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A178407
a(n+1) = a(n)*1000 + 101^n with a(0) = 0.
1
0, 1, 1101, 1111201, 1112231301, 1112335361401, 1112345871501501, 1112346933021651601, 1112347040235186811701, 1112347051063753867981801, 1112347052157439140666161901, 1112347052267901353207282352001, 1112347052279058036673935517552101
OFFSET
0,3
COMMENTS
Numerators of fractions in partial sums of sum_{n=0...infinity} 101^n/1000^(n+1)converging to fraction 1/899 or sum_{n=0...infinity} 100^n/999^(n+1) whose alternating sum converges to fraction 1/1099 [= alternating sum_99^n/1000^(n+1)].
FORMULA
From Colin Barker, Oct 02 2015: (Start)
a(n) = 1101*a(n-1) - 101000*a(n-2) for n>2.
G.f.: x / ((101*x-1)*(1000*x-1)).
(End)
EXAMPLE
As interlocking Pascal triangles starting a new triangle to the left of each row:
................1
............1.1.0.1
........1.1.1.1.2.0.1
....1.1.1.2.2.3.1.3.0.1
1.1.1.2.3.3.5.3.6.1.4.0.1
MATHEMATICA
RecurrenceTable[{a[0]==0, a[n]==1000*a[n-1]+101^(n-1)}, a, {n, 15}] (* Harvey P. Dale, Nov 18 2013 *)
PROG
(PARI) concat(0, Vec(x/((101*x-1)*(1000*x-1)) + O(x^30))) \\ Colin Barker, Oct 02 2015
CROSSREFS
Sequence in context: A283356 A283172 A283253 * A178348 A359149 A250796
KEYWORD
nonn,easy
AUTHOR
Mark Dols, May 27 2010
EXTENSIONS
Prepended zero (from definition), changed offset accordingly, and more terms from Harvey P. Dale, Nov 18 2013
STATUS
approved