kicking this tin can further: First argument of Mod: a[1+k*p] a[1+k*p^2] a[1+p^2*(-1+k*p)] a[1+p^2*(-1+p*(-1+k*p))] a[1+p^2*(-1+p*(-1+k*p^2))] a[1+p^2*(-1+p*(-1+k*p^3))] a[1+p^2*(-1+p*(-1+k*p^4))] a[1+p^2*(-1+p*(-1+p^4*(1+k*p)))] a[1+p^2*(-1+p*(-1+p^4*(1+k*p^2)))] a[1+p^2*(-1+p*(-1+p^4*(1+k*p^3)))] a[1+p^2*(-1+p*(-1+p^4*(1+p^3*(-1+k*p))))] Second argument of Mod: k*p^2 k*p^4 p^5*(-1+k*p) p^6*(-1+p*(-1+k*p)) p^7*(-1+p*(-1+k*p^2)) p^8*(-1+p*(-1+k*p^3)) p^9*(-1+p*(-1+k*p^4)) p^10*(-1+p*(-1+p^4*(1+k*p))) p^11*(-1+p*(-1+p^4*(1+k*p^2))) p^12*(-1+p*(-1+p^4*(1+k*p^3))) p^13*(-1+p*(-1+p^4*(1+p^3*(-1+k*p)))) for the first argument of Mod, these were generated by the following rules operating on a[1+k*p]: {k-> k*p, k-> k*p-1, k-> k*p-1, k-> k*p, k-> k*p, k-> k*p, k-> k*p+1, k-> k*p, k-> k*p, k-> k*p-1} for the second argument of Mod, the same rules operating on p*(previous), starting with previous=(k*p^2) Note that rule k-> k*p+1 is equivalent with k-> k*p-2 Symbolising the rules 'in absurdum' gives {0,-1,-1,0,0,0,1,0,0,-1,...}. No OEIS direct hits. Magnitude: we arrived at Mod[a[474544], 84063868821], and Mma burps at Round[474544! /E ]. Can't blame it. Misquoting William S. : 'there's a threefoldness in these derangements' W. ----- Original Message ----- From: "wouter meeussen" <wouter.meeussen@pandora.be> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Sunday, March 27, 2005 2:43 PM Subject: [math-fun] derangements divisible by 3 given the many alternatives of writing (!n), cfr. OEIS A000166, and considering there is a nice way to find the powers of p in n! by writing n in base p (if I remember correctly), it is funny to observe the powers of 3 in !n. I get, using a[ n ] for (!n) and p as representing 3 (to avoid immediate evaluation in Mathematica): {Mod[a[1 + k*p], k*p]==0, Mod[a[1 + k*p^2], k*p^3]==0, Mod[a[1 + k*p^2], k*p^4]==0, Mod[a[1 - p^2 + k*p^3], -p^5 + k*p^6]==0, Mod[a[1 - p^2 - p^3 + k*p^4], -p^6 - p^7 + k*p^8]==0, Mod[a[1 - p^2 - p^3 + k*p^5], -p^7 - p^8 + k*p^10]==0, Mod[a[1 - p^2 - p^3 + k*p^6], -p^8 - p^9 + k*p^12]==0, ......... } and I wonder where these lead to. W. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun