OFFSET
1,2
COMMENTS
Create a triangle having first column T(n,1) = 2*n-1 for n = 1,2,3... The remaining terms are set to T(r,c) = T(r,c-1) + T(r-1,c-1). The sum of the terms in row n is a(n). The first five rows of the triangle are 1; 3,4; 5,8,12; 7,12,20,32; 9,16,28,48,80. - J. M. Bergot, Jan 17 2013
Starting at n=1, a(n) = (n+1)*2^n - 2*n - 1. A001787(n) = n*2^n. - J. M. Bergot, Jan 27 2013
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (6,-13,12,-4).
FORMULA
a(n) = 6*a(n-1) - 13*a(n-2) + 12*a(n-3) - 4*a(n-4).
G.f.: x*(1 + x - 4*x^2)/( (1-2*x)^2*(1-x)^2 ).
a(n) = A001787(n+1)- 2*n - 1. - J. M. Bergot, Jan 22 2013
a(n) = Sum_{k=1..n} Sum_{i=0..n} (n-i) * C(k,i). - Wesley Ivan Hurt, Sep 19 2017
MAPLE
f:= gfun:-rectoproc({a(n) = 6*a(n-1) - 13*a(n-2) + 12*a(n-3) - 4*a(n-4),
a(1)=1, a(2)=7, a(3)=25, a(4)=71}, a(n), remember):
map(f, [$1..30]); # Robert Israel, Sep 19 2017
MATHEMATICA
(* First program *)
b[n_]:= 2^(n-1); c[n_]:= n;
t[n_, k_]:= Sum[b[k-i] c[n+i], {i, 0, k-1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n-k+1, k], {n, 12}, {k, n, 1, -1}]]
r[n_]:= Table[t[n, k], {k, 1, 60}] (* A213568 *)
d = Table[t[n, n], {n, 1, 40}] (* A213569 *)
s[n_]:= Sum[t[i, n+1-i], {i, 1, n}]
s1 = Table[s[n], {n, 1, 50}] (* A047520 *)
(* Additional programs *)
LinearRecurrence[{6, -13, 12, -4}, {1, 7, 25, 71}, 30] (* Harvey P. Dale, Jan 06 2015 *)
Table[2^n*(n+1) -(2*n+1), {n, 30}] (* G. C. Greubel, Jul 25 2019 *)
PROG
(PARI) my(x='x+O('x^30)); Vec(x*(1+x-4*x^2)/((1-2*x)^2*(1-x)^2)) \\ Altug Alkan, Sep 19 2017
(PARI) vector(30, n, 2^n*(n+1) -(2*n+1)) \\ G. C. Greubel, Jul 25 2019
(Magma) [2^n*(n+1) -(2*n+1): n in [1..30]]; // G. C. Greubel, Jul 25 2019
(Sage) [2^n*(n+1) -(2*n+1) for n in (1..30)] # G. C. Greubel, Jul 25 2019
(GAP) List([1..30], n-> 2^n*(n+1) -(2*n+1)); # G. C. Greubel, Jul 25 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 18 2012
STATUS
approved