A281362 -id:A281362 - cp's OEIS frontend

A109128 Triangle read by rows: T(n,k) = T(n-1,k-1) + T(n-1,k) + 1 for 0

1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 7, 11, 7, 1, 1, 9, 19, 19, 9, 1, 1, 11, 29, 39, 29, 11, 1, 1, 13, 41, 69, 69, 41, 13, 1, 1, 15, 55, 111, 139, 111, 55, 15, 1, 1, 17, 71, 167, 251, 251, 167, 71, 17, 1, 1, 19, 89, 239, 419, 503, 419, 239, 89, 19, 1, 1, 21, 109, 329, 659, 923, 923, 659, 329, 109, 21, 1

Offset: 0

Reinhard Zumkeller, Jun 20 2005

Eigensequence of the triangle = A001861. - Gary W. Adamson, Apr 17 2009

			Triangle begins as:
  1;
  1   1;
  1   3   1;
  1   5   5   1;
  1   7  11   7   1;
  1   9  19  19   9   1;
  1  11  29  39  29  11   1;
  1  13  41  69  69  41  13   1;
  1  15  55 111 139 111  55  15   1;
  1  17  71 167 251 251 167  71  17   1;
  1  19  89 239 419 503 419 239  89  19   1;

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Index entries for triangles and arrays related to Pascal's triangle

Cf. A000035, A000295, A001861, A005408, A014473, A028387, A030662.
Cf. A134760, A281362, A327767.
Cf. A000325 (row sums).
Sequence m*binomial(n,k) - (m-1): A007318 (m=1), this sequence (m=2), A131060 (m=3), A131061 (m=4), A131063 (m=5), A131065 (m=6), A131067 (m=7), A168625 (m=8).

a109128 n k = a109128_tabl !! n !! k
a109128_row n = a109128_tabl !! n
a109128_tabl = iterate (\row -> zipWith (+)
   ([0] ++ row) (1 : (map (+ 1) $ tail row) ++ [0])) [1]
-- Reinhard Zumkeller, Apr 10 2012

[2*Binomial(n,k) -1: k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 12 2020

A109128 := proc(n,k)
    2*binomial(n,k)-1 ;
end proc: # R. J. Mathar, Jul 12 2016

Table[2*Binomial[n,k] -1, {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 12 2020 *)

[[2*binomial(n,k) -1 for k in (0..n)] for n in (0..12)] # G. C. Greubel, Mar 12 2020

T(n,k) = T(n-1,k-1) + T(n-1,k) + 1 with T(n,0) = T(n,n) = 1.
Sum_{k=0..n} T(n, k) = A000325(n+1) (row sums).
T(n, k) = 2*binomial(n,k) - 1. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Sep 30 2007
T(n, 1) = 2*n - 1 = A005408(n+1) for n>0.
T(n, 2) = n^2 + n - 1 = A028387(n-2) for n>1.
T(n, k) = Sum_{j=0..n-k} C(n-k,j)*C(k,j)*(2 - 0^j) for k <= n. - Paul Barry, Apr 27 2006
T(n,k) = A014473(n,k) + A007318(n,k), 0 <= k <= n. - Reinhard Zumkeller, Apr 10 2012
From G. C. Greubel, Apr 06 2024: (Start)
T(n, n-k) = T(n, k).
T(2*n, n) = A134760(n).
T(2*n-1, n) = A030662(n), for n >= 1.
Sum_{k=0..n-1} T(n, k) = A000295(n+1), for n >= 1.
Sum_{k=0..n} (-1)^k*T(n, k) = 2*[n=0] - A000035(n+1).
Sum_{k=0..n-1} (-1)^k*T(n, k) = A327767(n), for n >= 1.
Sum_{k=0..floor(n/2)} T(n-k, k) = A281362(n).
Sum_{k=0..floor((n-1)/2)} T(n-k, k) = A281362(n-1) - (1+(-1)^n)/2 for n >= 1.
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = b(n), where b(n) is the repeating pattern {1,1,0,-2,-3,-1,2,2,-1,-3,-2,0} with b(n) = b(n-12). (End)

Offset corrected by Reinhard Zumkeller, Apr 10 2012

A020956 a(n) = Sum_{k>=1} floor(tau^(n-k)) where tau is A001622.

Original entry on oeis.org

1, 2, 4, 8, 14, 25, 42, 71, 117, 193, 315, 514, 835, 1356, 2198, 3562, 5768, 9339, 15116, 24465, 39591, 64067, 103669, 167748, 271429, 439190, 710632, 1149836, 1860482, 3010333, 4870830, 7881179, 12752025, 20633221, 33385263, 54018502, 87403783, 141422304

Offset: 1

Clark Kimberling

Colin Barker, Table of n, a(n) for n = 1..1000
Clark Kimberling, Problem 10520, Amer. Math. Mon. 103 (1996) p. 347.
Index entries for linear recurrences with constant coefficients, signature (2,1,-3,0,1).

Cf. A000032, A001622, A014217, A281362.

I:=[1,2,4,8,14]; [n le 5 select I[n] else 2*Self(n-1)+Self(n-2)-3*Self(n-3)+Self(n-5): n in [1..40]]; // Vincenzo Librandi, Nov 01 2016

[Lucas(n+1)-(2*n+5-(-1)^n)/4: n in [1..40]]; // G. C. Greubel, Apr 05 2024

LinearRecurrence[{2,1,-3,0,1}, {1,2,4,8,14}, 40] (* Vincenzo Librandi, Nov 01 2016 *)

Vec(x*(1-x^2+x^3)/((1-x-x^2)*(1+x)*(1-x)^2) + O(x^50)) \\ Michel Marcus, Nov 01 2016

prpr = 0
prev = 1
for n in range(2,100):
    print(prev, end=", ")
    curr = prpr+prev + n//2
    prpr = prev
    prev = curr
# Alex Ratushnyak, Jul 30 2012

[lucas_number2(n+1,1,-1) -(n+2+(n%2))//2 for n in range(1,41)] # G. C. Greubel, Apr 05 2024

G.f.: x*(1-x^2+x^3)/((1-x-x^2)*(1+x)*(1-x)^2). - Ralf Stephan, Apr 08 2004
a(n) = Lucas(n+1) - floor(n/2) - 1.
a(n) = Sum_{k=0..n-1} A014217(k).
a(n) = 2^(-2-n)*((-2)^n - 5*2^n + 2*(1-t)^(1+n) + 2*(1+t)^n + 2*t*(1+t)^n - 2^(1+n)*n) where t=sqrt(5). - Colin Barker, Feb 09 2017
From G. C. Greubel, Apr 05 2024: (Start)
a(n) = Lucas(n+1) - (1/4)*(2*n + 5 - (-1)^n).
E.g.f.: exp(x/2)*(cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2)) - (1/2)*((x+2)*cosh(x) + (x+3)*sinh(x)). (End)

More terms from Vladeta Jovovic, Apr 04 2002

A215004 a(0) = a(1) = 1; for n>1, a(n) = a(n-1) + a(n-2) + floor(n/2).

Original entry on oeis.org

1, 1, 3, 5, 10, 17, 30, 50, 84, 138, 227, 370, 603, 979, 1589, 2575, 4172, 6755, 10936, 17700, 28646, 46356, 75013, 121380, 196405, 317797, 514215, 832025, 1346254, 2178293, 3524562, 5702870, 9227448, 14930334, 24157799, 39088150, 63245967, 102334135, 165580121

Offset: 0

Alex Ratushnyak, Jul 31 2012

If the first two terms are {0,1}, we get A020956 except for the first term.

If the first two terms are {1,2}, we get A281362.

Colin Barker, Table of n, a(n) for n = 0..1000
Jean-Luc Baril, Sergey Kirgizov, and Armen Petrossian, Dyck Paths with catastrophes modulo the positions of a given pattern, Australasian J. Comb. (2022) Vol. 84, No. 2, 398-418.
Nathan Fox, Proof of formula for a(n).
Index entries for linear recurrences with constant coefficients, signature (2,1,-3,0,1).

Cf. A020956, except for first term: same formula, seed {0,1}.

Cf. A000045, A281362.

[Fibonacci(n+3)-(2*n+5-(-1)^n)/4: n in [0..40]]; // _G. C. Greubel, Feb 01 2018

Table[((-1)^n - 2 n + 8 Fibonacci[n] + 4 LucasL[n] - 5)/4, {n, 0, 20}] (* Vladimir Reshetnikov, May 18 2016 *)
RecurrenceTable[{a[0]==a[1]==1,a[n]==a[n-1]+a[n-2]+Floor[n/2]},a,{n,40}] (* or *) LinearRecurrence[{2,1,-3,0,1},{1,1,3,5,10},40] (* Harvey P. Dale, Jul 11 2020 *)

Vec(-(x^3-x+1)/((x-1)^2*(x+1)*(x^2+x-1)) + O(x^100)) \\ Colin Barker, Sep 16 2015

a(n)=([0,1,0,0,0;0,0,1,0,0;0,0,0,1,0;0,0,0,0,1;1,0,-3,1,2]^n* [1;1;3;5;10])[1,1] \\ Charles R Greathouse IV, Jan 16 2017

prpr = prev = 1
for n in range(2,100):
    print(prpr, end=', ')
    curr = prpr+prev + n//2
    prpr = prev
    prev = curr

[fibonacci(n+3) -(n+2+(n%2))//2 for n in range(41)] # G. C. Greubel, Apr 05 2024

From Colin Barker, Sep 16 2015: (Start)
a(n) = 2*a(n-1) + a(n-2) - 3*a(n-3) + a(n-5) for n>4.
G.f.: (1-x+x^3) / ((1-x)^2*(1+x)*(1-x-x^2)). (End)
a(n) = Fibonacci(n+3) - floor((n+3)/2). - Nathan Fox, Jan 27 2017
a(n) = (-3/4 + (-1)^n/4 + (2^(-n)*((1-t)^n*(-2+t) + (1+t)^n*(2+t)))/t + (-1-n)/2) where t=sqrt(5). - Colin Barker, Feb 09 2017
From G. C. Greubel, Apr 05 2024: (Start)
a(n) =  Fibonacci(n+3) - (1/4)*(2*n + 5 - (-1)^n).
E.g.f.: 2*exp(x/2)*( cosh(sqrt(5)*x/2) + (2/sqrt(5))*sinh(sqrt(5)*x/2) ) - (1/2)*( (x+2)*cosh(x) + (x+3)*sinh(x) ). (End)

cp's OEIS Frontend

A109128 Triangle read by rows: T(n,k) = T(n-1,k-1) + T(n-1,k) + 1 for 0

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A020956 a(n) = Sum_{k>=1} floor(tau^(n-k)) where tau is A001622.

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A215004 a(0) = a(1) = 1; for n>1, a(n) = a(n-1) + a(n-2) + floor(n/2).

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Magma

Mathematica

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