A225901 Write n in factorial base, then replace each nonzero digit d of radix k with k-d.
0, 1, 4, 5, 2, 3, 18, 19, 22, 23, 20, 21, 12, 13, 16, 17, 14, 15, 6, 7, 10, 11, 8, 9, 96, 97, 100, 101, 98, 99, 114, 115, 118, 119, 116, 117, 108, 109, 112, 113, 110, 111, 102, 103, 106, 107, 104, 105, 72, 73, 76, 77, 74, 75, 90, 91, 94, 95, 92, 93, 84, 85, 88, 89, 86, 87, 78, 79, 82, 83, 80, 81, 48, 49, 52, 53, 50, 51, 66, 67, 70, 71, 68
Offset: 0
Examples
a(1000) = a(1*6! + 2*5! + 1*4! + 2*3! + 2*2!) = (7-1)*6! + (6-2)*5! + (5-1)*4! + (4-2)*3! + (3-2)*2! = 4910. a(1397) = a(1*6! + 5*5! + 3*4! + 0*3! + 2*2! + 1*1!) = (7-1)*6! + (6-5)*5! + (5-3)*4! + (3-2)*2! + (2-1)*1! = 4491.
Links
Crossrefs
Cf. A000142, A007623, A004488, A048647, A001563, A007489, A257684, A257687, A276091, A275736, A276149.
Programs
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Mathematica
b = MixedRadix[Reverse@ Range[2, 12]]; Table[FromDigits[Map[Boole[# > 0] &, #] (Reverse@ Range[2, Length@ # + 1] - #), b] &@ IntegerDigits[n, b], {n, 0, 82}] (* Version 10.2, or *) f[n_] := Block[{a = {{0, n}}}, Do[AppendTo[a, {First@ #, Last@ #} &@ QuotientRemainder[a[[-1, -1]], Times @@ Range[# - i]]], {i, 0, #}] &@ NestWhile[# + 1 &, 0, Times @@ Range[# + 1] <= n &]; Most@ Rest[a][[All, 1]] /. {} -> {0}]; g[w_List] := Total[Times @@@ Transpose@ {Map[Times @@ # &, Range@ Range[0, Length@ w]], Reverse@ Append[w, 0]}]; Table[g[Map[Boole[# > 0] &, #] (Reverse@ Range[2, Length@ # + 1] - #)] &@ f@ n, {n, 0, 82}] (* Michael De Vlieger, Aug 29 2016 *)
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PARI
a(n)=my(s=0,d,k=2);while(n,d=n%k;n=n\k;if(d,s=s+(k-d)*(k-1)!);k=k+1);return(s)
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Python
from sympy import factorial as f def a(n): s=0 k=2 while(n): d=n%k n=(n//k) if d: s=s+(k - d)*f(k - 1) k+=1 return s print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 19 2017
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Scheme
(define (A225901 n) (let loop ((n n) (z 0) (m 2) (f 1)) (cond ((zero? n) z) (else (loop (quotient n m) (if (zero? (modulo n m)) z (+ z (* f (- m (modulo n m))))) (+ 1 m) (* f m)))))) ;; One implementing the first recurrence, with memoization-macro definec: (definec (A225901 n) (if (zero? n) n (+ (A276091 (A275736 n)) (A153880 (A225901 (A257684 n)))))) ;; Antti Karttunen, Aug 29 2016
Formula
From Antti Karttunen, Aug 29 2016: (Start)
(End)
Other identities. For n >= 0:
a(n!) = A001563(n).
a(n!-1) = A007489(n-1).
From Antti Karttunen, Aug 16 2016: (Start)
A153880(n) = a(A255411(a(n))) and A255411(n) = a(A153880(a(n))). [This involution conjugates between the two fundamental factorial base shifts.]
(End)
Comments