

A006072


Numbers with mirror symmetry about middle.
(Formerly M4481)


11



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
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OFFSET

1,3


COMMENTS

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 ndigit 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]


REFERENCES

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


LINKS



FORMULA



MATHEMATICA

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 *)


PROG

(PARI) {for(l=1, 5, u=vector((l+1)\2, i, 10^(i1)+(2*i1<l)*10^(li))~; forvec(v=vector((l+1)\2, i, [l>1&&i==1, 2]), print1((v+v\2*6)*u", ")))} \\ The nth 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
(Python)
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//21):
left = start + "".join(rest)
for mid in [[""], ["0", "1", "8"]][d%2]:
yield int(left + mid + left[::1])


CROSSREFS



KEYWORD

base,nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



