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A207361
Displacement under constant discrete unit surge.
4
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
0,3
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.
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
EXAMPLE
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.
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 *)
PROG
(Maxima) A207361(n) := block(
n*(1+n)*(2+n)*(1+11*n+8*n^2)/120
)$ /* R. J. Mathar, Mar 08 2012 */
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Jonathan Vos Post, Feb 18 2012
STATUS
approved