A276076 -id:A276076 - cp's OEIS frontend

A276075 a(1) = 0, a(n) = (e1i1! + e2i2! + ... + eziz!) for n = prime(i1)^e1 prime(i2)^e2 * ... * prime(iz)^ez, where prime(k) is the k-th prime, A000040(k).

0, 1, 2, 2, 6, 3, 24, 3, 4, 7, 120, 4, 720, 25, 8, 4, 5040, 5, 40320, 8, 26, 121, 362880, 5, 12, 721, 6, 26, 3628800, 9, 39916800, 5, 122, 5041, 30, 6, 479001600, 40321, 722, 9, 6227020800, 27, 87178291200, 122, 10, 362881, 1307674368000, 6, 48, 13, 5042, 722, 20922789888000, 7, 126, 27, 40322, 3628801, 355687428096000, 10, 6402373705728000, 39916801, 28, 6, 726, 123

Offset: 1

Antti Karttunen, Aug 18 2016

nonn

Additive with a(p^e) = e * (PrimePi(p)!), where PrimePi(n) = A000720(n).

a(3181) has 1001 decimal digits. - Michael De Vlieger, Dec 24 2017

Michael De Vlieger, Table of n, a(n) for n = 1..3180 (First 120 terms from Antti Karttunen).

Cf. A000040, A000142, A000720, A002110, A007489, A019565, A028234, A033312, A055396, A059590, A067029, A076954, A206296, A260443, A276080, A276081.
Left inverse of A276076.
Cf. also A048675.

Array[If[# == 1, 0, Total[FactorInteger[#] /. {p_, e_} /; p > 1 :> e PrimePi[p]!]] &, 66] (* Michael De Vlieger, Dec 24 2017 *)

from sympy import factorint, factorial as f, primepi
def a(n):
    F=factorint(n)
    return 0 if n==1 else sum(F[i]*f(primepi(i)) for i in F)
print([a(n) for n in range(1, 121)]) # Indranil Ghosh, Jun 21 2017

a(1) = 0; for n > 1, a(n) = a(A028234(n)) + (A067029(n) * A000142(A055396(n))).
Other identities.
For all n >= 0:
a(A276076(n)) = n.
a(A002110(n)) = A007489(n).
a(A019565(n)) = A059590(n).
a(A206296(n)) = A276080(n).
a(A260443(n)) = A276081(n).
For all n >= 1:
a(A000040(n)) = n! = A000142(n).
a(A076954(n-1)) = A033312(n).

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

Antti Karttunen, Aug 08 2016

These are prime-factorization representations of single-variable polynomials where the coefficient of term x^(k-1) (encoded as the exponent of prime(k) in the factorization of n) is equal to the number of nonzero digits that occur on the slope (k-1) levels below the "maximal slope" in the factorial base representation of n. See A275811 for the definition of the "digit slopes" in this context.

			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.

Antti Karttunen, Table of n, a(n) for n = 0..40320
Index entries for sequences related to factorial base representation

Cf. A001221, A001222, A002110, A007489, A007814, A048675, A051903, A056169, A056170, A060130, A060502, A225901.
Cf. A257684, A275732.
Cf. A275811.
Cf. A260736, A275728, A275808, A275812, A275946, A275947, A275962.
Cf. A275804 (indices of squarefree terms), A275805 (of terms not squarefree).
Cf. also A275725, A275733, A275735, A276076 for other such prime factorization encodings of A060117/A060118-related polynomials.

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

a(0) = 1; for n >= 1, a(n) = A275732(n) * a(A257684(n)).
Other identities and observations. For all n >= 0:
a(n) = A275735(A225901(n)).
a(A007489(n)) = A002110(n).
A001221(a(n)) = A060502(n).
A001222(a(n)) = A060130(n).
A007814(a(n)) = A260736(n).
A051903(a(n)) = A275811(n).
A048675(a(n)) = A275728(n).
A248663(a(n)) = A275808(n).
A056169(a(n)) = A275946(n).
A056170(a(n)) = A275947(n).
A275812(a(n)) = A275962(n).

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

Antti Karttunen, Aug 09 2016

These are prime-factorization representations of single-variable polynomials where the coefficient of term x^(k-1) (encoded as the exponent of prime(k) in the factorization of n) is equal to the number of times a nonzero digit k occurs in the factorial base representation of n. See the examples.

			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.

Antti Karttunen, Table of n, a(n) for n = 0..40320
Index entries for sequences related to factorial base representation

Cf. A000079, A001221, A001222, A003961, A007623, A008683, A181819, A225901, A257511, A257684, A265349, A265350, A264990, A275729, A275806, A351954.
Cf. also A275725, A275733, A275734 for other such prime factorization encodings of A060117/A060118-related polynomials, and also A276076.
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

a(0) = 1; for n >= 1, a(n) = 2^A257511(n) * A003961(a(A257684(n))).
Other identities and observations. For all n >= 0:
a(n) = A275734(A225901(n)).
A001221(a(n)) = A275806(n).
A001222(a(n)) = A060130(n).
A048675(a(n)) = A275729(n).
A051903(a(n)) = A264990(n).
A008683(a(A265349(n))) = -1 or +1 for all n >= 0.
A008683(a(A265350(n))) = 0 for all n >= 1.
From Antti Karttunen, Apr 03 2022: (Start)
A342001(a(n)) = A351954(n).
a(n) = A181819(A276076(n)). (End)

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

Antti Karttunen, Apr 27 2015

Antti Karttunen, Table of n, a(n) for n = 0..10080
Index entries for sequences related to factorial base representation

Cf. A001221, A007623, A007814, A034968, A056169, A060130, A225901, A257687, A265333, A275732, A275735, A260736, A276076.
Cf. A255411 (numbers n such that a(n) = 0), A255341 (such that a(n) = 1), A255342 (such that a(n) = 2), A255343 (such that a(n) = 3).
Positions of records: A007489.
Cf. also A257510.

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)))))))

a(0) = 0; for n >= 1, a(n) = A265333(n) + a(A257687(n)). - Antti Karttunen, Aug 29 2016
Other identities and observations. For all n >= 0:
a(n) = A260736(A225901(n)).
a(n) = A001221(A275732(n)) = A001222(A275732(n)).
a(n) = A007814(A275735(n)).
a(n) = A056169(A276076(n)).
a(A007489(n)) = n. [Particularly, A007489(n) gives the position where n first appears.]
a(n) <= A060130(n) <= A034968(n).

A054842 If n = a + 10 * b + 100 * c + 1000 * d + ... then a(n) = (2^a) * (3^b) * (5^c) * (7^d) * ...

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 3, 6, 12, 24, 48, 96, 192, 384, 768, 1536, 9, 18, 36, 72, 144, 288, 576, 1152, 2304, 4608, 27, 54, 108, 216, 432, 864, 1728, 3456, 6912, 13824, 81, 162, 324, 648, 1296, 2592, 5184, 10368, 20736, 41472, 243, 486, 972, 1944

Offset: 0

Henry Bottomley, Apr 11 2000

a((10^k-1)/9) = Primorial(k)= A061509((10^k-1)/9). This is a rearrangement of whole numbers. a(m) = a(n) iff m = n. (Unlike A061509, in which a(n) = a(n*10^k).) - Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), Jul 14 2003

Part of the previous comment is incorrect: as a set, this sequence consists of numbers n such that the largest exponent appearing in the prime factorization of n is 9. So this cannot be a rearrangement (or permutation) of the natural numbers. - Tom Edgar, Oct 20 2015

			a(15)=96 because 3^1 * 2^5 = 3*32 = 96.

Reinhard Zumkeller, Table of n, a(n) for n = 0..9999

Cf. A054841, A069877, A085840.

Cf. analogous sequences for other bases: A019565 (base 2), A101278 (base 3), A101942 (base 4), A101943 (base 5), A276076 (factorial base), A276086 (primorial base).

a054842 = f a000040_list 1 where
   f _      y 0 = y
   f (p:ps) y x = f ps (y * p ^ d) x'  where (x', d) = divMod x 10
-- Reinhard Zumkeller, Aug 03 2015

A054842[n_] := Times @@ (Prime[Range[Length[#], 1, -1]]^#) & [IntegerDigits[n]];
Array[A054842, 100, 0] (* Paolo Xausa, Nov 25 2024 *)

a(n)= my(d=Vecrev(digits(n))); factorback(primes(#d), d); \\ Ruud H.G. van Tol, Nov 28 2024

a(n) = f(n, 1, 1) with f(x, y, z) = if x > 0 then f(floor(x/10), y*prime(z)^(x mod 10), z+1) else y. - Reinhard Zumkeller, Mar 13 2010

A047257 Numbers that are congruent to {4, 5} mod 6.

Original entry on oeis.org

4, 5, 10, 11, 16, 17, 22, 23, 28, 29, 34, 35, 40, 41, 46, 47, 52, 53, 58, 59, 64, 65, 70, 71, 76, 77, 82, 83, 88, 89, 94, 95, 100, 101, 106, 107, 112, 113, 118, 119, 124, 125, 130, 131, 136, 137, 142, 143, 148, 149

Offset: 1

N. J. A. Sloane

Equivalently, numbers m such that 2^m - m is divisible by 3. Indeed, for every prime p, there are infinitely many numbers m such that 2^m - m (A000325) is divisible by p, here are numbers m corresponding to p = 3. - Bernard Schott, Dec 10 2021

Numbers k for which A276076(k) and A276086(k) are multiples of nine. For a simple proof, consider the penultimate digit in the factorial and primorial base expansions of n, A007623 and A049345. - Antti Karttunen, Feb 08 2024

Michael Doob, The Canadian Mathematical Olympiad & L'Olympiade Mathématique du Canada 1969-1993, Canadian Mathematical Society & Société Mathématique du Canada, Problem 4, 1983, page 158, 1993.

David Lovler, Table of n, a(n) for n = 1..1000
The IMO Compendium, Problem 4, 15th Canadian Mathematical Olympiad 1983.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).
Index to sequences related to Olympiads.

Cf. A000325.
Similar with: A299174 (p = 2), this sequence (p = 3), A349767 (p = 5).
Cf. also A047227, A047235, A047246, A047247, A047270.
Cf. A007623, A049345, A276076, A276086.

Select[Range@ 150, 4 <= Mod[#, 6] <= 5 &] (* Michael De Vlieger, Mar 20 2015 *)
LinearRecurrence[{1,1,-1},{4,5,10},50] (* Harvey P. Dale, Oct 16 2017 *)

A047257(n):=4 + 6*floor(n/2) + mod(n,2)$ akelist(A047257(n),n,0,40); /* Martin Ettl, Oct 24 2012 */

a(n) = 3*n - (-1)^n \\ David Lovler, Aug 25 2022

a(n) = 4 + 6*floor(n/2) + n mod 2.
a(n) = 6*n-a(n-1)-3, with a(1)=4. - Vincenzo Librandi, Aug 05 2010
G.f.: ( x*(4+x+x^2) ) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Oct 08 2011
a(n) = 3*n - (-1)^n. - Wesley Ivan Hurt, Mar 20 2015
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(6*sqrt(3)) - log(2)/3. - Amiram Eldar, Dec 14 2021
E.g.f.: 1 + 3*x*exp(x) - exp(-x). - David Lovler, Aug 25 2022

A246359 Maximum digit in the factorial base expansion of n (A007623).

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4

Offset: 0

Antti Karttunen, Oct 20 2014

Maximum entry in n-th row of A108731.

			Factorial base representation of 46 is "1320" as 46 = 1*4! + 3*3! + 2*2! + 0*1!, and the largest of these digits is 3, thus a(46) = 3.

Antti Karttunen, Table of n, a(n) for n = 0..10080
Index entries for sequences related to factorial base representation

Cf. A007623, A034968, A051903, A060130, A084558, A099563, A257684, A257687, A276076.

Cf. also A249070.

nn = 96; m = 1; While[Factorial@ m < nn, m++]; m; Table[Max@ IntegerDigits[n, MixedRadix[Reverse@ Range[2, m]]], {n, 0, nn}] (* 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}]; Table[Max@ f@ n, {n, 0, 96}] (* Michael De Vlieger, Aug 29 2016 *)

def a007623(n, p=2): return n if nIndranil Ghosh, Jun 21 2017

From Antti Karttunen, Aug 29 2016: (Start)
a(0) = 0; for n >= 1, a(n) = 1 + a(A257684(n)).
a(0) = 0; for n >= 1, a(n) = max(A099563(n), a(A257687(n))).
a(n) = A051903(A276076(n)).
(End)

A276077 Number of distinct prime factors p of n such that p^(1+A000720(p)) is a divisor of n, where A000720(p) gives the index of prime p, 1 for 2, 2 for 3, 3 for 5, and so on.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1

Offset: 1

Antti Karttunen, Aug 18 2016

			For n = 2 (= prime(1)), 2 is not divisible by 2^(1+1), thus a(2) = 0.
For n = 3 (= prime(3)), 3 is not divisible by 3^(2+1), thus a(3) = 0.
For n = 4 (= prime(1)^2), 4 is divisible by 2^(1+1), and there are no other prime factors apart from 2, thus a(4) = 1.
For n = 108 = 2^2 * 3^3, it is divisible both by 2^(1+1) and 3^(2+1), thus a(108) = 2.
For n = 625 = prime(3)^4, it is divisible by 5^(3+1), thus a(625) = 1.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A000720, A028234, A055396, A067029.
Cf. A276078 (positions of zeros), A276079 (nonzeros), also A276076.
Differs from A129251 for the first time at n=625, where a(625) = 1, while A129251(625) = 0.

f[p_, e_] := If[PrimePi[p] < e, 1, 0]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 30 2023 *)

a(n) = {my(f = factor(n)); sum(i = 1, #f~, primepi(f[i,1]) < f[i,2]);} \\ Amiram Eldar, Sep 30 2023

from sympy import primepi, isprime, primefactors, factorint
def a028234(n):
    f=factorint(n)
    return 1 if n==1 else n//(min(f)**f[min(f)])
def a067029(n):
    f=factorint(n)
    return 0 if n==1 else f[min(f)]
def a049084(n): return primepi(n)*(isprime(n))
def a055396(n): return 0 if n==1 else a049084(min(primefactors(n)))
def a(n):
    if n==1: return 0
    val = a(a028234(n))
    if a067029(n) > a055396(n):
        val += 1
    return val
print([a(n) for n in range(1, 201)]) # Indranil Ghosh, Jun 21 2017

(define (A276077 n) (if (= 1 n) 0 (+ (A276077 (A028234 n)) (if (> (A067029 n) (A055396 n)) 1 0))))

This formula uses Iverson bracket, which gives 1 as its value if the condition given inside [ ] is true and 0 otherwise:
a(1) = 0, for n > 1, a(n) = a(A028234(n)) + [A067029(n) > A055396(n)].
Other identities. For all n >= 1:
a(A276076(n)) = 0.
From Amiram Eldar, Sep 30 2023: (Start)
Additive with a(p^e) = 1 if primepi(p) < e, and 0 otherwise.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=1} 1/prime(k)^(k+1) = 0.2886971166123417096098... . (End)

A277012 Factorial base representation of n is rewritten as a base-2 number with each nonzero digit k replaced by a run of k 1's (followed by one extra zero if not the rightmost run of 1's) and with each 0 kept as 0.

Original entry on oeis.org

0, 1, 2, 5, 6, 13, 4, 9, 10, 21, 22, 45, 12, 25, 26, 53, 54, 109, 28, 57, 58, 117, 118, 237, 8, 17, 18, 37, 38, 77, 20, 41, 42, 85, 86, 173, 44, 89, 90, 181, 182, 365, 92, 185, 186, 373, 374, 749, 24, 49, 50, 101, 102, 205, 52, 105, 106, 213, 214, 429, 108, 217, 218, 437, 438, 877, 220, 441, 442, 885, 886, 1773, 56, 113, 114, 229, 230, 461, 116, 233

Offset: 0

Antti Karttunen, Sep 25 2016

			9 = "111" in factorial base (3! + 2! + 1! = 9) is converted to three 1-bits with separating zeros between, in binary as "10101" = A007088(21), thus a(9) = 21.
91 = "3301" in factorial base (91 = 3*4! + 3*3! + 1!) is converted to binary number "1110111001" = A007088(953), thus a(91) = 953. Between the rightmost 1-runs the other zero comes from the factorial base representation, while the other zero is an extra separating zero inserted after each run of 1-bits apart from the rightmost 1-run. The single zero between the two leftmost 1-runs is similarly used to separate the two "unary representations" of 3's.

Cf. A000035, A000079, A000120, A000225, A007088, A007623, A034968, A060130, A069010, A156552, A227153, A227349.
Cf. A277008 (terms sorted into ascending order).
Cf. A277011 (a left inverse).
Differs from analogous A277022 for the first time at n=24, where a(24) = 8, while A277022(24) = 60.

(define (A277012 n) (let loop ((n n) (z 0) (i 2) (j 0)) (if (zero? n) z (let ((d (remainder n i))) (loop (quotient n i) (+ z (* (A000225 d) (A000079 j))) (+ 1 i) (+ 1 j d))))))

a(n) = A156552(A276076(n)).
Other identities. For all n >= 0:
A277011(a(n)) = n.
A005940(1+a(n)) = A276076(n).
A000035(a(n)) = A000035(n). [Preserves the parity of n.]
A000120(a(n)) = A034968(n).
A069010(a(n)) = A060130(n).
A227349(a(n)) = A227153(n).

A351576 Factorial base expansion of n reinterpreted as a primorial base expansion, then converted back to decimal.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100

Offset: 0

Antti Karttunen, Apr 01 2022

			n = 313 has factorial base representation (see A007623) "23001" because 2*5! + 3*4! + 1*1! = 240+72+1 = 313. When this is reinterpreted as a primorial base expansion (see A049345), we obtain 2*A002110(4) + 3*A002110(3) + 1*A002110(0) = 511, therefore a(313) = 511.

Cf. A000142, A002110, A007623, A049345, A276076, A276085.

Cf. also A276156.

a[n_] := Module[{k = n, m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0|| r != 0, AppendTo[s, r]; m++]; FromDigits[Reverse[s], MixedRadix[Reverse@ Prime@ Range@ Length[s]]]]; Array[a, 100, 0] (* Amiram Eldar, Feb 07 2024 *)

A002110(n) = prod(i=1,n,prime(i));
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; };
A276085(n) = { my(f = factor(n)); sum(k=1, #f~, f[k, 2]*A002110(primepi(f[k, 1])-1)); };
A351576(n) = A276085(A276076(n));

a(n) = A276085(A276076(n)).

cp's OEIS Frontend

A276075 a(1) = 0, a(n) = (e1*i1! + e2*i2! + ... + ez*iz!) for n = prime(i1)^e1 * prime(i2)^e2 * ... * prime(iz)^ez, where prime(k) is the k-th prime, A000040(k).

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A275734 Prime-factorization representations of "factorial base slope polynomials": a(0) = 1; for n >= 1, a(n) = A275732(n) * a(A257684(n)).

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A275735 Prime-factorization representations of "factorial base level polynomials": a(0) = 1; for n >= 1, a(n) = 2^A257511(n) * A003961(a(A257684(n))).

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A257511 Number of 1's in factorial base representation of n (A007623).

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A054842 If n = a + 10 * b + 100 * c + 1000 * d + ... then a(n) = (2^a) * (3^b) * (5^c) * (7^d) * ...

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A047257 Numbers that are congruent to {4, 5} mod 6.

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A246359 Maximum digit in the factorial base expansion of n (A007623).

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A276075 a(1) = 0, a(n) = (e1i1! + e2i2! + ... + eziz!) for n = prime(i1)^e1 prime(i2)^e2 * ... * prime(iz)^ez, where prime(k) is the k-th prime, A000040(k).