This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
In this table each row consists of A001563[n] permutations of (n+1) terms; i.e., we have (1/) 2,1/ 1,3,2; 3,1,2; 3,2,1; 2,3,1/ 1,2,4,3; 2,1,4,3; Append to each an infinite number of fixed terms and we get a list of rearrangements of natural numbers, but with only a finite number of terms permuted: 1/2,3,4,5,6,7,8,9,... 2,1/3,4,5,6,7,8,9,... 1,3,2/4,5,6,7,8,9,... 3,1,2/4,5,6,7,8,9,... 3,2,1/4,5,6,7,8,9,... 2,3,1/4,5,6,7,8,9,... 1,2,4,3/5,6,7,8,9,... 2,1,4,3/5,6,7,8,9,...
with(group); permul := (a,b) -> mulperms(b,a); PermUnrank3R := proc(r) local n; n := nops(factorial_base(r)); convert(PermUnrank3Raux(n+1,r,[]),'permlist',1+(((r+2) mod (r+1))*n)); end; PermUnrank3Raux := proc(n,r,p) local s; if(0 = r) then RETURN(p); else s := floor(r/((n-1)!)); RETURN(PermUnrank3Raux(n-1, r-(s*((n-1)!)), permul(p,[[n,n-s]]))); fi; end;
19 = 3*(3!) + 0*(2!) + 1*(1!), thus it is written as "301" in factorial base (A007623). The count of nonzero digits in that representation is 2, so a(19) = 2.
A060130(n) = count_nonfixed(convert(PermUnrank3R(n), 'disjcyc'))-nops(convert(PermUnrank3R(n), 'disjcyc')) or nops(fac_base(n))-nops(positions(0, fac_base(n))) fac_base := n -> fac_base_aux(n, 2); fac_base_aux := proc(n, i) if(0 = n) then RETURN([]); else RETURN([op(fac_base_aux(floor(n/i), i+1)), (n mod i)]); fi; end; count_nonfixed := l -> convert(map(nops, l), `+`); positions := proc(e, ll) local a, k, l, m; l := ll; m := 1; a := []; while(member(e, l[m..nops(l)], 'k')) do a := [op(a), (k+m-1)]; m := k+m; od; RETURN(a); end; # For procedure PermUnrank3R see A060117
Block[{nn = 105, r}, r = MixedRadix[Reverse@ Range[2, -1 + SelectFirst[Range@ 12, #! > nn &]]]; Array[Count[IntegerDigits[#, r], k_ /; k > 0] &, nn, 0]] (* Michael De Vlieger, Dec 30 2017 *)
(define (A060130 n) (let loop ((n n) (i 2) (s 0)) (cond ((zero? n) s) (else (loop (quotient n i) (+ 1 i) (+ s (if (zero? (remainder n i)) 0 1))))))) ;; Two other implementations, that use memoization-macro definec: (definec (A060130 n) (if (zero? n) n (+ 1 (A060130 (A257687 n))))) (definec (A060130 n) (if (zero? n) n (+ (A257511 n) (A060130 (A257684 n))))) ;; Antti Karttunen, Dec 30 2017
n = 288 = 2*2*2*2*2*3*3; lcm(5,2) = 10; Product(5,2) = 10, max(5,2) = 5; n = 180 = 2*2*3*3*5; lcm(2,2,1) = 2; Product(2,2,1) = 4; max(2,2,1) = 2; it deviates both from maximum of exponents (A051903, for the first time at n=72), and product of exponents (A005361, for the first time at n=36). For n = 36 = 2*2*3*3 = 2^2 * 3^2 we have a(36) = lcm(2,2) = 2. For n = 72 = 2*2*2*3*3 = 2^3 * 3^2 we have a(72) = lcm(2,3) = 6. For n = 144 = 2^4 * 3^2 we have a(144) = lcm(2,4) = 4. For n = 360 = 2^3 * 3^2 * 5^1 we have a(360) = lcm(1,2,3) = 6.
Table[LCM @@ Last /@ FactorInteger[n], {n, 2, 100}] (* Ray Chandler, Jan 24 2006 *)
a(n) = lcm(factor(n)[,2]); \\ Michel Marcus, Mar 25 2017
from sympy import lcm, factorint
def a(n):
l=[]
f=factorint(n)
for i in f: l+=[f[i],]
return lcm(l)
print([a(n) for n in range(1, 151)]) # Indranil Ghosh, Mar 25 2017
Consider the first eight permutations (indices 0-7) listed in A060117: 1 [Only the first 1-cycle explicitly listed thus a(0) = 2^1 = 2] 2,1 [One transposition (2-cycle) in beginning, thus a(1) = 2^2 = 4] 1,3,2 [One fixed element in beginning, then transposition, thus a(2) = 2^1 * 3^2 = 18] 3,1,2 [One 3-cycle, thus a(3) = 2^3 = 8] 3,2,1 [One transposition jumping over a fixed element, a(4) = 2^2 * 3^1 = 12] 2,3,1 [One 3-cycle, thus a(5) = 2^3 = 8] 1,2,4,3 [Two 1-cycles, then a 2-cycle, thus a(6) = 2^1 * 3^1 * 5^2 = 150]. 2,1,4,3 [Two 2-cycles, not crossed, thus a(7) = 2^2 * 5^2 = 100] and also the seventeenth one at n=16 [A007623(16)=220] where we have: 3,4,1,2 [Two 2-cycles crossed, thus a(16) = 2^2 * 3^2 = 36].
For n=27, which in factorial base (A007623) is "1011" and encodes (in A060118-order) permutation "23154" with one 3-cycle and one 2-cycle, the longest cycle has three elements, thus a(27) = 3.
Comments