A100631 -id:A100631 - cp's OEIS frontend

A341344 a(n) = A100631(n, floor(n/2)).

1, 1, 2, 4, 12, 32, 104, 304, 1008, 3072, 10272, 32064, 107712, 341504, 1150592, 3688192, 12451584, 40239104, 136053248, 442442752, 1497664512, 4894728192, 16583583744, 54419632128, 184511361024, 607524225024, 2061074178048, 6805625192448, 23100352413696, 76462341095424, 259648659554304

Offset: 0

Petros Hadjicostas, Feb 09 2021

nonn

"Middle" diagonal of Reinhard Zumkeller's symmetric (Pascal-like) triangular array A100631.

Cf. A084773, A100631, A152254.

a(n) = {my(m=matrix(n+1, n+1)); for (i=1, n+1, for (j=1, n+1, if ((j==1) || (j==i), m[i, j] = 1, if (j<=n, m[i,j] = 2*(if (i>1, m[i-1,j-1] + m[i-1,j], 0) - if (i>2, m[i-2,j-1], 0) ))););); m[n+1, (n+2)\2];} \\ Michel Marcus, Feb 10 2021

a(n) = A100631(n, floor(n/2)) = A100631(n, ceiling(n/2)).
a(2*n) = A152254(n-1) = 2*A084773(n-1) for n >= 1.
a(n) = 2^ceiling(n/2)*hypergeom([-floor(n/2) + 1, ceiling(n/2)], [1], -1); see the comments for A100631. - Petros Hadjicostas, Feb 10 2021

A087161 Records in A087159, i.e., A087159(a(n)) = n, and satisfies the recurrence a(n+3) = 5a(n+2) - 6 a(n+1) + 2*a(n) with a(1) = 1, a(2) = 2, and a(3) = 4.

Original entry on oeis.org

1, 2, 4, 10, 30, 98, 330, 1122, 3826, 13058, 44578, 152194, 519618, 1774082, 6057090, 20680194, 70606594, 241065986, 823050754, 2810071042, 9594182658, 32756588546, 111837988866, 381838778370, 1303679135746, 4451038986242

Offset: 1

Paul D. Hanna, Aug 22 2003

nonn

Binomial transform of A001333 (which, with an extra leading 1, is the expansion of (1 - x - 2*x^2)/(1 - 2*x - x^2)). - Paul Barry, Aug 26 2003

Partial sums of the binomial transform of Pell(n-1). - Paul Barry, Apr 24 2004

Index entries for linear recurrences with constant coefficients, signature (5,-6,2).

Cf. A000129, A007070, A087159, A087160, A100631.

CoefficientList[Series[(1-3x)/(1-5x+6x^2-2x^3),{x,0,30}],x] (* or *) LinearRecurrence[{5,-6,2},{1,2,4},30] (* Harvey P. Dale, Oct 12 2015 *)

G.f.: x*(1 - 3*x)/(1 - 5*x + 6*x^2 - 2*x^3).
a(n) = 2 + 2*A007070(n-3) for n > 2.
a(n) = ((2 - sqrt(2))^(n)/(1 - sqrt(2)) + (2 + sqrt(2))^(n)/(1 + sqrt(2)))/2 + 2 (offset 0) - Paul Barry, Aug 26 2003
a(n+1) - a(n) = A006012(n-1) for n >= 2. - Philippe Deléham, Feb 01 2012
a(1) = 1, a(2) = 2, a(3) = 4, a(n) = 5*a(n-1) - 6*a(n-2) + 2*a(n-3) for n >= 4. - Harvey P. Dale, Oct 12 2015
a(n+1) = Sum_{k=0..n} A100631(n,k) for n >= 0. - Petros Hadjicostas, Feb 09 2021

More terms from Paul Barry, Apr 24 2004

cp's OEIS Frontend

A341344 a(n) = A100631(n, floor(n/2)).

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A087161 Records in A087159, i.e., A087159(a(n)) = n, and satisfies the recurrence a(n+3) = 5a(n+2) - 6 a(n+1) + 2*a(n) with a(1) = 1, a(2) = 2, and a(3) = 4.

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Programs

Mathematica

Formula

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cp's OEIS Frontend

A341344 a(n) = A100631(n, floor(n/2)).

Views

Author

Keywords

Comments

Crossrefs

Programs

PARI

Formula

A087161 Records in A087159, i.e., A087159(a(n)) = n, and satisfies the recurrence a(n+3) = 5*a(n+2) - 6* a(n+1) + 2*a(n) with a(1) = 1, a(2) = 2, and a(3) = 4.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A087161 Records in A087159, i.e., A087159(a(n)) = n, and satisfies the recurrence a(n+3) = 5a(n+2) - 6 a(n+1) + 2*a(n) with a(1) = 1, a(2) = 2, and a(3) = 4.