OFFSET
0,2
COMMENTS
Form an array having the first column all 1's and the first row the squares 1, 4, 9, ..., so m(n,1) = 1 and m(1,n) = n^2 for n = 1, 2, 3, ..., and let the interior terms be m(i,j) = m(i,j-1) + m(i-1,j-1) + m(i-1,j). Then the sums of the terms in the antidiagonals are the terms of this sequence. - J. M. Bergot, Nov 16 2012
Define a triangle with T(n,n)=n+1 and T(n,0)=n*(n+1)+1 for n >= 0. Define the interior terms via T(r,c) = T(r-2,c-1) + T(r-1,c-1) + T(r-1,c). Then the row sums are a(n) = Sum_{k=0..n} T(n,k). - J. M. Bergot, Feb 27 2013
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-4,0,1).
FORMULA
a(n) = 2*a(n-1) + a(n-2) + 2*n+1 with a(0)=1, a(1)=5.
a(n) = ( {(5+(7/2)*sqrt(2))*(1+sqrt(2))^n - (5-(7/2)*sqrt(2))*(1-sqrt(2))^n}/2*sqrt(2) ) - (2*n+5)/2.
a(n) = (1/2)*( Pell(n+3) + Pell(n+2) -2*n -5 ), with Pell(n) = A000129(n). - Ralf Stephan, May 15 2007
From Colin Barker, Sep 20 2012: (Start)
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-4).
G.f.: (1+x)/((1-x)^2*(1-2*x-x^2)). (End)
a(n) = (A002203(n+3) - 2*(2*n+5))/4. - G. C. Greubel, Apr 23 2021
E.g.f.: exp(x)*(7*cosh(sqrt(2)*x) + 5*sqrt(2)*sinh(sqrt(2)*x) - 2*x - 5)/2. - Stefano Spezia, May 13 2023
MATHEMATICA
LinearRecurrence[{4, -4, 0, 1}, {1, 5, 16, 44}, 40] (* Harvey P. Dale, Nov 12 2017 *)
Table[(LucasL[n+3, 2] -2*(2n+5))/4, {n, 0, 35}] (* G. C. Greubel, Apr 23 2021 *)
PROG
(Magma) I:=[1, 5, 16, 44]; [n le 4 select I[n] else 4*Self(n-1) -4*Self(n-2) +Self(n-4): n in [1..36]]; // G. C. Greubel, Apr 23 2021
(Sage) [(lucas_number2(n+3, 2, -1) -2*(2*n+5))/4 for n in (0..35)] # G. C. Greubel, Apr 23 2021
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
EXTENSIONS
More terms from Harvey P. Dale, Nov 12 2017
STATUS
approved