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A048777 First partial sums of A005409; second partial sums of A001333. 1
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=0,1,2..  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

Table of n, a(n) for n=0..29.

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; 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.

(1/2) [Pell(n+3) - Pell(n+2) - 2n - 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)). - Colin Barker, Sep 20 2012

a(n) = A048776(n-1)+A048776(n). - R. J. Mathar, Feb 28 2013

MATHEMATICA

LinearRecurrence[{4, -4, 0, 1}, {1, 5, 16, 44}, 40] (* Harvey P. Dale, Nov 12 2017 *)

CROSSREFS

Cf. A001333, A000129, A005409.

Sequence in context: A079094 A144952 A053220 * A300961 A270134 A269754

Adjacent sequences:  A048774 A048775 A048776 * A048778 A048779 A048780

KEYWORD

easy,nonn

AUTHOR

Barry E. Williams

EXTENSIONS

More terms from Harvey P. Dale, Nov 12 2017

STATUS

approved

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Last modified October 19 02:41 EDT 2019. Contains 328211 sequences. (Running on oeis4.)