A230403 -id:A230403 - cp's OEIS frontend

A339012 Written in factorial base, n ends in a(n) consecutive non-0 digits.

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

Offset: 0

Kevin Ryde, Nov 19 2020

Also, a(n) is the least p for which n mod (p+2)! < (p+1)!.  A small remainder like this means a 0 digit at position p in the factorial base representation of n, where the least significant digit is position p=0.  The least such p means only nonzero digits below.
Those n with a(n)=p are characterized by remainders n mod (p+2)!, per the formula below.  These remainders are terms of A227157 which is factorial base digits all nonzero.  A227157 can be taken by rows where row p lists the terms having p digits in factorial base.  Each digit ranges from 1 up to 1,2,3,... respectively so there are p! values in a row, and so the asymptotic density of terms p here is p!/(p+2)! = 1/((p+2)*(p+1)) = 1/A002378(p+1) = 1/2, 1/6, etc.
The smallest n with a(n)=p is the factorial base repunit n = 11..11 with p 1's = A007489(p).

			n = 10571 written in factorial base is 2,0,4,0,1,2,1.  It ends in 3 consecutive nonzero digits (1,2,1) so a(10571) = 3.  Remainder 10571 mod (3+2)! = 11 is in A227157 row 3.

Kevin Ryde, Table of n, a(n) for n = 0..10080

Cf. A108731 (factorial base digits), A016921 (where a(n)=1), A339013 (+2), A230403 (ending 0's).

In other bases: A215887, A328570.

a[n_] := Module[{k = n, m = 2, r}, While[{k, r} = QuotientRemainder[k, m]; r != 0, m++]; m - 2]; Array[a, 30, 0] (* Amiram Eldar, Feb 15 2021 after Kevin Ryde's PARI code *)

a(n) = my(b=2,r); while([n,r]=divrem(n,b);r!=0, b++); b-2;

a(n)=p iff n mod (p+2)! is a term in row p of A227157 (row p terms having p digits), including p=0 by reckoning an initial A227157(0) = 0 as no digits.
a(n)=0 iff n mod 2 = 0.
a(n)=1 iff n mod 6 = 1, which is A016921.
a(n)=2 iff n mod 24 = 3 or 5.

A233286 Number of trailing zeros in the factorial base representation of n-th Fibonacci number.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 5, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 1

Offset: 1

Antti Karttunen, Dec 07 2013

A233285 is the main entry for this topic, see comments there.

			The factorial base representation (A007623(A000045(n))) of Fibonacci numbers look like this, from n=1 onward: 1, 1, 10, 11, 21, 110, 201, 311, 1120, 2101, 3221, 11000, 14221, 30221, 50120, 121011, 211201, 332220, ...
When we count the trailing zeros of each, we get 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, ..., the first terms of this sequence.

Antti Karttunen, Table of n, a(n) for n = 1..100000
Index entries for sequences related to factorial base representation.

One less than A233285. Cf. also A007623, A000045, A230403.

a[n_] := Module[{k = Fibonacci[n], m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0|| r != 0, AppendTo[s, r]; m++]; FirstPosition[s, ?(# > 0&)][[1]] - 1]; Array[a, 100] (* _Amiram Eldar, Feb 21 2024 *)

A233286(n) = if(n<=2, 0, my(f=fibonacci(n)); for(k=2,oo,if(f%(k!), return(k-2)))); \\ Antti Karttunen, Jan 18 2025

(define (A233286 n) (A230403 (A000045 n)))

a(n) = A230403(A000045(n)) = A233285(n)-1.

Data section extended to a(105) by Antti Karttunen, Jan 18 2025

A352427 a(n) is the number of trailing 0's in the minimal representation of n in terms of the positive Pell numbers (A317204).

Original entry on oeis.org

0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 3, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 3, 0, 1, 0, 1, 4, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 3, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 3, 0, 1, 0, 1, 4, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 5, 0, 1, 0, 1, 2, 0, 1, 0, 1, 2, 0, 3, 0, 1, 0, 1

Offset: 1

Amiram Eldar, Mar 16 2022

The asymptotic density of the occurrences of 0 is sqrt(2)-1 and of the occurrences of k = 1, 2, ... is 2*(sqrt(2)-1)^(k+1).

The asymptotic mean of this sequence is 1 and its asymptotic variance is sqrt(2).

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000129, A003151, A003152, A286666, A286667, A317204.

Similar sequences: A003849 (dual Zeckendorf), A035614 (Zeckendorf), A230403 (factorial), A276084 (primorial), A278045 (tribonacci).

pell[1] = 1; pell[2] = 2; pell[n_] := pell[n] = 2*pell[n - 1] + pell[n - 2]; a[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[pell[k] <= m, k++]; k--; AppendTo[s, k]; m -= pell[k]; k = 1]; IntegerExponent[Total[3^(s - 1)], 3]]; Array[a, 100]

a(A000129(n)) = n-1 for n>=1.
a(n) = 0 if and only if n is in A286666.
a(n) > 0 if and only if n is in A286667.
a(n) == 0 (mod 2) if and only if n is in A003152.
a(n) == 1 (mod 2) if and only if n is in A003151.

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A339012 Written in factorial base, n ends in a(n) consecutive non-0 digits.

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A233286 Number of trailing zeros in the factorial base representation of n-th Fibonacci number.

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A352427 a(n) is the number of trailing 0's in the minimal representation of n in terms of the positive Pell numbers (A317204).

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