + 10 = 21 more digits '6' in { 16, 26, ..., 56, 60, ..., 69, 76, 86, ..., 116 }, where 66 accounts for two '6's.

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

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

T[int_Integer, {bndsLow_Integer, bndsUpp_Integer}] := Table[
   Count[
    Flatten[Table[
      IntegerDigits[m],
      {m, 1, Sum[
         10^i - 1,
         {i, n}
         ]/9
       }
      ]],
    int
    ],
   {n, bndsLow, bndsUpp}
   ];
T[6, {0, 7}](* Robert P. P. McKone, Jan 01 2021 *)

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==6,digits(k)))))

A277836(n,m=6)=if(n>m,A277836(n,m+1)+(m+2)*10^(n-m-1),A277830(n)-(m>n)) \\ M. F. Hasler, Nov 02 2016

a(n) = A277839(n) =  A083449(n) = A277830(n) - 1 for n < 6,
a(n) = A277835(n) - 7*10^(n-6) for n >= 6,
a(n) = A277837(n) + 8*10^(n-7) for n >= 7.

More terms from Lars Blomberg, Nov 05 2016

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

A277834 Number of '4' 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, 344, 4671, 59053, 713985, 8374417, 96089849, 1084355281, 12078120713, 133126886145, 1454725651577, 15781824417015, 170163923182508, 1825096021948551, 19485528120720094, 207200960219546637, 2195466392318923180, 23189231824423799723

Offset: 0

M. F. Hasler, Nov 01 2016

			For n=2 there is only one digit '4' in the sequence 0, 1, 2, ..., 12.
For n=3 there are 11 + 10 = 21 more digits '4' in { 14, 24, 34, 40, ..., 49, 54, ..., 114 }, where 44 accounts for two '4'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==4,digits(k)))))

A277834(n,m=4)=if(n>m,A277833(n,m+1)+(m+2)*10^(n-m-1),A277830(n)-(m>n))

a(n) = A277849(n) = A083449(n) = A277830(n) - 1 for n < 4,
a(n) = A277833(n) - 5*10^(n-4) for n >= 4, a(n) = A277835(n) + 6*10^(n-5) for n >= 5.
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

A014822 Numbers k such that k divides s(k), where s(1)=1, s(j)=10*s(j-1)+j (A014824).

Original entry on oeis.org

1, 2, 3, 5, 6, 9, 10, 15, 18, 22, 27, 30, 42, 45, 54, 66, 78, 81, 90, 110, 111, 126, 135, 162, 198, 205, 210, 222, 234, 242, 243, 270, 294, 330, 333, 342, 378, 390, 405, 410, 462, 465, 486, 506, 546, 555, 594, 615, 630, 666, 702, 726, 729, 810, 858, 882, 930

Offset: 1

N. J. A. Sloane

nonn

Robert Israel, Table of n, a(n) for n = 1..1000

s:= proc(n) option remember; 10*procname(n-1)+n end proc: s(1):= 1:
select(t -> s(t) mod t = 0, [$1..10000]); # Robert Israel, Feb 28 2025

Transpose[Select[NestList[{#[[1]]+1,10#[[2]]+#[[1]]+1}&,{1,1},1000],Divisible[ #[[2]],#[[1]]]&]][[1]] (* Harvey P. Dale, Sep 24 2012 *)

A095761 a(n) = A014824(2*n-1).

Original entry on oeis.org

1, 123, 12345, 1234567, 123456789, 12345679011, 1234567901233, 123456790123455, 12345679012345677, 1234567901234567899, 123456790123456790121, 12345679012345679012343

Offset: 1

Michael Joseph Halm, Jul 10 2004

Previous name was: Numbers whose square root shows strings of seemingly rational and irrational strings.

Muniru A Asiru, Table of n, a(n) for n = 1..101
K. S. Brown, Schizophrenic numbers

Cf. A014824.

a:=[1];; for n in [2..20] do a[n]:=100*a[n-1]+22*n-21; od; a; # Muniru A Asiru, Mar 30 2018

[-19/81-(2/9)*(n-1)+(100/81)*100^(n-1): n in [1..20]]; // Vincenzo Librandi, Apr 03 2018

RecurrenceTable[{a[1]==1, a[n]==100 a[n-1] + 22 n - 21}, a, {n, 20}] (* Vincenzo Librandi, Apr 03 2018 *)

a=vector(10^3); a[1]=1; for(n=2, #a, a[n]=100*a[n-1]+22*n-21); a \\ Altug Alkan, Mar 30 2018

a(1) = 1; for n > 1, a(n) = 100*a(n - 1) + 22*n - 21.

O.g.f.: x(1+21x)/((-1+x)^2*(1-100x)) = -17/(81(-1+x)) - 1/(81(-1+100*x)) - 2/(9(-1+x)^2). - R. J. Mathar, Feb 01 2008

Better name from Georg Fischer, Mar 30 2018

A179477 Antonym of A014824: each term is 10 times the previous term minus n.

Original entry on oeis.org

10, 99, 988, 9877, 98766, 987655, 9876544, 98765433, 987654322, 9876543211, 98765432100, 987654320989, 9876543209878, 98765432098767, 987654320987656, 9876543209876545, 98765432098765434, 987654320987654323, 9876543209876543212, 98765432098765432101

Offset: 0

Mark Dols, Jul 16 2010

Colin Barker, Table of n, a(n) for n = 0..998
Index entries for linear recurrences with constant coefficients, signature (12,-21,10).

Cf. A014824.

Join[{a=10,b=99},Table[c=11*b-10*a-1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 20 2011*)
Join[{a=10}, Table[a=10a-n, {n,60}]] (* T. D. Noe, Jan 20 2011 *)

Vec(-(10*x^2-21*x+10)/((x-1)^2*(10*x-1)) + O(x^30)) \\ Colin Barker, Oct 03 2015

a(n) + A014824(n) = 10^(n+1).
From Colin Barker, Oct 03 2015: (Start)
a(n) = 12*a(n-1) - 21*a(n-2) + 10*a(n-3) for n>2.
G.f.: -(10*x^2-21*x+10) / ((x-1)^2*(10*x-1)). (End)
a(n) = (10 + 2^(5+n)*5^(2+n) + 9*n)/81. - Colin Barker, Mar 25 2017

A083449 a(n) = A019566(n)/9, where A019566(n) = concat(n,...,1) - concat(1,...,n).

Original entry on oeis.org

0, 1, 22, 343, 4664, 58985, 713306, 8367627, 96021948, -150891621, -13731137410, -260644605199, 86159119727012, 19839246664059223, 3106259112208391434, 422859356777752723645, 53509280234443297055856, 6473262479112108841388067, 759559693477989774385720278

Offset: 1

Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 01 2003

Are there any palindromes > 58985?

This sequence also gives the number of occurrences of any digit d > n (thus n < 9) in the list of all numbers from 1 to concatenation(1,...,n) = A007908(n) = A014824(n) = sum_{i=1..n} i*10^(n-i). See A277849, A061217, A277830 etc. - M. F. Hasler, Nov 01 2016, edited Nov 07 2020

Cf. A061217.

a:= n-> (parse(cat((n+1-i)$i=1..n))-parse(cat($1..n)))/9:
seq(a(n), n=1..20);  # Alois P. Heinz, Nov 09 2020

Array[(FromDigits@ Apply[Join, Reverse@ #] - FromDigits@ Apply[Join, #])/9 &@ Map[IntegerDigits, Range[#]] &, 19] (* Michael De Vlieger, Nov 12 2020 *)

apply( {A083449(n)=A019566(n)\9}, [1..20]) \\ - M. F. Hasler, Nov 07 2020

For n < 10, a(n) = ceiling((9*n-11)*(10^n+1)/729). - M. F. Hasler, Nov 07 2020

More terms from David Wasserman, Nov 09 2004

A057137 Concatenate next digit at right hand end (where the next digit after 9 is again 0).

Original entry on oeis.org

0, 1, 12, 123, 1234, 12345, 123456, 1234567, 12345678, 123456789, 1234567890, 12345678901, 123456789012, 1234567890123, 12345678901234, 123456789012345, 1234567890123456, 12345678901234567, 123456789012345678, 1234567890123456789, 12345678901234567890, 123456789012345678901

Offset: 0

Henry Bottomley, Aug 12 2000

Also called the triangle of the gods (see Pickover link).

See A037610 for a general formula. - Hieronymus Fischer, Jan 03 2013

Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 61.

T. D. Noe and Hieronymus Fischer, Table of n, a(n) for n = 0..200 (terms up to 100 from T. D. Noe)
Clifford Pickover, Triangle of the Gods
Index entries for linear recurrences with constant coefficients, signature (10,0,0,0,0,0,0,0,0,1,-10).

Alternative progression for n >= 10 compared with A007908 and A014824.
Cf. A057138 for reverse. Cf. A010879 (decimal digits).
For primes see A120819.

A057137:=n->floor((137174210/1111111111)*10^n); seq(A057137(n), n=0..20); # Wesley Ivan Hurt, Apr 18 2014

a[n_]:=Floor[137174210/1111111111*10^n]; Array[a,19,0] (* Robert G. Wilson v, Apr 18 2014 *)

A057137(n)=sum(i=1,n,i%10*10^(n-i)) \\ M. F. Hasler, Jan 13 2013

A057137(n)=137174210*10^n\1111111111 \\ M. F. Hasler, Jan 13 2013

def A057137(n): s = '0123456789'; return int((n+1)//10*s + s[:(n+1)%10]) # Ya-Ping Lu, Apr 08 2025

a(n) = 10*(a(n-1)-floor[n/10]) + n = floor[A057139(n)/10^(n-1)].

a(n) = floor((137174210/1111111111)*10^n). - Hieronymus Fischer, Jan 03 2013, corrected by M. F. Hasler, Jan 13 2013

A057932 a(n) = floor(10^(n+1)/81).

Original entry on oeis.org

0, 1, 12, 123, 1234, 12345, 123456, 1234567, 12345679, 123456790, 1234567901, 12345679012, 123456790123, 1234567901234, 12345679012345, 123456790123456, 1234567901234567, 12345679012345679, 123456790123456790, 1234567901234567901, 12345679012345679012, 123456790123456790123

Offset: 0

Henry Bottomley, Oct 04 2000

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Index entries for linear recurrences with constant coefficients, signature (10,0,0,0,0,0,0,0,1,-10).

Cf. A014824, A057933.

[Floor(10^(n+1)/81): n in [1..20]]; // Vincenzo Librandi, Jul 07 2011

Floor[10^(Range[0, 25] + 1)/81] (* Paolo Xausa, Jan 29 2025 *)

a(n)=10^(n+1)\81 \\ Charles R Greathouse IV, Jul 07 2011

A277635 Number of 7's appearing in the sequence of consecutive natural numbers from 1 to A007908(n), where A007908 = (1, 12, 123, 1234, ...).

Original entry on oeis.org

0, 1, 22, 343, 4664, 58985, 713307, 8367637, 96022049

Offset: 1

Alexander R. Povolotsky, Oct 24 2016

First 6 terms are the same as in A083449, also see A272525. [See the OEIS wiki page for more details. - M. F. Hasler, Dec 29 2020]
a(n) gives the number of times the digit 7 occurs in all terms of A000027 in the interval [A000027(1), A007908(n)]. - Felix Fröhlich, Oct 28 2016
The sequence was initially defined only up to n = 9 and then extended using A007908 = concat(1..n); see A277837 for the extension using A014824 (a(n) = 10 a(n-1) + n) leading to a smoother growth, in particular at powers of 10. - M. F. Hasler, Nov 01 2016, edited Dec 29 2020

			22 is the third term of the sequence because there are 22 occurrences of the digit '7' contained in numbers within the range of 1 to 123.
96022049 is the 9th term of the sequence because there are 96022049 occurrences of the digit '7' contained in numbers within the range of 1 to 123456789.

Puzzling Stack Exchange, How many sevens?
M. F. Hasler, Digits d in 0 through 123...n, OEIS Wiki, Nov. 2016.

Cf. A083449, A272525, A014824.

Cf. A277830 - A277838 and A277849: analog for digits 0 .. 9, but based on A014824 instead of A083449.

Table[a[n] = Count[Flatten@ Map[IntegerDigits, Range@ FromDigits@ Range@ n], k_ /; k == 8]; Print@ a@ n; an = a[n]; an, {n, 0, 9}] (* Michael De Vlieger, Oct 30 2016 *)

print1(c=0);N=1;for(n=2,8,print1(","c+=sum(k=N+1,N=eval(Str(N,n)),#select(d->d==7,digits(k))))) \\ For illustration; more efficient code below. - M. F. Hasler, Oct 31 2016

A277635(n, m=7)=if(n>m,A277635(n, m+1)+(m+2)*10^(n-m-1),A277830(n)-(m>n)) \\ Valid only for n <= 9. - M. F. Hasler, Nov 02 2016

cp's OEIS Frontend

A277835 Number of '5' 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, ...).

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A277836 Number of '6' 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, ...).

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A277834 Number of '4' 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, ...).

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A014822 Numbers k such that k divides s(k), where s(1)=1, s(j)=10*s(j-1)+j (A014824).

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A095761 a(n) = A014824(2*n-1).

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A179477 Antonym of A014824: each term is 10 times the previous term minus n.

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A083449 a(n) = A019566(n)/9, where A019566(n) = concat(n,...,1) - concat(1,...,n).

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