A134311 Row sums of triangle A134310.
1, 2, 7, 20, 51, 122, 281, 632, 1399, 3062, 6645, 14324, 30707, 65522, 139249, 294896, 622575, 1310702, 2752493, 5767148, 12058603, 25165802, 52428777, 109051880, 226492391, 469762022, 973078501, 2013265892, 4160749539
Offset: 0
Examples
a(3) = 20 = sum of row 4 terms of triangle A134310: (4 + 4 + 5 + 7). a(3) = 20 = (1, 3, 3, 1) dot (1, 1, 4, 4) = (1 + 3 + 12 + 4).
Links
- M. F. Hasler, Table of n, a(n) for n = 0..1000 (terms a(0..999) from Gennady Eremin), May 03 2022
- Index entries for linear recurrences with constant coefficients, signature (6,-13,12,-4).
Crossrefs
Cf. A134310.
Programs
-
Mathematica
Join[{1},LinearRecurrence[{6,-13,12,-4},{2,7,20,51},30]] (* Harvey P. Dale, Apr 16 2013 *) -
PARI
apply( {A134311(n)=max(n+3,4)<<(n-1)-n-1}, [0..33]) \\ M. F. Hasler, Mar 29 2022 -
Python
a = lambda n: (n+3)*2**(n-1)-n-1 if n > 0 else 1 print([a(n) for n in range(40)]) # Gennady Eremin, Mar 26 2022
Formula
Binomial transform of [1, 1, 4, 4, 6, 6, 8, 8, 10, 10, 12, 12, ...].
For n > 0, a(n) = (n+3)*2^(n-1) - n - 1. - R. J. Mathar, Apr 04 2012, edited by M. F. Hasler, Mar 29 2022
G.f.: (2*x^4-8*x^3+8*x^2-4*x+1)/((x-1)^2*(2*x-1)^2). - Colin Barker, Aug 13 2012
Extensions
Offset corrected to 0 by M. F. Hasler, Mar 29 2022