|
| |
|
|
A048777
|
|
First partial sums of A005409; second partial sums of A001333.
|
|
2
|
|
|
|
1, 5, 16, 44, 113, 281, 688, 1672, 4049, 9789, 23648, 57108, 137889, 332913, 803744, 1940432, 4684641, 11309749, 27304176, 65918140, 159140497, 384199177, 927538896, 2239277016, 5406092977, 13051463021, 31509019072, 76069501220, 183648021569, 443365544417
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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
|
|
|
|
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
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)
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
|
|
|
|
STATUS
|
approved
|
| |
|
|
