A019303 "Pascal sweep" for k=2: draw a horizontal line through the 1 at C(k,0) in Pascal's triangle; rotate this line and record the sum of the numbers on it (excluding the initial 1).

1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 1, 10, 1, 11, 1, 12, 1, 13, 1, 14, 1, 15, 1, 16, 1, 17, 1, 18, 1, 19, 1, 20, 1, 21, 1, 22, 1, 23, 1, 24, 1, 25, 1, 26, 1, 27, 1, 28, 1, 29, 1, 30, 1, 31, 1, 32, 1, 33, 1, 34, 1, 35, 1, 36, 1, 37, 1, 38, 1, 39, 1, 40, 1, 41, 1, 42, 1, 43, 1, 44, 1, 45, 1, 46, 1

Offset

0,2

Comments

The line is never horizontal, but slopes down: For any n >= 0, a(n) is the sum of all C(m,p) where (m,p) are the integer points of the line going through (2,0) and (3+n,3+n). It does not matter whether the triangle is written with left-aligned or centered lines. - M. F. Hasler, Oct 12 2018

It might have been more natural to start with a horizontal line for n=0 (which would have prefixed a 3 to the sequence), or even, to start with the line going through (0,0), resulting in (2, 1, 3, 1, 4, ...). - M. F. Hasler, Oct 13 2018

Links

Table of n, a(n) for n=0..86.

N. J. A. Sloane, Illustration of initial terms

Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Formula

For all n >= 0: a(2n) = 1, a(2n+1) = n + 4.

G.f.: ( 1+4*x-x^2-3*x^3 ) / ( (x-1)^2*(1+x)^2 ). - R. J. Mathar, Oct 15 2018

Prog

(PARI) a(n, k=2)=sum(i=1, n+=1, if((n+k)*i%n==0, binomial(k+i, i*(n+k)/n)) \\ For illustration. - M. F. Hasler, Oct 13 2018
(PARI) A019303(n)=if(bittest(n, 0), n\2+4, 1) \\ M. F. Hasler, Oct 13 2018

Crossrefs

Sequence in context: A046785 A060044 A323413 * A339967 A357311 A107463

Adjacent sequences: A019300 A019301 A019302 * A019304 A019305 A019306

Keyword

nonn,easy

Author

Jonas Wallgren

Extensions

More terms from David W. Wilson

Status

approved