A115549 -id:A115549 - cp's OEIS frontend

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

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

Offset: 0

Clark Kimberling

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

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

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

Cf. A002281, A062397.

[(10^n + 8)/9: n in [0..40]]; // G. C. Greubel, Jan 11 2025

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

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 *)

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

def A047855(n): return (pow(10,n) +8)//9
print([A047855(n) for n in range(41)]) # G. C. Greubel, Jan 11 2025

[gaussian_binomial(n,1,10)+1 for n in range(17)] # Zerinvary Lajos, May 29 2009

(List.fill(20)(10: BigInt)).scanLeft(1: BigInt)( * ).map(n => (n + 8)/9) // Alonso del Arte, Apr 12 2020

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)

More terms from Harvey P. Dale, Apr 17 2013

A380428 Numbers k for which nonnegative integers x and y exist such that k is the concatenation of x and y as well as k = (x + y)^2.

Original entry on oeis.org

81, 100, 2025, 3025, 88209, 494209, 4941729, 7441984, 24502500, 25502500, 52881984, 60481729, 300814336, 493817284, 6049417284, 6832014336, 20408122449, 21948126201, 33058148761, 35010152100, 43470165025, 101558217124, 108878221089, 123448227904, 127194229449, 152344237969

Offset: 1

Felix Huber, Jan 25 2025

Subsequence of A000290.
From David A. Corneth, Apr 26 2025: (Start)
If y has q digits then a term m is of the form (x + y) = 10^q * x + y. Choosing some y we can solve for x (the equation is a quadratic with respect to x) and see if it produces a term.
y comes from A238712.
The sequence is infinite; it contains (25*100^i +- 5*10^i)^2 = concat(25*100^i +- 5*10^i, 25*100^i) for all i >= 0.
Neither x nor y can have a leading 0. (End)

			2025 is in the sequence because (20 + 25)^2 = 2025.
100 is in the sequence because (10 + 0)^2 = 100.
88209 is in the sequence because (88 + 209)^2 = 88209.
From _David A. Corneth_, Apr 26 2025: (Start)
9801 is not in the sequence even though (98 + 01)^2 = 9801 but 01 has a leading 0 which is disallowed.
If a term m ends in y = 209 where y has three digits we have 10^3*x + y = (x + y)^2. Solving for x gives x = 88 or x = 494 corresponding to terms 88209 and 494209. (End)

David A. Corneth, Table of n, a(n) for n = 1..96 (terms <= 10^20)

Cf. A000290, A238712.

Cf. A115527, A115528, A115529, A115530, A115531, A115532, A115533, A115534, A115535, A115536, A115537, A115538, A115539, A115540, A115541, A115542, A115543, A115544, A115545, A115546, A115547, A115548, A115549, A115550, A115551, A115552, A115553, A115554, A115555, A115556.

A380428:=proc(n)
    option remember;
    local a,i,k,x,y;
    if n=1 then
        81
    elif n=2 then
        100
    else
        for a from isqrt(procname(n-1))+1 do
            k:=length(a^2);
            for i to k-1 do
                x:=floor(a^2/10^i);
                y:=a^2-x*10^i;
                if x+y=a and length(x)+length(y)=k then
                    return a^2
                fi
            od
        od
    fi;
end proc;
seq(A380428(n),n=1..26);

cp's OEIS Frontend

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

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