A093137 -id:A093137 - cp's OEIS frontend

A098210 a(n) = -1 + A093137(n)^2.

0, 15, 1155, 111555, 11115555, 1111155555, 111111555555, 11111115555555, 1111111155555555, 111111111555555555, 11111111115555555555, 1111111111155555555555, 111111111111555555555555

Offset: 0

Labos Elemer, Oct 20 2004

			Like other -1 + square of near-repdigits give special digit patterns: (33333334^2) - 1 = 11111115555555.

Vincenzo Librandi, Table of n, a(n) for n = 0..100

Cf. A093137.

[(1/9)*(10^n-1)*(5+10^n): n in [0..15]]; // Vincenzo Librandi, Jun 02 2013

Table[(1/9) (10^n - 1) (5 + 10^n), {n, 0, 25}] (* Vincenzo Librandi, Jun 02 2013 *)

a(n) = (1/9)*(10^n-1)*(5+10^n). - Vincenzo Librandi, Jun 02 2013

A067275 Number of Fibonacci numbers A000045(k), k <= 10^n, which end in 4.

Original entry on oeis.org

0, 1, 7, 67, 667, 6667, 66667, 666667, 6666667, 66666667, 666666667, 6666666667, 66666666667, 666666666667, 6666666666667, 66666666666667, 666666666666667, 6666666666666667, 66666666666666667, 666666666666666667, 6666666666666666667, 66666666666666666667, 666666666666666666667

Offset: 0

Joseph L. Pe, Feb 21 2002

Numbers n such that the digits of A000326(n), the n-th pentagonal number, begin with n. [original definition of this sequence]
The sequence 1,7,67.... has a(n) = 6*10^n/9+3/9. It is the second binomial transform of 6*A001045(3n)/3 + (-1)^n. In general the second binomial transform of k*Jacobsthal(3n)/3+(-1)^n is k*10^n/9 + (1-k/9) = 1, 1+k, 1+11k, 1+111k, ... - Paul Barry, Mar 24 2004
Except for the first two terms, these are the 3-automorphic numbers ending in 7. - Eric M. Schmidt, Aug 28 2012
From Wolfdieter Lang, Feb 08 2017: (Start)
This sequence appears in curious identities based on the Armstrong numbers 370 = A005188(11), 371 = A005188(12) and 407 = A005188(13).
For such identities based on 153 = A005188(10) see a comment in A246057 with the van der Poorten et al. reference.
For 370 these identities are A002277(n)^3 + a(n+1)^3  + 0(n)^3 = A002277(n)*10^(2*n) + a(n+1)*10^n + 0(n) = A281858(n), where 0(n) means n 0's.
For 371 these identities are A002277(n)^3 + a(n+1)^3 + (0(n-1)1)^3 = A002277(n)*10^(2*n) + a(n+1)*10^n + 0(n-1)1 = A281860(n), where 0(n-1)1 means n-1 0's followed by a 1.
For 407 these identities are A093137(n)^3 + 0(n)^3 + a(n+1)^3 = concatenation(A093137(n), 0(n), a(n+1)) = A281859(n). (End)

			a(2) = 7 because 7 of the first 10^2 Fibonacci numbers end in 4.
From _Wolfdieter Lang_, Feb 08 2017: (Start)
Curious cubic identities:
3^3 + 7^3 + 0^3 = 370, 33^3 + 67^3 + (00)^3 = 336700, 333^3 + 667^3 + (000)^3 = 333667000, ...
3^3 + 7^3 + 1^3 = 371, 33^3 + 67^3 + (01)^3 = 336701, 333^3 + 667^3 + (001)^3 = 333667001, ...
4^3 + 0^3 + 7^3 = 407, 34^3 + (00)^3 + 67^3 = 340067 , 334^3 + (000)^3 + 677^3 = 334000677, ... (End)

Seiichi Manyama, Table of n, a(n) for n = 0..1001
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A072702, A093137, A109344, A246057, A281858, A281859, A281860.

Cf. A000045, A000326, A001045, A002277, A005188, A073553, A199682.

s = Fibonacci@ Range[10^5]; Table[Count[Take[s, 10^n], m_ /; Mod[m, 10] == 4], {n, 0, Floor@ Log10@ Length@ s}] (* or *) Table[Boole[n > 0] Ceiling[10^n/15], {n, 0, 20}] (* or *) CoefficientList[Series[x (1 - 4 x)/((1 - x) (1 - 10 x)), {x, 0, 20}], x] (* Michael De Vlieger, Feb 08 2017 *)

a(n)=(10^n+13)\15 \\ Charles R Greathouse IV, Jun 05 2011

a(n) = ceiling((2/30)*10^n) - Benoit Cloitre, Aug 27 2002
From Paul Barry, Mar 24 2004: (Start)
G.f.: x*(1 - 4*x)/((1 - x)*(1 - 10*x)).
a(n) = 10^n/15 + 1/3 for n>0. (End)
a(n) = 10*a(n-1) - 3 for n>1. - Vincenzo Librandi, Dec 07 2010 [Immediate consequence of the previous formula, R. J. Mathar]
From Eric M. Schmidt, Oct 28 2012: (Start)
For n>0, a(n) = A199682(n-1)/3 = (2*10^(n-1) + 1)/3.
For n>=2, a(n+1) = a(n) + 6*10^n. (End)
From Elmo R. Oliveira, Jul 22 2025: (Start)
E.g.f.: (-6 + 5*exp(x) + exp(10*x))/15.
a(n) = 11*a(n-1) - 10*a(n-2) for n >= 3.
a(n) = A073553(n)/2 for n >= 1. (End)

A073552 merged into this sequence by Eric M. Schmidt, Oct 28 2012

A102807 a(n) is the square of one plus the number consisting of n 3's.

Original entry on oeis.org

1, 16, 1156, 111556, 11115556, 1111155556, 111111555556, 11111115555556, 1111111155555556, 111111111555555556, 11111111115555555556, 1111111111155555555556, 111111111111555555555556, 11111111111115555555555556, 1111111111111155555555555556, 111111111111111555555555555556

Offset: 0

Ron Knott, Feb 27 2005

Old name was: The number (333...334)^2.
An infinite sequence of squares with no zeros in base 10.
a(n) = A104265(2n) for n > 0. - Chai Wah Wu, Mar 24 2020

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 31 at p. 61.
Italo Ghersi, Matematica dilettevole e curiosa, pp. 111-112, Hoepli, Milano, 1967. [Vincenzo Librandi, Dec 31 2008]

Seiichi Manyama, Table of n, a(n) for n = 0..500
Emile Fourrey, Récréations arithmétiques, Vuibert, 1899 and after, Paris, pages 72-73.
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A052041, A093137, A075415, A109344, A104265.

a:= n-> (1+parse(cat(0, 3$n)))^2:
seq(a(n), n=0..20);  # Alois P. Heinz, Sep 03 2018

Table[(10^n + 2)^2/9, {n, 0, 20}] (* Paolo Xausa, Jun 26 2024 *)

From R. J. Mathar, Jan 06 2009: (Start)
a(n) = (100^n + 4*10^n + 4)/9.
G.f.: (1 - 95*x + 490*x^2)/((1-x)*(100*x-1)*(10*x-1)). (End)
E.g.f.: exp(x)*(4 + 4*exp(9*x) + exp(99*x))/9. - Stefano Spezia, Jul 31 2024

New name from Alois P. Heinz, Sep 03 2018

A237424 Numbers of the form (10^a + 10^b + 1)/3.

Original entry on oeis.org

1, 4, 7, 34, 37, 67, 334, 337, 367, 667, 3334, 3337, 3367, 3667, 6667, 33334, 33337, 33367, 33667, 36667, 66667, 333334, 333337, 333367, 333667, 336667, 366667, 666667, 3333334, 3333337, 3333367, 3333667, 3336667

Offset: 1

Ahmad J. Masad, Feb 07 2014

nonn

Has the property that the product of any two (not necessarily distinct) terms has digits in nondecreasing order.
Conjecture: This sequence is in a sense the maximally dense sequence with this nondecreasing products property. That is, it appears that every maximal sequence is either (i) A237424, (ii) a finite set of "extra" terms plus A237424 restricted to b=0 (which is A093137), or (iii) a finite set of "extra" terms plus A237424 restricted to a=b (which is A067275). (There might be one more case, not yet identified.) - David Applegate, Feb 12 2014
See A254143 and link for products a(i)*a(j) in natural order. - Reinhard Zumkeller, Jan 28 2015

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 (first 1035 terms from Robert G. Wilson v)
Reinhard Zumkeller, First 10000 products of any two terms of A237424

Cf. A028819, A028820, A028821, A052216, A067275, A093137, A234841, A254143.

a237424 = flip div 3 . (+ 1) . a052216
-- Reinhard Zumkeller, Jan 28 2015

A052216:=[10^(n-1) + 10^(k-1): k in [1..n], n in [1..100]];
A237424:= func< n | (A052216[n]+1)/3 >;
[A237424(n): n in [1..100]]; // G. C. Greubel, Feb 22 2024

Union@ Flatten@ Table[(10^a + 10^b + 1)/3, {a, 0, 8}, {b, a, 8}] (* Robert G. Wilson v, Jan 26 2015 *)
(10^#[[1]]+10^#[[2]]+1)/3&/@Tuples[Range[0,8],2]//Union (* Harvey P. Dale, May 28 2019 *)

list(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++) \\ Charles R Greathouse IV, May 13 2015

from math import isqrt
def A237424(n): return (10**(a:=(k:=isqrt(m:=n<<1))+(m>k*(k+1))-1)+10**(n-1-(a*(a+1)>>1))+1)//3 # Chai Wah Wu, Apr 08 2025

A052216=flatten([[10^(n-1) + 10^(k-1) for k in range(1,n+1)] for n in range(1,101)])
def A237424(n): return (A052216[n-1]+1)//3
[A237424(n) for n in range(1,101)] # G. C. Greubel, Feb 22 2024

a(n) = (A052216(n) + 1)/3. - Reinhard Zumkeller, Jan 28 2015

Edited by David Applegate, Feb 07 2014

A093138 Expansion of (1-7x)/(1-10x).

Original entry on oeis.org

1, 3, 30, 300, 3000, 30000, 300000, 3000000, 30000000, 300000000, 3000000000, 30000000000, 300000000000, 3000000000000, 30000000000000, 300000000000000, 3000000000000000, 30000000000000000, 300000000000000000, 3000000000000000000, 30000000000000000000, 300000000000000000000

Offset: 0

Paul Barry, Mar 24 2004

Partial sums are A093137. A convex combination of 10^n and 0^n.

M. H. Cilasun, An Analytical Approach to Exponent-Restricted Multiple Counting Sequences, arXiv preprint arXiv:1412.3265 [math.NT], 2014.
M. H. Cilasun, Generalized Multiple Counting Jacobsthal Sequences of Fermat Pseudoprimes, Journal of Integer Sequences, Vol. 19, 2016, #16.2.3.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Index entries for linear recurrences with constant coefficients, signature (10).

Cf. A093137.

Vec((1-7*x)/(1-10*x) + O(x^30)) \\ Michel Marcus, Sep 07 2015

a(n) = 3*10^n/10 + 7*0^n/10.
a(n) = 3*10^(n-1) with a(0)=1. - Vincenzo Librandi, Aug 02 2010
E.g.f.: (3*exp(10*x) + 7)/10. - Elmo R. Oliveira, Aug 13 2024

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

A325907 a(n) = ( (-1)^n * Sum_{k=0..n-2} (-1)^k*10^(2^k) + 10^(2^(n-1)) - ((-1)^n+3)/2 )/3.

Original entry on oeis.org

3, 36, 3363, 33336636, 3333333366663363, 33333333333333336666666633336636, 3333333333333333333333333333333366666666666666663333333366663363

Offset: 1

Seiichi Manyama, Sep 08 2019

nonn

All terms are elements of A213517.

			              36 =        -3 - 1 +        4 * 10^1.
            3363 =       -36 - 1 +       34 * 10^2.
        33336636 =     -3363 - 1 +     3334 * 10^4.
3333333366663363 = -33336636 - 1 + 33333334 * 10^8.
------------------------------------------------------
T(n) = n*(n+1)/2.
               T(3) =                               6.
              T(36) =                             666.
            T(3363) =                         5656566.
        T(33336636) =                 555665666566566.
T(3333333366663363) = 5555555666655656666556566566566.

Seiichi Manyama, Table of n, a(n) for n = 1..10

Cf. A000217, A062691, A093137, A119219, A213517, A309597, A325906.

a[n_] := ((-1)^n * Sum[(-1)^k * 10^(2^k), {k, 0, n - 2}] + 10^(2^(n - 1)) - ((-1)^n + 3)/2)/3; Array[a, 7] (* Amiram Eldar, May 07 2021 *)

{a(n) = ((-1)^n*sum(k=0, n-2, (-1)^k*10^2^k)+10^2^(n-1)-((-1)^n+3)/2)/3}

a(n) = 3 * A325906(n).

a(n) = -a(n-1) - 1 + A093137(2^(n-2)) * 10^(2^(n-2)).

A281859 Curious identities based on the Armstrong number 407 = A005188(13).

Original entry on oeis.org

407, 340067, 334000667, 333400006667, 333340000066667, 333334000000666667, 333333400000006666667, 333333340000000066666667, 333333334000000000666666667, 333333333400000000006666666667, 333333333340000000000066666666667, 333333333334000000000000666666666667

Offset: 1

Wolfdieter Lang, Feb 08 2017

See a comment in A093137.

			Curious cubic identities: 407 = 4^3 + 0^3 + 7^3, 340067 = 34^3 + (00)^3 + 67^3, 334000677 = 334^3 + (000)^3 + 677^3, ...

Colin Barker, Table of n, a(n) for n = 1..333
Index entries for linear recurrences with constant coefficients, signature (1111,-112110,1111000,-1000000).

Table[FromDigits@ Join[ReplacePart[ConstantArray[3, n], -1 -> 4], ConstantArray[0, n], ReplacePart[ConstantArray[6, n], -1 -> 7]], {n, 12}] (* Michael De Vlieger, Feb 08 2017 *)
LinearRecurrence[{1111,-112110,1111000,-1000000},{407,340067,334000667,333400006667},20] (* Harvey P. Dale, May 10 2018 *)

Vec(x*(407 - 112110*x + 1815000*x^2 - 2000000*x^3) / ((1 - x)*(1 - 10*x)*(1 - 100*x)*(1 - 1000*x)) + O(x^30)) \\ Colin Barker, Feb 08 2017

From Colin Barker, Feb 08 2017: (Start)
G.f.: x*(407 - 112110*x + 1815000*x^2 - 2000000*x^3) / ((1 - x)*(1 - 10*x)*(1 - 100*x)*(1 - 1000*x)).
a(n) = (1 + 2^(1+n)*5^n + 2^(1+2*n)*25^n + 1000^n) / 3.
a(n) = 1111*a(n-1) - 112110*a(n-2) + 1111000*a(n-3) - 1000000*a(n-4) for n>4. (End)

A034978 a(n)^2 is smallest square starting with a string of n 1's.

Original entry on oeis.org

1, 34, 334, 3334, 10541, 333334, 3333334, 33333334, 333333334, 3333333334, 10540925534, 105409255339, 3333333333334, 10540925533895, 105409255338946, 105409255338946, 10540925533894598, 105409255338945978, 3333333333333333334, 33333333333333333334, 333333333333333333334, 1054092553389459777333

Offset: 1

Patrick De Geest, Nov 15 1998

			a(5)^2 = 10541^2 = {11111}2681.

Robert Israel, Table of n, a(n) for n = 1..996

Cf. A034979, A093137.

f:= proc(n) local x, k, s;
  x:= (10^n-1)/9;
  for k from 0 do
    s:= ceil(sqrt(10^k*x));
    if s^2 < (x+1)*10^k then return s fi
  od
end proc:
map(f, [$1..20]); # Robert Israel, May 31 2023

a(n) <= A093137(n). - Robert Israel, May 31 2023

More terms from Francisco Salinas (franciscodesalinas(AT)hotmail.com), Dec 23 2001

A073555 Number of Fibonacci numbers F(k), k <= 10^n, which end in 8.

Original entry on oeis.org

1, 8, 68, 668, 6668, 66668, 666668, 6666668, 66666668, 666666668, 6666666668, 66666666668, 666666666668, 6666666666668, 66666666666668, 666666666666668, 6666666666666668, 66666666666666668, 666666666666666668, 6666666666666666668, 66666666666666666668, 666666666666666666668, 6666666666666666666668, 66666666666666666666668

Offset: 1

Shyam Sunder Gupta, Aug 15 2002

			a(2) = 8 because there are 8 Fibonacci numbers up to 10^2 which end in 8.

Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A000045, A093137.

If n>1 then a(n) = (10^n + 20)/15. - Robert Gerbicz, Sep 06 2002
From Elmo R. Oliveira, Jul 22 2025: (Start)
G.f.: x*(1 - 3*x - 10*x^2)/((1-x)*(1-10*x)).
E.g.f.: (-21 - 15*x + 20*exp(x) + exp(10*x))/15.
a(n) = 2*A093137(n-1) for n >= 2.
a(n) = 11*a(n-1) - 10*a(n-2) for n > 3. (End)

More terms from Robert Gerbicz, Sep 06 2002

cp's OEIS Frontend

A098210 a(n) = -1 + A093137(n)^2.

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A067275 Number of Fibonacci numbers A000045(k), k <= 10^n, which end in 4.

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A102807 a(n) is the square of one plus the number consisting of n 3's.

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A237424 Numbers of the form (10^a + 10^b + 1)/3.

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A093138 Expansion of (1-7x)/(1-10x).

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A254338 Initial digits of A254143 in decimal representation.

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A325907 a(n) = ( (-1)^n * Sum_{k=0..n-2} (-1)^k*10^(2^k) + 10^(2^(n-1)) - ((-1)^n+3)/2 )/3.

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