cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A276086 Primorial base exp-function: digits in primorial base representation of n become the exponents of successive prime factors whose product a(n) is.

Original entry on oeis.org

1, 2, 3, 6, 9, 18, 5, 10, 15, 30, 45, 90, 25, 50, 75, 150, 225, 450, 125, 250, 375, 750, 1125, 2250, 625, 1250, 1875, 3750, 5625, 11250, 7, 14, 21, 42, 63, 126, 35, 70, 105, 210, 315, 630, 175, 350, 525, 1050, 1575, 3150, 875, 1750, 2625, 5250, 7875, 15750, 4375, 8750, 13125, 26250, 39375, 78750, 49, 98, 147, 294, 441, 882, 245, 490, 735, 1470, 2205, 4410, 1225, 2450
Offset: 0

Views

Author

Antti Karttunen, Aug 21 2016

Keywords

Comments

Prime product form of primorial base expansion of n.
Sequence is a permutation of A048103. It maps the smallest prime not dividing n to the smallest prime dividing n, that is, A020639(a(n)) = A053669(n) holds for all n >= 1.
The sequence satisfies the exponential function identity, a(x + y) = a(x) * a(y), whenever A329041(x,y) = 1, that is, when adding x and y together will not generate any carries in the primorial base. Examples of such pairs of x and y are A328841(n) & A328842(n), and also A328770(n) (when added with itself). - Antti Karttunen, Oct 31 2019
From Antti Karttunen, Feb 18 2022: (Start)
The conjecture given in A327969 asks whether applying this function together with the arithmetic derivative (A003415) in some combination or another can eventually transform every positive integer into zero.
Another related open question asks whether there are any other numbers than n=6 such that when starting from that n and by iterating with A003415, one eventually reaches a(n). See comments in A351088.
This sequence is used in A351255 to list the terms of A099308 in a different order, by the increasing exponents of the successive primes in their prime factorization. (End)
From Bill McEachen, Oct 15 2022: (Start)
From inspection, the least significant decimal digits of a(n) terms form continuous chains of 30 as follows. For n == i (mod 30), i=0..5, there are 6 ordered elements of these 8 {1,2,3,6,9,8,7,4}. Then for n == i (mod 30), i=6..29, there are 12 repeated pairs = {5,0}.
Moreover, when the individual elements of any of the possible groups of 6 are transformed via (7*digit) (mod 10), the result matches one of the other 7 groupings (not all 7 may be seen). As example, {1,2,3,6,9,8} transforms to {7,4,1,2,3,6}. (End)
The least significant digit of a(n) in base 4 is given by A353486, and in base 6 by A358840. - Antti Karttunen, Oct 25 2022, Feb 17 2024

Examples

			For n = 24, which has primorial base representation (see A049345) "400" as 24 = 4*A002110(2) + 0*A002110(1) + 0*A002110(0) = 4*6 + 0*2 + 0*1, thus a(24) = prime(3)^4 * prime(2)^0 * prime(1)^0 = 5^4 = 625.
For n = 35 = "1021" as 35 = 1*A002110(3) + 0*A002110(2) + 2*A002110(1) + 1*A002110(0) = 1*30 + 0*6 + 2*2 + 1*1, thus a(35) = prime(4)^1 * prime(2)^2 * prime(1) = 7 * 3*3 * 2 = 126.
		

Crossrefs

Cf. A276085 (a left inverse) and also A276087, A328403.
Cf. A048103 (terms sorted into ascending order), A100716 (natural numbers not present in this sequence).
Cf. A278226 (associated filter-sequence), A286626 (and its rgs-version), A328477.
Cf. A328316 (iterates started from zero).
Cf. A327858, A327859, A327860, A327963, A328097, A328098, A328099, A328110, A328112, A328382 for various combinations with arithmetic derivative (A003415).
Cf. also A327167, A329037.
Cf. A019565 and A054842 for base-2 and base-10 analogs and A276076 for the analogous "factorial base exp-function", from which this differs for the first time at n=24, where a(24)=625 while A276076(24)=7.
Cf. A327969, A351088, A351458 for sequences with conjectures involving this sequence.

Programs

  • Mathematica
    b = MixedRadix[Reverse@ Prime@ Range@ 12]; Table[Function[k, Times @@ Power @@@ # &@ Transpose@ {Prime@ Range@ Length@ k, Reverse@ k}]@ IntegerDigits[n, b], {n, 0, 51}] (* Michael De Vlieger, Aug 23 2016, Version 10.2 *)
    f[n_] := Block[{a = {{0, n}}}, Do[AppendTo[a, {First@ #, Last@ #} &@ QuotientRemainder[a[[-1, -1]], Times @@ Prime@ Range[# - i]]], {i, 0, #}] &@ NestWhile[# + 1 &, 0, Times @@ Prime@ Range[# + 1] <= n &]; Rest[a][[All, 1]]]; Table[Times @@ Flatten@ MapIndexed[Prime[#2]^#1 &, Reverse@ f@ n], {n, 0, 73}] (* Michael De Vlieger, Aug 30 2016, Pre-Version 10 *)
    a[n0_] := Module[{m = 1, i = 1, n = n0, p}, While[n > 0, p = Prime[i]; m *= p^Mod[n, p]; n = Quotient[n, p]; i++]; m];
    Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Dec 01 2021, after Antti Karttunen's Sage code *)
  • PARI
    A276086(n) = { my(i=0,m=1,pr=1,nextpr); while((n>0),i=i+1; nextpr = prime(i)*pr; if((n%nextpr),m*=(prime(i)^((n%nextpr)/pr));n-=(n%nextpr));pr=nextpr); m; }; \\ Antti Karttunen, May 12 2017
    
  • PARI
    A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); }; \\ (Better than above one, avoids unnecessary construction of primorials). - Antti Karttunen, Oct 14 2019
    
  • Python
    from sympy import prime
    def a(n):
        i=0
        m=pr=1
        while n>0:
            i+=1
            N=prime(i)*pr
            if n%N!=0:
                m*=(prime(i)**((n%N)/pr))
                n-=n%N
            pr=N
        return m # Indranil Ghosh, May 12 2017, after Antti Karttunen's PARI code
    
  • Python
    from sympy import nextprime
    def a(n):
        m, p = 1, 2
        while n > 0:
            n, r = divmod(n, p)
            m *= p**r
            p = nextprime(p)
        return m
    print([a(n) for n in range(74)])  # Peter Luschny, Apr 20 2024
  • Sage
    def A276086(n):
        m=1
        i=1
        while n>0:
            p = sloane.A000040(i)
            m *= (p**(n%p))
            n = floor(n/p)
            i += 1
        return (m)
    # Antti Karttunen, Oct 14 2019, after Indranil Ghosh's Python code above, and my own leaner PARI code from Oct 14 2019. This avoids unnecessary construction of primorials.
    
  • Scheme
    (define (A276086 n) (let loop ((n n) (t 1) (i 1)) (if (zero? n) t (let* ((p (A000040 i)) (d (modulo n p))) (loop (/ (- n d) p) (* t (expt p d)) (+ 1 i))))))
    
  • Scheme
    (definec (A276086 n) (if (zero? n) 1 (* (expt (A053669 n) (A276088 n)) (A276086 (A276093 n))))) ;; Needs macro definec from http://oeis.org/wiki/Memoization#Scheme
    
  • Scheme
    (definec (A276086 n) (if (zero? n) 1 (* (A053669 n) (A276086 (- n (A002110 (A276084 n))))))) ;; Needs macro definec from http://oeis.org/wiki/Memoization#Scheme
    

Formula

a(0) = 1; for n >= 1, a(n) = A053669(n) * a(A276151(n)) = A053669(n) * a(n-A002110(A276084(n))).
a(0) = 1; for n >= 1, a(n) = A053669(n)^A276088(n) * a(A276093(n)).
a(n) = A328841(a(n)) + A328842(a(n)) = A328843(n) + A328844(n).
a(n) = a(A328841(n)) * a(A328842(n)) = A328571(n) * A328572(n).
a(n) = A328475(n) * A328580(n) = A328476(n) + A328580(n).
a(A002110(n)) = A000040(n+1). [Maps primorials to primes]
a(A143293(n)) = A002110(n+1). [Maps partial sums of primorials to primorials]
a(A057588(n)) = A276092(n).
a(A276156(n)) = A019565(n).
a(A283477(n)) = A324289(n).
a(A003415(n)) = A327859(n).
Here the text in brackets shows how the right hand side sequence is a function of the primorial base expansion of n:
A001221(a(n)) = A267263(n). [Number of nonzero digits]
A001222(a(n)) = A276150(n). [Sum of digits]
A067029(a(n)) = A276088(n). [The least significant nonzero digit]
A071178(a(n)) = A276153(n). [The most significant digit]
A061395(a(n)) = A235224(n). [Number of significant digits]
A051903(a(n)) = A328114(n). [Largest digit]
A055396(a(n)) = A257993(n). [Number of trailing zeros + 1]
A257993(a(n)) = A328570(n). [Index of the least significant zero digit]
A079067(a(n)) = A328620(n). [Number of nonleading zeros]
A056169(a(n)) = A328614(n). [Number of 1-digits]
A056170(a(n)) = A328615(n). [Number of digits larger than 1]
A277885(a(n)) = A328828(n). [Index of the least significant digit > 1]
A134193(a(n)) = A329028(n). [The least missing nonzero digit]
A005361(a(n)) = A328581(n). [Product of nonzero digits]
A072411(a(n)) = A328582(n). [LCM of nonzero digits]
A001055(a(n)) = A317836(n). [Number of carry-free partitions of n in primorial base]
Various number theoretical functions applied:
A000005(a(n)) = A324655(n). [Number of divisors of a(n)]
A000203(a(n)) = A324653(n). [Sum of divisors of a(n)]
A000010(a(n)) = A324650(n). [Euler phi applied to a(n)]
A023900(a(n)) = A328583(n). [Dirichlet inverse of Euler phi applied to a(n)]
A069359(a(n)) = A329029(n). [Sum a(n)/p over primes p dividing a(n)]
A003415(a(n)) = A327860(n). [Arithmetic derivative of a(n)]
Other identities:
A276085(a(n)) = n. [A276085 is a left inverse]
A020639(a(n)) = A053669(n). [The smallest prime not dividing n -> the smallest prime dividing n]
A046523(a(n)) = A278226(n). [Least number with the same prime signature as a(n)]
A246277(a(n)) = A329038(n).
A181819(a(n)) = A328835(n).
A053669(a(n)) = A326810(n), A326810(a(n)) = A328579(n).
A257993(a(n)) = A328570(n), A328570(a(n)) = A328578(n).
A328613(a(n)) = A328763(n), A328620(a(n)) = A328766(n).
A328828(a(n)) = A328829(n).
A053589(a(n)) = A328580(n). [Greatest primorial number which divides a(n)]
A276151(a(n)) = A328476(n). [... and that primorial subtracted from a(n)]
A111701(a(n)) = A328475(n).
A328114(a(n)) = A328389(n). [Greatest digit of primorial base expansion of a(n)]
A328389(a(n)) = A328394(n), A328394(a(n)) = A328398(n).
A235224(a(n)) = A328404(n), A328405(a(n)) = A328406(n).
a(A328625(n)) = A328624(n), a(A328626(n)) = A328627(n). ["Twisted" variants]
a(A108951(n)) = A324886(n).
a(n) mod n = A328386(n).
a(a(n)) = A276087(n), a(a(a(n))) = A328403(n). [2- and 3-fold applications]
a(2n+1) = 2 * a(2n). - Antti Karttunen, Feb 17 2022

Extensions

Name edited and new link-formulas added by Antti Karttunen, Oct 29 2019
Name changed again by Antti Karttunen, Feb 05 2022

A327969 The length of a shortest path from n to zero when using the transitions x -> A003415(x) and x -> A276086(x), or -1 if no zero can ever be reached from n.

Original entry on oeis.org

0, 1, 2, 2, 5, 2, 3, 2, 6, 4, 3, 2, 5, 2, 5, 6, 6, 2, 5, 2, 7, 4, 3, 2
Offset: 0

Views

Author

Antti Karttunen, Oct 07 2019

Keywords

Comments

The terms of this sequence are currently known only up to n=23, with the value of a(24) still being uncertain. For the tentative values of the later terms, see sequence A328324 which gives upper bounds for these terms, many of which are very likely also exact values for them.
As A051903(A003415(n)) >= A051903(n)-1, it means that it takes always at least A051903(n) steps to a prime if iterating solely with A003415.
Some known values and upper bounds from n=24 onward:
a(24) <= 11.
a(25) = 4.
a(26) = 7.
a(27) <= 22.
a(33) = 4.
a(39) = 4.
a(40) = 5.
a(42) = 3.
a(44) <= 10.
a(45) = 5.
a(46) = 5.
a(48) = 9.
a(49) = 6.
a(50) = 6.
a(55) = 7.
a(74) = 5.
a(77) = 6.
a(80) <= 18.
a(111) = 6.
a(112) = 8.
a(125) <= 9.
a(240) = 7.
a(625) <= 10.
a(875) = 8.
From Antti Karttunen, Feb 20 2022: (Start)
a(2556) <= 20.
a(5005) <= 19.
What is the value of a(128), and is A328324(128) well-defined?
When I created this sequence, I conjectured that by applying two simple arithmetic operations "arithmetic derivative" (A003415) and "primorial base exp-function" (A276086) in some combination, and starting from any positive integer, we could always reach zero (via a prime and 1).
At the first sight it seems almost certain that the conjecture holds, as it is always possible at every step to choose from two options (which very rarely meet, see A351088), leading to an exponentially growing search tree, and also because A276086 always jumps out of any dead-end path with p^p-factors (dead-end from the arithmetic derivative's point of view). However, it should be realized that one can reach the terms of either A157037 or A327978 with a single step of A003415 only from squarefree numbers (or respectively, cubefree numbers that are not multiples of 4, see A328234), and in general, because A003415 decreases the maximal exponent of the prime factorization (A051903) at most by one, if the maximal exponent in the prime factorization of n is large, there is a correspondingly long path to traverse if we take only A003415-steps in the iteration, and any step could always lead with certain probability to a p^p-number. Note that the antiderivatives of primorials with a square factor seem quite rare, see A351029.
And although taking a A276086-step will always land us to a p^p-free number (which a priori is not in the obvious dead-end path of A003415, although of course it might eventually lead to one), it (in most cases) also increases the magnitude of number considerably, that tends to make the escape even harder. Particularly, in the majority of cases A276086 increases the maximal exponent (which in the preimage is A328114, "maximal digit value used when n is written in primorial base"), so there will be even a longer journey down to squarefree numbers when using A003415. See the sequences A351067 and A351071 for the diminishing ratios suggesting rapidly diminishing chances of successfully reaching zero from larger terms of A276086. Also, the asymptotic density of A276156 is zero, even though A351073 may contain a few larger values.
On the other hand, if we could prove that by (for example) continuing upwards with any p^p-path of A003415 we could eventually reach with a near certainty a region of numbers with low values of A328114 (i.e., numbers with smallish digits in primorial base, like A276156), then the situation might change (see also A351089). However, a few empirical runs seemed to indicate otherwise.
For all of the above reasons, I now conjecture that there are natural numbers from which it is not possible to reach zero with any combination of steps. For example 128 or 5^5 = 3125.
(End)

Examples

			Let -A> stand for an application of A003415 and -B> for an application of A276086, then, we have for example:
a(8) = 6 as we have 8 -A>  12 -B>  25 -A> 10 -A>  7 -A> 1 -A> 0, six transitions in total (and there are no shorter paths).
a(15) = 6 as we have 15 -B> 150 -A> 185 -A> 42 -A> 41 -A> 1 -A> 0, six transitions in total (and there are no shorter paths).
a(20) = 7, as 20 -B> 375 -A> 350 -A> 365 -A> 78 -A> 71 -A> 1 -A> 0, and there are no shorter paths.
For n=112, we know that a(112) cannot be larger than eight, as A328099^(8)(112) = 0, so we have a path of length 8 as 112 -A> 240 -B> 77 -A> 18 -A> 21 -A> 10 -A> 7 -A> 1 -A> 0. Checking all 32 combinations of the paths of lengths of 5 starting from 112 shows that none of them or their prefixes ends with a prime, thus there cannot be any shorter path, and indeed a(112) = 8.
a(24) <= 11 as A328099^(11)(24) = 0, i.e., we have 24 -A> 44 -A> 48 -A> 112 -A> 240 -B> 77 -A> 18 -A> 21 -A> 10 -A> 7 -A> 1 -A> 0. On the other hand, 24 -B> 625 -B> 17794411250 -A> 41620434625 -A> 58507928150 -A> 86090357185 -A> 54113940517 -A> 19982203325 -A> 12038411230 -A> 8426887871 -A> 1 -A> 0, thus offering another path of length 11.
		

Crossrefs

Cf. A328324 (a sequence giving upper bounds, computed with restricted search space).
Sequences for whose terms k, value a(k) has a guaranteed constant upper bound: A000040, A002110, A143293, A157037, A192192, A327978, A328232, A328233, A328239, A328240, A328243, A328249, A328313.
Sequences for whose terms k, it is guaranteed that a(k) has finite value > 0, even if not bound by a constant: A099308, A328116.

Programs

  • PARI
    A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
    A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
    A327969(n,searchlim=0) = if(!n,n,my(xs=Set([n]),newxs,a,b,u); for(k=1,oo, print("n=", n, " k=", k, " xs=", xs); newxs=Set([]); for(i=1,#xs,u = xs[i]; a = A003415(u); if(0==a, return(k)); if(isprime(a), return(k+2)); b = A276086(u); if(isprime(b), return(k+1+(u>2))); newxs = setunion([a],newxs); if(!searchlim || (b<=searchlim),newxs = setunion([b],newxs))); xs = newxs));

Formula

a(0) = 0, a(p^p) = 1 + a(A276086(p^p)) for primes p, and for other numbers, a(n) = 1+min(a(A003415(n)), a(A276086(n))).
a(p) = 2 for all primes p.
For all n, a(n) <= A328324(n).
Let A stand the transition x -> A003415(x), and B stand for x -> A276086(x). The following sequences give some constant upper limits, because it is guaranteed that the combination given in brackets (the leftmost A or B is applied first) will always lead to a prime:
For all n, a(A157037(n)) = 3. [A]
For n > 1, a(A002110(n)) = 3. [B]
For all n, a(A192192(n)) <= 4. [AA]
For all n, a(A327978(n)) = 4. [AB]
For all n, a(A328233(n)) <= 4. [BA]
For all n, a(A143293(n)) <= 4. [BB]
For all n, a(A328239(n)) <= 5. [AAA]
For all n, a(A328240(n)) <= 5. [BAA]
For all n, a(A328243(n)) <= 5. [ABB]
For all n, a(A328313(n)) <= 5. [BBB]
For all n, a(A328249(n)) <= 6. [BAAA]
For all k in A046099, a(k) >= 4, and if A328114(k) > 1, then certainly a(k) > 4.

A328099 a(n) = min(A003415(n), A276086(n)).

Original entry on oeis.org

0, 1, 1, 4, 1, 5, 1, 12, 6, 7, 1, 16, 1, 9, 8, 32, 1, 21, 1, 24, 10, 13, 1, 44, 10, 15, 27, 32, 1, 7, 1, 21, 14, 19, 12, 35, 1, 21, 16, 68, 1, 41, 1, 48, 39, 25, 1, 112, 14, 45, 20, 56, 1, 81, 16, 92, 22, 31, 1, 49, 1, 33, 51, 192, 18, 61, 1, 72, 26, 59, 1, 156, 1, 39, 55, 80, 18, 71, 1, 176, 108, 43, 1, 124, 22, 45, 32, 140, 1, 123, 20, 96, 34, 49, 24
Offset: 1

Views

Author

Antti Karttunen, Oct 07 2019

Keywords

Comments

Always choosing the lesser of A003415(n) and A276086(n) is often a good heuristic when trying to find the shortest path to zero. However, this doesn't always guarantee the optimal result. E.g., if we define b(0) = 0; and for n > 0, b(n) = 1+(a(n)), then we have b(8) = 8 > A327969(8) = 6, b(12) = 7 > A327969(12) = 5, and b(15) = 9 > A327969(15) = 6.

Crossrefs

Programs

  • PARI
    A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
    A276086(n) = { my(i=0,m=1,pr=1,nextpr); while((n>0),i=i+1; nextpr = prime(i)*pr; if((n%nextpr),m*=(prime(i)^((n%nextpr)/pr));n-=(n%nextpr));pr=nextpr); m; };
    A328099(n) = min(A003415(n),A276086(n));
Showing 1-3 of 3 results.