A325983 Row sums of the triangle A325982.

1, 1, 2, 2, 5, 5, 18, 21, 77, 102, 337, 480, 1449, 2155, 6107, 9348, 25355, 39639, 104188, 165596, 425156, 684926, 1726737, 2813582, 6990175, 11501905, 28232753, 46854161, 113841632, 190362483, 458480128, 771855377, 1844765161, 3124639626, 7417428613, 12633074088

Offset: 1

Author

Stefano Spezia, May 29 2019

Keywords

• nonn

Links

• Stefano Spezia, Table of n, a(n) for n = 1..2000

Crossrefs

Cf. A004526, A007318, A325982.

Programs

GAP

List([1..40], n->Sum([0..Int((n-1)/2)], k->Binomial(n-1, k-1)-Binomial(n-k-1, k-1)+1));

Magma

[(&+[Binomial(n-1, k-1)-Binomial(n-k-1, k-1)+1: k in [0..Floor((n-1)/2)]]): n in [1..40]];

Maple

a := n -> add(binomial(n-1, k-1)-binomial(n-k-1, k-1)+1, k = 0 .. floor((n-1)/2)): seq(a(n), n = 1 .. 40);

Mathematica

T[n_,k_]:=Binomial[n-1,k-1]-Binomial[n-k-1,k-1]+1; a[n_]:=Sum[T[n,k],{k,0,Floor[(n-1)/2]}]; Array[a,40]
Table[If[EvenQ[n], 2^(n - 2) + n/2 + 1 - Binomial[n, n/2]/2 + Fibonacci[n]/2 - LucasL[n]/2, 2^(n - 2) + (n + 1)/2 - Binomial[n - 1, (n - 1)/2]/2 - Fibonacci[n - 3] - 3*Fibonacci[n]/2 + LucasL[n]/2], {n, 1, 40}] (* Vaclav Kotesovec, Jun 20 2021 *)

PARI

a(n) = sum(k=0, floor((n-1)/2), binomial(n - 1, k - 1) - binomial(n - k - 1, k - 1) + 1);

Formula

a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n - 1, k - 1) - binomial(n - k - 1, k - 1) + 1.

a(n) = Sum_{k=0..A004526(n-1)} A007318(n - 1, k - 1) - A007318(n - k - 1, k - 1) + 1.

Conjecture: a(n) ~ 2^(n-2). - Stefano Spezia, Jun 16 2021

This conjecture is true. Its proof follows from Vaclav Kotesovec's formula used in the 2nd Mathematica code. - Stefano Spezia, Jun 27 2021

Recurrence: n*(15*n^6 - 600*n^5 + 9874*n^4 - 85078*n^3 + 403791*n^2 - 1000762*n + 1013720)*a(n) = (45*n^7 - 1815*n^6 + 30507*n^5 - 273677*n^4 + 1393992*n^3 - 3930412*n^2 + 5362320*n - 2240000)*a(n-1) + 2*(30*n^7 - 1230*n^6 + 20813*n^5 - 186314*n^4 + 944533*n^3 - 2692476*n^2 + 3985604*n - 2395680)*a(n-2) - (255*n^7 - 10470*n^6 + 180608*n^5 - 1685204*n^4 + 9136801*n^3 - 28646510*n^2 + 47833000*n - 32569600)*a(n-3) + (15*n^7 - 585*n^6 + 11089*n^5 - 129179*n^4 + 931040*n^3 - 3910812*n^2 + 8540192*n - 7271040)*a(n-4) + 2*(165*n^7 - 6810*n^6 + 118529*n^5 - 1121692*n^4 + 6211866*n^3 - 20094734*n^2 + 35150196*n - 25693920)*a(n-5) - 4*(15*n^7 - 600*n^6 + 10309*n^5 - 98168*n^4 + 555848*n^3 - 1857428*n^2 + 3364504*n - 2524480)*a(n-6) - 8*(n-7)*(15*n^6 - 510*n^5 + 7099*n^4 - 51282*n^3 + 202026*n^2 - 411828*n + 340960)*a(n-7). - Vaclav Kotesovec, Jun 20 2021