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.

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A329377 Number of iterations done when n is divided by its divisors starting from the smallest one in increasing order until one no longer gets an integer, or until divisors are exhausted.

Original entry on oeis.org

1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 3, 3, 2, 3, 2, 2, 3, 3, 2, 4, 2, 3, 3, 2, 2, 4, 2, 3, 3, 3, 3, 3, 2, 3, 3, 4, 2, 3, 2, 2, 3, 3, 2, 4, 2, 3, 3, 2, 2, 3, 3, 4, 3, 3, 2, 3, 2, 3, 3, 4, 3, 3, 2, 2, 3, 4, 2, 4, 2, 3, 3, 2, 3, 3, 2, 4, 3, 3, 2, 3, 3, 3, 3, 3, 2, 4, 3, 2, 3, 3, 3, 4, 2, 3, 2, 2, 2, 3, 2, 3, 4
Offset: 1

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Author

Antti Karttunen, Nov 17 2019

Keywords

Examples

			For n = 12, its divisors are [1, 2, 3, 4, 6, 12]. We can divide only three times so that the quotient remains an integer: 12/1 = 12, 12/2 = 6, 6/3 = 2 (but 2/4 = 1/2, a fraction). Thus a(12) = 3.
For n = 24, its divisors are [1, 2, 3, 4, 6, 8, 12, 24]. We can divide only four times so that the quotient remains an integer: 24/1 = 24, 24/2 = 12, 12/3 = 4, 4/4 = 1, but on the fifth time 1/6 would be a rational, thus a(24) = 4.
		

Crossrefs

Cf. A000142, A076933 (final integer reached), A240694.

Programs

  • PARI
    A329377(n) = { my(k=n,i=0); fordiv(k, d, if(n%d, return(i)); n /= d; i++); (i); };

Formula

a(A000142(n)) = n.

A329549 Numbers 4*k such that 1 is the last integer obtained when 4*k is successively divided by its divisors in increasing order.

Original entry on oeis.org

8, 24, 40, 56, 64, 120, 144, 280, 320, 448, 704, 720, 832, 1008, 1024, 1152, 2240, 3200, 4928, 5040, 5760, 5824, 6272, 8064, 9152, 10368, 11264, 13312, 17408, 19456, 22400, 23552, 29696, 31744, 32768, 35200, 40320, 41600, 51200, 51840, 64064, 68992, 72576, 81536, 100352, 114048
Offset: 1

Views

Author

David A. Corneth, Nov 16 2019

Keywords

Comments

At sequence A076933, the question is asked: "What is the longest string of ones in this sequence?" As A076933(4*n) is rarely 1, such a string is not very long. The longest starting below 4*10^8 has length 6 and starts at 141. Checking multiples of 4 may help in finding longer such strings.
Terms are also a multiple of 8. Proof: If m = 8*k + 4 then its divisors are 1, 2, 4 (and maybe 3). After dividing by 4 we have a fraction with denominator 2. Before that we did not see 1.

Examples

			The divisors of 8 are 1, 2, 4 and 8. Dividing from left to right gives 8/1 = 8, 8/2 = 4, 4/4 = 1, and then 1/8 isn't an integer so as the last integer we see is 1, 8 is in the sequence.
		

Crossrefs

Cf. A076933, A240694 (partial products of divisors of n).
Subsequence of A008586 (multiples of 4) and of A008590 (multiples of 8).
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