Mathematics Magazine for Grades 1-12  

Smarandacheials (1)

1, 2

Edited by J. Dezert

Let n>k≥1 be two integers.  Then the Smarandacheial is defined as:

!n!_k =  ∏(n-k·i)

              0<|n-k·i|≤n  i=0, 1, 2, … .

For examples:

                         0<|n-i|≤n   i=0, 1, 2, … .

Thus !5!=5(5-1)(5-2)(5-3)(5-4)(5-6)(5-7)(5-8)(5-9)(5-10)=5·4·3·2·1·(-1)·(-2)·(-3)·(-4)·(-5) =-14400.

The sequence is: 4, -36, 576, -14400, 518400, -25401600, 1625702400, -131681894400, 13168189440000, -1593350922240000, 229442532802560000, -38775788043632640000, 7600054456551997440000, -1710012252724199424000000, … .

2) In the case k=2:

      a) If n is odd, then

          !n!₂ =  ∏(n-2i) = n(n-2)(n-4)…(3)(1)(-1)(-3)…(-n+4)(-n+2)(-n) = (-1)^(n+1)/2(n!!)².

              0<|n-2i|≤n      i=0, 1, 2, … .

      b) If n is even, then

          !n!₂ =  ∏(n-2i) = n(n-2)(n-4)…(4)(2)(-2)(-4)…(-n+4)(-n+2)(-n) = (-1)^n/2(n!!)².

              0<|n-2i|≤n      i=0, 1, 2, … .

         Thus: !3!₂ = 3(3-2)(3-4)(3-6) = 9  and !4!₂ = 4(4-2)(4-6)(4-8) = 64.

         The sequence is: 9, 64, -225, -2304, 11025, 147456, -893025, -14745600, 108056025,        2123366400, … 

3) In the case k=3:      !n!₃ =  ∏(n-3i) = n(n-3)(n-6)… .

              0<|n-3i|≤n        i=0, 1, 2, … .

          Thus !7!₃ = 7(7-3)(7-6)(7-9)(7-12) = 7(4)(1)(-2)(-5) = 280.

          The sequence is: -8, 40, 324, 280, -2240, -26244, -22400, 246400, 3779136, 3203200,       -44844800,..