A276088 The least significant nonzero digit in primorial base representation of n: a(n) = A276094(n) / A002110(A276084(n)) (with a(0) = 0).

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

Offset: 0

Author

Antti Karttunen, Aug 22 2016

Keywords

• nonn
• base

Comments

For any n >= 1, start from k = n and repeatedly try to divide as many successive primes as possible out of k, iterating as k/2 -> k, k/3 -> k, k/5 -> k, until a nonzero remainder is encountered, which is then the value of a(n). (See the last example).

Note that the sequence has been defined so that it will eventually include also "digits" (actually: value holders) > 9 that occur as the least significant nonzero digits in primorial base representation. Thus any eventual decimal corruption of A049345 will not affect these values.

The sums of the first 10^k terms (starting from n=1), for k = 1, 2, ..., are 12, 138, 1441, 14565, 145950, 1459992, 14600211, 146002438, 1460025336, 14600254674, ... . Apparently, the asymptotic mean of this sequence is limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1.460025... . - Amiram Eldar, Sep 10 2022

The asymptotic density of the occurrences of k = 1, 2, ..., is d(1) = c = A064648 for k = 1, and d(k) = c - Sum_{i = 1..PrimePi(k)} 1/primorial(i) for k >= 2, where primorial(i) = A002110(i). The asymptotic mean of this sequence is Sum_{k>=1} k * d(k) = c + Sum_{i>=1} (prime(i) * gap(i) + t(gap(i)-1)) * (c - Sum_{j = 1..i} 1/primorial(i)) = 1.460025488658067356046281458556..., where t(i) = A000217(i) and gap(i) = A001223(i). - Amiram Eldar, Feb 20 2025

Examples

			   n   A049345  the rightmost nonzero = a(n)
---------------------------------------------------------
   0       0             0
   1       1             1
   2      10             1
   3      11             1
   4      20             2
   5      21             1
   6     100             1
   7     101             1
   8     110             1
   9     111             1
  10     120             2
  11     121             1
  12     200             2
  13     201             1
  14     210             1
  15     211             1
  16     220             2
.
For n=48 according to the iteration interpretation, we obtain first 48/2 = 24, and the remainder is zero, so we continue: 24/3 = 8 and here the remainder is zero as well, so we try next 8/5, but this gives the nonzero remainder 3, thus a(48)=3.
For n=2100, which could be written "A0000" in primorial base (where A stands for digit "ten", as 2100 = 10*A002110(4)), the least significant nonzero value holder (also the most significant) is thus 10 and a(2100) = 10. (The first point where this sequence attains a value larger than 9).
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..2310
• Index entries for sequences related to primorial base.

Crossrefs

Cf. A000040, A002110, A049345, A067029, A276084, A276086, A276094, A328569.

Cf. A000217, A000720, A001223, A064648.

Programs

Mathematica

nn = 120; b = MixedRadix[Reverse@ Prime@ Range@ PrimePi[nn + 1]]; Table[Last[IntegerDigits[n, b] /. 0 -> Nothing, 0], {n, 0, nn}] (* Version 11, or *)
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]]]; {0}~Join~Table[Last@ DeleteCases[f@ n, d_ /; d == 0], {n, 120}] (* Michael De Vlieger, Aug 30 2016 *)

PARI

A276088(n) = { my(e=0, p=2); while(n && !(e=(n%p)), n = n/p; p = nextprime(1+p)); (e); }; \\ Antti Karttunen, Oct 29 2019

Python

from sympy import nextprime, primepi, primorial
def a053669(n):
    p = 2
    while True:
        if n%p!=0: return p
        else: p=nextprime(p)
def a257993(n): return primepi(a053669(n))
def a002110(n): return 1 if n<1 else primorial(n)
def a276094(n): return 0 if n==0 else n%a002110(a257993(n))
def a(n): return 0 if n==0 else a276094(n)//a002110(a257993(n) - 1)
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 22 2017

Scheme

(define (A276088 n) (if (zero? n) n (let loop ((n n) (i 1)) (let* ((p (A000040 i)) (d (modulo n p))) (if (not (zero? d)) d (loop (/ (- n d) p) (+ 1 i)))))))

Scheme

(define (A276088 n) (if (zero? n) n (/ (A276094 n) (A002110 (A276084 n)))))

Formula

a(0) = 0, and for n >= 1, a(n) = A276094(n) / A002110(A276084(n)).

From Antti Karttunen, Oct 29 2019: (Start)

a(n) = A067029(A276086(n)).

a(A276086(n)) = A328569(n).

(End).