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.

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A081139 9th binomial transform of (0,0,1,0,0,0,...).

Original entry on oeis.org

0, 0, 1, 27, 486, 7290, 98415, 1240029, 14880348, 172186884, 1937102445, 21308126895, 230127770466, 2447722649502, 25701087819771, 266895911974545, 2745215094595320, 28001193964872264, 283512088894331673
Offset: 0

Views

Author

Paul Barry, Mar 08 2003

Keywords

Comments

Starting at 1, the three-fold convolution of A001019 (powers of 9).

Crossrefs

Sequences similar to the form q^(n-2)*binomial(n, 2): A000217 (q=1), A001788 (q=2), A027472 (q=3), A038845 (q=4), A081135 (q=5), A081136 (q=6), A027474 (q=7), A081138 (q=8), this sequence (q=9), A081140 (q=10), A081141 (q=11), A081142 (q=12), A027476 (q=15).
Cf. A001019.

Programs

  • Magma
    [9^n* Binomial(n+2, 2): n in [-2..20]]; // Vincenzo Librandi, Oct 16 2011
  • Mathematica
    LinearRecurrence[{27,-243,729},{0,0,1},30] (* Harvey P. Dale, Jan 30 2018 *)

Formula

a(n) = 27*a(n-1) - 243*a(n-2) + 729*a(n-3), a(0)=a(1)=0, a(2)=1.
a(n) = 9^(n-2)*binomial(n, 2).
G.f.: x^2/(1-9*x)^3.
E.g.f.: (x^2/2)*exp(9*x). - G. C. Greubel, May 13 2021
From Amiram Eldar, Jan 06 2022: (Start)
Sum_{n>=2} 1/a(n) = 18 - 144*log(9/8).
Sum_{n>=2} (-1)^n/a(n) = 180*log(10/9) - 18. (End)

A017629 a(n) = 12*n + 9.

Original entry on oeis.org

9, 21, 33, 45, 57, 69, 81, 93, 105, 117, 129, 141, 153, 165, 177, 189, 201, 213, 225, 237, 249, 261, 273, 285, 297, 309, 321, 333, 345, 357, 369, 381, 393, 405, 417, 429, 441, 453, 465, 477, 489, 501, 513, 525, 537, 549, 561, 573, 585, 597, 609, 621, 633
Offset: 0

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Author

Keywords

Comments

Numbers k such that k mod 2 = (k+1) mod 3 = 1 and (k+2) mod 4 != 1. - Klaus Brockhaus, Jun 15 2004
For n > 3, the number of squares on the infinite 3-column chessboard at <= n knight moves from any fixed point. - Ralf Stephan, Sep 15 2004
A016946 is the subsequence of squares (for n = 3*k*(k+1) = A028896(k), then a(n) = (6k+3)^2 = A016946(k)). - Bernard Schott, Apr 05 2021

Crossrefs

Programs

Formula

a(n) = 6*(4*n+1) - a(n-1) (with a(0)=9). - Vincenzo Librandi, Dec 17 2010
A089911(2*a(n)) = 4. - Reinhard Zumkeller, Jul 05 2013
G.f.: (9 + 3*x)/(1 - x)^2. - Alejandro J. Becerra Jr., Jul 08 2020
Sum_{n>=0} (-1)^n/a(n) = (Pi + log(3-2*sqrt(2)))/(12*sqrt(2)). - Amiram Eldar, Dec 12 2021
E.g.f.: 3*exp(x)*(3 + 4*x). - Stefano Spezia, Feb 25 2023

A018913 a(n) = 9*a(n - 1) - a(n - 2) for n>1, a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 9, 80, 711, 6319, 56160, 499121, 4435929, 39424240, 350382231, 3114015839, 27675760320, 245967827041, 2186034683049, 19428344320400, 172669064200551, 1534593233484559, 13638670037160480, 121213437100959761
Offset: 0

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Author

Keywords

Comments

Define the sequence L(a_0,a_1) by a_{n+2} is the greatest integer such that a_{n+2}/a_{n+1}= 0. This is L(1,9).
For n>=2, a(n) equals the permanent of the (n-1)X(n-1) tridiagonal matrix with 9's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,8}. - Milan Janjic, Jan 25 2015
Not to be confused with the Pisot L(1,9) sequence, which is A001019. - R. J. Mathar, Feb 13 2016
Lim_{n->oo} a(n+1)/a(n) = (9 + sqrt(77))/2 = A092290 + 1 = 8.887482... - Wolfdieter Lang, Nov 16 2023

Examples

			G.f. = x + 9*x^2 + 80*x^3 + 711*x^4 + 6319*x^5 + 56160*x^6 + 499121*x^7 + ...
		

Crossrefs

Cf. A056918(n)=sqrt{77*(a(n))^2 +4}, that is, a(n)=sqrt((A056918(n)^2 - 4)/77).
Cf. A092290 + 1.

Programs

  • Magma
    I:=[0, 1]; [n le 2 select I[n] else 9*Self(n-1) - Self(n-2): n in [1..30]]; // Vincenzo Librandi, Dec 23 2012
    
  • Mathematica
    CoefficientList[Series[x/(1 - 9*x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 23 2012 *)
  • PARI
    concat(0, Vec(x/(1-9*x+x^2) + O(x^30))) \\ Michel Marcus, Sep 06 2017
  • Sage
    [lucas_number1(n,9,1) for n in range(22)] # Zerinvary Lajos, Jun 25 2008
    

Formula

G.f.: x/(1-9*x+x^2).
a(n) = S(2*n-1, sqrt(11))/sqrt(11) = S(n-1, 9); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(-1, x) := 0.
a(n) = (((9+sqrt(77))/2)^n - ((9-sqrt(77))/2)^n)/sqrt(77). - Barry E. Williams, Aug 21 2000
a(n+1) = Sum_{k, 0<=k<=n} A101950(n,k)*8^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 1} (1 + 1/a(n)) = 1/7*(7 + sqrt(77)).
Product {n >= 2} (1 - 1/a(n)) = 1/18*(7 + sqrt(77)). (End)
a(n) = Sum_{k = 0..n-1} binomial(n+k, 2*k+1)*7^k = Sum_{k = 0..n-1} (-1)^(n+k+1)* binomial(n+k, 2*k+1)*11^k. - Peter Bala, Jul 17 2023
E.g.f.: 2*exp(9*x/2)*sinh(sqrt(77)*x/2)/sqrt(77). - Stefano Spezia, Feb 23 2025
Product_{n >= 1} (a(2*n) + 1)/(a(2*n) - 1) = sqrt(11/7) [telescoping product: ((a(2*n) + 1)/(a(2*n) - 1))^2 = (11 - 4/(a(n+1) - a(n))^2)/(11 - 4/(a(n) - a(n-1))^2), leading to 11 - 7*Product_{k = 1..n} ((a(2*k) + 1)/(a(2*k) - 1))^2 = 4/A070998(n)^2]. - Peter Bala, May 18 2025

Extensions

G.f. adapted to the offset by Vincenzo Librandi, Dec 23 2012

A047855 a(n) = A047848(7,n).

Original entry on oeis.org

1, 2, 12, 112, 1112, 11112, 111112, 1111112, 11111112, 111111112, 1111111112, 11111111112, 111111111112, 1111111111112, 11111111111112, 111111111111112, 1111111111111112, 11111111111111112, 111111111111111112, 1111111111111111112, 11111111111111111112, 111111111111111111112
Offset: 0

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Author

Keywords

Comments

n-th difference of a(n), a(n-1), ..., a(0) is A001019(n-1) for n >= 1.
Range of A164898, apart from first term. - Reinhard Zumkeller, Aug 30 2009
a(n) is the number of integers less than or equal to 10^n, whose initial digit is 1. - Michel Marcus, Jul 04 2019
a(n) is 2^n represented in bijective base-2 numeration. - Alois P. Heinz, Aug 26 2019
This sequence proves both A028842 (numbers with prime product of digits) and A028843 (numbers with prime iterated product of digits) are infinite. Proof: Suppose either of those sequences is finite. Label as omega the supposed last term. Compute n = ceiling(log_10 omega) + 1. Then a(n) > omega. The product of digits of a(n) is 2, contradicting the assumption that omega is the final term of either A028842 or A028843. - Alonso del Arte, Apr 14 2020
For n >= 2, the concatenation of a(n) with 8*a(n) equals (3*R_n+3)^2, where R_n = A002275(n) is the repunit with n 1's; hence this sequence, except for {1,2}, is a subsequence of A115549. - Bernard Schott, Apr 30 2022

Crossrefs

Programs

  • Magma
    [(10^n + 8)/9: n in [0..40]]; // G. C. Greubel, Jan 11 2025
    
  • Maple
    a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=10*a[n-1]+1 od: seq(a[n]+1, n=0..18); # Zerinvary Lajos, Mar 20 2008
  • Mathematica
    Join[{1}, Table[FromDigits[PadLeft[{2}, n, 1]], {n, 30}]] (* Harvey P. Dale, Apr 17 2013 *)
    (10^Range[0, 29] + 8)/9 (* Alonso del Arte, Apr 12 2020 *)
  • PARI
    a(n)=if(n==0,1,if(n==1,2,11*a(n-1)-10*a(n-2)))
    for(i=0,10,print1(a(i),",")) \\ Lambert Klasen, Jan 28 2005
    
  • Python
    def A047855(n): return (pow(10,n) +8)//9
    print([A047855(n) for n in range(41)]) # G. C. Greubel, Jan 11 2025
  • Sage
    [gaussian_binomial(n,1,10)+1 for n in range(17)] # Zerinvary Lajos, May 29 2009
    
  • Scala
    (List.fill(20)(10: BigInt)).scanLeft(1: BigInt)( * ).map(n => (n + 8)/9) // Alonso del Arte, Apr 12 2020
    

Formula

a(n) = (10^n + 8)/9. - Ralf Stephan, Feb 14 2004
a(0) = 1, a(1) = 2, a(n) = 11*a(n-1) - 10*a(n-2) for n > 1. - Lambert Klasen (lambert.klasen(AT)gmx.net), Jan 28 2005
G.f.: (1 - 9*x)/(1 - 11*x + 10*x^2). - Philippe Deléham, Oct 05 2009
a(n) = 10*a(n-1) - 8 (with a(0) = 1). - Vincenzo Librandi, Aug 06 2010
From Elmo R. Oliveira, Apr 03 2025: (Start)
E.g.f.: exp(x)*(8 + exp(9*x))/9.
a(n) = (A062397(n) - A002281(n))/2. (End)

Extensions

More terms from Harvey P. Dale, Apr 17 2013

A016185 a(n) = 9^n - 8^n.

Original entry on oeis.org

0, 1, 17, 217, 2465, 26281, 269297, 2685817, 26269505, 253202761, 2413042577, 22791125017, 213710059745, 1992110014441, 18478745943857, 170706760005817, 1571545212141185, 14425381885981321, 132080236787517137, 1206736529597136217, 11004743954450081825
Offset: 0

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Author

Keywords

Comments

a(n) is also the number of n-digit numbers whose smallest decimal digit is 1. - Stefano Spezia, Nov 15 2023

Crossrefs

Programs

Formula

G.f.: x/((1-8*x)*(1-9*x)).
E.g.f.: e^(9*x) - e^(8*x). - Mohammad K. Azarian, Jan 14 2009
a(n) = 9*a(n-1) + 8^(n-1), a(0)=0. - Vincenzo Librandi, Feb 09 2011
a(n) = 17*a(n-1) - 72*a(n-2), a(0)=0, a(1)=1. - Vincenzo Librandi, Feb 09 2011
a(n) = A001019(n) - A001018(n). - Alois P. Heinz, Aug 19 2021

A102518 a(n) = Sum_{k=0..n} binomial(n, k) * Sum_{j=0..k} binomial(3k, 3j).

Original entry on oeis.org

1, 3, 27, 243, 2187, 19683, 177147, 1594323, 14348907, 129140163, 1162261467, 10460353203, 94143178827, 847288609443, 7625597484987, 68630377364883, 617673396283947, 5559060566555523, 50031545098999707, 450283905890997363, 4052555153018976267, 36472996377170786403
Offset: 0

Views

Author

Paul Barry, Jan 13 2005

Keywords

Comments

Binomial transform of A007613.
a(n+1) is the smallest number with a reciprocal with repeating decimal of period a(n). - Matthew Goers, Nov 09 2017
a(n) is the number of walks of 2n steps on the utility graph that start and end at the same vertex (excursions). A001019(n) is the number of 2n+1-step walks on the utility graph that end at one of the 3 adjacent vertices. A013708(n) is the number of 2n+2-step walks that end at one of the 2 remote vertices (at distance 2). The number of n-step walks on the utility (3-regular) graph, summed over all 3 types of final vertices, is 3^n. - R. J. Mathar, Nov 03 2020

Crossrefs

Programs

  • Mathematica
    Join[{1},NestList[9#&,3,20]] (* Harvey P. Dale, Feb 03 2021 *)
  • PARI
    Vec((1-6*x)/(1-9*x) + O(x^30)) \\ Colin Barker, Mar 17 2016

Formula

a(n) = 3^(2*n-1) + 2*0^k/3; a(n+1) = A013708(n).
G.f.: (1-6*x) / (1-9*x). - Colin Barker, Mar 17 2016
E.g.f.: (exp(9*x) + 2)/3. - Stefano Spezia, Jul 09 2024

A084175 Jacobsthal oblong numbers.

Original entry on oeis.org

0, 1, 3, 15, 55, 231, 903, 3655, 14535, 58311, 232903, 932295, 3727815, 14913991, 59650503, 238612935, 954429895, 3817763271, 15270965703, 61084037575, 244335800775, 977343902151, 3909374210503, 15637499638215, 62549992960455
Offset: 0

Views

Author

Paul Barry, May 18 2003

Keywords

Comments

Inverse binomial transform is A001019 doubled up.
Binomial transform is A084177.
Partial sums of A003683.

Crossrefs

Except for initial terms, same as A015249 and A084152.

Programs

Formula

a(n) = A001045(n)*A001045(n+1).
a(n) = (2*4^n - (-2)^n - 1)/9;
a(n) = 3*a(n-1) + 6*a(n-2) - 8*a(n-3), a(0)=0, a(1)=1, a(2)=3.
G.f.: x/((1+2*x)*(1-x)*(1-4*x)).
E.g.f.: (2*exp(4*x) - exp(x) - exp(-2*x))/9.
a(n+1) - 4*a(n) = 1, -1, 3, -5, 11, ... = A001045(n+1) signed. - Paul Curtz, May 19 2008
a(n) = round(2^n/3) * round(2^(n+1)/3). - Gary Detlefs, Feb 10 2010
From Peter Bala, Mar 30 2015: (Start)
The shifted o.g.f. A(x) := 1/( (1 + 2*x)*(1 - x)*(1 - 4*x) ) = 1/(1 - 3*x - 6*x^2 + 8*x^3). Hence A(x) == 1/(1 - 3*x + 3*x^2 - x^3) (mod 9) == 1/(1 - x)^3 (mod 9). It follows by Theorem 1 of Heninger et al. that (A(x))^(1/3) = 1 + x + 4*x^2 + 10*x^3 + ... has integral coefficients.
Sum_{n >= 0} a(n+1)*x^n = exp( Sum_{n >= 1} J(3*n)/J(n)*x^n/n ), where J(n) = A001045(n) are the Jacobsthal numbers. Cf. A001656, A099930. (End)

A241010 Numbers n with the property that the number of parts in the symmetric representation of sigma(n) is odd, and that all parts have width 1.

Original entry on oeis.org

1, 2, 4, 8, 9, 16, 25, 32, 49, 50, 64, 81, 98, 121, 128, 169, 242, 256, 289, 338, 361, 484, 512, 529, 578, 625, 676, 722, 729, 841, 961, 1024, 1058, 1156, 1250, 1369, 1444, 1681, 1682, 1849, 1922, 2048, 2116, 2209, 2312, 2401, 2738, 2809, 2888, 3025, 3249, 3362, 3364, 3481, 3698, 3721, 3844
Offset: 1

Views

Author

Hartmut F. W. Hoft, Aug 07 2014

Keywords

Comments

The first eight entries in A071562 but not in this sequence are 6, 12, 15, 18, 20, 24, 28 & 30.
The first eight entries in A238443 but not in this sequence are 6, 12, 18, 20, 24, 28, 30 & 36.
The union of A241008 and of this sequence equals A174905 (for a proof see link in A174905).
Let n = 2^m * product(p_i^e_i, i=1,...,k) = 2^m * q with m >= 0, k >= 0, 2 < p_1, ...< p_k primes and e_i >= 1, for all 1 <= i <= k. For each number n in this sequence all e_i are even, and for any two odd divisors f < g of n, 2^(m+1) * f < g. The sum of the areas of the regions r(n, z) equals sigma(n). For a proof of the characterization and the formula see the theorem in the link below.
Numbers 3025 = 5^2 * 11^2 and 510050 = 2^1 * 5^2 * 101^2 are the smallest odd and even numbers, respectively, in the sequence with two distinct odd prime factors.
Among the 706 numbers in the sequence less than 1000000 (see link to the table) there are 143 that have two different odd prime factors, but none with three. All numbers with three different odd prime factors are larger than 500000000. Number 4450891225 = 5^2 * 11^2 * 1213^2 is in the sequence, but may not be the smallest one with three different odd prime factors. Note that 1213 is the first prime that extends the list of divisors of 3025 while preserving the property for numbers in this sequence.
The subsequence of numbers n = 2^(k-1) * p^2 satisfying the constraints above is A247687.
n = 3^(2*h) = 9^h = A001019(h), h>=0, is the smallest number for which the symmetric representation of sigma(n) has 2*h+1 regions of width one, for example for h = 1, 2, 3 and 5, but not for h = 4 in which case 3025 = 5^2 * 11^2 < 3^8 = 6561 is the smallest (see A318843). [Comment corrected by Hartmut F. W. Hoft, Sep 04 2018]
Computations using this characterization are more efficient than those of Dyck paths for the symmetric representations of sigma(n), e.g., the Mathematica code below.

Examples

			This irregular triangle presents in each column those elements of the sequence that have the same factor of a power of 2.
  row/col      2^0    2^1   2^2   2^3    2^4    2^5  ...
   2^k:          1      2     4     8     16     32  ...
   3^2:          9
   5^2:         25     50
   7^2:         49     98
   3^4:         81
  11^2:        121    242   484
  13^2:        169    338   676
  17^2:        289    578  1156  2312
  19^2:        361    722  1444  2888
  23^2:        529   1058  2116  4232
   5^4:        625   1250
   3^6:        729
  29^2:        841   1682  3364  6728
  31^2:        961   1922  3844  7688
  37^2:       1369   2738  5476 10952 21904
  41^2:       1681   3362  6724 13448 26896
  43^2:       1849   3698  7396 14792 29584
  47^2:       2209   4418  8836 17672 35344
   7^4:       2401   4802
  53^2:       2809   5618 11236 22472 44944
  5^2*11^2:   3025
  3^2*19^2:   3249
  59^2:       3481   6962 13924 27848 55696
  61^2:       3721   7442 14884 29768 59536
  67^2:       4489   8978 17956 35912 71824 143648
  3^2*23^2:   4761
  71^2:       5041
  ...
  5^2*101^2:225025 510050
  ...
Number 3025 = 5^2 * 11^2 is in the sequence since its divisors are 1, 5, 11, 25, 55, 121, 275, 605 and 3025. Number 6050 = 2^1 * 5^2 * 11^2 is not in the sequence since 2^2 * 5 > 11 while 5 < 11.
Number 510050 = 2^1 * 5^2 * 101^2 is in the sequence since its 9 odd divisors 1, 5, 25, 101, 505, 2525, 10201, 51005 and 225025 are separated by factors larger than 2^2. The areas of its 9 regions are 382539, 76515, 15339, 3939, 1515, 3939, 15339, 76515 and 382539. However, 2^2 * 5^2 * 101^2 is not in the sequence.
The first row is A000079.
The rows, except the first, are indexed by products of even powers of the odd primes satisfying the property, sorted in increasing order.
The first column is a subsequence of A244579.
A row labeled p^(2*h), h>=1 and p>=3 with p = A000040(n), has A098388(n) entries.
Starting with the second column, dividing the entries of a column by 2 creates a proper subsequence of the prior column.
See A259417 for references to other sequences of even powers of odd primes that are subsequences of column 1.
The first entry greater than 16 in column labeled 2^4 is 21904 since 37 is the first prime larger than 2^5. The rightmost entry in the row labeled 19^2 is 2888 in the column labeled 2^3 since 2^4 < 19 < 2^5.
		

Crossrefs

Cf. A000203, A174905, A236104, A237270 (symmetric representation of sigma(n)), A237271, A237593, A238443, A241008, A071562, A246955, A247687, A250068, A250070, A250071.

Programs

  • Mathematica
    (* path[n] and a237270[n] are defined in A237270 *)
    atmostOneDiagonalsQ[n_] := SubsetQ[{0, 1}, Union[Flatten[Drop[Drop[path[n], 1], -1] - path[n-1], 1]]]
    Select[Range[1000], atmostOneDiagonalsQ[#] && OddQ[Length[a237270[#]]]&] (* data *)
    (* more efficient code based on numeric characterization *)
    divisorPairsQ[m_, q_] := Module[{d = Divisors[q]}, Select[2^(m + 1)*Most[d] - Rest[d], # >= 0 &] == {}]
    a241010AltQ[n_] := Module[{m, q, p, e}, m=IntegerExponent[n, 2]; q=n/2^m; {p, e} = Transpose[FactorInteger[q]]; q==1||(Select[e, EvenQ]==e && divisorPairsQ[m, q])]
    a241010Alt[m_,n_] := Select[Range[m, n], a241010AltQ]
    a241010Alt[1,4000] (* data *)

Formula

Formula for the z-th region in the symmetric representation of n = 2^m * q in this sequence, 1 <= z <= sigma_0(q) and q odd: r(n, z) = 1/2 * (2^(m+1) - 1) * (d_z + d_(2*x+2-z)) where 1 = d_1 < ... < d_(2*x+1) = q are the odd divisors of n.

Extensions

More terms and further edited by Hartmut F. W. Hoft, Jun 26 2015 and Jul 02 2015 and corrected Oct 11 2015

A065999 Sum of digits of 9^n.

Original entry on oeis.org

1, 9, 9, 18, 18, 27, 18, 45, 27, 45, 45, 45, 54, 63, 72, 63, 63, 99, 81, 90, 90, 90, 90, 108, 117, 144, 117, 108, 90, 126, 99, 153, 144, 117, 153, 144, 162, 171, 153, 153, 153, 198, 162, 171, 198, 216, 171, 198, 198, 225, 153, 252, 216, 234, 207
Offset: 0

Views

Author

N. J. A. Sloane, Dec 11 2001

Keywords

Comments

a(n) mod 9 = 0 for n > 0. - Reinhard Zumkeller, May 14 2011

Crossrefs

Cf. sum of digits of k^n: A001370 (k=2), A004166 (k=3), A065713 (k=4), A066001 (k=5), A066002 (k=6), A066003(k=7), A066004 (k=8), this sequence (k=9), A066005 (k=11), A066006 (k=12), A175527 (k=13).
Cf. also A056888, A001019.

Programs

  • Mathematica
    Table[Total[IntegerDigits[9^n]], {n, 0, 60}] (* Vincenzo Librandi, Oct 08 2013 *)
  • PARI
    a(n) = sumdigits(9^n); \\ Michel Marcus, Nov 01 2013

Formula

a(n) = A007953(A001019(n)). - Michel Marcus, Nov 01 2013

A035023 One ninth of 9-factorial numbers.

Original entry on oeis.org

1, 18, 486, 17496, 787320, 42515280, 2678462640, 192849310080, 15620794116480, 1405871470483200, 139181275577836800, 15031577762406374400, 1758694598201545804800, 221595519373394771404800, 29915395115408294139648000, 4307816896618794356109312000
Offset: 1

Views

Author

Keywords

Comments

E.g.f. is g.f. for A001019(n-1) (powers of nine).

Crossrefs

Programs

  • Magma
    [9^(n-1)*Factorial(n): n in [1..40]]; // G. C. Greubel, Oct 19 2022
    
  • Mathematica
    With[{nn=20},Rest[CoefficientList[Series[(-1+1/(1-9*x))/9,{x,0,nn}],x] Range[ 0,nn]!]] (* Harvey P. Dale, Apr 07 2019 *)
    Table[9^(n-1)*n!, {n, 40}] (* G. C. Greubel, Oct 19 2022 *)
  • SageMath
    [9^(n-1)*factorial(n) for n in range(1,40)] # G. C. Greubel, Oct 19 2022

Formula

9*a(n) = (9*n)(!^9) = Product_{j=1..n} 9*j = 9^n*n!.
E.g.f.: (-1+1/(1-9*x))/9.
D-finite with recurrence: a(n) - 9*n*a(n-1) = 0. - R. J. Mathar, Jan 28 2020
From Amiram Eldar, Jan 08 2022: (Start)
Sum_{n>=1} 1/a(n) = 9*(exp(1/9)-1).
Sum_{n>=1} (-1)^(n+1)/a(n) = 9*(1-exp(-1/9)). (End)
a(n) = A001019(n-1) * A000142(n). - G. C. Greubel, Oct 19 2022
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