A346375 a(n) = Sum_{k=0..n} (2^k + 2) * (2^k + 3) / 2.

6, 16, 37, 92, 263, 858, 3069, 11584, 44995, 177350, 704201, 2806476, 11205327, 44780242, 179038933, 715991768, 2863639259, 11453901534, 45814295265, 183254559460, 733012994791, 2932041493226, 11728145001197, 46912538061552, 187650068359923, 750600105667318, 3002400087124729

Offset: 0

Author

Paul Weisenhorn, Jul 14 2021

Keywords

• nonn
• easy

Links

• Roger B. Nelson, Proof without Words: A Triangular Sum, Mathematics Magazine Vol. 78, No. 5 (December 2005), p. 395.
• Index entries for linear recurrences with constant coefficients, signature (8,-21,22,-8).

Crossrefs

Cf. A006095, A076024.

Programs

Maple

a:= proc(n) option remember:
     if n=0 then 6 else procname(n-1)+(2^n+3)*(2^n+2)/2 fi:
    end proc:
seq(a(n), n=0..26);

Mathematica

a[n_]:=Sum[(2^k+2)*(2^k+3)/2,{k,0,n}];Array[a,30,0] (* Giorgos Kalogeropoulos, Jul 27 2021 *)

PARI

a(n) = sum(k=0, n, (2^k+2)*(2^k+3)/2); \\ Michel Marcus, Jul 28 2021

Formula

a(n) = Sum_{k=0..n} (2^k + 2) * (2^k + 3) / 2.

a(n) = (2^(n+1) + 7) * (2^(n+1) + 8)/6 - 9 + 3*n.

More generally: let f(n, b) = Sum_{k=0..n} (2^k + b) * (2^k + b + 1)/2 then f(n, b) = (2^(n+1) + 3*b + 1) * (2^(n+1) + 3*b + 2) / 6 - (b + 1)^2 + b*(b + 1)*n/2.

G.f.: ((b^2+3*b+2)/2 - (3*b^2+8*b+4)*x + (4*b^2+8*b+3)*x^2) / ((4*x-1) * (2*x-1) * (x-1)^2).

E.g.f.: exp(x)*((6*b+3)*exp(x) + 2*exp(3*x) + 3*(b^2+b)*x/2 +(3*b^2-3*b-4) / 2) / 3.

Then b = -1 gives A006095, b = 0 gives A076024, b = 1 gives A346295, b = 2 gives A346375.

a(n) = 8*a(n-1) - 21*a(n-2) + 22*a(n-3) - 8*a(n-4) with n > 3.

This recurrence is valid for all sequences f(n, b).

G.f.: (35*x^2 - 32*x + 6) / ((4*x - 1) * (2*x - 1) * (x - 1)^2).

E.g.f.: exp(x) * (1 + 15*exp(x) + 2*exp(3*x) + 9*x)/3. - Stefano Spezia, Aug 15 2021