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a(n) = A047848(7, n).
20

%I #71 Jan 14 2025 10:30:48

%S 1,2,12,112,1112,11112,111112,1111112,11111112,111111112,1111111112,

%T 11111111112,111111111112,1111111111112,11111111111112,

%U 111111111111112,1111111111111112,11111111111111112,111111111111111112,1111111111111111112,11111111111111111112

%N a(n) = A047848(7, n).

%C n-th difference of a(n), a(n-1), ..., a(0) is A001019(n-1) for n >= 1.

%C Range of A164898, apart from first term. - _Reinhard Zumkeller_, Aug 30 2009

%C a(n) is the number of integers less than or equal to 10^n, whose initial digit is 1. - _Michel Marcus_, Jul 04 2019

%C a(n) is 2^n represented in bijective base-2 numeration. - _Alois P. Heinz_, Aug 26 2019

%C 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

%C 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

%H Ivan Panchenko, <a href="/A047855/b047855.txt">Table of n, a(n) for n = 0..200</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Bijective_numeration">Bijective numeration</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11,-10).

%F a(n) = (10^n + 8)/9. - _Ralf Stephan_, Feb 14 2004

%F 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

%F G.f.: (1 - 9*x)/(1 - 11*x + 10*x^2). - _Philippe Deléham_, Oct 05 2009

%F a(n) = 10*a(n-1) - 8 (with a(0) = 1). - _Vincenzo Librandi_, Aug 06 2010

%p 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

%t Join[{1}, Table[FromDigits[PadLeft[{2}, n, 1]], {n, 30}]] (* _Harvey P. Dale_, Apr 17 2013 *)

%t (10^Range[0, 29] + 8)/9 (* _Alonso del Arte_, Apr 12 2020 *)

%o (PARI) a(n)=if(n==0,1,if(n==1,2,11*a(n-1)-10*a(n-2)))

%o for(i=0,10,print1(a(i),",")) \\ Lambert Klasen, Jan 28 2005

%o (Sage) [gaussian_binomial(n,1,10)+1 for n in range(17)] # _Zerinvary Lajos_, May 29 2009

%o (Scala) (List.fill(20)(10: BigInt)).scanLeft(1: BigInt)(_ * _).map(n => (n + 8)/9) // _Alonso del Arte_, Apr 12 2020

%o (Magma) [(10^n + 8)/9: n in [0..40]]; // _G. C. Greubel_, Jan 11 2025

%o (Python)

%o def A047855(n): return (pow(10,n) +8)//9

%o print([A047855(n) for n in range(41)]) # _G. C. Greubel_, Jan 11 2025

%Y Cf. A001019, A002275, A028842, A028843, A047848, A115549, A164898.

%K nonn,easy,changed

%O 0,2

%A _Clark Kimberling_

%E More terms from _Harvey P. Dale_, Apr 17 2013

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