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
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
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 *)
-
(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)))))))
A276091
Numbers obtained by reinterpreting base-2 representation of n in A001563-base (A276326): a(n) = Sum_{k>=0} A030308(n,k)*A001563(k+1).
Original entry on oeis.org
0, 1, 4, 5, 18, 19, 22, 23, 96, 97, 100, 101, 114, 115, 118, 119, 600, 601, 604, 605, 618, 619, 622, 623, 696, 697, 700, 701, 714, 715, 718, 719, 4320, 4321, 4324, 4325, 4338, 4339, 4342, 4343, 4416, 4417, 4420, 4421, 4434, 4435, 4438, 4439, 4920, 4921, 4924, 4925, 4938, 4939, 4942, 4943, 5016, 5017, 5020, 5021, 5034, 5035, 5038, 5039, 35280, 35281
Offset: 0
Cf.
A000120,
A000142,
A001563,
A030308,
A059590,
A060130,
A260736,
A225901,
A255411,
A275959,
A276082,
A276083,
A276090,
A276326,
A276328,
A276336,
A276337,
A276339,
A276340.
-
Table[Total[Times @@@ Transpose@ {Map[# #! &, Range@ Length@ #], Reverse@ #}] &@ IntegerDigits[n, 2], {n, 64}] (* Michael De Vlieger, Aug 31 2016 *)
-
from sympy import factorial as f
def a007623(n, p=2): return n if n0 else '0' for i in x)[::-1]
return 0 if n==0 else sum(int(y[i])*f(i + 1) for i in range(len(y)))
def a(n): return 0 if n==0 else a255411(a(n//2)) if n%2==0 else 1 + a255411(a((n - 1)//2))
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 20 2017
-
;; This is a standalone program:
(define (A276091 n) (let loop ((n n) (s 0) (f 1) (i 2)) (cond ((zero? n) s) ((even? n) (loop (/ n 2) s (* i f) (+ 1 i))) (else (loop (/ (- n 1) 2) (+ s (* (- i 1) f)) (* i f) (+ 1 i))))))
;; This implements one of the given recurrences:
(definec (A276091 n) (cond ((zero? n) n) ((even? n) (A255411 (A276091 (/ n 2)))) (else (+ 1 (A255411 (A276091 (/ (- n 1) 2)))))))
;; Alternatively, we can use A276340 in place of A255411:
(definec (A276091 n) (cond ((zero? n) n) ((even? n) (A276340 (A276091 (/ n 2)))) (else (+ 1 (A276340 (A276091 (/ (- n 1) 2)))))))
Name changed (to emphasize the functional nature of the sequence) with the original definition moved to the comments by
Antti Karttunen, Sep 01 2016
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