A111065 -id:A111065 - cp's OEIS frontend

A000787 Strobogrammatic numbers: the same upside down.

0, 1, 8, 11, 69, 88, 96, 101, 111, 181, 609, 619, 689, 808, 818, 888, 906, 916, 986, 1001, 1111, 1691, 1881, 1961, 6009, 6119, 6699, 6889, 6969, 8008, 8118, 8698, 8888, 8968, 9006, 9116, 9696, 9886, 9966, 10001, 10101, 10801, 11011, 11111, 11811, 16091, 16191

Offset: 1

N. J. A. Sloane

Strobogrammatic numbers are a kind of ambigrams that retain the same meaning when viewed upside down. - Daniel Mondot, Sep 27 2016

"Upside down" here means rotated by 180 degrees (i.e., central symmetry), NOT "vertically flipped" (symmetry w.r.t. horizontal line, which are in A045574). - M. F. Hasler, May 04 2012

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)
J. M. Howell, Strobogrammatic years, Math. Mag., 34 (1961), p. 182 and 184.
N. J. A. Sloane, "A Handbook of Integer Sequences" Fifty Years Later, arXiv:2301.03149 [math.NT], 2023, p. 5.

Cf. A007597 (Primes in this sequence), A057770, A111065, A169731 (another version).

Subsequence of A045574. - M. F. Hasler, May 04 2012

fQ[n_] := Block[{s = {0, 1, 6, 8, 9}, id = IntegerDigits[n]}, If[ Union[ Join[s, id]] == s && (id /. {6 -> 9, 9 -> 6}) == Reverse[id], True, False]]; Select[ Range[0, 16190], fQ[ # ] &] (* Robert G. Wilson v, Oct 11 2005 *)

from itertools import count, islice, product
def ud(s): return s[::-1].translate({ord('6'):ord('9'), ord('9'):ord('6')})
def agen():
    yield from [0, 1, 8]
    for d in count(2):
        for start in "1689":
            for rest in product("01689", repeat=d//2-1):
                left = start + "".join(rest)
                right = ud(left)
                for mid in [[""], ["0", "1", "8"]][d%2]:
                    yield int(left + mid + right)
print(list(islice(agen(), 47))) # Michael S. Branicky, Mar 29 2022

More terms from Robert G. Wilson v, Oct 11 2005

A006072 Numbers with mirror symmetry about middle.

Original entry on oeis.org

0, 1, 8, 11, 88, 101, 111, 181, 808, 818, 888, 1001, 1111, 1881, 8008, 8118, 8888, 10001, 10101, 10801, 11011, 11111, 11811, 18081, 18181, 18881, 80008, 80108, 80808, 81018, 81118, 81818, 88088, 88188, 88888, 100001, 101101, 108801, 110011, 111111, 118811, 180081

Offset: 1

N. J. A. Sloane

Apparently this sequence and A111065 have the same parity. - Jeremy Gardiner, Oct 15 2005
Obviously, terms of this sequence also have the same parity (and also the same digital sum mod 6) as those of A118594, see below. - M. F. Hasler, May 08 2013
The number of n-digit terms is given by A225367 -- which counts palindromes in base 3, A118594. The terms here are the base 3 palindromes considered there, with 2 replaced by 8 (which means this sequence A006072 arises from A118594 not only by taking the 3rd power of each digit, but also by superposing the number with its horizontal or vertical reflection, somehow remarkably given the symmetry of numbers considered here). - M. F. Hasler, May 05 2013 [Part of the comment moved from A225367 to here on May 08 2013]

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..1450 from Vincenzo Librandi)
Eric Weisstein's World of Mathematics, Tetradic Number

Subsequence of A000787. Cf. A045574.

NextPalindrome[n_] := Block[{l = Floor[Log[10, n] + 1], idn = IntegerDigits[n]}, If[ Union[idn] == {9}, Return[n + 2], If[l < 2, Return[n + 1], If[ FromDigits[ Reverse[ Take[idn, Ceiling[l/2]]]] > FromDigits[ Take[idn, -Ceiling[l/2]]], FromDigits[ Join[ Take[idn, Ceiling[l/2]], Reverse[ Take[idn, Floor[l/2]]]]], idfhn = FromDigits[ Take[idn, Ceiling[l/2]]] + 1; idp = FromDigits[ Join[ IntegerDigits[ idfhn], Drop[ Reverse[ IntegerDigits[ idfhn]], Mod[l, 2]]]]]]]]; np = 0; t = {0}; Do[np = NextPalindrome[np]; If[Union[Join[{0, 1, 8}, IntegerDigits[np]]] == {0, 1, 8}, AppendTo[t, np]], {n, 1150}]; t (* Robert G. Wilson v *)
TetrNumsUpTo10powerK[k_]:= Select[FromDigits/@ Tuples[{0, 1, 8}, k],IntegerDigits[#] == Reverse[IntegerDigits[#]] &]; TetrNumsUpTo10powerK[7] (* Mikk Heidemaa, May 21 2017 *)

{for(l=1,5,u=vector((l+1)\2,i,10^(i-1)+(2*i-11&&i==1,2]), print1((v+v\2*6)*u", ")))} \\ The n-th term could be produced by using (partial sums of) A225367 to skip all shorter terms, and then skipping the adequate number of vectors v until n is reached. - M. F. Hasler, May 05 2013

from itertools import count, islice, product
def agen():
    yield from [0, 1, 8]
    for d in count(2):
        for start in "18":
            for rest in product("018", repeat=d//2-1):
                left = start + "".join(rest)
                for mid in [[""], ["0", "1", "8"]][d%2]:
                    yield int(left + mid + left[::-1])
print(list(islice(agen(), 42))) # Michael S. Branicky, Mar 29 2022

a(n) = digit-wise application of A000578 to A118594(n). - M. F. Hasler, May 08 2013

More terms from Robert G. Wilson v, Nov 16 2005

A111575 Powers of 3 repeated four times.

Original entry on oeis.org

1, 1, 1, 1, 3, 3, 3, 3, 9, 9, 9, 9, 27, 27, 27, 27, 81, 81, 81, 81, 243, 243, 243, 243, 729, 729, 729, 729, 2187, 2187, 2187, 2187, 6561, 6561, 6561, 6561, 19683, 19683, 19683, 19683, 59049, 59049, 59049, 59049, 177147, 177147, 177147, 177147, 531441

Offset: 0

Jeremy Gardiner, Nov 17 2005

Generating sequence for the number of 0's and 1's (run lengths) in the parity of A006072, A111065 and A118594.

			a(10) = 3^floor(10/4) = 3^2 = 9.

Vincenzo Librandi, Table of n, a(n) for n = 0..8000
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 3).

Cf. A006072, A111065, A108411, A118594, A127975.

[3^(Floor(n/4)):n in [0..50]]; // Vincenzo Librandi, Sep 20 2011

With[{pt=3^Range[0,15]},Sort[Join[pt,pt,pt,pt]]] (* Harvey P. Dale, Sep 17 2011 *)
Table[PadRight[{},4,3^n],{n,0,15}]//Flatten (* Harvey P. Dale, Jan 17 2019 *)

a(n)=3^(n\4) \\ Charles R Greathouse IV, Oct 03 2016

a(n) = 3^floor(n/4).

O.g.f.: -(1+x)*(1+x^2)/(-1+3*x^4). - R. J. Mathar, Jan 08 2008

A111156 Numbers that look the same when printed upside down.

Original entry on oeis.org

0, 8, 69, 88, 96, 609, 689, 808, 888, 906, 986, 6009, 6699, 6889, 6969, 8008, 8698, 8888, 8968, 9006, 9696, 9886, 9966, 60009, 60809, 66099, 66899, 68089, 68889, 69069, 69869, 80008, 80808, 86098, 86898, 88088, 88888, 89068, 89868, 90006, 90806, 96096, 96896, 98086, 98886, 99066, 99866, 600009

Offset: 1

Paul Stoeber (pstoeber(AT)uni-potsdam.de), Oct 09 2005

Numbers with 1's are excluded.

Numbers written with digits 0,6,8,9, with 6 and 9 interchanged when reversed. - Robert Israel, Jul 02 2018

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. strobogrammatic numbers A000787. If 8's are excluded we get A111065.

main=print$"0":concat[concat[[reverse(reverse(map f x)++z++x)|x<-y]|z<-["","0"]]|y<-s(iterate i"6")];f '0'='0';f '6'='9';f '8'='8';f '9'='6';i('0':x)='6':x;i('6':x)='8':x;i('8':x)='9':x;i('9':x)='0':i x;i""="6";s(x:y@(z:_))=let w:v=s y in if length x==length z then(x:w):v else[x]:w:v

f:= proc(n) local L,Lp,nl;
   L:= subs(1=6,2=8,3=9,convert(n,base,4));
   nl:= nops(L);
   Lp:= subs([6=9,9=6],L);
   add(Lp[-i]*10^(i-1),i=1..nl)+add(L[i]*10^(nl+i-1),i=1..nl);
end proc:
g:= proc(n) local L,Lp,nl;
   L:= subs(1=6,2=8,3=9,convert(n,base,4));
   nl:= nops(L);
   Lp:= subs([6=9,9=6],L);
   seq(add(Lp[-i]*10^(i-1),i=1..nl)+x*10^nl+add(L[i]*10^(nl+i),i=1..nl),x=[0,8]);
end proc:
0,8,seq(op([seq(f(n),n=4^i..4^(i+1)-1),seq(g(n),n=4^i..4^(i+1)-1)]),i=0..2); # Robert Israel, Jul 02 2018

Select[Range[0,600010],ContainsOnly[IntegerDigits[#],{0,6,8,9}]&&IntegerReverse[FromDigits[IntegerDigits[#]/.{6->9,9->6}]]==#&] (* James C. McMahon, Apr 30 2024 *)

from itertools import count, islice, product
def ud(s): return s[::-1].translate({ord('6'):ord('9'), ord('9'):ord('6')})
def agen():
    yield from [0, 8]
    for d in count(2):
        for start in "689":
            for rest in product("0689", repeat=d//2-1):
                left = start + "".join(rest)
                right = ud(left)
                for mid in [[""], ["0", "8"]][d%2]:
                    yield int(left + mid + right)
print(list(islice(agen(), 48))) # Michael S. Branicky, Mar 29 2022

Corrected by Robert Israel, Jul 02 2018

A246880 6((10^n-1)/9)(10^(n+1))+9*(10^n-1)/9.

Original entry on oeis.org

0, 609, 66099, 6660999, 666609999, 66666099999, 6666660999999, 666666609999999, 66666666099999999, 6666666660999999999, 666666666609999999999, 66666666666099999999999, 6666666666660999999999999, 666666666666609999999999999, 66666666666666099999999999999

Offset: 0

Felix Fröhlich, Sep 06 2014

Numbers of the form 6...609...9 (i.e., consisting of an odd number of digits with the middle digit 0, all digits to the left of the middle digit 6 and all digits to the right of the middle digit 9).

Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Subsequence of A001743, A039434, A111065, A111156, A153806, A169731 and A227793.

[(6*((10^n - 1)/9))*(10^(n + 1)) + (9*(10^n - 1)/9) : n in [0..15]]; // Wesley Ivan Hurt, Sep 15 2014

A246880:=n->(6*((10^n - 1)/9))*(10^(n + 1)) + (9*(10^n - 1)/9): seq(A246880(n),n=0..15); # Wesley Ivan Hurt, Sep 15 2014

Table[(6*((10^n - 1)/9))*(10^(n + 1)) + (9*(10^n - 1)/9), {n, 15}] (* Wesley Ivan Hurt, Sep 15 2014 *)
Join[{0}, CoefficientList[Series[20/(3 x - 300 x^2) + 1/(x^2 - x) + 17/(30 x^2 - 3 x), {x, 0, 30}], x]] (* Wesley Ivan Hurt, Sep 15 2014 *)

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

G.f.: 20/(3-300*x)+1/(x-1)+17/(30*x-3); a(n) = 111*a(n-1)-1110*a(n-2)+1000*a(n-3). - Wesley Ivan Hurt, Sep 15 2014

cp's OEIS Frontend

A000787 Strobogrammatic numbers: the same upside down.

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A006072 Numbers with mirror symmetry about middle.

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A111575 Powers of 3 repeated four times.

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Magma

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A111156 Numbers that look the same when printed upside down.

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A246880 6*((10^n-1)/9)*(10^(n+1))+9*(10^n-1)/9.

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Author

Keywords

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Programs

Magma

Maple

Mathematica

PARI

Formula

A246880 6((10^n-1)/9)(10^(n+1))+9*(10^n-1)/9.