OFFSET
0,1
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).
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.
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
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 *)
PROG
(PARI) a(n) = sum(k=0, n, (2^k+2)*(2^k+3)/2); \\ Michel Marcus, Jul 28 2021
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul Weisenhorn, Jul 14 2021
STATUS
approved