A019303 - OEIS

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

(list; graph; refs; listen; history; text; internal format)

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

Paolo Xausa, Table of n, a(n) for n = 0..10000

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

MATHEMATICA

Riffle[Range[4, 50], 1, {1, -2, 2}] (* Paolo Xausa, Sep 21 2024 *)

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