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.
A275735
Prime-factorization representations of "factorial base level polynomials": a(0) = 1; for n >= 1, a(n) = 2^A257511(n) * A003961(a(A257684(n))).
Original entry on oeis.org
1, 2, 2, 4, 3, 6, 2, 4, 4, 8, 6, 12, 3, 6, 6, 12, 9, 18, 5, 10, 10, 20, 15, 30, 2, 4, 4, 8, 6, 12, 4, 8, 8, 16, 12, 24, 6, 12, 12, 24, 18, 36, 10, 20, 20, 40, 30, 60, 3, 6, 6, 12, 9, 18, 6, 12, 12, 24, 18, 36, 9, 18, 18, 36, 27, 54, 15, 30, 30, 60, 45, 90, 5, 10, 10, 20, 15, 30, 10, 20, 20, 40, 30, 60, 15, 30, 30, 60, 45, 90, 25, 50, 50, 100, 75
Offset: 0
For n = 0 whose factorial base representation (A007623) is also 0, there are no nonzero digits at all, thus there cannot be any prime present in the encoding, and a(0) = 1.
For n = 1 there is just one 1, thus a(1) = prime(1) = 2.
For n = 2 ("10"), there is just one 1-digit, thus a(2) = prime(1) = 2.
For n = 3 ("11") there are two 1-digits, thus a(3) = prime(1)^2 = 4.
For n = 18 ("300") there is just one 3, thus a(18) = prime(3) = 5.
For n = 19 ("301") there is one 1 and one 3, thus a(19) = prime(1)*prime(3) = 2*5 = 10.
For n = 141 ("10311") there are three 1's and one 3, thus a(141) = prime(1)^3 * prime(3) = 2^3 * 5^1 = 40.
Cf.
A000079,
A001221,
A001222,
A003961,
A007623,
A008683,
A181819,
A225901,
A257511,
A257684,
A265349,
A265350,
A264990,
A275729,
A275806,
A351954.
Differs from
A227154 for the first time at n=18, where a(18) = 5, while
A227154(18) = 4.
-
A276076(n) = { my(i=0,m=1,f=1,nextf); while((n>0),i=i+1; nextf = (i+1)*f; if((n%nextf),m*=(prime(i)^((n%nextf)/f));n-=(n%nextf));f=nextf); m; };
A181819(n) = factorback(apply(e->prime(e),(factor(n)[,2])));
A275735(n) = A181819(A276076(n)); \\ Antti Karttunen, Apr 03 2022
-
from sympy import prime
from operator import mul
import collections
def a007623(n, p=2): return n if nIndranil Ghosh, Jun 19 2017
A275725
a(n) = A275723(A002110(1+A084558(n)), n); prime factorization encodings of cycle-polynomials computed for finite permutations listed in the order that is used in tables A060117 / A060118.
Original entry on oeis.org
2, 4, 18, 8, 12, 8, 150, 100, 54, 16, 24, 16, 90, 40, 54, 16, 36, 16, 60, 40, 36, 16, 24, 16, 1470, 980, 882, 392, 588, 392, 750, 500, 162, 32, 48, 32, 270, 80, 162, 32, 108, 32, 120, 80, 72, 32, 48, 32, 1050, 700, 378, 112, 168, 112, 750, 500, 162, 32, 48, 32, 450, 200, 162, 32, 72, 32, 300, 200, 108, 32, 48, 32, 630, 280, 378, 112, 252, 112, 450, 200
Offset: 0
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].
Cf.
A000040,
A001222,
A001221,
A002110,
A007814,
A046660,
A048675,
A051903,
A056169,
A056170,
A084558,
A243054,
A257510,
A275723,
A275803,
A275832.
A275804
Numbers with at most one nonzero digit on each digit slope of the factorial base representation of n.
Original entry on oeis.org
0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 12, 13, 16, 18, 20, 24, 25, 26, 27, 28, 30, 31, 32, 33, 34, 36, 37, 40, 42, 44, 48, 49, 50, 51, 52, 60, 61, 64, 66, 68, 72, 73, 76, 78, 79, 82, 90, 96, 98, 102, 104, 108, 120, 121, 122, 123, 124, 126, 127, 128, 129, 130, 132, 133, 136, 138, 140, 144, 145, 146, 147, 148, 150, 151, 152, 153, 154, 156, 157, 160
Offset: 0
-
from operator import mul
from sympy import prime, factorial as f
from sympy.ntheory.factor_ import core
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))
def ok(n): return 1 if n==0 else core(a(n))==a(n)
print([n for n in range(201) if ok(n)]) # Indranil Ghosh, Jun 19 2017
A275732
One-based positions of 1-digits in the factorial base representation of n are converted to primes with those indices, then multiplied together.
Original entry on oeis.org
1, 2, 3, 6, 1, 2, 5, 10, 15, 30, 5, 10, 1, 2, 3, 6, 1, 2, 1, 2, 3, 6, 1, 2, 7, 14, 21, 42, 7, 14, 35, 70, 105, 210, 35, 70, 7, 14, 21, 42, 7, 14, 7, 14, 21, 42, 7, 14, 1, 2, 3, 6, 1, 2, 5, 10, 15, 30, 5, 10, 1, 2, 3, 6, 1, 2, 1, 2, 3, 6, 1, 2, 1, 2, 3, 6, 1, 2, 5, 10, 15, 30, 5, 10, 1, 2, 3, 6, 1, 2, 1, 2, 3, 6, 1, 2, 1, 2, 3, 6, 1, 2, 5, 10, 15, 30
Offset: 0
22 has factorial base representation "320" (= A007623(22)), which does not contain any "1". Thus a(22) = 1, as the empty product is 1.
35 has factorial base representation "1121" (= A007623(35)). 1's occur in the following positions, when counted from right, starting with 1: 1, 3 and 4. Thus a(35) = prime(1)*prime(3)*prime(4) = 2*5*7 = 70.
Cf.
A000040,
A001221,
A001222,
A002110,
A005117,
A007623,
A007489,
A048675,
A257261,
A257511,
A275730.
-
nn = 105; m = 1; While[Factorial@ m < nn, m++]; m; Map[Times @@ Map[Prime, Flatten@ Position[#, 1]] &@ Reverse@ IntegerDigits[#, MixedRadix[Reverse@ Range[2, m]]] &, Range[0, nn]] (* Michael De Vlieger, Aug 11 2016, Version 10.2 *)
-
from operator import mul
from sympy import prime
def a007623(n, p=2): return n if nIndranil Ghosh, Jun 19 2017
-
;; Recursive definition using memoizing definec-macro:
(definec (A275732 n) (cond ((zero? (A257261 n)) 1) (else (* (A000040 (A257261 n)) (A275732 (A275730bi n (- (A257261 n) 1)))))))
(define (A275732 n) (let loop ((z 1) (n n)) (let ((y (A257261 n))) (cond ((zero? y) z) (else (loop (* z (A000040 y)) (A275730bi n (- y 1))))))))
;; Code for A275730bi given in A275730.
Original entry on oeis.org
1, 2, 3, 6, 3, 6, 5, 10, 15, 30, 15, 30, 5, 10, 15, 30, 15, 30, 5, 10, 15, 30, 15, 30, 7, 14, 21, 42, 21, 42, 35, 70, 105, 210, 105, 210, 35, 70, 105, 210, 105, 210, 35, 70, 105, 210, 105, 210, 7, 14, 21, 42, 21, 42, 35, 70, 105, 210, 105, 210, 35, 70, 105, 210, 105, 210, 35, 70, 105, 210, 105, 210, 7, 14, 21, 42, 21, 42
Offset: 0
For n=19, A007623(19) = 301, thus a(19) = prime(3)*prime(1) = 5*2 = 10.
For n=52, A007623(52) = 2020, thus a(52) = prime(2)*prime(4) = 3*7 = 21.
Cf.
A001221,
A002110,
A003961,
A005117,
A007489,
A007623,
A048675,
A060130,
A061395,
A084558,
A257684,
A275732.
A290095
a(n) = A275725(A060126(n)); prime factorization encodings of cycle-polynomials computed for finite permutations listed in reversed colexicographic ordering.
Original entry on oeis.org
2, 4, 18, 8, 8, 12, 150, 100, 54, 16, 16, 24, 54, 16, 90, 40, 36, 16, 16, 24, 40, 60, 16, 36, 1470, 980, 882, 392, 392, 588, 750, 500, 162, 32, 32, 48, 162, 32, 270, 80, 108, 32, 32, 48, 80, 120, 32, 72, 750, 500, 162, 32, 32, 48, 1050, 700, 378, 112, 112, 168, 450, 200, 162, 32, 32, 72, 200, 300, 32, 48, 108, 32, 162, 32, 270, 80, 108, 32, 378, 112, 630, 280
Offset: 0
Consider the first eight permutations (indices 0-7) listed in A055089:
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]
2,3,1 [One 3-cycle, thus a(4) = 2^3 = 8]
3,2,1 [One transposition jumping over a fixed element, a(5) = 2^2 * 3^1 = 12]
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].
A275808
a(0) = 0, for n >= 1, a(n) = A275736(n) XOR a(A257684(n)), where XOR is given by A003987.
Original entry on oeis.org
0, 1, 2, 3, 1, 0, 4, 5, 6, 7, 5, 4, 2, 3, 0, 1, 3, 2, 1, 0, 3, 2, 0, 1, 8, 9, 10, 11, 9, 8, 12, 13, 14, 15, 13, 12, 10, 11, 8, 9, 11, 10, 9, 8, 11, 10, 8, 9, 4, 5, 6, 7, 5, 4, 0, 1, 2, 3, 1, 0, 6, 7, 4, 5, 7, 6, 5, 4, 7, 6, 4, 5, 2, 3, 0, 1, 3, 2, 6, 7, 4, 5, 7, 6, 0, 1, 2, 3, 1, 0, 3, 2, 1, 0, 2, 3, 1, 0, 3, 2, 0
Offset: 0
Cf.
A275809 (positions of zeros),
A275810 (and their first differences).
A276010
a(0) = 0, for n >= 1, a(n) = A275736(n) OR a(A257684(n)), where OR is given by A003986.
Original entry on oeis.org
0, 1, 2, 3, 1, 1, 4, 5, 6, 7, 5, 5, 2, 3, 2, 3, 3, 3, 1, 1, 3, 3, 1, 1, 8, 9, 10, 11, 9, 9, 12, 13, 14, 15, 13, 13, 10, 11, 10, 11, 11, 11, 9, 9, 11, 11, 9, 9, 4, 5, 6, 7, 5, 5, 4, 5, 6, 7, 5, 5, 6, 7, 6, 7, 7, 7, 5, 5, 7, 7, 5, 5, 2, 3, 2, 3, 3, 3, 6, 7, 6, 7, 7, 7, 2, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 1, 1, 3, 3, 1, 1, 5, 5, 7, 7, 5, 5, 3
Offset: 0
Comments