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
A060112
Sums of nonconsecutive factorial numbers.
Original entry on oeis.org
0, 1, 2, 6, 7, 24, 25, 26, 120, 121, 122, 126, 127, 720, 721, 722, 726, 727, 744, 745, 746, 5040, 5041, 5042, 5046, 5047, 5064, 5065, 5066, 5160, 5161, 5162, 5166, 5167, 40320, 40321, 40322, 40326, 40327, 40344, 40345, 40346, 40440, 40441, 40442
Offset: 1
Zeckendorf Expansions of first few natural numbers and the corresponding values when interpreted as factorial expansions: 0 = 0 = 0, 1 = 1 = 1, 2 = 10 = 2, 3 = 100 = 6, 4 = 101 = 7, 5 = 1000 = 24, 6 = 1001 = 25, 7 = 1010 = 26, 8 = 10000 = 120, etc.,
-
CampanoPerm := proc(n) local z,p,i; p := []; z := fibbinary(n); i := 1; while(z > 0) do if(1 = (z mod 2)) then p := permul(p,[[i,i+1]]); fi; i := i+1; z := floor(z/2); od; RETURN(convert(p,'permlist',i)); end;
-
With[{b = MixedRadix[Range[12, 2, -1]]}, FromDigits[#, b] & /@ Select[Tuples[{0, 1}, 8], SequenceCount[#, {1, 1}] == 0 &]] (* Michael De Vlieger, Jun 26 2017 *)
-
fill(lim,k,val)=if(k>#f, return); my(t=val+f[k]); if(t<=lim, listput(v,t); fill(lim,k+2,t)); fill(lim,k+1,val)
list(lim)=my(k,t=1); local(f=List(),v=List([0])); while((t*=k++)<=lim, listput(f,t)); f=Vecrev(f); fill(lim,1,0); Set(v) \\ Charles R Greathouse IV, Jun 25 2017
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first(n) = my(res = [0, 1], k = 1, t = 1, p = 1); while(#res < n, k++; t++; p *= t; res = concat(res, vector(fibonacci(k), i, res[i]+p))); vector(n, i, res[i]) \\ David A. Corneth, Jun 26 2017
A227130
Numbers k for which there is an even number of nonzero digits when k is written in the factorial base (A007623).
Original entry on oeis.org
0, 3, 5, 7, 8, 10, 13, 14, 16, 19, 20, 22, 25, 26, 28, 30, 33, 35, 36, 39, 41, 42, 45, 47, 49, 50, 52, 54, 57, 59, 60, 63, 65, 66, 69, 71, 73, 74, 76, 78, 81, 83, 84, 87, 89, 90, 93, 95, 97, 98, 100, 102, 105, 107, 108, 111, 113, 114, 117, 119, 121, 122, 124
Offset: 1
-
q[n_] := Module[{k = n, m = 2, c = 0, r}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, If[r != 0, c++]; m++]; EvenQ[c]]; Select[Range[0, 150], q] (* Amiram Eldar, Jan 23 2024 *)
A227132
Numbers k for which there is an odd number of nonzero digits when k is written in the factorial base (A007623).
Original entry on oeis.org
1, 2, 4, 6, 9, 11, 12, 15, 17, 18, 21, 23, 24, 27, 29, 31, 32, 34, 37, 38, 40, 43, 44, 46, 48, 51, 53, 55, 56, 58, 61, 62, 64, 67, 68, 70, 72, 75, 77, 79, 80, 82, 85, 86, 88, 91, 92, 94, 96, 99, 101, 103, 104, 106, 109, 110, 112, 115, 116, 118, 120, 123, 125
Offset: 1
-
q[n_] := Module[{k = n, m = 2, c = 0, r}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, If[r != 0, c++]; m++]; OddQ[c]]; Select[Range[150], q] (* Amiram Eldar, Jan 24 2024 *)
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