Octagonal and Triangular Numbers (3)

We have seen that the nth term of the octagonal series, i.e.

1, 8, 21, 40, 65, …,

is given by 3n² – 2n where n = 1, 2, 3, …

However what is interesting is that a “shadow” series (comprising positive valued terms) can be obtained by letting n = 0, – 1, – 2, …, i.e.

0, 5, 16, 33, 56, …

And again the same neat numerical features that characterise the octagonal series, also characterise its “shadow” referred to in OEIS as A045944 (The Rhombic Matchstick Numbers).

So when we multiply each of the numbers of this sequence by 3, we obtain a number that is 1 less than a perfect square.

So 0 * 3 = 0 = 1² – 1; 5 * 3 = 15 = 4² – 1; 16 * 3 = 48 = 7² – 1 and so on.

Likewise each of these numbers can be expressed as the difference of two squares.

So 0 = 1² – 1²; 5 = 3² – 2² ; 16 = 5² – 3² and so on.

And                     Also in the case of the octagonal series

0 =   0 * 2            1 =  1 * 1

5 =   1 * 5            8 =  2 * 4

16 = 2 * 8           21 = 3 * 7

33 = 3 * 11         40 = 4 * 10

     …                         …

So we can see that the number bases that were so relevant in the case of self generating numbers and the octagonal series i.e. 2, 5, 8, 11, …, now naturally appear with this “shadow” sequence.

Likewise in complementary fashion the number bases applicable to self generating numbers in the “shadow” sequence i.e. 1, 4, 7, 10, …naturally appear in the octagonal series.

In fact because the value of the respective numbers bases is directly related to the value of k it is the negative number bases that are relevant with respect to the “shadow” sequence.

In other words the self generating numbers that arise in the negative number bases i.e.

– 1, – 4, – 7, – 10,…, in fact are the respective terms 0, 5, 16, 33, … of the shadow sequence.

The 1st case is trivial for 00 in base – 1  is 00 and 00 – 00 = 00.

However it is somewhat trickier in subsequent number bases.

In fact in base – 4, 5 would be written as – 23.

Now if we look at the right hand digit this indeed represent 3 units. However the 2 in base – 4 = 2 * (– 4) = – 8.

Therefore 23 in base – 4 =  – 8 + 3 =   – 5 (in denary terms).

Therefore 5 (in base 10) = – 23 (in base – 4).

And the reverse of – 23 = – 32 (in base – 4).

The difference of the two = – 32 – (– 23) = – 32 + 23

And  32 = 3 * (– 4) + 2 = – 12 + 2 = – 10 (in denary terms)

Therefore – 32 = 10 and 23 = – 5.

So – 32  + 23 = 10 – 5 = 5.

Therefore when we subtract – 23  from its reverse in base – 4,  we get – 23.

So – 23 is self-generating (in both cyclical and linear non-hierarchical terms) in this base.

Then next case arises in base – 7, where – 35 (= 16 in base 10) is now the relevant number.

And – 53 – (– 35 ) in base – 7,  =  – 53 + 35 = 32 – 16 = 16 (in base 10).

So – 35 is self-generating in base – 7.

And just as the two digits of the self-generating numbers in the positive bases 2, 5, 8, 11, … represent the unique digit sequence of the reciprocal of the cyclic prime 3 in these bases, likewise the two digits in the negative bases – 1, – 4, – 7, – 10 again represent the unique digit sequence of the reciprocal of 3.

Now if we line up the octagonal and its corresponding “shadow” sequence as follows, we have,

1, 8, 21, 40, 65, …

0, 5, 16, 33, 56, …

Then subtracting each term of the latter “shadow” from the corresponding term of the former octagonal sequence we get,

1, 3, 5, 7, 9, …

In other words, we have the sequence of odd-numbered natural integers.

This for example using Leibniz’s well- known formula, we can express π in terms of the octagonal and shadow sequences.

So π/4 = 1/(1 – 0) – 1/(8 – 5) + 1/(21 – 16) – 1/(40 – 33) + …

It is also fascinating how we can derive the natural number sequence, through dividing each successive term of one sequence by the corresponding successive number base of the complementary sequence.

Thus with respect to the “shadow” sequence the terms are 0, 5, 16, 33, ..

and the corresponding number bases (related to self-generating numbers) of the octagonal sequence are 2, 5, 8, 11, …

And 0/2 = 0; 5/5 = 1; 16/8 = 2; 33/11 = 3 and so on.

Then with respect to the octagonal sequence, the terms are 1, 8, 21, 40, … and the corresponding number bases (related to self-generating numbers) of the “shadow” sequence are – 1, – 4, – 7, – 10, …

And 1/– 1 = – 1; 8/– 4 = – 2; 21/– 7 = – 3; 40/– 10 = – 4 and so on.

If we now combine the positive number bases i.e. 2, 5, 8, 11, … directly with the corresponding terms of the octagonal series, a fascinating pattern emerges.

So  2 *  1  =  2

  5 *  8  =  40

  8 *  21 = 168 

 11 * 40 = 440

      …

Now the sum of the first (i.e. one) term = 2              =  1 * 2

The sum of the first two terms  = 2 + 40 = 42           =  6 * 7

The sum of the first 3 terms = 42 + 168 = 210          = 14 * 15

The sum of the first 4 terms = 210 = 440 = 650        = 25 * 26

So the cumulative total of terms always can be expressed as the product of two consecutive numbers.

This also applies to the shadow sequence, multiplied by the number bases,  1, 4, 7, 10, …

So  1 *  0 =      0

    4  * 5 =     20

    7 * 16 =   112

  10 * 33 =   330.

So the cumulative total of 1 = 0 =     0 *  1

The cumulative total of 2 =    20   =  4 *  5 

The cumulative total of 3 =    132 = 11 * 12

The cumulative total of 4 =    462 = 21 * 22

Notice how the increase of each number in the consecutive sequence is related to the number bases. So each number increases by 4, then 7, then 10 and so on whereas with octagonal numbers each number (in the consecutive sequence).

Finally, we can again make a connection with the triangular nos. in the following way.

 

Again, if we line up the octagonal and its corresponding “shadow” sequence as follows, we have,

1, 8, 21, 40, 65, …

0, 5, 16, 33, 56, …

Then adding each respective term we get,

1, 13, 37, 73, 121, …

Then subtracting 1 we get

0, 12, 36, 72, 120, …

= 12 (0, 1, 3, 6, 10, …) 

Octagonal and Triangular Numbers (2)

The octagonal numbers i.e.

1, 8, 21, 40, 65, 96, …have built in to their numerical structure some very interesting relationships.

1 =   1 * 1,

8 =   2 * 4,

21 = 3 * 7,

40 = 4 * 10

65 = 5 * 13

96 = 6 * 16

      …

So the kth term i.e. t_k = k(3k – 2) = 3k² – 2k

Also when each term is multiplied by 3, it results in a number 1 less than a perfect square.

So 1 * 3 =  3  =  2² – 1

     8 * 3  = 24 = 5²  – 1

   21 * 3  = 63 = 8² – 1

             …

So once again we see the relationship between 3 and the the sequence 2, 5, 8, … (in which bases we have the unique digit sequence of the reciprocal of 3).

So 3 * t_k  = (3k – 1)² – 1; so t_k = {(3k – 1)² – 1}/3 = 3k² – 2k

Also each term can be expressed as the difference of two squares.

So 1 =  1² – 0²

       8  = 3² – 1²

   21  = 5² – 2²

    40  = 7² – 3²

                 …

So again t_k = (2k – 1)² – (k – 1)²  = k(3k – 2) = 3k² – 2k.

We have already shown the connection between the octagonal self-generating numbers (with both non-hierarchical linear and circular features).

Now the coincidence of both linear and circular self-generating features can only apply when numbers have two digits.

However the non-hierarchical linear aspect can be continued into numbers with more than 2 digits.

Quite simply when a 2-digit number is self-generating in this non-hierarchical sense then a n-digit number (where n > 2) is likewise self-generating where the additional digits are 1 less than the base in question and inserted between the 1st and 2^nd digits (of the 2-digit number).

So for the simplest case, in base 2, 01 is self-generating (as 10 – 01 = 01).

This therefore entails that 3-digit number 011 is also self-generating (where the additional digit inserted is 1 less than the corresponding number base 2).

So 110 – 011 = 011  i.e. 3 in denary terms. 

So in the next relevant number base 5, 143 is self generating, where 341 – 143 = 143, i.e. 48 in denary terms.

In the next relevant number base 8, 275 is self-generating, where 572 – 275 = 275, i.e. 189 in denary terms.

Then in the next relevant number base 11, 3A7 is self-generating as 7A3 – 3A7 = 3A7, i.e. 480 in denary terms.

Now if we look at the denary nature of the sequence generated 3, 48, 189, 480, …, we can again show its links to the octagonal numbers

So, 3 = (1 * 3) *  1

    48 = (2 * 3) *  8

  189 = (3 * 3) * 21

  480 = (4 * 3) * 40  

              …

And the sum of the 1^st n terms of the sequence 1, 2, 3, 4, … is the nth triangular number.

So the sum of the first 4 terms i.e. 1 + 2 + 3 + 4 = 10 i.e. the 4^th triangular no.

Then when we move into 4-digit numbers 0111 is self-generating with respect to the base 2 i.e. 7 in denary terms.

Next 1443 is self-generating with respect to base 5 i.e. 248 in denary terms.

Then 2775 is self-generating with respect to base 8 i.e. 1533 in denary terms.

Finally to illustrate 3AA7 is self-generating with respect to base 11 i.e. 5320 in denary terms.

Again a fascinating connection can be shown with the octagonal numbers.

So   7     =     7  * 1

248   =   31 * 8

1533 =   73 * 21

5320 = 133 * 40

                 …

Now the other sequence of numbers 7, 31, 73, 133, … is fascinating in its own right

7 = 6 + 1, 31 = 30 + 1, 73 = 72 + 1 133 = 132 + 1, …

And 6 =  6 *  1

      30 =  6 *  5

      72 =  6 * 12

     132 = 6 * 22

           …

And 1, 5, 12, 22, … are the pentagonal numbers, which in turn are related to the triangular numbers.

The average of the 1^st n pentagonal numbers is the nth triangular no.

So the average for example of the first four pentagonal numbers = 40/4 = 10 (i.e. the 4^th triangular number).

Octagonal and Triangular Numbers (1)

Some time ago when I was exploring the use of holistic (circular) notions for deriving perhaps novel new insights with respect to simple number relationships, I became fascinated with - what I referred to at the time as - self generating numbers.

So a self generating number refers to a number that repeats itself according to some well-defined procedure based on subtraction.

And I basically defined two types of such relationships linear and circular respectively.

So one linear type example (with hierarchy) results from the ordering of the digits in descending (largest digit 1^st) and ascending (smallest digit 1^st) respectively and then subtracting the latter from the former number. So when the original number results from this operation, we thereby have a self-generating number.

For example if we take the number 495 to illustrate when arranging its digits in descending order, we obtain 954 and then in ascending order 459.

And then when we obtain 954 – 459, the result is 495 (i.e. the original number).

Now in fact 495 represents the only example of a 3-digit number (in the denary system) that is self-generating in this sense, with 6174, the only example of a corresponding 4-digit number.

One other interesting feature of such numbers is that when divided by 9 that a palindrome (or close palindrome) frequently results.

So 495/9 = 55; 6174/9 = 686.

I also looked at the circular equivalent, which is especially well represented through cyclic primes.

Self generation is this sense results when the same circular sequence of digits is preserved through subtraction.

For example 142857 represents the well-known unique cyclic system of digits associated with the reciprocal of 7.

Then if we obtain the difference of any 6-digit numbers, that preserves the same cyclic order of these digits that the result will be a number (that likewise preserves the same cyclic order).

So for example, 428571 – 142857 = 285714 (which preserves the same cyclic order).

Note that if we adopt the initial linear order (already mentioned) that self-generation will not occur

So 875421 – 124578 = 750483 (which is neither self-generating in a linear or circular sense).

I then examined the simpler linear case (without hierarchy) where subtraction takes place with the order of digits simply reversed.

So for example taking 6174 again to illustrate its reverse is 4716

And the difference = 6174 – 4716 = 1458. So self-generation does not take place.

Now, perhaps surprisingly there does not appear to be any numbers (in base 10) that are self-generating (in this non-hierarchical sense).

So we have to look to other number bases for this feature to arise.

Fro example in base 8, if we take the 3-digit number 275 and then subtract it from its reverse we get 572 – 275 = 275.  So this clearly then is a self-generating number in a linear (non-hierarchical) manner.

I then realised that a particularly interesting result would arise here in the 2-digit case as - by definition - both linear and circular self-generation would be involved. And also the hierarchical and non-hierarchical definitions would likewise coincide.

So the quest was on therefore to find all 2-digit self generating numbers that satisfied these requirements (in all relevant number bases).

And I found that a fascinating solution existed. So in general terms if n represents the number base, the relevant bases where self-generation occurs is given by 3k + 2  (where k = 0, 1, 2, 3, …)

And if a represents the 1^st digit of the corresponding self-generating number, its value is given as k with the second digit = 2k + 1.

Therefore when k = 0 the relevant number base = 2. The 1st digit = 0 and the second digit = 1

Therefore in base 2, 01 is a self generating number (in both a linear and cyclical sense).

So 10 – 01 = 01.

Then when k = 1, the next relevant number base = 5. Therefore the 1^st digit = 1, and the 2^nd = 3.

So in base 5, 13 is a self-generating number in the manner defined.

So 31 – 13 = 13.

When k = 2, the next number base = 8. The 1^st digit = 2, and the 2^nd digit = 5.

So 52 – 25 = 25.

And giving just one more example to illustrate, the next number base is 11 with the 1^st digit = 3 and the 2^nd digit = 7.

So 73 – 37 = 37 (in base 11).   

Now if one converts all these self-generating numbers into conventional denary (base 10) format, the resulting sequence arise

i.e. 1, 8, 21, 40, …, which represents the well-known octagonal numbers.

Thus the octagonal numbers in this sense are uniquely related to all these 2-digit self-generating numbers (in other number bases).

One other fascinating feature of these numbers struck me at the time i.e. that they represent the cyclic digit sequence (in their respective number bases) of the reciprocal of 3.

So for example if we were to express 1/3 in base 8, we would obtain .252525…

Thus the uniquely recurring digits here are 25.

In one way, this is a remarkable feature, which connects the octagonal and triangular numbers 1, 3, 6, 10, … in a way that is perhaps not commonly realised.

And when we look more closely at the triangular, striking connections can indeed be found with the octagonal numbers.

Now the triangular numbers repeat naturally groups of 3 where the 1st member of the group leaves a remainder of 1 (when divided by 3) while 3 is always a factor of the other two members of the group.

So taking the 1^st 3 members 1, 3 and 6 if we now add the 3 members of the group and subtract 1, the total will then be divisible by 3. So {(1 + 3 + 6) – 1}/3 = 9/3 = 3 (which is the 2^nd triangular number).

Then we treat the next group of 3 in the same manner, we obtain {(10 + 15 + 21) – 1}/3 = 45/3 = 15 (which is the 5^th triangular number).

Then with the next group of 3 we have {(28 + 36 + 45) – 1}/3 = 108/3 = 36 (which is the 8^th triangular number).

So the ordinal ranking of the triangular numbers obtained in this manner exactly match the corresponding number bases (where the self generating numbers arise).

And again the octagonal numbers are then the representation of these numbers in base 10.

And a more direct connection can be made.

If we start with the 3^rd triangular number and then divide by 3 we obtain 6/3 = 2.

If we now keep increasing the ordinal rank by 6 and then dividing (the number arising) by the corresponding cardinal number we will generate the sequence 2, 5, 8, 11, … which matches in cardinal fashion the ordinal rankings of the number bases in which the self generating numbers arise.

So the 9^th triangular number = 45 and 45/9 = 5; the 15^th cardinal number = 120 and 120/15 = 8; the 21^st triangular number = 231 and 231/21 = 11 and so on.

Therefore in a way it is very striking how such connections can be shown as between number (with respect to both cardinal and ordinal usage) both within the denary base and across other relevant number bases.

And this in turn relates to the very nature of the self-generating numbers (on which these connections are based) where both linear and circular notions coincide.


Again the octagonal numbers are:

1, 8, 21, 40, 65, 96, 133, ....

And the triangular numbers are:

1, 3,  6, 10, 15, 21,  28, ...

And if we subtract each successive term in the triangular from the corresponding term in the octagonal, we obtain,

0, 5, 15, 30, 50, 75, 105

= 5 (1, 3, 6, 10, 15, 21, ...)

And the sequence within the brackets is once again that of the triangular numbers.

Interesting Log Relationships (3)

In previous entries, Interesting Log Relationships (1) and Interesting Log Relationships (2), I showed how the (infinite) reciprocal sum of the unique number sequences associated with (x – k)² =  0 and 
(x – k)³ = 0, (with k = 2, 3, 4, …) together with (x + k)² = 0 and (x + k)³ = 0 (with k = 1, 2, 3, …), result from the application of generally applicable formulae resulting in answers of the form a log k + b (where a and b are rational numbers that can take on positive or negative values).

So once again for example for the general case where (x + k)³ = 0, the general formula is

2k{(k + 1)[log(k + 1/k)] – 1}.

Thus when k = 1, we have the polynomial equation (x + 1)³  = 0 i.e. x³ + 3x² + 3x + 1 = 0, with the unique number sequence 1, – 3, 6, – 10, 15, …

Therefore the corresponding (infinite) sum of reciprocals (i.e. the alternating sum of reciprocals of the triangular nos.) = 1 – 1/3 + 1/6 – 1/10 + 1/15 … 

= 2{2(log 2) – 1}  = 4 log 2 – 2.

However one can continue on both for higher powers of (x – k)ⁿ =  0 and (x + k)ⁿ = 0, to generate corresponding results for the associated (infinite) sums of reciprocals relating to the unique number sequences occurring.

Now while it appears that the general form of these results does not change (remaining as a log k + b, it is not yet clear to me whether for any given power of n, that a general formula expression can be given to cover all possible results.

However, I did manage to calculate some specific results (which, not surprisingly, become somewhat more difficult to establish as the power of n and value of k increases).

For example in the simplest case where (x – k)² =  0 (with k = 2), we have (x – 2)⁴ =  0,

i.e. x⁴ – 8x³ + 24x² –  32x + 16 = 0 with the corresponding unique number sequence

1, 8. 40, 160, 560, … (A001789 in OEIS).

And the corresponding (infinite) sum of reciprocals is,

1 + 1/8 + 1/40 + 1/160 + 1/560 + …,  =  1.15888 …   = 6 log 2 – 3.

In the next case where (x – 3)⁴ =  0, i.e. x⁴ – 12x³ + 54x² – 108x + 81 = 0, the unique numbers associated are 1, 12, 90, 540, 2835, … (A036216 in OEIS).   

Thus the corresponding (infinite) sum of reciprocals is,

1 + 1/12 + 1/90  + 1/540 + 1/2835 + …,   = 1.09674… = {72 log (3/2) – 27}/2.

Then in the next case, where (x – 4)⁴ =  0, i.e. x⁴ – 16x³ + 96x² – 256x + 256 = 0,

the unique associated number sequence is

1, 16, 160, 1280, 8960, … (i.e. A038846 in OEIS).

Thus the corresponding sum of reciprocals

= 1 + 1/16 + 1/160 + 1/1280 + 1/8960 + … =  1.06966…  = 108 log (4/3) – 30

Then in the case where (x + 1)⁴ = 0, i.e. x⁴ + 4x³ + 6x² + 4x + 1 = 0, the unique associated number sequence is 1, – 4, 10, – 20, 35, … 

So the corresponding (infinite) sum of reciprocals (i.e. sum of alternating reciprocal terms of the tetrahedral numbers) is,

1 – 1/4 + 1/10 – 1/20 + 1/35, …,  = .817766…    = (24 log 2 – 15)/2.   

Then in the case where (x + 2)⁴ = 0, i.e. x⁴ + 8x³ + 24x² + 32x + 16 = 0, the unique associated number sequence is 1, – 8, 40, – 160, 560, …

So the corresponding (infinite) alternating sum of reciprocals,

= 1 – 1/8 + 1/40 – 1/160 + 1/560 – … = .89511…,   = 54 log (3/2) – 21.

And in the case where (x – 5)⁴ = 0 i.e. x⁴ – 20x³ + 150x² – 500x + 625 = 0, the unique digit sequence is 0, 0, 0, 1, 20, 250, 2500, ...

So the sum of reciprocals = 1 + 1/20 + 1/250 + 1/2500 + ... , (A081143 in OEIS).

= {480 log (5/4)  – 105)}/2.

Interesting Log Relationships (2)

In a former entry, “Interesting Log Relationships 1” I showed how the infinite sums of reciprocals of the unique numbers associated with simple polynomial equations of the form (x – k)² = 0, where k = 2, 3, 4, … are related to simple log type relationships.

So, again for example, when k = 2, the equation is (x – 2)² = 0, i.e. x² – 4x + 4 = 0.

Then by the same process by which the unique numbers forming the Fibonacci sequence is obtained from the equation x² – x – 1 = 0, in like manner a unique number sequence can be obtained for x² – 4x + 4 = 0, i.e. 1, 4, 12, 32, 80, … (listed as A001787 in the On-line Encyclopedia of Integer Sequences).

Then the infinite sum of reciprocals of this number sequence,

= 1 + 1/4 + 1/12 + 1/32 + 1/80 + … = 2 log 2.

And in that piece, I provided a general formula to cover the infinite sum of reciprocals for all cases entailing unique number sequences of form (x – k)²= 0,

i.e. k{log k/(k – 1)}.

So for example when k = 6 we have (x – 6)²= x² – 12x + 36 = 0.

And the unique sequence of numbers associated with this equation – again derived in a similar manner to the Fibonacci sequence – is 1, 12, 108, 864, (i.e. A053469 in the OEIS).

Therefore the infinite sum of reciprocals of this sequence

= 1 + 1/12 + 1/108 + 1/864 + …  = 6{log (6 – 1)/5} = 6 log 6/5 = 1.0939…

I also provided in that entry a general formula to cover all cases where (x + k)² = 0, with k = 2, 3, 4, …

i.e. k{log (k + 1)/k}

In this case the same sequence of numbers is obtained as for (x – k)² = 0, except that they alternate as between positive and negative values.

So the corresponding infinite sum of associated reciprocals

= 1 – 1/12 + 1/108 – 1/864 + …   = 6{log (6 + 1)/6} = 6 log 7/6 = .9249…

So interestingly the sum of the alternating sum of reciprocals can be directly obtained from the (positive) sum of reciprocals by increasing both numerator and denominator of the log expression in the latter formula by 1.

We can however proceed to obtain further log expressions for the infinite sums of reciprocals associated with the unique numbers associated with the polynomial equations of the form (x – k)³ = 0, where k = 1, 2, 3, 4, …    

For example, when k = 2, we have (x – 2)³ = x³ – 6x²  + 12x – 8 = 0.

And the unique sequence of numbers associated with this equation is 1, 8, 40, 160, 560, … (A001789 in OEIS).

So the corresponding sum of reciprocals of this sequence is given as

1 + 1/8 + 1/40 + 1/160 + 1/560 + …  = 4(1 – log 2).

Once again a general formula can be given to cover all such cases,

i.e. 2k{1 – (k – 1)[log k/(k – 1)]}

So for example when k = 3, we have (x – 3)³ = x³ – 9x² + 27x – 27 = 0.

And the unique sequence of digits associated with this equation is 1, 9, 54, 270, 1215, …, (A027472  in OEIS).

Therefore the corresponding infinite sum of reciprocals

= 1 + 1/9 + 1/54 + 1/270 + 1/1215 + …

= 6{1 – [(3 – 1)log 3/(3 – 1)]} = 6{1 – 2 log (3/2)} = 1.1344 …

And as before, a related general formula can be given to cover the infinite sums of reciprocals of the unique number sequences associated with (x + k)³ = 0,

i.e. 2k{(k + 1)[log(k + 1/k)] – 1}.

So once again this provides the alternating case to the corresponding positive sum of reciprocals

Thus again for k = 3, the infinite sum of reciprocals of the unique number sequence associated with (x + 3)³

 = 1 – 1/9 + 1/54 – 1/270 + 1/1215 – …

Therefore from the formula the required sum = 6{[(3 + 1)log 3 + 1/(3)] – 1}

= 6{4 log (4/3) – 1} = .9043 …

A special case with respect to (x – k)³ = 0, occurs where k = 1 i.e. (x – 1)³

= x³ – 3x² + 3x – 1 = 0.

And the unique sequence of numbers associated with this sequence is the set of triangular nos. i.e. 1, 3, 6, 10, 15, …

So the infinite sum of reciprocals of this sequence, i.e.

1 + 1/3 + 1/6 + 1/10 + 1/15 + … = 2{1 – (1 – 1)[log 1/(1 – 1)]} = 2{1 – 0[log (1/0)]}

= 2.

And the alternating sum of reciprocals, corresponding to the unique sequence of numbers associated with (x + 1)³ = 0, i.e.

1 – 1/3 + 1/6 – 1/10 + 1/15 – …

= 2{(1 + 1)[log(1 + 1/1)] – 1} = 2(2 log 2 – 1) = .772588 …