0, 1, 1, 2, 1, 1, 1, 2, 2, 3, 2, 2, 1, 2, 1, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 3, 2, 2, 2, 3, 3, 4, 3, 3, 2, 3, 2, 3, 3, 3, 2, 2, 3, 3, 2, 2, 1, 2, 2, 3, 2, 2, 1, 2, 2, 3, 2, 2, 2, 3, 2, 3, 3, 3, 2, 2, 3, 3, 2, 2, 1, 2, 1, 2, 2, 2, 2, 3, 2, 3, 3, 3, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1
Offset: 0
For n=23 ("321" in factorial base representation, A007623), all the digits are maximal for their positions (they occur on the "maximal slope"), thus there is only one distinct digit slope present and a(23)=1. Also, for the 23rd permutation in the ordering A060117, [2341], there is just one drop, as p[4] = 1 < 4.
For n=29 ("1021"), there are three nonzero digits, where both 2 and the rightmost 1 are on the maximal slope, while the most significant 1 is on the "sub-sub-sub-maximal", thus there are two occupied slopes in total, and a(29) = 2. In the 29th permutation of A060117, [23154], there are two drops as p[3] = 1 < 3 and p[5] = 4 < 5.
For n=37 ("1201"), there are three nonzero digits, where the rightmost 1 is on the maximal slope, 2 is on the submaximal, and the most significant 1 is on the "sub-sub-sub-maximal", thus there are three occupied slopes in total, and a(37) = 3. In the 37th permutation of A060117, [51324], there are three drops at indices 2, 4 and 5.
A275734
Prime-factorization representations of "factorial base slope polynomials": a(0) = 1; for n >= 1, a(n) = A275732(n) * a(A257684(n)).
Original entry on oeis.org
1, 2, 3, 6, 2, 4, 5, 10, 15, 30, 10, 20, 3, 6, 9, 18, 6, 12, 2, 4, 6, 12, 4, 8, 7, 14, 21, 42, 14, 28, 35, 70, 105, 210, 70, 140, 21, 42, 63, 126, 42, 84, 14, 28, 42, 84, 28, 56, 5, 10, 15, 30, 10, 20, 25, 50, 75, 150, 50, 100, 15, 30, 45, 90, 30, 60, 10, 20, 30, 60, 20, 40, 3, 6, 9, 18, 6, 12, 15, 30, 45, 90, 30, 60, 9, 18, 27
Offset: 0
For n=23 ("321" in factorial base representation, A007623), all three nonzero digits are maximal for their positions (they all occur on "maximal slope"), thus a(23) = prime(1)^3 = 2^3 = 8.
For n=29 ("1021"), there are three nonzero digits, where both 2 and the rightmost 1 are on the "maximal slope", while the most significant 1 is on the "sub-sub-sub-maximal", thus a(29) = prime(1)^2 * prime(4)^1 = 2*7 = 28.
For n=37 ("1201"), there are three nonzero digits, where the rightmost 1 is on the maximal slope, 2 is on the sub-maximal, and the most significant 1 is on the "sub-sub-sub-maximal", thus a(37) = prime(1) * prime(2) * prime(4) = 2*3*7 = 42.
For n=55 ("2101"), the least significant 1 is on the maximal slope, and the digits "21" at the beginning are together on the sub-sub-maximal slope (as they are both two less than the maximal digit values 4 and 3 allowed in those positions), thus a(55) = prime(1)^1 * prime(3)^2 = 2*25 = 50.
Cf.
A001221,
A001222,
A002110,
A007489,
A007814,
A048675,
A051903,
A056169,
A056170,
A060130,
A060502,
A225901.
Cf.
A275804 (indices of squarefree terms),
A275805 (of terms not squarefree).
-
from operator import mul
from sympy import prime, factorial as f
def a007623(n, p=2): return n if n0 else '0' for i in x)[::-1]
return 0 if n==1 else sum(int(y[i])*f(i + 1) for i in range(len(y)))
def a(n): return 1 if n==0 else a275732(n)*a(a257684(n))
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 19 2017
A134640
Permutational numbers (numbers with k different digits in k-positional system).
Original entry on oeis.org
0, 1, 2, 5, 7, 11, 15, 19, 21, 27, 30, 39, 45, 54, 57, 75, 78, 99, 108, 114, 120, 135, 141, 147, 156, 177, 180, 198, 201, 210, 216, 225, 228, 194, 198, 214, 222, 238, 242, 294, 298, 334, 346, 358, 366, 414, 422, 434, 446, 482, 486, 538, 542, 558, 566, 582, 586
Offset: 1
We build permutational numbers:
a(1)=0 in unitary positional system we have only one digit 0
a(2)=1 because in binary positional system smaller number with two different digits is 01 = 1
a(3)=2 because in binary positional system bigger number with two different digits is 10 = 2 (binary system is over)
a(4)=5 because smallest number in ternary system with 3 different digits is 012=5
a(5)=7 second number in ternary system with 3 different digits is 021=7
a(6)=11 third number in ternary system with 3 different digits is 102=11
a(7)=15 120=15
etc.
-
import Data.List (permutations, sort)
a134640 n k = a134640_tabf !! (n-1) !! (k-1)
a134640_row n = sort $
map (foldr (\dig val -> val * n + dig) 0) $ permutations [0 .. n - 1]
a134640_tabf = map a134640_row [1..]
a134640_list = concat a134640_tabf
-- Reinhard Zumkeller, Aug 29 2014
-
a = {}; b = {}; Do[AppendTo[b, n]; w = Permutations[b]; Do[j = FromDigits[w[[m]], n + 1]; AppendTo[a, j], {m, 1, Length[w]}], {n, 0, 5}]; a (*Artur Jasinski*)
Flatten[Table[FromDigits[#,n]&/@Permutations[Range[0,n-1]],{n,5}]] (* Harvey P. Dale, Dec 09 2014 *)
-
from itertools import permutations
def fd(d, b): return sum(di*b**i for i, di in enumerate(d[::-1]))
def row(n): return [fd(d, n) for d in permutations(range(n))]
print([an for r in range(1, 6) for an in row(r)]) # Michael S. Branicky, Oct 21 2022
Corrected indices in examples. Replaced dashes in comments by the word "to" -
R. J. Mathar, Aug 26 2009
A257511
Number of 1's in factorial base representation of n (A007623).
Original entry on oeis.org
0, 1, 1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 2, 3, 3, 4, 2, 3, 1, 2, 2, 3, 1, 2, 1, 2, 2, 3, 1, 2, 0, 1, 1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 1, 2, 2, 3, 1, 2, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 1, 1
Offset: 0
Cf.
A001221,
A007623,
A007814,
A034968,
A056169,
A060130,
A225901,
A257687,
A265333,
A275732,
A275735,
A260736,
A276076.
-
factBaseIntDs[n_] := Module[{m, i, len, dList, currDigit}, i = 1; While[n > i!, i++]; m = n; len = i; dList = Table[0, {len}]; Do[currDigit = 0; While[m >= j!, m = m - j!; currDigit++]; dList[[len - j + 1]] = currDigit, {j, i, 1, -1}]; If[dList[[1]] == 0, dList = Drop[dList, 1]]; dList]; s = Table[FromDigits[factBaseIntDs@ n], {n, 0, 120}];
First@ DigitCount[#] & /@ s (* Michael De Vlieger, Apr 27 2015, after Alonso del Arte at A007623 *)
nn = 120; b = Module[{m = 1}, While[Factorial@ m < nn, m++]; MixedRadix[Reverse@ Range[2, m]]]; Table[Count[IntegerDigits[n, b], 1], {n, 0, nn}] (* Michael De Vlieger, Aug 29 2016, Version 10.2 *)
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(define (A257511 n) (let loop ((n n) (i 2) (s 0)) (cond ((zero? n) s) (else (loop (floor->exact (/ n i)) (+ 1 i) (+ s (if (= 1 (modulo n i)) 1 0)))))))
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