OFFSET
1,2
COMMENTS
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)
E.g.f.: 2*exp(2*x) - (2+x)*exp(x). - G. C. Greubel, Jul 26 2019
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}, 40] (* 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<<n-n-2 \\ M. F. Hasler, Jul 30 2015
(PARI) {a(n) = n++; if( n<0, 0, 2^n - (1+n))}; /* Michael Somos, Aug 17 2015 */
(Sage) [2^(n+1) -n-2 for n in (1..40)] # G. C. Greubel, Jul 26 2019
(GAP) List([1..40], n-> 2^(n+1) -n-2); # G. C. Greubel, Jul 26 2019
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, Nov 22 2006
EXTENSIONS
Edited by M. F. Hasler, Jul 30 2015
STATUS
approved