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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

%I

%S 1,4,11,26,57,120,247,502,1013,2036,4083,8178,16369,32752,65519,

%T 131054,262125,524268,1048555,2097130,4194281,8388584,16777191,

%U 33554406,67108837,134217700,268435427,536870882,1073741793,2147483616,4294967263,8589934558

%N a(n) = 2^(n+1) - n - 2, or partial sums of main diagonal of array A125127 of k-step Lucas numbers.

%C Essentially a duplicate of A000295: a(n) = A000295(n+1).

%C Partial sums of main diagonal of array A125127 = L(k,n): k-step Lucas numbers, read by antidiagonals.

%C Equals row sums of triangle A130128. - _Gary W. Adamson_, May 11 2007

%C 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

%C Row sums of triangle A131768. - _Gary W. Adamson_, Jul 13 2007

%C Convolution A130321 * (1, 2, 3, ...). Binomial transform of (1, 3, 4, 4, 4, ...). - _Gary W. Adamson_, Jul 27 2007

%C Row sums of triangle A131816. - _Gary W. Adamson_, Jul 30 2007

%C A000975 convolved with [1, 2, 2, 2, ...]. - _Gary W. Adamson_, Jun 02 2009

%C The eigensequence of a triangle with the triangular series as the left border and the rest 1's. - _Gary W. Adamson_, Jul 24 2010

%H Vincenzo Librandi, <a href="/A125128/b125128.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,-5,2).

%F 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]

%F From _Colin Barker_, Jun 17 2012: (Start)

%F a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3).

%F G.f.: x/((1-x)^2*(1-2*x)). (End)

%F a(n) = A000225(n) + A000325(n) - 1. - _Miquel Cerda_, Aug 07 2016

%F a(n) = A095151(n) - A000225(n). - _Miquel Cerda_, Aug 12 2016

%F E.g.f.: 2*exp(2*x) - (2+x)*exp(x). - _G. C. Greubel_, Jul 26 2019

%e a(1) = 1 because "1-step Lucas number"(1) = 1.

%e a(2) = 4 = a(1) + [2-step] Lucas number(2) = 1 + 3.

%e a(3) = 11 = a(2) + 3-step Lucas number(3) = 1 + 3 + 7.

%e a(4) = 26 = a(3) + 4-step Lucas number(4) = 1 + 3 + 7 + 15.

%e a(5) = 57 = a(4) + 5-step Lucas number(5) = 1 + 3 + 7 + 15 + 31.

%e a(6) = 120 = a(5) + 6-step Lucas number(6) = 1 + 3 + 7 + 15 + 31 + 63.

%e G.f. = x + 4*x^2 + 11*x^3 + 26*x^4 + 57*x^5 + 120*x^6 + 247*x^7 + 502*x^8 + ...

%t CoefficientList[Series[1/((1-x)^2*(1-2*x)),{x,0,40}],x] (* _Vincenzo Librandi_, Jun 28 2012 *)

%t LinearRecurrence[{4,-5,2},{1,4,11},40] (* _Harvey P. Dale_, Nov 16 2014 *)

%t a[ n_] := With[{m = n + 1}, If[ m < 0, 0, 2^m - (1 + m)]]; (* _Michael Somos_, Aug 17 2015 *)

%o (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

%o (PARI) A125128(n)=2<<n-n-2 \\ _M. F. Hasler_, Jul 30 2015

%o (PARI) {a(n) = n++; if( n<0, 0, 2^n - (1+n))}; /* _Michael Somos_, Aug 17 2015 */

%o (Sage) [2^(n+1) -n-2 for n in (1..40)] # _G. C. Greubel_, Jul 26 2019

%o (GAP) List([1..40], n-> 2^(n+1) -n-2); # _G. C. Greubel_, Jul 26 2019

%Y Cf. A000295, A125127, A125129, A130103.

%K easy,nonn

%O 1,2

%A _Jonathan Vos Post_, Nov 22 2006

%E Edited by _M. F. Hasler_, Jul 30 2015

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Last modified March 31 01:53 EDT 2020. Contains 333133 sequences. (Running on oeis4.)