A133384 -id:A133384 - cp's OEIS frontend

A199682 a(n) = 2*10^n + 1.

3, 21, 201, 2001, 20001, 200001, 2000001, 20000001, 200000001, 2000000001, 20000000001, 200000000001, 2000000000001, 20000000000001, 200000000000001, 2000000000000001, 20000000000000001, 200000000000000001, 2000000000000000001

Offset: 0

Vincenzo Librandi, Nov 09 2011

Numbers k such that (R(k) - 1)/(k + 1) = 1/2, where R(k) denotes the digit reversal of k (cf. A004086). - Stefano Spezia, Nov 25 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A004086, A011557, A133384.

a199682 = (+ 1) . (* 2) . (10 ^)  -- Reinhard Zumkeller, Jan 30 2015

Magma
```
[2*10^n+1: n in [0..30]];
```

NestList[10#-9&,3,20] (* or *) LinearRecurrence[{11,-10},{3,21},20] (* Harvey P. Dale, Sep 30 2017 *)

a(n)=2*10^n+1 \\ Charles R Greathouse IV, Oct 16 2015

a(n) = 10*a(n-1)-9.
a(n) = 11*a(n-1)-10*a(n-2).
G.f.: 3*(1-4*x)/((1-x)*(1-10*x)).
E.g.f.: 2*exp(10*x) + exp(x). - Stefano Spezia, Nov 25 2023

A349194 a(n) is the product of the sum of the first i digits of n, as i goes from 1 to the total number of digits of n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 49, 56, 63, 70, 77

Offset: 1

Malo David, Nov 10 2021

The only primes in the sequence are 2, 3, 5 and 7. - Bernard Schott, Nov 23 2021

			For n=256, a(256) = 2*(2+5)*(2+5+6) = 182.

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Malo David, About the Product Sum Digit Sequence

Cf. A055642, A284001 (binary analog), A349190 (fixed points).
Cf. A007953 (sum of digits), A059995 (floor(n/10)).
Cf. A349278 (similar, with the last digits).
Cf. A349898, A349899.

f:=func; [f(n):n in [1..100]]; // Marius A. Burtea, Nov 23 2021

Table[Product[Sum[Part[IntegerDigits[n],j],{j,i}],{i,Length[IntegerDigits[n]]}],{n,74}] (* Stefano Spezia, Nov 10 2021 *)

a(n) = my(d=digits(n)); prod(i=1, #d, sum(j=1, i, d[j])); \\ Michel Marcus, Nov 10 2021

first(n)=if(n<9,return([1..n])); my(v=vector(n)); for(i=1,9,v[i]=i); for(i=10,n, v[i]=sumdigits(i)*v[i\10]); v \\ Charles R Greathouse IV, Dec 04 2021

from math import prod
from itertools import accumulate
def a(n): return prod(accumulate(map(int, str(n))))
print([a(n) for n in range(1, 100)]) # Michael S. Branicky, Nov 10 2021

For n>10: a(n) = a(A059995(n))*A007953(n) where A059995(n) = floor(n/10).
In particular, for n<100: a(n) = floor(n/10)*A007953(n)
From Bernard Schott, Nov 23 2021: (Start)
a(n) = 1 iff n = 10^k, k >= 0 (A011557).
a(n) = 2 iff n = 10^k + 1, k >= 0 (A000533 \ {1}).
a(n) = 3 iff n = 10^k + 2, k >= 0 (A133384).
a(n) = 5 iff n = 10^k + 4, k >= 0.
a(n) = 7 iff n = 10^k + 6, k >= 0. (End)
From Marius A. Burtea, Nov 23 2021: (Start)
a(A002275(n)) = n! = A000142(n), n >= 1.
a(A090843(n - 1)) = (2*n - 1)!! = A001147(n), n >= 1.
a(A097166(n)) = (3*n - 2)!!! = A007559(n).
a(A093136(n)) = 2^n = A000079(n).
a(A093138(n)) = 3^n = A000244(n). (End)

A254338 Initial digits of A254143 in decimal representation.

Original entry on oeis.org

1, 4, 7, 1, 2, 3, 3, 4, 6, 1, 1, 2, 2, 2, 3, 3, 3, 4, 6, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2

Offset: 1

Reinhard Zumkeller, Feb 27 2015

a(n) = A000030(A254143(n));
also initial digits of A254323: a(n) = A000030(A254323(n)).
all terms are of the form u*v mod 10, where u <= v and belonging to {1,3,4,6,7}, the distinct elements of A254397:
length of k-th run of consecutive 1s = A005993(k-2), k > 1;
length of k-th run of consecutive 2s = k*(k+1)/2 = A000217(k), k >= 1;
length of k-th run of consecutive 3s = k+1, k >= 1;
length of k-th run of consecutive 4s = A065033(k-1);
n with a(n) = 4: A237424(n) = (10^a+10^b+1)/3 with b = 0, see also A093137, A133384;
n with a(n) = 6: A237424(n) = (10^a+10^b+1)/3 with a = b; A005994(a(n)) = 6 for n > 1; see also A199682;

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A254143, A254323, A000030, A254339, A005993, A000217, A065033.

Haskell
```
a254338 = a000030 . a254143
```

listA237424(lim)=my(v=List(),a,t); while(1, for(b=0,a, t=(10^a+10^b+1)/3; if(t>lim, return(Set(v))); listput(v, t)); a++)
do(lim)=my(v=List(),u=listA237424(lim),t); for(i=1,#u, for(j=1,i, t=u[i]*u[j]; if(t>lim,break); listput(v,t))); apply(n->digits(n)[1], Set(v)) \\ Charles R Greathouse IV, May 13 2015

A126109 a(n) = (5*10^n + 1)/3.

Original entry on oeis.org

2, 17, 167, 1667, 16667, 166667, 1666667, 16666667, 166666667, 1666666667, 16666666667, 166666666667, 1666666666667, 16666666666667, 166666666666667, 1666666666666667, 16666666666666667, 166666666666666667, 1666666666666666667, 16666666666666666667, 166666666666666666667

Offset: 0

Alex P. Lamoreux (Chamale_ic(AT)hotmail.com), Mar 05 2007

a(n+1)*(10^n)*(10^n+1) is the sum of squares of the positive even numbers not exceeding 10^n. - Graeme McRae, Aug 22 2021

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A011557.

One sixth of A133384.

[(5*10^n+1)/3: n in [0..30]]; // Vincenzo Librandi, Nov 09 2011

A126109:=n->(5*10^n+1)/3: seq(A126109(n), n=0..30); # Wesley Ivan Hurt, Jan 24 2017

NestList[10#-3&,2,20] (* or *) LinearRecurrence[{11,-10},{2,17},20] (* Harvey P. Dale, Jan 14 2012 *)

From Vincenzo Librandi, Nov 09 2011: (Start)
a(n) = 10*a(n-1) - 3.
a(n) = 11*a(n-1) - 10*a(n-2).
G.f.: (2-5*x)/((1-x)*(1-10*x)). (End)
E.g.f.: exp(x)*(1 + 5*exp(9*x))/3. - Stefano Spezia, Nov 30 2023

Edited by Don Reble, Mar 09 2007

A133472 a(n) = 10^n + 5.

Original entry on oeis.org

6, 15, 105, 1005, 10005, 100005, 1000005, 10000005, 100000005, 1000000005, 10000000005, 100000000005, 1000000000005, 10000000000005, 100000000000005, 1000000000000005, 10000000000000005, 100000000000000005, 1000000000000000005, 10000000000000000005, 100000000000000000005

Offset: 0

Paul Curtz, Nov 29 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A068128, A133384, A133473.

[10^n+5: n in [0..20]]; // Vincenzo Librandi, Aug 10 2011

Join[{6},Table[10*FromDigits[PadRight[{1},n,0]]+5,{n,20}]] (* or *) LinearRecurrence[{11,-10},{6,15},20] (* Harvey P. Dale, Sep 06 2020 *)

apply( {A133472(n)=10^n+5}, [0..20]) \\ M. F. Hasler, Jun 15 2025

From R. J. Mathar, Nov 30 2007: (Start)
O.g.f.: -3*(-2+17*x)/((-1+x)*(-1+10*x)) = -5/(-1+x) - 1/(-1+10*x).
a(n) = 10*a(n-1) - 45. (End)
a(n) = 10^n + 5. - Vincenzo Librandi, Aug 10 2011
From Elmo R. Oliveira, Jun 09 2025: (Start)
E.g.f.: exp(x)*(5 + exp(9*x)).
a(n) = 11*a(n-1) - 10*a(n-2).
a(n) = 3*A133473(n). (End)

Renamed by editors, Jun 16 2025

A356349 Primitive Niven numbers: terms of A005349 that are not ten times another term of A005349.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 18, 21, 24, 27, 36, 42, 45, 48, 54, 63, 72, 81, 84, 102, 108, 110, 111, 112, 114, 117, 126, 132, 133, 135, 140, 144, 150, 152, 153, 156, 162, 171, 190, 192, 195, 198, 201, 204, 207, 209, 216, 220, 222, 224, 225, 228, 230, 234

Offset: 1

Bernard Schott and Rémy Sigrist, Oct 15 2022

A005349(k) belongs to this sequence iff A113315(k) is not a multiple of 10.
This sequence is infinite as it contains A133384 and A199682.
Each Niven number can be uniquely written as a(m)*10^z for some m > 0 and z >= 0.
This sequence contains numbers with k trailing zeros for any k >= 0; for example R(2^k) * 10^k (where R = A002275).

			190 is a term as 190 is a Niven number and 19 is not a Niven number.
192 is a term as 192 is a Niven number and 192 is not divisible by 10.

Index entries for sequences related to decimal expansion of n

Cf. A002275, A005349, A113315, A133384, A199682.

is(n, base=10) = my (s=sumdigits(n, base)); n%s==0 && (n%base || (n/base)%s)

def ok(n):
    sd = sum(map(int, str(n)))
    return sd and not n%sd and (n%10 or (n//10)%sd)
print([k for k in range(235) if ok(k)]) # Michael S. Branicky, Oct 16 2022

A254398 Final digits of A237424 in decimal representation.

Original entry on oeis.org

1, 4, 7, 4, 7, 7, 4, 7, 7, 7, 4, 7, 7, 7, 7, 4, 7, 7, 7, 7, 7, 4, 7, 7, 7, 7, 7, 7, 4, 7, 7, 7, 7, 7, 7, 7, 4, 7, 7, 7, 7, 7, 7, 7, 7, 4, 7, 7, 7, 7, 7, 7, 7, 7, 7, 4, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 4, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 4, 7, 7, 7, 7, 7, 7, 7

Offset: 1

Reinhard Zumkeller, Feb 23 2015

a(n) = A237424(n) mod 10;
n with a(n) = 4: A237424(n) = (10^a+10^b+1)/3 with b = 0, see also A093137, A133384;
n with a(n) = 7: A237424(n) = (10^a+10^b+1)/3 with 0 < b <= a;
length of k-th run of consecutive 7s = k;
digits 0, 2, 3, 5, 6, 8 and 9 do not occur.

Index entries for sequences related to final digits of numbers

Cf. A237424, A010879, A254339, A093137, A133384.

Haskell
```
a254398 = flip mod 10 . a237424
```

A260702 Numbers n such that 3*n and n^2 have the same digit sum.

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 18, 21, 30, 33, 39, 45, 48, 51, 60, 66, 90, 96, 99, 102, 105, 111, 120, 123, 129, 132, 150, 153, 156, 159, 162, 165, 180, 189, 195, 198, 201, 210, 225, 231, 246, 252, 255, 261, 285, 300, 330, 333, 348, 351, 390, 399, 429, 450, 453, 459, 462

Offset: 1

Vincenzo Librandi, Nov 17 2015

All terms are multiple of 3.

If n is in the sequence, then so is 10*n. - Robert Israel, Apr 05 2020

			159 is in the sequence because 159^2 = 25281 and 3*159 = 477 have the same digit sum: 18.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000290, A007953, A008585, A049343, A058369.

Contains A133384 and A199682.

[n: n in [0..500] | &+Intseq(3*n) eq &+Intseq(n^2)];

select(n -> convert(convert(3*n,base,10),`+`)=convert(convert(n^2,base,10),`+`), [seq(i,i=0..1000,3)]); # Robert Israel, Apr 05 2020

Select[Range[0, 500], Total[IntegerDigits[3 #]] == Total[IntegerDigits[#^2]] &]

isok(n) = sumdigits(3*n) == sumdigits(n^2); \\ Michel Marcus, Nov 17 2015

[n for n in (0..500) if sum((3*n).digits())==sum((n^2).digits())] # Bruno Berselli, Nov 17 2015

A007953(A008585(a(n))) = A007953(A000290(a(n))).

A348589 a(n) = (10^n+2)^2 / 6.

Original entry on oeis.org

24, 1734, 167334, 16673334, 1666733334, 166667333334, 16666673333334, 1666666733333334, 166666667333333334, 16666666673333333334, 1666666666733333333334, 166666666667333333333334, 16666666666673333333333334, 1666666666666733333333333334

Offset: 1

Bernard Schott, Oct 24 2021

Numbers q.r such that q.r = 3*q*r, when q and r have the same number of digits, "." means concatenation, r = 2q and r may not begin with 0.
We must solve the Diophantine equation q.r = q*10^m+r = 3 * q*r where m = length(q) = length(r).
The number of solutions is infinite with (r, q) = ((10^n+2)/3, (10^n+2)/6) and n >= 1.
Note that 15 satisfies also q.r = 3*q*r, 15 = 3*1*5 with here r = 5*q.
For further information about the general equation q.r = k * q*r, see A347541.
Problem proposed on the French website Diophante (see link).

			a(1) = 12^2 / 6 = 24 and 2.4 = 3 * 2*4.
a(2) = 102^2 / 6 = 1734 and 17.34 = 3 * 17*34.

Diophante, A1945 - Concaténations en tous genres (in French).
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A102807, A133384, A147553.

Subsequence of A347541.

Maple
```
seq((10^n+2)^2 / 6, n=1..14);
```

Table[(10^n + 2)^2/6, {n, 1, 14}] (* Amiram Eldar, Oct 24 2021 *)
LinearRecurrence[{111,-1110,1000},{24,1734,167334},20] (* Harvey P. Dale, Sep 05 2025 *)

def a(n): return (10**n+2)**2//6
print([a(n) for n in range(1, 15)]) # Michael S. Branicky, Oct 24 2021

a(n) = (10^n+2)^2 / 6.
a(n) = A133384(n-1)^2/6.
G.f.: 6*x*(4-155*x+250*x^2)/((1-x)*(1-10*x)*(1-100*x)). - Stefano Spezia, Oct 25 2021
a(n) = 3*A102807(n)/2. - Hugo Pfoertner, Oct 30 2021

A350382 a(n) = 9 + 4 * 10^n.

Original entry on oeis.org

49, 409, 4009, 40009, 400009, 4000009, 40000009, 400000009, 4000000009, 40000000009, 400000000009, 4000000000009, 40000000000009, 400000000000009, 4000000000000009, 40000000000000009, 400000000000000009, 4000000000000000009, 40000000000000000009, 400000000000000000009, 4000000000000000000009

Offset: 1

Bernard Schott, Dec 28 2021

The 4th problem of 16th Tournament of Towns in 1994-1995, Spring tour 1995, 8-9 grades, Training option, asked for a proof that the number 400...009 with at least one zero is not a perfect square (see link).
Indeed, the first few squares whose digits are 0, 4 and 9 are 4900, 9409, 490000, 940900, 994009, ... (comes from A019544).
Generalization: the 4th problem of 16th Tournament of Towns in 1994-1995, Spring tour 1995, 10-11 grades, Training option, asked for a proof that the number d00...009 with at least one zero is not a perfect square, when d is a digit with 1 <= d <= 9 (see link).

			a(3) = 9 + 4 * 10^3 = 4009 = 19 * 211 is not a square.

Steve Dinh, The Hard Mathematical Olympiad Problems And Their Solutions, AuthorHouse, 2011, Problem 1 (in fact, it is Problem 4) of Tournament of Towns 1995, page 301.

Tournament of Towns 1994-1995, Spring tour Problem 4, 8-9 grades, Training option & Problem 4, 10-11 grades, Training option (in Russian and English, problems in red).
Index entries for linear recurrences with constant coefficients, signature (11,-10).
Index to sequences related to Olympiads and other Mathematical competitions.

Cf. A019544, A126109, A133384, A199684.

Maple
```
Data := [seq(9 + 4*10^n,  n = 1..20)];
```

a[n_] := 9 + 4*10^n; Array[a, 20] (* Amiram Eldar, Dec 28 2021 *)

a(n) = 9 + 4*10^n = 4*A133384(n-1) + 1.
a(n) = 24*A126109(n-1) + 1 = 10*A199684(n-1) - 1. - Hugo Pfoertner, Dec 28 2021
From Stefano Spezia, Dec 28 2021: (Start)
G.f.: x*(49 - 130*x)/((1 - x)*(1 - 10*x)).
a(n) = 11*a(n-1) - 10*a(n-2) for n > 2. (End)

cp's OEIS Frontend

A199682 a(n) = 2*10^n + 1.

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A349194 a(n) is the product of the sum of the first i digits of n, as i goes from 1 to the total number of digits of n.

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A254338 Initial digits of A254143 in decimal representation.

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A126109 a(n) = (5*10^n + 1)/3.

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A133472 a(n) = 10^n + 5.

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A356349 Primitive Niven numbers: terms of A005349 that are not ten times another term of A005349.

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A254398 Final digits of A237424 in decimal representation.

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