A002283 -id:A002283 - cp's OEIS frontend

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

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

Offset: 0

N. J. A. Sloane

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

Ivan Panchenko, Table of n, a(n) for n = 0..200
Eric Weisstein's World of Mathematics, Repdigit.
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A002283, A178632, A010785.

Cf. A002275, A002276, A002277, A002278, A002280, A002281, A002282, A075415.

Table[FromDigits[PadRight[{},n,5]],{n,0,20}] (* Harvey P. Dale, Oct 05 2013 *)

A002279(n):=5*(10^n-1)/9$
makelist(A002279(n),n,0,30); /* Martin Ettl, Nov 08 2012 */

a(n)=5*(10^n-1)/9 \\ Charles R Greathouse IV, Sep 24 2015

def A002279(n): return 5*(10**n-1)//9 # Karl-Heinz Hofmann, Nov 28 2023

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)

A002280 a(n) = 6*(10^n - 1)/9.

Original entry on oeis.org

0, 6, 66, 666, 6666, 66666, 666666, 6666666, 66666666, 666666666, 6666666666, 66666666666, 666666666666, 6666666666666, 66666666666666, 666666666666666, 6666666666666666, 66666666666666666, 666666666666666666, 6666666666666666666, 66666666666666666666, 666666666666666666666

Offset: 0

N. J. A. Sloane

a(n-1) = number of Fibonacci numbers F(k), k <= 10^n, which end in 0. a(1)=6 because there are 6 Fibonacci numbers up to 10^2 which end in 0. - Shyam Sunder Gupta and Benoit Cloitre, Aug 15 2002

a(n) is the total number of holes in a certain triangle fractal (start with 10 triangles, 6 holes) after n iterations. See illustration in links. - Kival Ngaokrajang, Feb 21 2015

Ivan Panchenko, Table of n, a(n) for n = 0..200
Kival Ngaokrajang, Illustration of initial terms.
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Different from A072912. Cf. A073548, etc.
Cf. A002275, A002276, A002277, A002278, A002279, A002281, A002282, A002283, A075415, A178631, A178633.
Cf. A010785, A017233, A073551.

LinearRecurrence[{11,-10},{0,6},20] (* Harvey P. Dale, Dec 20 2012 *)

a(n)=6*(10^n-1)/9 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 6*A002275(n).
From Jaume Oliver Lafont, Feb 03 2009: (Start)
G.f.: 6*x/((1-x)*(1-10*x)).
a(n) = 11*a(n-1) - 10*a(n-2) with a(0)=0, a(1)=6. (End)
a(n) = A178633(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
a(n) = a(n-1) + 6*10^(n-1) with a(0)=0. - Vincenzo Librandi, Jul 22 2010
E.g.f.: 2*exp(x)*(exp(9*x) - 1)/3. - Stefano Spezia, Sep 13 2023
From Elmo R. Oliveira, Jul 21 2025: (Start)
a(n) = A073551(n+1)/2 for n >= 1.
a(n) = A010785(A017233(n-1)) for n >= 1. (End)

A002281 a(n) = 7*(10^n - 1)/9.

Original entry on oeis.org

0, 7, 77, 777, 7777, 77777, 777777, 7777777, 77777777, 777777777, 7777777777, 77777777777, 777777777777, 7777777777777, 77777777777777, 777777777777777, 7777777777777777, 77777777777777777, 777777777777777777, 7777777777777777777, 77777777777777777777, 777777777777777777777

Offset: 0

N. J. A. Sloane

Ivan Panchenko, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A002275, A002276, A002277, A002278, A002279, A002280, A002282, A002283, A178630, A178634.

Cf. A010785, A017245, A099915.

[7*(10^n-1)/9 : n in [0..30]]; // Wesley Ivan Hurt, Mar 24 2015

A002281:=n->7*(10^n-1)/9: seq(A002281(n), n=0..30); # Wesley Ivan Hurt, Mar 24 2015

LinearRecurrence[{11,-10},{0,7},25] (* Robert G. Wilson v, Jul 06 2013 *)

a(n)=7*(10^n-1)/9 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = A178634(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
From Vincenzo Librandi, Jul 22 2010: (Start)
a(n) = a(n-1) + 7*10^(n-1) with n>0, a(0)=0.
a(n) = 11*a(n-1) - 10*a(n-2) with n>1, a(0)=0, a(1)=7. (End)
G.f.: 7*x/((x-1)*(10*x-1)). - Colin Barker, Jan 24 2013
a(n) = 7*A002275(n). - Wesley Ivan Hurt, Mar 24 2015
E.g.f.: 7*exp(x)*(exp(9*x) - 1)/9. - Stefano Spezia, Sep 13 2023
From Elmo R. Oliveira, Jul 20 2025: (Start)
a(n) = (A099915(n) - 1)/2.
a(n) = A010785(A017245(n-1)) for n >= 1. (End)

A033819 Trimorphic numbers: n^3 ends with n. Also m-morphic numbers for all m > 5 such that m-1 is not divisible by 10 and m == 3 (mod 4).

Original entry on oeis.org

0, 1, 4, 5, 6, 9, 24, 25, 49, 51, 75, 76, 99, 125, 249, 251, 375, 376, 499, 501, 624, 625, 749, 751, 875, 999, 1249, 3751, 4375, 4999, 5001, 5625, 6249, 8751, 9375, 9376, 9999, 18751, 31249, 40625, 49999, 50001, 59375, 68751, 81249, 90624, 90625

Offset: 1

David W. Wilson

n is in this sequence iff it occurs in one of A002283, A007185, A016090, A198971, A199685, A216092, A216093, A224473, A224474, A224475, A224476, A224477, and A224478. - Eric M. Schmidt, Apr 08 2013

Let q(n) = floor(a(n)^3 / 10^A055642(a(n))), where A055642(n) is the number of digits in the decimal expansion of n. As well, let na and nb denote the indices of the preceding and next terms that begin with a 9. Then (1/q(n)) * (a(n)^4 - a(n)^3 - a(n)^2 + a(n)) - 2*a(n)^2 + a(n) + q(n) + 1 = a(na+nb-n)^2 - a(na+nb-n) - q(na+nb-n). - Christopher Hohl, Apr 08 2019

			376^3 = 53157376 which ends with 376.

S. Premchaud, A class of numbers, Math. Student, 48 (1980), 293-300.

Eric M. Schmidt, Table of n, a(n) for n = 1..1000
Robert Dawson, On Some Sequences Related to Sums of Powers, J. Int. Seq., Vol. 21 (2018), Article 18.7.6.
Shyam Sunder Gupta, Elegance of Squares, Cubes, and Higher Powers, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 2, 29-81.
Eric Weisstein's World of Mathematics, Trimorphic Number
Index entries for sequences related to automorphic numbers

Cf. A074194, A215558 (cubes of the terms).

[n: n in [0..10^5] | Intseq(n^3)[1..#Intseq(n)] eq Intseq(n)]; // Bruno Berselli, Apr 04 2013

Do[x=Floor[N[Log[10, n], 25]]+1; If[Mod[n^3, 10^x] == n, Print[n]], {n, 1, 10000}]
Select[Range[100000],PowerMod[#,3,10^IntegerLength[#]]==#&](* Harvey P. Dale, Nov 04 2011 *)
Select[Range[0, 10^5], 10^IntegerExponent[#^3-#, 10]>#&] (* Jean-François Alcover, Apr 04 2013 *)

A002278 a(n) = 4*(10^n - 1)/9.

Original entry on oeis.org

0, 4, 44, 444, 4444, 44444, 444444, 4444444, 44444444, 444444444, 4444444444, 44444444444, 444444444444, 4444444444444, 44444444444444, 444444444444444, 4444444444444444, 44444444444444444, 444444444444444444, 4444444444444444444, 44444444444444444444, 444444444444444444444

Offset: 0

N. J. A. Sloane

Ivan Panchenko, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A002275, A002276, A002277, A002279, A002280, A002281, A002282, A002283, A075415, A178632.

Cf. A007091, A010785, A017209, A024049.

LinearRecurrence[{11, -10}, {0, 4}, 20] (* Robert G. Wilson v, Jul 06 2013 *)

a(n)=4*(10^n-1)/9 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = A075415(n)/A002283(n). - Reinhard Zumkeller, May 31 2010
From Vincenzo Librandi, Jul 22 2010: (Start)
a(n) = a(n-1) + 4*10^(n-1) with a(0)=0;
a(n) = 11*a(n-1) - 10*a(n-2) with a(0)=0, a(1)=4. (End)
G.f.: 4*x/((1 - x)*(1 - 10*x)). - Ilya Gutkovskiy, Feb 24 2017
E.g.f.: 4*exp(x)*(exp(9*x) - 1)/9. - Stefano Spezia, Sep 13 2023
a(n) = A007091(A024049(n)). - Michel Marcus, Jun 16 2024
From Elmo R. Oliveira, Jul 19 2025: (Start)
a(n) = 4*A002275(n).
a(n) = A010785(A017209(n-1)) for n >= 1. (End)

A001232 Numbers k such that 9*k = (k written backwards), k > 0.

Original entry on oeis.org

1089, 10989, 109989, 1099989, 10891089, 10999989, 108901089, 109999989, 1089001089, 1098910989, 1099999989, 10890001089, 10989010989, 10999999989, 108900001089, 108910891089, 109890010989, 109989109989, 109999999989, 1089000001089, 1089109891089

Offset: 1

N. J. A. Sloane and Simon Plouffe

This sequence contains the least n-digit non-palindromic number which is a factor of its reversal. Quotient is always 9. - Lekraj Beedassy, Jun 11 2004. (But it contains many other numbers as well. - N. J. A. Sloane, Jul 02 2013)
Nonzero fixed points of the map which sends x to x - reverse(x) if that is nonnegative, otherwise to x + reverse(x). - Sébastien Dumortier, Nov 05 2006. (Clarified comment, see A124074. - Ray Chandler, Oct 11 2017)
Numbers k such that reversal(k)=reversal(k+reversal(k)). Also numbers k such that reversal(k)=reversal(10*k-reversal(k)). - Farideh Firoozbakht, Jun 11 2010
From M. F. Hasler, Oct 04 2022: (Start)
(1) The first digit of any term must be 1, otherwise multiplication by 9 yields one more digit. For the same reason, no "overflow" must occur from the second to the first digit, so the last digit must be 9.
(2) Continuing the reasoning "from right to left" implies that the trailing nonzero digits must be ...9*89, where 9* means any nonnegative number of consecutive digits 9, preceded by a digit 0, which must be preceded by a digit 1. This implies that the initial and also final digits of any term must be 109*89. We might call a term of this form a "primitive" term. So there is exactly one primitive term b(k) = 11*10^(k-2)-11 with k digits, for all k >= 4.
(3) All terms of the sequence are a "symmetric" concatenation of such b(k)'s, "spaced out" with any number of digits 0, also in a symmetrical way: For any n >= 1, let k = (k[1], ..., k[n]) with k[n+1-j] = k[j] >= 4, and m = (m[1], ..., m[n-1]) (possibly of length 0) with m[n-j] = m[j] >= 0, then N = concat(b(k[j])*10^m[j], 1 <= j < n; k[n]) is a term of the sequence, and this yields all terms of the sequence. (For example, with 1089 we also have 1089{0...0}1089 and 1089,001089,001089, etc.) (End)

			1089*9 = 9801.

H. Camous, Jouer Avec Les Maths, "Cardinaux Réversibles", Section I, Problem 6, pp. 27, 37-38; Les Editions d'Organisation, Paris, 1984.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 41.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, under #1089.

Ray Chandler, Table of n, a(n) for n = 1..10000
C. A. Van Cott, The Integer Hokey Pokey, Math Horizons, Vol. 28, pp. 24-27, November 2020.
L. H. Kendrick, Young Graphs: 1089 et al., J. Int. Seq. 18 (2015) 15.9.7.
N. J. A. Sloane, 2178 And All That, arXiv:1307.0453 [math.NT], 2013; Fib. Quart., 52 (2014), 99-120.
N. J. A. Sloane, 2178 And All That [Local copy]
Simon Weisgerber, Value Judgments in Mathematics: GH Hardy and the (Non-)seriousness of Mathematical Theorems, Global Phil. (2024) Vol. 34, Art. No. 1. See p. 8.

Cf. A008918, A008919, A193434, A222814, A222815, A031877, A124074.

Cf. A002275, A002283, A004086, A094707.

Rest@Select[FromDigits /@ Tuples[{0, 99}, 11], IntegerDigits[9*#] == Reverse@IntegerDigits[#] &] (* Arkadiusz Wesolowski, Aug 14 2012 *)
okQ[t_]:=t==Reverse[t]&&First[t]!=0&&Min[Length/@Split[t]]>1; 99#&/@Flatten[Table[ FromDigits/@ Select[Tuples[{0,1},n],okQ],{n,20}]] (* Harvey P. Dale, Jul 03 2013 *)

isok(n) = 9*n == eval(concat(Vecrev(Str(n)))); \\ Michel Marcus, Feb 21 2015

{A001232_row(n, L(v, s=0)=for(i=1, #v, s*=10^v[i]; i%2 && s+=10^v[i]\900); s)=if(n<4, [], L, Set(apply(L, self()(n, 0)))*99, L=List([[n]]); for(k=4, n\2, listput(L,[k,n-2*k,k]); for(p=0, n\2-k, foreach(self()(n-(k+p)*2, 0), M, listput(L, concat([[k, p], M, [p, k]]))))); L)} \\ List of n-digit terms. - M. F. Hasler, Oct 04 2022
concat(apply(A001232_row, [1..14]))

def A001232_row(n, r=11): # list of n-digit terms
    L = [] if n<4 else [[n]]
    for L1 in range(4, n//2+1):
        L.append([L1, n-2*L1, L1])
        L.extend([L1,L2]+M+[L2,L1] for L2 in range(n//2-1-L1)
                                     for M in A001232_row(n-(L1+L2)*2, 0))
    if not r: return L
    def f(L, s=0):
        for k,L in enumerate(L):
            s *= 10**L
            if not k%2: s += 10**(L-2)-1
        return r*s
    return sorted(map(f, A001232_row(n, 0))) # M. F. Hasler, Oct 04 2022

Theorem: Terms in this sequence have the form 99*m, where the decimal representation of m contains only 1's and 0's, is palindromic and contains no singleton 1's or 0's. Hence contains Fib(floor(k/2)-1) k-digit terms, k >= 4. - David W. Wilson, Dec 15 1997
a(A094707(n)) = 11*(10^n - 1) = 11*A002283(n) = 99*A002275(n), for n>1. - Lekraj Beedassy, Jun 11 2004. (Restored from history and corrected. - Ray Chandler, Oct 11 2017)
a(n) = 99*A061851(n) = A008918(n)/2. - M. F. Hasler, Oct 06 2022

Corrected and extended by David W. Wilson, Aug 15 1996, Dec 15 1997
a(20)-a(21) from Arkadiusz Wesolowski, Aug 14 2012
a(1..10^4) in b-file double-checked with independent code by M. F. Hasler, Oct 04 2022

A075415 Squares of A002280 or numbers (666...6)^2.

Original entry on oeis.org

0, 36, 4356, 443556, 44435556, 4444355556, 444443555556, 44444435555556, 4444444355555556, 444444443555555556, 44444444435555555556, 4444444444355555555556, 444444444443555555555556, 44444444444435555555555556, 4444444444444355555555555556, 444444444444443555555555555556

Offset: 0

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

A transformation of the Wonderful Demlo numbers (A002477).

			a(2) = 66^2 = 4356.
From _Reinhard Zumkeller_, May 31 2010: (Start)
n=1: ..................... 36 = 9 * 4;
n=2: ................... 4356 = 99 * 44;
n=3: ................. 443556 = 999 * 444;
n=4: ............... 44435556 = 9999 * 4444;
n=5: ............. 4444355556 = 99999 * 44444;
n=6: ........... 444443555556 = 999999 * 444444;
n=7: ......... 44444435555556 = 9999999 * 4444444;
n=8: ....... 4444444355555556 = 99999999 * 44444444;
n=9: ..... 444444443555555556 = 999999999 * 444444444. (End)

Seiichi Manyama, Table of n, a(n) for n = 0..500
Gérard Villemin, Variations sur les carrés.
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A075411, A075412, A075413, A075414, A075415, A075416, A075417.
Cf. A059988, A178630, A178631, A178632, A178633, A178634, A178635. [From Reinhard Zumkeller, May 31 2010]
Cf. A002275, A002278, A002279,  A002280, A002283, A002477, A052041, A102807, A309827.

Table[FromDigits[PadRight[{},n,6]]^2,{n,0,20}] (* or *) LinearRecurrence[ {111,-1110,1000},{0,36,4356},20] (* Harvey P. Dale, May 20 2021 *)

a(n) = A002280(n)^2 = (6*A002275(n))^2 = 36*A002275(n)^2.
a(n) = (6*(10^n-1)/9)^2 = (4/9)*(10^(2*n) - 2*10^n + 1), which is n-1 4's, followed by a 3, n-1 5's and a 6. - Ignacio Larrosa Cañestro, Feb 26 2005
From Reinhard Zumkeller, May 31 2010: (Start)
a(n) = ((A002278(n-1)*10 + 3)*10^(n-1) + A002279(n-1))*10 + 6 for n>0.
a(n) = A002283(n)*A002278(n). (End)
G.f.: 36*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)). - Arkadiusz Wesolowski, Dec 26 2011
From Elmo R. Oliveira, Jul 27 2025: (Start)
E.g.f.: 4*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) = 36*A002477(n). (End)

Edited by Alois P. Heinz, Aug 21 2019 (merged with A102794, submitted by Richard C. Schroeppel, Feb 26 2005)

A186669 Total number of positive integers below 10^n requiring 12 positive biquadrates in their representation as sum of biquadrates.

Original entry on oeis.org

0, 6, 68, 724, 7126, 67537, 667927, 6668357, 66668807, 666669258, 6666669682

Offset: 1

Martin Renner, Feb 25 2011

A114322(n) + A186649(n) + A186651(n) + A186653(n) + A186655(n) + A186657(n) + A186659(n) + A186661(n) + A186663(n) + A186665(n) + A186667(n) + a(n) + A186671(n) + A186673(n) + A186675(n) + A186677(n) + A186680(n) + A186682(n) + A186684(n) = A002283(n).

Eric Weisstein's World of Mathematics, Waring's Problem.

Cf. A003346, A186670.

a(5)-a(7) from Lars Blomberg, May 08 2011
a(8)-a(9) from Hiroaki Yamanouchi, Oct 13 2014
a(10)-a(11) from Giovanni Resta, Apr 29 2016

A048379 Apply the transformation 0->1->2->3->4->5->6->7->8->9->0 to digits of n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 21, 22, 23, 24, 25, 26, 27, 28, 29, 20, 31, 32, 33, 34, 35, 36, 37, 38, 39, 30, 41, 42, 43, 44, 45, 46, 47, 48, 49, 40, 51, 52, 53, 54, 55, 56, 57, 58, 59, 50, 61, 62, 63, 64, 65, 66, 67, 68, 69, 60, 71, 72, 73, 74, 75, 76, 77, 78, 79, 70, 81, 82

Offset: 0

Patrick De Geest, Mar 15 1999

This is the same as a(n) = 1*n in the arithmetic defined in A169918 (cf. A169930). - M. F. Hasler, Mar 25 2015

			a(8) = 9.
a(9) = 0.
a(10) = 21 because the original 1 is changed to a 2 and the 0 is changed to a 1.

Reinhard Zumkeller, Table of n, a(n) for n = 0..90000

a048379 n = if n == 0 then 1 else x n where
   x m = if m == 0 then 0 else 10 * x m' + (d + 1) `mod` 10
         where (m',d) = divMod m 10
-- Reinhard Zumkeller, Feb 21 2014

Table[FromDigits[ReplaceAll[IntegerDigits[n] + 1, 10 -> 0]], {n, 0, 79}] (* Alonso del Arte, Feb 27 2014 *)

A048379(n)=n+sum(i=1, #n=digits(n), if(n[i]<9, 10^(i-1), -9*10^(i-1))) \\ M. F. Hasler, Mar 21 2015

A048379(n)=!n+apply(t->(t+1)%10, n=digits(n))*vector(#n, i, 10^(#n-i))~ \\ M. F. Hasler, Mar 21 2015

d = {ord(str(i)):ord(str((i+1)%10)) for i in range(10)}
def a(n): return int(str(n).translate(d))
print([a(n) for n in range(72)]) # Michael S. Branicky, Dec 20 2022

a(A002283(n)) = 0. - Reinhard Zumkeller, Feb 21 2014

A057951 Number of prime factors of 10^n - 1 (counted with multiplicity).

Original entry on oeis.org

2, 3, 4, 4, 4, 7, 4, 6, 6, 6, 4, 9, 5, 6, 8, 8, 4, 11, 3, 9, 9, 9, 3, 12, 7, 8, 9, 10, 7, 15, 5, 13, 8, 8, 9, 14, 5, 5, 8, 13, 6, 17, 6, 13, 12, 8, 4, 15, 6, 12, 10, 11, 6, 16, 10, 14, 8, 10, 4, 22, 9, 7, 16, 17, 9, 17, 5, 12, 8, 14, 4, 20, 5, 9, 14, 8, 10, 18

Offset: 1

Patrick De Geest, Nov 15 2000

nonn

Max Alekseyev, Table of n, a(n) for n = 1..352 (first 322 terms from Ray Chandler)
Makoto Kamada, Factorizations of 11...11 (Repunit).
S. S. Wagstaff, Jr., Main Tables from the Cunningham Project.
S. S. Wagstaff, Jr., The Cunningham Project

bigomega(b^n-1): this sequence (b=10), A057952 (b=9), A057953 (b=8), A057954 (b=7), A057955 (b=6), A057956 (b=5), A057957 (b=4), A057958 (b=3), A046051 (b=2).

Cf. A001222, A002283, A046053, A085035, A003020, A046107, A005422, A061075, A102380, A095370, A147556, A081317, A081318, A102347, A112505.

PrimeOmega[10^Range[70]-1] (* Jayanta Basu, May 29 2013 *)

a(n)=bigomega(10^n-1) \\ Charles R Greathouse IV, Sep 14 2015

Mobius transform of A085035 - T. D. Noe, Jun 19 2003
a(n) = Omega(10^n -1) = Omega(R_n) + 2 = A046053(n) + 2 {where Omega(n) = A001222(n) and R_n = (10^n - 1)/9 = A002275(n)}. - Lekraj Beedassy, Jun 09 2006
a(n) = A001222(A002283(n)). - Ray Chandler, Apr 22 2017

Erroneous b-file replaced by Ray Chandler, Apr 26 2017

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Maxima

PARI

Python

Formula

A002280 a(n) = 6*(10^n - 1)/9.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A002281 a(n) = 7*(10^n - 1)/9.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A033819 Trimorphic numbers: n^3 ends with n. Also m-morphic numbers for all m > 5 such that m-1 is not divisible by 10 and m == 3 (mod 4).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Mathematica

A002278 a(n) = 4*(10^n - 1)/9.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A001232 Numbers k such that 9*k = (k written backwards), k > 0.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Python

Formula

Extensions

A075415 Squares of A002280 or numbers (666...6)^2.

Views

Author

Keywords

Comments