cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 50 results. Next

A075412 Squares of A002277.

Original entry on oeis.org

0, 9, 1089, 110889, 11108889, 1111088889, 111110888889, 11111108888889, 1111111088888889, 111111110888888889, 11111111108888888889, 1111111111088888888889, 111111111110888888888889, 11111111111108888888888889, 1111111111111088888888888889, 111111111111110888888888888889
Offset: 0

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Author

Michael Taylor (michael.taylor(AT)vf.vodafone.co.uk), Sep 14 2002

Keywords

Comments

A transformation of the Wonderful Demlo numbers (A002477).

Examples

			a(2) = 33^2 = 1089.
Contribution from _Reinhard Zumkeller_, May 31 2010: (Start)
n=1: ...................... 9 = 9 * 1;
n=2: ................... 1089 = 99 * 11;
n=3: ................. 110889 = 999 * 111;
n=4: ............... 11108889 = 9999 * 1111;
n=5: ............. 1111088889 = 99999 * 11111;
n=6: ........... 111110888889 = 999999 * 111111;
n=7: ......... 11111108888889 = 9999999 * 1111111;
n=8: ....... 1111111088888889 = 99999999 * 11111111;
n=9: ..... 111111110888888889 = 999999999 * 111111111. (End)
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{11, -10}, {0, 3}, 20]^2 (* Vincenzo Librandi, Mar 20 2014 *)
    Table[FromDigits[PadRight[{},n,9]]FromDigits[PadRight[{},n,1]],{n,0,15}] (* Harvey P. Dale, Feb 12 2023 *)

Formula

a(n) = A002277(n)^2 = (3*A002275(n))^2 = 9*A002275(n)^2.
a(n) = {111111... (2n times)} - 2*{ 111... (n times)} a(n) = A000042(2*n) - 2*A000042(n). - Amarnath Murthy, Jul 21 2003
a(n) = {333... (n times)}^2 = {111...(n times)}{000... (n times)} - {111... (n times)}. For example, 333^2 = 111000 - 111 = 110889. - Kyle D. Balliet, Mar 07 2009
From Reinhard Zumkeller, May 31 2010: (Start)
a(n) = A002283(n)*A002275(n).
For n>0, a(n) = (A002275(n-1)*10^n + A002282(n-1))*10 + 9. (End)
a(n) = (10^(n+1)-10)^2/900. - José de Jesús Camacho Medina, Apr 01 2016
From Elmo R. Oliveira, Jul 27 2025: (Start)
G.f.: 9*x*(1+10*x)/((1-x)*(1-10*x)*(1-100*x)).
E.g.f.: exp(x)*(1 - 2*exp(9*x) + exp(99*x))/9.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3).
a(n) = 9*A002477(n). (End)

A095967 Numbers n such that r3(k) * 2^n + 1 is prime, where r3() = A002277 and k is the number of decimal digits of 2^n.

Original entry on oeis.org

1, 2, 6, 9, 20, 46, 58, 64, 69, 110, 158, 178, 186, 268, 424, 624, 641, 1236, 1593, 2264, 2870, 5797, 7518, 7688, 9300
Offset: 1

Views

Author

Jason Earls, Jul 15 2004

Keywords

Comments

a(1) through a(25) have been proved to be prime with WinPFGW. a(25) has 5600 digits. No more terms up to 12800.
Results were computed using the PrimeFormGW (PFGW) primality-testing program. - Hugo Pfoertner, Nov 14 2019

Examples

			a(4)=9 because 333 * 2^9 + 1 = 170497, a prime.
		

A099657 a(n) is the least prime following A002277(n) repdigits.

Original entry on oeis.org

2, 5, 37, 337, 3343, 33343, 333337, 3333373, 33333347, 333333349, 3333333403, 33333333343, 333333333367, 3333333333347, 33333333333437, 333333333333389, 3333333333333343, 33333333333333391, 333333333333333391
Offset: 0

Views

Author

Labos Elemer, Nov 17 2004

Keywords

Examples

			n=3: 33 is followed by 37.
		

Crossrefs

Programs

  • Mathematica
    Table[NextPrime[3*(10^n-1)/9], {n, 0, 35}]

A099670 Partial sums of repdigits of A002277.

Original entry on oeis.org

3, 36, 369, 3702, 37035, 370368, 3703701, 37037034, 370370367, 3703703700, 37037037033, 370370370366, 3703703703699, 37037037037032, 370370370370365, 3703703703703698, 37037037037037031, 370370370370370364, 3703703703703703697, 37037037037037037030, 370370370370370370363
Offset: 1

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Author

Labos Elemer, Nov 17 2004

Keywords

Examples

			3 + 33 + 333 + 3333 = a(4) = 3702.
		

Crossrefs

Programs

  • Maple
    a:=n->sum((10^(n-j)-1^(n-j))/3,j=0..n): seq(a(n), n=1..18); # Zerinvary Lajos, Jan 15 2007
  • Mathematica
    <Robert G. Wilson v, Nov 20 2004 *)

Formula

a(n) = (3/81)*(10^(n+1) - 9*n - 10). - R. Piyo (nagoya314(AT)yahoo.com), Dec 10 2004
From Elmo R. Oliveira, Apr 02 2025: (Start)
G.f.: 3*x/((1 - x)^2*(1 - 10*x)).
E.g.f.: 3*exp(x)*(10*exp(9*x) - 9*x - 10)/81.
a(n) = 3*A014824(n).
a(n) = 12*a(n-1) - 21*a(n-2) + 10*a(n-3) for n > 3. (End)

Extensions

More terms from Elmo R. Oliveira, Apr 02 2025

A246881 a(n)=A246880(n) where no k exists such that A246880(n)=A002277(k)*(2*10^(A081677(k-1))+3).

Original entry on oeis.org

609, 6660999, 666666609999999, 6666666660999999999, 66666666666666666099999999999999999, 666666666666666666666666609999999999999999999999999, 6666666666666666666666666660999999999999999999999999999
Offset: 1

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Author

Felix Fröhlich, Sep 06 2014

Keywords

Comments

For any n there exists a k such that A246880(n)=A002277(k-1)*A173041(k).

Crossrefs

Subsequence of A246880.

A002275 Repunits: (10^n - 1)/9. Often denoted by R_n.

Original entry on oeis.org

0, 1, 11, 111, 1111, 11111, 111111, 1111111, 11111111, 111111111, 1111111111, 11111111111, 111111111111, 1111111111111, 11111111111111, 111111111111111, 1111111111111111, 11111111111111111, 111111111111111111, 1111111111111111111, 11111111111111111111
Offset: 0

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Author

Keywords

Comments

R_n is a string of n 1's.
Base-4 representation of Jacobsthal bisection sequence A002450. E.g., a(4)= 1111 because A002450(4)= 85 (in base 10) = 64 + 16 + 4 + 1 = 1*(4^3) + 1*(4^2) + 1*(4^1) + 1. - Paul Barry, Mar 12 2004
Except for the first two terms, these numbers cannot be perfect squares, because x^2 != 11 (mod 100). - Zak Seidov, Dec 05 2008
For n >= 0: a(n) = (A000225(n) written in base 2). - Jaroslav Krizek, Jul 27 2009, edited by M. F. Hasler, Jul 03 2020
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=10, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010
Except 0, 1 and 11, all these integers are Brazilian numbers, A125134. - Bernard Schott, Dec 24 2012
Numbers n such that 11...111 = R_n = (10^n - 1)/9 is prime are in A004023. - Bernard Schott, Dec 24 2012
The terms 0 and 1 are the only squares in this sequence, as a(n) == 3 (mod 4) for n>=2. - Nehul Yadav, Sep 26 2013
For n>=2 the multiplicative order of 10 modulo the a(n) is n. - Robert G. Wilson v, Aug 20 2014
The above is a special case of the statement that the order of z modulo (z^n-1)/(z-1) is n, here for z=10. - Joerg Arndt, Aug 21 2014
From Peter Bala, Sep 20 2015: (Start)
Let d be a divisor of a(n). Let m*d be any multiple of d. Split the decimal expansion of m*d into 2 blocks of contiguous digits a and b, so we have m*d = 10^k*a + b for some k, where 0 <= k < number of decimal digits of m*d. Then d divides a^n - (-b)^n (see McGough). For example, 271 divides a(5) and we find 2^5 + 71^5 = 11*73*271*8291 and 27^5 + 1^5 = 2^2*7*31*61*271 are both divisible by 271. Similarly, 4*271 = 1084 and 10^5 + 84^5 = 2^5*31*47*271*331 while 108^5 + 4^5 = 2^12*7*31*61*271 are again both divisible by 271. (End)
Starting with the second term this sequence is the binary representation of the n-th iteration of the Rule 220 and 252 elementary cellular automaton starting with a single ON (black) cell. - Robert Price, Feb 21 2016
If p > 5 is a prime, then p divides a(p-1). - Thomas Ordowski, Apr 10 2016
0, 1 and 11 are only terms that are of the form x^2 + y^2 + z^2 where x, y, z are integers. In other words, a(n) is a member of A004215 for all n > 2. - Altug Alkan, May 08 2016
Except for the initial terms, the binary representation of the x-axis, from the left edge to the origin, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 737", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. - Robert Price, Mar 17 2017
The term "repunit" was coined by Albert H. Beiler in 1964. - Amiram Eldar, Nov 13 2020
q-integers for q = 10. - John Keith, Apr 12 2021
Binomial transform of A001019 with leading zero. - Jules Beauchamp, Jan 04 2022

References

  • Albert H. Beiler, Recreations in the Theory of Numbers: The Queen of Mathematics Entertains, New York: Dover Publications, 1964, chapter XI, p. 83.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 235-237.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 197-198.
  • Samuel Yates, Peculiar Properties of Repunits, J. Recr. Math. 2, 139-146, 1969.
  • Samuel Yates, Prime Divisors of Repunits, J. Recr. Math. 8, 33-38, 1975.

Crossrefs

Programs

  • Haskell
    a002275 = (`div` 9) . subtract 1 . (10 ^)
    a002275_list = iterate ((+ 1) . (* 10)) 0
    -- Reinhard Zumkeller, Jul 05 2013, Feb 05 2012
    
  • Magma
    [(10^n-1)/9: n in [0..25]]; // Vincenzo Librandi, Nov 06 2014
    
  • Maple
    seq((10^k - 1)/9, k=0..30); # Wesley Ivan Hurt, Sep 28 2013
  • Mathematica
    Table[(10^n - 1)/9, {n, 0, 19}] (* Alonso del Arte, Nov 15 2011 *)
    Join[{0},Table[FromDigits[PadRight[{},n,1]],{n,20}]] (* Harvey P. Dale, Mar 04 2012 *)
  • Maxima
    a[0]:0$
    a[1]:1$
    a[n]:=11*a[n-1]-10*a[n-2]$
    A002275(n):=a[n]$
    makelist(A002275(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
    
  • PARI
    a(n)=(10^n-1)/9; \\ Michael B. Porter, Oct 26 2009
    
  • PARI
    my(x='x+O('x^30)); concat(0, Vec(x/((1-10*x)*(1-x)))) \\ Altug Alkan, Apr 10 2016
    
  • Python
    print([(10**n-1)//9 for n in range(100)]) # Michael S. Branicky, Apr 30 2022
  • Sage
    [lucas_number1(n, 11, 10) for n in range(21)]  # Zerinvary Lajos, Apr 27 2009
    

Formula

a(n) = 10*a(n-1) + 1, a(0)=0.
a(n) = A000042(n) for n >= 1.
Second binomial transform of Jacobsthal trisection A001045(3n)/3 (A015565). - Paul Barry, Mar 24 2004
G.f.: x/((1-10*x)*(1-x)). Regarded as base b numbers, g.f. x/((1-b*x)*(1-x)). - Franklin T. Adams-Watters, Jun 15 2006
a(n) = 11*a(n-1) - 10*a(n-2), a(0)=0, a(1)=1. - Lekraj Beedassy, Jun 07 2006
a(n) = A125118(n,9) for n>8. - Reinhard Zumkeller, Nov 21 2006
a(n) = A075412(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
a(n) = a(n-1) + 10^(n-1) with a(0)=0. - Vincenzo Librandi, Jul 22 2010
a(n) = A242614(n,A242622(n)). - Reinhard Zumkeller, Jul 17 2014
E.g.f.: (exp(9*x) - 1)*exp(x)/9. - Ilya Gutkovskiy, May 11 2016
a(n) = Sum_{k=0..n-1} 10^k. - Torlach Rush, Nov 03 2020
Sum_{n>=1} 1/a(n) = A065444. - Amiram Eldar, Nov 13 2020
From Elmo R. Oliveira, Aug 02 2025: (Start)
a(n) = A002283(n)/9 = A105279(n)/10.
a(n) = A010785(A017173(n-1)) for n >= 1. (End)

A002283 a(n) = 10^n - 1.

Original entry on oeis.org

0, 9, 99, 999, 9999, 99999, 999999, 9999999, 99999999, 999999999, 9999999999, 99999999999, 999999999999, 9999999999999, 99999999999999, 999999999999999, 9999999999999999, 99999999999999999, 999999999999999999, 9999999999999999999, 99999999999999999999, 999999999999999999999, 9999999999999999999999
Offset: 0

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Author

Keywords

Comments

A friend from Germany remarks that the sequence 9, 99, 999, 9999, 99999, 999999, ... might be called the grumpy German sequence: nein!, nein! nein!, nein! nein! nein!, ...
The Regan link shows that integers of the form 10^n -1 have binary representations with exactly n trailing 1 bits. Also those integers have quinary expressions with exactly n trailing 4's. For example, 10^4 -1 = (304444)5. The first digits in quinary correspond to the number 2^n -1, in our example (30)5 = 2^4 -1. A similar pattern occurs in the binary case. Consider 9 = (1001)2. - Washington Bomfim Dec 23 2010
a(n) is the number of positive integers with less than n+1 digits. - Bui Quang Tuan, Mar 09 2015
From Peter Bala, Sep 27 2015: (Start)
For n >= 1, the simple continued fraction expansion of sqrt(a(2*n)) = [10^n - 1; 1, 2*(10^n - 1), 1, 2*(10^n - 1), ...] has period 2. The simple continued fraction expansion of sqrt(a(2*n))/a(n) = [1; 10^n - 1, 2, 10^n - 1, 2, ...] also has period 2. Note the occurrence of large partial quotients in both expansions.
A theorem of Kuzmin in the measure theory of continued fractions says that large partial quotients are the exception in continued fraction expansions.
Empirically, we also see the presence of unexpectedly large partial quotients early in the continued fraction expansions of the m-th roots of the numbers a(m*n) for m >= 3. Some typical examples are given below. (End)
For n > 0, numbers whose smallest decimal digit is 9. - Stefano Spezia, Nov 16 2023

Examples

			From _Peter Bala_, Sep 27 2015: (Start)
Continued fraction expansions showing large partial quotients:
a(12)^(1/3) = [9999; 1, 299999998, 1, 9998, 1, 449999998, 1, 7998, 1, 535714284, 1, 2, 2, 142, 2, 2, 1, 599999999, 3, 1, 1,...].
Compare with a(30)^(1/3) = [9999999999; 1, 299999999999999999998, 1, 9999999998, 1, 449999999999999999998, 1, 7999999998, 1, 535714285714285714284, 1, 2, 2, 142857142, 2, 2, 1, 599999999999999999999, 3, 1, 1,...].
a(24)^(1/4) = [999999; 1, 3999999999999999998, 1, 666665, 1, 1, 1, 799999999999999999, 3, 476190, 7, 190476190476190476, 21, 43289, 1, 229, 1, 1864801864801863, 1, 4, 6,...].
Compare with a(48)^(1/4) = [999999999999; 1, 3999999999999999999999999999999999998, 1, 666666666665, 1, 1, 1, 799999999999999999999999999999999999, 3, 476190476190, 7, 190476190476190476190476190476190476, 21, 43290043289, 1, 229, 1, 1864801864801864801864801864801863, 1, 4, 6,...].
a(25)^(1/5) = [99999, 1, 499999999999999999998, 1, 49998, 1, 999999999999999999998, 1, 33332, 3, 151515151515151515151, 5, 1, 1, 1947, 1, 1, 38, 3787878787878787878, 1, 3, 5,...].
(End)
		

Crossrefs

Programs

Formula

From Mohammad K. Azarian, Jan 14 2009: (Start)
G.f.: 1/(1-10*x)-1/(1-x).
E.g.f.: e^(10*x)-e^x. (End)
a(n) = A075412(n)/A002275(n) = A178630(n)/A002276(n) = A178631(n)/A002277(n) = A075415(n)/A002278(n) = A178632(n)/A002279(n) = A178633(n)/A002280(n) = A178634(n)/A002281(n) = A178635(n)/A002282(n). - Reinhard Zumkeller, May 31 2010
a(n) = a(n-1) + 9*10^(n-1) with a(0)=0; Also: a(n) = 11*a(n-1) - 10*a(n-2) with a(0)=0, a(1)=9. - Vincenzo Librandi, Jul 22 2010
For n>0, A007953(a(n)) = A008591(n) and A010888(a(n)) = 9. - Reinhard Zumkeller, Aug 06 2010
A048379(a(n)) = 0. - Reinhard Zumkeller, Feb 21 2014
a(n) = Sum_{k=1..n} 9*10^k. - Carauleanu Marc, Sep 03 2016
Sum_{n>=1} 1/a(n) = A073668. - Amiram Eldar, Nov 13 2020
From Elmo R. Oliveira, Jul 19 2025: (Start)
a(n) = 9*A002275(n).
a(n) = A010785(A008591(n)). (End)

Extensions

More terms from Michael De Vlieger, Sep 27 2015

A002276 a(n) = 2*(10^n - 1)/9.

Original entry on oeis.org

0, 2, 22, 222, 2222, 22222, 222222, 2222222, 22222222, 222222222, 2222222222, 22222222222, 222222222222, 2222222222222, 22222222222222, 222222222222222, 2222222222222222, 22222222222222222, 222222222222222222, 2222222222222222222, 22222222222222222222, 222222222222222222222
Offset: 0

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Keywords

Comments

a(n) is also the total number of holes in a variation of a box fractal as in illustration. - Kival Ngaokrajang, May 23 2014 [As observed by Hans Havermann, this seems to be incorrect: e.g., for n = 2 the illustration shows 28 small holes plus two larger holes. - M. F. Hasler, Oct 05 2020]

Crossrefs

Programs

Formula

a(n) = A178630(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
From Vincenzo Librandi, Jul 22 2010: (Start)
a(n) = a(n-1) + 2*10^(n-1) with a(0) = 0.
a(n) = 11*a(n-1) - 10*a(n-2) with a(0) = 0, a(1) = 2. (End)
G.f.: 2*x/((1 - x)*(1 - 10*x)). - Ilya Gutkovskiy, Feb 24 2017
E.g.f.: 2*exp(x)*(exp(9*x) - 1)/9. - Stefano Spezia, Sep 13 2023
From Elmo R. Oliveira, Jul 19 2025: (Start)
a(n) = 2*A002275(n).
a(n) = A010785(A017185(n-1)) for n >= 1. (End)

A002282 a(n) = 8*(10^n - 1)/9.

Original entry on oeis.org

0, 8, 88, 888, 8888, 88888, 888888, 8888888, 88888888, 888888888, 8888888888, 88888888888, 888888888888, 8888888888888, 88888888888888, 888888888888888, 8888888888888888, 88888888888888888, 888888888888888888, 8888888888888888888, 88888888888888888888, 888888888888888888888
Offset: 0

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Author

Keywords

Comments

If the initial term is omitted, might be called eightful (or hateful) numbers!

Examples

			Curious multiplications:
9*9 + 7 = 88;
98*9 + 6 = 888;
987*9 + 5 = 8888;
9876*9 + 4 = 88888;
98765*9 + 3 = 888888;
987654*9 + 2 = 8888888;
9876543*9 + 1 = 88888888;
98765432*9 + 0 = 888888888;
987654321*9 - 1 = 8888888888;
9876543210*9 - 2 = 88888888888. - _Philippe Deléham_, Mar 09 2014
		

References

  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 32.

Crossrefs

Programs

  • Maple
    A002282:=n->8*(10^n - 1)/9; seq(A002282(n), n=0..20); # Wesley Ivan Hurt, Mar 10 2014
  • Mathematica
    LinearRecurrence[{11,-10}, {0,8}, 20] (* Harvey P. Dale, May 30 2013 *)
  • PARI
    { a=-4/5; for (n = 0, 200, a+=8*10^(n - 1); write("b002282.txt", n, " ", a); ) } \\ Harry J. Smith, Jun 27 2009
    
  • Python
    def a(n): return 8*(10**n - 1)//9 # Martin Gergov, Oct 19 2022

Formula

From Jaume Oliver Lafont, Feb 03 2009: (Start)
a(n) = 11*a(n-1) - 10*a(n-2), with a(0)=0, a(1)=8.
G.f.: 8*x/((1-x)*(1-10*x)). (End)
a(n) = A178635(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
a(n) = a(n-1) + 8*10^(n-1), with a(0)=0. - Vincenzo Librandi, Jul 22 2010
a(n) = 8*A002275(n) = A002283(n) - A002275(n). - Carauleanu Marc, Sep 03 2016
From Ilya Gutkovskiy, Sep 03 2016: (Start)
E.g.f.: 8*(exp(9*x) - 1)*exp(x)/9.
a(n) = floor(8*10^n/9). (End)
From Elmo R. Oliveira, Jul 20 2025: (Start)
a(n) = (A246058(n) - 1)/2.
a(n) = A010785(A017257(n-1)) for n >= 1. (End)

A002279 a(n) = 5*(10^n - 1)/9.

Original entry on oeis.org

0, 5, 55, 555, 5555, 55555, 555555, 5555555, 55555555, 555555555, 5555555555, 55555555555, 555555555555, 5555555555555, 55555555555555, 555555555555555, 5555555555555555, 55555555555555555, 555555555555555555, 5555555555555555555, 55555555555555555555, 555555555555555555555
Offset: 0

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Author

Keywords

Comments

Arithmetic mean of all n-digit odd numbers. E.g., a(2) = arithmetic mean of {11,13,15,...,97,99} = (11+99)/2 = 55. - Amarnath Murthy, Aug 02 2005

Crossrefs

Programs

Formula

a(n) = A178632(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
From Vincenzo Librandi, Jul 22 2010: (Start)
a(n) = a(n-1) + 5*10^(n-1) with a(0)=0.
a(n) = 11*a(n-1) - 10*a(n-2) with a(0)=0, a(1)=5. (End)
G.f.: 5*x/((1 - x)*(1 - 10*x)). - Ilya Gutkovskiy, Feb 24 2017
E.g.f.: 5*exp(x)*(exp(9*x) - 1)/9. - Stefano Spezia, Sep 13 2023
From Karl-Heinz Hofmann, Nov 28 2023: (Start)
a(n) = A010785(9*n-4) for n > 0.
a(n) = 5 * A002275(n).
a(n) = 5 * A002283(n) / 9. (End)
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