A207361 Displacement under constant discrete unit surge.

0, 1, 11, 53, 173, 448, 994, 1974, 3606, 6171, 10021, 15587, 23387, 34034, 48244, 66844, 90780, 121125, 159087, 206017, 263417, 332948, 416438, 515890, 633490, 771615, 932841, 1119951, 1335943, 1584038, 1867688

Offset: 0

Author

Jonathan Vos Post, Feb 18 2012

Keywords

• nonn
• easy

Comments

Assume discrete times 0, 1, 2, 3, ...

Assume constant discrete unit surge (= jerk = rate of change of acceleration) starting surge(0) = 0.

Also assume acceleration(0) = velocity(0) = displacement(0) = 0.

So at t = 0, 1, 2, 3, 4, ... the acceleration = 0, 1, 2, 3, 4, ...

Then the velocity v(t) = v(t-1) + a(t)*t.

So the displacement = s(t) = s(t-1) + v(t)*t.

v(0,1,2,3,4,...) = 0, 1, 5, 14, 30, 55, 91, 140, ... = A000330(n).

The subsequence of primes is finite with three terms 11, 53, and 173.

Examples

			s(4) = s(3) + v(4)*4 =  53 +  30*4 =  53 + 120 =  173;
s(5) = s(4) + v(5)*5 = 173 +  55*5 = 173 + 275 =  448;
s(6) = s(5) + v(6)*6 = 448 +  91*6 = 448 + 546 =  994;
s(7) = s(6) + v(7)*7 = 994 + 140*7 = 994 + 980 = 1974.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Crossrefs

Cf. A000330, A101856, A108678.

Programs

Maple

a:=n->sum(sum(i^2*j,j=i..n),i=0..n): seq(a(n),n=0..30); # Robert FERREOL, May 24 2022

Mathematica

a[0] = 0; a[n_] := a[n] = a[n-1] + n^2*(n+1)*(2*n+1)/6; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Oct 22 2015 *)

Maxima

A207361(n) := block(
        n*(1+n)*(2+n)*(1+11*n+8*n^2)/120
)$ /* R. J. Mathar, Mar 08 2012 */

Formula

a(0) = 0; for n>0, a(n) = a(n-1) + n*A000330(n) = a(n-1) + n*(0^2 + 1^2 + 2^2 + ... + n^2) = a(n-1) + n^2*(n+1)*(2*n+1)/6 = n*(1+n)*(2+n)*(1 + 11*n + 8*n^2)/120 = (2*n + 25*n^2 + 50*n^3 + 35*n^4 + 8*n^5)/120.

G.f.: x*(2*x^2+5*x+1) / (x-1)^6. - Colin Barker, May 06 2013

a(n) = Sum_{i=0..n-1} A108678(i). - J. M. Bergot, May 02 2018

a(n) = Sum_{0<=i<=j<=n} i^2*j. - Robert FERREOL, May 24 2022