A319209 a(n) = 1*2*3*4*5*6*7*8 + 9*10*11*12*13*14*15*16 + 17*18*19*20*21*22*23*24 + ... + (up to n).

1, 2, 6, 24, 120, 720, 5040, 40320, 40329, 40410, 41310, 52200, 194760, 2202480, 32472720, 518958720, 518958737, 518959026, 518964534, 519075000, 521400600, 572680080, 1754550000, 30173149440, 30173149465, 30173150090, 30173166990, 30173640840, 30187400040

Offset: 1

Author

Wesley Ivan Hurt, Sep 13 2018

Keywords

• nonn
• easy

Comments

In general, for sequences that multiply the first k natural numbers, and then add the product of the next k natural numbers (preserving the order of operations up to n), we have a(n) = Sum_{i=1..floor(n/k)} (k*i)!/(k*i-k)! + Sum_{j=1..k-1} (1-sign((n-j) mod k)) * (Product_{i=1..j} n-i+1). Here, k=8.

Examples

			a(1) = 1;
a(2) = 1*2 = 2;
a(3) = 1*2*3 = 6;
a(4) = 1*2*3*4 = 24;
a(5) = 1*2*3*4*5 = 120;
a(6) = 1*2*3*4*5*6 = 720;
a(7) = 1*2*3*4*5*6*7 = 5040;
a(8) = 1*2*3*4*5*6*7*8 = 40320;
a(9) = 1*2*3*4*5*6*7*8 + 9 = 40329;
a(10) = 1*2*3*4*5*6*7*8 + 9*10 = 40410;
a(11) = 1*2*3*4*5*6*7*8 + 9*10*11 = 41310;
a(12) = 1*2*3*4*5*6*7*8 + 9*10*11*12 = 52200;
a(13) = 1*2*3*4*5*6*7*8 + 9*10*11*12*13 = 194760;
a(14) = 1*2*3*4*5*6*7*8 + 9*10*11*12*13*14 = 2202480;
a(15) = 1*2*3*4*5*6*7*8 + 9*10*11*12*13*14*15 = 32472720;
a(16) = 1*2*3*4*5*6*7*8 + 9*10*11*12*13*14*15*16 = 518958720;
a(17) = 1*2*3*4*5*6*7*8 + 9*10*11*12*13*14*15*16 + 17 = 518958737;
a(18) = 1*2*3*4*5*6*7*8 + 9*10*11*12*13*14*15*16 + 17*18 = 518959026; etc.
		

Links

• Colin Barker, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A093361, (k=1) A000217, (k=2) A228958, (k=3) A319014, (k=4) A319205, (k=5) A319206, (k=6) A319207, (k=7) A319208, (k=8) this sequence, (k=9) A319211, (k=10) A319212.

Programs

Mathematica

a[n_]:=Sum[(8*i)!/(8*i-8)!, {i, 1, Floor[n/8] }] + Sum[(1-Sign[Mod[n-j, 8]])*Product[n-i+1, {i, 1, j}], {j, 1, 7}] ; Array[a, 29] (* Stefano Spezia, Apr 18 2023 *)

Formula

a(n) = Sum_{i=1..floor(n/8)} (8*i)!/(8*i-8)! + Sum_{j=1..7} (1-sign((n-j) mod 8)) * (Product_{i=1..j} n-i+1).