A193345 Digits occurring in A173616.

1, 1, 7, 8, 7, 0, 1, 9, 7, 2, 3, 0, 8, 5, 5, 1, 9, 6, 5, 4, 6, 3, 8, 8, 0, 5, 5, 0, 3, 2, 7, 9, 6, 8, 6, 7, 5, 0, 4, 9, 5, 0, 5, 9, 9, 0, 5, 2, 5, 3, 3, 6, 6, 3, 4, 8, 2, 7, 8, 0, 0, 9, 0, 9, 4, 8, 5, 0, 3, 4, 4, 4, 8, 7, 2, 2, 9, 7, 9, 3, 7, 7, 7, 3, 8, 4

Offset

1,3

Comments

a(n) = A173616(n) - 10*A173616(n-1).

This is the 10-adic integer x such that x^9 == (10^n-9) mod 10^n for all n. It is the 10's complement of A225458. - Aswini Vaidyanathan, May 11 2013

Links

Table of n, a(n) for n=1..86.

Example

1111^1111=.........8711; 111^111=........711;
10^(1-4)(8711-711)=8 ==> a(4)=8
Comment from Aswini Vaidyanathan, May 11 2013:
1^9 == 1 (mod 10).
11^9 == 91 (mod 100).
711^9 == 991 (mod 1000).
8711^9 == 9991 (mod 10000).
78711^9 == 99991 (mod 100000).
78711^9 == 999991 (mod 1000000).

Mathematica

repunit[n_] := Sum[10^i, {i, 0, n-1}]; a[n_] := 10^(1-n)(PowerMod[repunit[n], repunit[n], 10^n] - PowerMod[repunit[n-1], repunit[n-1], 10^(n-1)]); Table[a[n], {n, 200}]

Prog

(PARI) n=0; for(i=1, 100, m=(10^i-9); for(x=0, 9, if(((n+(x*10^(i-1)))^9)%(10^i)==m, n=n+(x*10^(i-1)); print1(x", "); break))) (From Aswini Vaidyanathan, May 11 2013)

Crossrefs

Cf. A173616, A225458.

Sequence in context: A244263 A288023 A020506 * A197822 A055060 A010515

Adjacent sequences: A193342 A193343 A193344 * A193346 A193347 A193348

Keyword

nonn,base,easy

Author

José María Grau Ribas, Jul 23 2011

Extensions

Edited by N. J. A. Sloane, May 12 2013

Status

approved