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 A125128 a(n) = 2^(n+1) - n - 2, or partial sums of main diagonal of array A125127 of k-step Lucas numbers. 14
 1, 4, 11, 26, 57, 120, 247, 502, 1013, 2036, 4083, 8178, 16369, 32752, 65519, 131054, 262125, 524268, 1048555, 2097130, 4194281, 8388584, 16777191, 33554406, 67108837, 134217700, 268435427, 536870882, 1073741793, 2147483616, 4294967263, 8589934558 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Essentially a duplicate of A000295: a(n) = A000295(n+1). Partial sums of main diagonal of array A125127 = L(k,n): k-step Lucas numbers, read by antidiagonals. Equals row sums of triangle A130128. - Gary W. Adamson, May 11 2007 Row sums of triangle A130330 which is composed of (1,3,7,15,...) in every column, thus: row sums of (1; 3,1; 7,3,1; ...). - Gary W. Adamson, May 24 2007 Row sums of triangle A131768. - Gary W. Adamson, Jul 13 2007 Convolution A130321 * (1, 2, 3, ...). Binomial transform of (1, 3, 4, 4, 4, ...). - Gary W. Adamson, Jul 27 2007 Row sums of triangle A131816. - Gary W. Adamson, Jul 30 2007 A000975 convolved with [1, 2, 2, 2, ...]. - Gary W. Adamson, Jun 02 2009 The eigensequence of a triangle with the triangular series as the left border and the rest 1's. - Gary W. Adamson, Jul 24 2010 LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..1000 Index entries for linear recurrences with constant coefficients, signature (4,-5,2). FORMULA a(n) = A000295(n+1) = 2^(n+1) - n - 2 = Sum_{i=1..n} A125127(i,i) = Sum_{i=1..n} ((2^i)-1). [Edited by M. F. Hasler, Jul 30 2015] From Colin Barker, Jun 17 2012: (Start) a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3). G.f.: x/((1-x)^2*(1-2*x)). (End) a(n) = A000225(n) + A000325(n) - 1. - Miquel Cerda, Aug 07 2016 a(n) = A095151(n) - A000225(n). - Miquel Cerda, Aug 12 2016 EXAMPLE a(1) = 1 because "1-step Lucas number"(1) = 1. a(2) = 4 = a(1) + [2-step] Lucas number(2) = 1 + 3. a(3) = 11 = a(2) + 3-step Lucas number(3) = 1 + 3 + 7. a(4) = 26 = a(3) + 4-step Lucas number(4) = 1 + 3 + 7 + 15. a(5) = 57 = a(4) + 5-step Lucas number(5) = 1 + 3 + 7 + 15 + 31. a(6) = 120 = a(5) + 6-step Lucas number(6) = 1 + 3 + 7 + 15 + 31 + 63. G.f. = x + 4*x^2 + 11*x^3 + 26*x^4 + 57*x^5 + 120*x^6 + 247*x^7 + 502*x^8 + ... MATHEMATICA CoefficientList[Series[1/((1-x)^2*(1-2*x)), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 28 2012 *) LinearRecurrence[{4, -5, 2}, {1, 4, 11}, 30] (* Harvey P. Dale, Nov 16 2014 *) a[ n_] := With[{m = n + 1}, If[ m < 0, 0, 2^m - (1 + m)]]; (* Michael Somos, Aug 17 2015 *) PROG (MAGMA) I:=[1, 4, 11]; [n le 3 select I[n] else 4*Self(n-1)-5*Self(n-2)+2*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jun 28 2012 (PARI) A125128(n)=2<

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Last modified July 22 20:51 EDT 2019. Contains 325226 sequences. (Running on oeis4.)