A new number family related to the fibonaccis.

[Leo Hennig]

Hi, 

I have been looking at the divisor spectrum of numbers, so simply the divisors of a number n, with all proper divisors from smallest to greatest: 

For 12, this would be 2,3,4,6. 

Then it occured to me to do the following. At most know, the Fibonacci numbers come to exist, by adding the first two numbers in the series (1,1) and then going on to add just the last two.
For me, I wanted to check how Fibonacci like the divisors of a given number are. 

For twelve, we can add 2,3 makes 5, and 3 and 4 makes seven. 
The actual divisor that comes next is 4 and 6, thus we compute: 4-5 and 6-7 makes -2. 



Now it further occured to me, what if all the residuals sum up to 0? THis would give as a certain family, quite rare number members, that have Fibonacci like properties "on average". 

Example: 

20 has divisors 2,4,5,10.

2+4 = 6 - 5 (next divisor) is 1

4+5 = 9 -10 (next divisor) is -1

sum of 1 -1 = 0.

105 has divsors 3, 5, 7, 15, 35.

3+5 = 8 - 7 = 1

5+7 = 12 -15 = -3

7+15 = 22 -21 = 1

15+21= 36 -35 = 1

sum of 1-3+1+1 = 0.

This is the series: 20, 42, 105, 294, 693, 735, 2058, 5145, 14406, 36015

Now what is even more surprising, is that if we take the second order Fibonaccis, thus three divisors summed together, then sum together the residuals, we get for the zero property:  78, 1014, 13182, 171366 see  A121057, for m = 6, and this is quite easy to compute, just a power of 6 multiplied by n -1 divided by two. 

That we found a relatively "easy", "exponential" way for the second order to find these numbers for the second order that have very specific divisor properties,  makes me hopeful that we find an easy way for first order and higher order and very much reminds me of the perfect numbers!

I hope everyone has a nice day. 

LAH

[Allan Wechsler]

This is very peculiar. I haven't checked Leo's numbers, but the ones he lists are, with two exceptions, of the form 6*7^n or 15*7^n, with all numbers of this form for n in the range 1..4 being present. Leo, you might want to check 100842 and 252105. The exceptions are 20 and 693.

Second, I notice that Leo has excluded the number itself from his lists of divisors (that is, he is only considering aliquot divisors), and I wonder whether things would change much if all divisors were included.

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[Leo Hennig]

If we take all numbers, including the number itself, we get, for groups (degree of fibonacciness) of 

size 2: 12, 

size 3: 6, 78, 1014, 

size 4: 114, 765, 2166, 

size 6: 28 

size 10: 496, 

size 14: 8128, 

thus we get interesting numbers including 12 a sublime number, and for higher iterations we get the perfect numbers!

[Allan Wechsler]

Now I'm not following, Leo. You've made some sort of generalization involving a quantity called "size", where size 2 corresponds to your original definition (but including the original number in the calculation). Hence, for 12, the divisors, preceded by the "Fibonacci prediction" in parentheses, are:

1, 2, (3) 3, (5) 4, (7) 6, (10) 12

so the aggregate discrepancy is 0 + 1 + 1 - 2 = 0. I'm with you up to here.

But, for example, 78 gives

1, 2, (3) 3, (5) 6, (9) 13, (19) 26, (39) 39, (58) 78

for a discrepancy of 0 - 1 - 4 - 7 + 0 - 20 = -32. Can you spell out the property of 78 that led you to list it as a "size 3" example?

-- Allan

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[Leo Hennig]

Yes it is like this, you take three instead of just the last two. 

78:
 Divisors: [1, 2, 3, 6, 13, 26, 39, 78]

    (1, 2, 3) ->next divisor  = 6, Abstand = 0
    (2, 3, 6) -> next divisor = 13, Abstand = 2
    (3, 6, 13) -> next divisor  = 26, Abstand = 4
    (6, 13, 26) -> next divisor  = 39, Abstand = -6
    (13, 26, 39) -> next divisor  = 78, Abstand = 0


Sums to zero. 

[Allan Wechsler]

Leo, thank you. That clears it up.

-- Allan

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[Gareth McCaughan]

On 29/04/2026 13:03, Leo Hennig wrote:

Hi, 

I have been looking at the divisor spectrum of numbers, so simply the divisors of a number n, with all proper divisors from smallest to greatest: 

For 12, this would be 2,3,4,6. 

Then it occured to me to do the following. At most know, the Fibonacci numbers come to exist, by adding the first two numbers in the series (1,1) and then going on to add just the last two.
For me, I wanted to check how Fibonacci like the divisors of a given number are. 

For twelve, we can add 2,3 makes 5, and 3 and 4 makes seven. 
The actual divisor that comes next is 4 and 6, thus we compute: 4-5 and 6-7 makes -2. 



Now it further occured to me, what if all the residuals sum up to 0? THis would give as a certain family, quite rare number members, that have Fibonacci like properties "on average". 

...

This is the series: 20, 42, 105, 294, 693, 735, 2058, 5145, 14406, 36015
Now what is even more surprising, is that if we take the second order Fibonaccis, thus three divisors summed together, then sum together the residuals, we get for the zero property:  78, 1014, 13182, 171366 see  A121057, for m = 6, and this is quite easy to compute, just a power of 6 multiplied by n -1 divided by two.

If we call the proper divisors d1, ..., dk then your residual is d1 + ... + d(k-2) + d2 - dk, so this is zero precisely when dk = d1 + ... + d(k-2) + d2.

With your "second order Fibonaccis" the condition is instead d1 + 2d2 + 3d3 + 2(d4+...+d(k-2)) - dk.

Allan Wechsler observes that Leo's numbers (for the ordinary "second-order" differentials) mostly look like 6.7^k or 15.7^k. Both of these do indeed work.

6.7^k: the divisors, in increasing order, are

[1],2,3,6 . 7^0
1,2,3,6 . 7^1
...
1,2,3,[6] . 7^k

where the square brackets indicate the not-proper divisors that we are ignoring. So what we need is that

(2+3+6) + (1+2+3+6).7 + ... + (1+2+3+6).7^(k-1) + (1).7^k + 3 = 3.7^k

or 12.(1+7+...+7^(k-1)) - 1 + 7^k + 3 = 3.7^k

or 12.(7^k-1)/(7-1) - 1 + 7^k + 3 = 3.7^k

or 2.(7^k-1) - 1 + 7^k + 3 = 3.7^k

and at this point even I can see that the LHS and RHS are indeed equal.

15.7^k: all the divisors, _not_ yet in increasing order, are

1,3,5,15 . 7^0
1,3,5,15 . 7^1
...
1.3.5.15 . 7^k

and so the divisors in increasing order are

[1],3,5 . 7^0
1,15/7,3,5 . 7^1
...
1,15/7,3,5 . 7^k
[15 . 7^k]

and now what we need is that

(3+5).7^0 + (1+15/7+3+5).7^1 + ... + (1+15/7+3+5).7^(k-1) + (1+15/7).7^k + 5 = 5.7^k

or (1+15/7+3+5).(7^0+...+7^(k-1)) - (1+15/7) + (1+15/7).7^k + 5 = 5.7^k

or (1+15/7+3+5).(7^k-1)/(7-1) + (1+15/7).(7^k-1) + 5 = 5.7^k

or (1+15/7+3+5).(7^k-1)/(7-1) + (1+15/7).(7^k-1) = 5.7^k - 5

or (1+15/7+3+5)/(7-1) + (1+15/7) = 5

or 13/7 + 22/7 = 5

and again at this point even I can see that it's true.

So indeed numbers of the form 6.7^k and 15.7^k do have the required property. Leo also finds that 20 and 693 have the property. Are these perhaps all of them? Well ...

That we found a relatively "easy", "exponential" way for the second order to find these numbers for the second order that have very specific divisor properties,  makes me hopeful that we find an easy way for first order and higher order and very much reminds me of the perfect numbers!

It reminds me of the perfect numbers, too. With the perfect numbers, we have an easily verified sequence of perfect numbers and an apparently incredibly difficult conjecture that that's them all. I am not very optimistic about the prospects of proving any conjectures we might make about Leo's zero-discrepancy numbers.

The above is all about what Leo calls "first order" though I would be inclined to call them "second order" since the recurrence whose residuals we're summing is a second-order one. Leo made a conjecture about what I would call the "third order" equivalent, and there again I would not be surprised if 1. it were true but 2. it were incredibly difficult to prove.

(It might turn out that these things are not-too-painfully provable after all, despite the apparent analogy to perfect numbers. I haven't thought about it very hard. But it's not the way I'd bet.)

-- 
g