A129696 Antidiagonal sums of triangular array T defined in A014430: T(j,k) = binomial(j+1, k) - 1 for 1 <= k <= j.
1, 2, 5, 9, 17, 29, 50, 83, 138, 226, 370, 602, 979, 1588, 2575, 4171, 6755, 10935, 17700, 28645, 46356, 75012, 121380, 196404, 317797, 514214, 832025, 1346253, 2178293, 3524561, 5702870, 9227447, 14930334, 24157798, 39088150, 63245966
Offset: 1
References
- Paul Curtz, Intégration numérique des systèmes différentiels à conditions initiales. Note no. 12 du Centre de Calcul Scientifique de l'Armement, 1969.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..200
- Daniel F. Checa, Arndt Compositions: Connections with Fibonacci Numbers, Statistics, and Generalizations, 2023, p. 31.
- Larry Ericksen, Problem 7, Problem Section, The Fibonacci Quarterly, Vol. 57, No. 5 (2019), pp. 172, 175-183.
- Index entries for linear recurrences with constant coefficients, signature (2,1,-3,0,1).
Crossrefs
Programs
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Magma
m:=36; M:=ZeroMatrix(IntegerRing(), m, m); for j:=1 to m do for k:=1 to j do M[j, k]:=Binomial(j+1, k)-1; end for; end for; [ &+[ M[j-k+1, k]: k in [1..(j+1) div 2] ]: j in [1..m] ]; // Klaus Brockhaus, Jun 11 2007
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Magma
[Fibonacci(n+3)-2-Floor(n/2): n in [1..40]]; // Vincenzo Librandi, Nov 23 2014
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Maple
with(combinat): a := proc (n) options operator, arrow: fibonacci(n+3)-2-floor((1/2)*n) end proc: seq(a(n), n = 1 .. 34); # Emeric Deutsch, Nov 22 2014
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Mathematica
a[n_]:= a[n]= If[n<3, n, a[n-1] + a[n-2] + (n + Mod[n, 2])/2]; Table[a[n], {n,40}] (* Jean-François Alcover, Mar 04 2013 *) Table[Fibonacci[n+3] -2 -Floor[n/2], {n, 100}] (* Vincenzo Librandi, Nov 23 2014 *)
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Python
prpr = 1 prev = 2 for n in range(2,100): print(prpr, end=", ") curr = prpr+prev + 1 + n//2 prpr = prev prev = curr # Alex Ratushnyak, Jul 30 2012
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SageMath
[fibonacci(n+3) -2 -(n//2) for n in range(1,41)] # G. C. Greubel, Mar 17 2023
Formula
From Paul Barry, Jan 18 2009: (Start)
a(n) = Sum_{k=0..floor(n/2)} A000071(n-2*k+3).
a(n) = Sum_{k=0..floor(n/2)} (Sum_{j=0..n-2*k} Fibonacci(j+1)). (End)
a(n+1) = a(n-1) + a(n) + 1 + floor(n/2) for n>1, a(0)=1, a(1)=2. - Alex Ratushnyak, Jul 30 2012
From R. J. Mathar, Jul 25 2013: (Start)
G.f.: x/((1 + x)*(1 - x)^2*(1 - x - x^2)).
a(n) + a(n+1) = A001924(n+1). (End)
a(n) = Fibonacci(n+3) - 2 - floor(n/2). - Emeric Deutsch, Nov 22 2014
a(n) = (-5/4 - (-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
E.g.f.: (4*exp(x/2)*(5*cosh(sqrt(5)*x/2) + 2*sqrt(5)*sinh(sqrt(5)*x/2)) - 5*(4 + x)*cosh(x) - 5*(3 + x)*sinh(x))/10. - Stefano Spezia, Apr 06 2024
a(n) = max_{k = 2^(n+1)..2^(n+2)-1} (A002487(k) - A000120(k)) (Ericksen, 2019). - Amiram Eldar, Jan 30 2025
Extensions
Edited and extended by Klaus Brockhaus, Jun 11 2007
Comments