A319874 a(n) = 10*9*8*7*6*5*4*3*2*1 + 20*19*18*17*16*15*14*13*12*11 + ... + (up to the n-th term).

10, 90, 720, 5040, 30240, 151200, 604800, 1814400, 3628800, 3628800, 3628820, 3629180, 3635640, 3745080, 5489280, 31536000, 394329600, 5082739200, 60952953600, 670446201600, 670446201630, 670446202470, 670446225960, 670446859320, 670463302320, 670873719600

Offset: 1

Author

Wesley Ivan Hurt, Sep 30 2018

Keywords

• nonn
• easy

Comments

For similar multiply/add sequences in descending blocks of k natural numbers, we have: a(n) = Sum_{j=1..k-1} (floor((n-j)/k)-floor((n-j-1)/k)) * (Product_{i=1..j} n-i-j+k+1) + Sum_{j=1..n} (floor(j/k)-floor((j-1)/k)) * (Product_{i=1..k} j-i+1). Here, k=10.

Examples

			a(1) = 10;
a(2) = 10*9 = 90;
a(3) = 10*9*8 = 720;
a(4) = 10*9*8*7 = 5040;
a(5) = 10*9*8*7*6 = 30240;
a(6) = 10*9*8*7*6*5 = 151200;
a(7) = 10*9*8*7*6*5*4 = 604800;
a(8) = 10*9*8*7*6*5*4*3 = 1814400;
a(9) = 10*9*8*7*6*5*4*3*2 = 3628800;
a(10) = 10*9*8*7*6*5*4*3*2*1 = 3628800;
a(11) = 10*9*8*7*6*5*4*3*2*1 + 20 = 3628820;
a(12) = 10*9*8*7*6*5*4*3*2*1 + 20*19 = 3629180;
a(13) = 10*9*8*7*6*5*4*3*2*1 + 20*19*18 = 3635640;
a(14) = 10*9*8*7*6*5*4*3*2*1 + 20*19*18*17 = 3745080;
a(15) = 10*9*8*7*6*5*4*3*2*1 + 20*19*18*17*16 = 5489280;
a(16) = 10*9*8*7*6*5*4*3*2*1 + 20*19*18*17*16*15 = 31536000;
a(17) = 10*9*8*7*6*5*4*3*2*1 + 20*19*18*17*16*15*14 = 394329600;
a(18) = 10*9*8*7*6*5*4*3*2*1 + 20*19*18*17*16*15*14*13 = 5082739200;
a(19) = 10*9*8*7*6*5*4*3*2*1 + 20*19*18*17*16*15*14*13*12 = 60952953600;
etc.
		

Links

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

Crossrefs

For similar sequences, see: A000217 (k=1), A319866 (k=2), A319867 (k=3), A319868 (k=4), A319869 (k=5), A319870 (k=6), A319871 (k=7), A319872 (k=8), A319873 (k=9), this sequence (k=10).

Programs

Maple

a:=(n,k)->add((floor((n-j)/k)-floor((n-j-1)/k))*(mul(n-i-j+k+1,i=1..j)),j=1..k-1) + add((floor(j/k)-floor((j-1)/k))*(mul(j-i+1,i=1..k)),j=1..n): seq(a(n,10),n=1..25); # Muniru A Asiru, Sep 30 2018

Mathematica

k:=10; a[n_]:=Sum[(Floor[(n-j)/k]-Floor[(n-j-1)/k])* Product[n-i-j+k+1, {i,1,j }], {j,1,k-1} ] + Sum[(Floor[j/k]-Floor[(j-1)/k])* Product[j-i+1, {i,1,k} ], {j,1,n}]; Array[a, 50] (* Stefano Spezia, Sep 30 2018 *)