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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 (list; graph; refs; listen; history; text; internal format)
 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)]. LINKS Colin Barker, Table of n, a(n) for n = 0..334 Index entries for linear recurrences with constant coefficients, signature (1101,-101000). FORMULA a(n) = (1/899)*(1000^n-101^n), with n>=1. - Paolo P. Lava, Jun 10 2010 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 Cf. A000930, A007318. Sequence in context: A283356 A283172 A283253 * A178348 A359149 A250796 Adjacent sequences: A178404 A178405 A178406 * A178408 A178409 A178410 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

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Last modified November 29 14:38 EST 2023. Contains 367445 sequences. (Running on oeis4.)