A319873 a(n) = 9*8*7*6*5*4*3*2*1 + 18*17*16*15*14*13*12*11*10 + ... + (up to the n-th term).

9, 72, 504, 3024, 15120, 60480, 181440, 362880, 362880, 362898, 363186, 367776, 436320, 1391040, 13728960, 160755840, 1764685440, 17643588480, 17643588507, 17643589182, 17643606030, 17644009680, 17653276080, 17856715680, 22119259680, 107157012480

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=9.

Examples

			a(1) = 9;
a(2) = 9*8 = 72;
a(3) = 9*8*7 = 504;
a(4) = 9*8*7*6 = 3024;
a(5) = 9*8*7*6*5 = 15120;
a(6) = 9*8*7*6*5*4 = 60480;
a(7) = 9*8*7*6*5*4*3 = 181440;
a(8) = 9*8*7*6*5*4*3*2 = 362880;
a(9) = 9*8*7*6*5*4*3*2*1 = 362880;
a(10) = 9*8*7*6*5*4*3*2*1 + 18 = 362898;
a(11) = 9*8*7*6*5*4*3*2*1 + 18*17 = 363186;
a(12) = 9*8*7*6*5*4*3*2*1 + 18*17*16 = 367776
a(13) = 9*8*7*6*5*4*3*2*1 + 18*17*16*15 = 436320;
a(14) = 9*8*7*6*5*4*3*2*1 + 18*17*16*15*14 = 1391040;
a(15) = 9*8*7*6*5*4*3*2*1 + 18*17*16*15*14*13 = 13728960;
a(16) = 9*8*7*6*5*4*3*2*1 + 18*17*16*15*14*13*12 = 160755840;
a(17) = 9*8*7*6*5*4*3*2*1 + 18*17*16*15*14*13*12*11 = 1764685440;
a(18) = 9*8*7*6*5*4*3*2*1 + 18*17*16*15*14*13*12*11*10 = 17643588480;
a(19) = 9*8*7*6*5*4*3*2*1 + 18*17*16*15*14*13*12*11*10 + 27 = 17643588507;
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), this sequence (k=9), A319874 (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,9),n=1..30); # Muniru A Asiru, Sep 30 2018

Mathematica

k:=9; 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 *)