The Additive Sequence Hidden in Ramanujan's Pi Formula: Public Notebook

Additive Sequences Hidden in Ramanujan's Pi Formula - Public Notes -

Overview

This article demonstrates that the two large integers, 1103 and 26390, provided by Ramanujan in his 1914 pi formula, are not unrelated coefficients but rather consecutive terms in a single additive recurrence relation.

Ramanujan's Pi Formula

$$\frac{1}{\pi} = \frac{2\sqrt{2}}{9801}\sum_{k=0}^{\infty}\frac{(4k)!\,(1103+26390k)}{(k!)^4\,396^{4k}}$$

Specifically, the values $82,\ 1103,\ 26390,\ \dots$ appear as three consecutive terms in the second-order recurrence relation $a_{n+1}=24\,a_n-a_{n-1}$. The coefficient $24=11+13$ is the sum of twin primes.

Furthermore, an algebraic identity holds: the discriminant of an additive recurrence relation whose coefficient sum is a pair of twin primes $(p,q)$ is $4pq$ if and only if $|p-q|=2$. While the ratio of consecutive terms of this recurrence converges to $12+\sqrt{143}=12+\sqrt{11\cdot13}$, it never reaches this value, nor the integer $24$, at any finite stage, approaching it while carrying a residue.

This "convergence without reaching" motion is the same structure by which the Fibonacci sequence fails to reach the golden ratio $\varphi$, and the partial sums of Ramanujan's formula fail to reach $1/\pi$. The three differ only in the reduction rate of their residues.

As a starting point, I also investigated the world of imaginary quadratic fields (Eisenstein series and $j$-invariants), but 1103 and 26390 did not appear in the integer structure of $j(i\sqrt{58})$. I record this as a negative result.

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1. Introduction: Why these two integers?

There are countless formulas for pi, but Ramanujan's 1914 formula is known for achieving extreme precision with very few terms. Just the 0-th term, for $k=0$, gives:

$$S_0=\frac{2\sqrt{2}}{9801}\cdot 1103 = 0.31830987844\ldots$$

This matches $1/\pi=0.31830988618\ldots$ to 8 decimal places. By simply dividing the single integer 1103 by $9801=99^2$ and multiplying by $2\sqrt{2}$, we reach within 8 digits of the reciprocal of pi.

This naturally raises a question: Why 1103? Why 26390? These two values form a linear expression within the formula:

$$a_k = 1103 + 26390\,k$$

With a constant term of 1103 and a slope of 26390, were these two numbers given separately, or are they linked to each other?

This note is an attempt to trace the thread connecting these two numbers using only the language of integers. I use no trigonometric functions, no calculus, and no continuous limit operations. All that is used is addition and its repetition.

---## 2. A Detour and Its Record: Eisenstein series were not the wayMy first approach was an attempt from the world of imaginary quadratics. The origin of Ramanujan's formula is a special point in the upper half-plane, $\tau=i\sqrt{58}$, where Eisenstein series $E_4, E_6$ and the $j$-invariant take algebraic values. I hypothesized that 1103 or 26390 might be embedded within the integer structure of these values.As a result of numerical calculation, $\tau=i\sqrt{58}$ is a CM point with discriminant $-232$ and class number 2. The $j$-invariant $j(i\sqrt{58})$, along with its conjugate, becomes the root of a quadratic equation (Hilbert class polynomial) with rational integer coefficients. Factoring those coefficients gives:- Sum of the two roots: $604729957849891344000 = 2^7\cdot3^3\cdot5^3\cdot140989\cdot9928702703$- Product of the two roots: $2^{12}\cdot3^6\cdot5^9\cdot53^3\cdot149^3\cdot173^3$**Neither 1103, nor 26390, nor 9801 appear here.**This is a negative result. The implication is simple: the coefficients 1103 and 26390 of Ramanujan's formula are not integer values of the $j$-invariant itself, but rather quantities that appear as ratios of special values of $E_4/E_6/\Delta$, or in the process of converting to elliptic integrals or hypergeometric functions.At the same time, this is also a record of having "taken the wrong path." Eisenstein series and Hilbert class polynomials are tools of continuous mathematics. The moment I introduced complex machinery, I moved away from the starting point of building the world solely through addition. When things become complex, it is evidence that one has taken the wrong path. Back to the origin.

3. Connecting 1103 and 26390 with Addition Alone

3.1 Putting the Two Numbers into a Single Formula

Returning to the origin, I will write the relationship between the two numbers directly. Dividing $26390$ by $1103$ yields:

$$\frac{26390}{1103}=23.9256\ldots$$

This is close to the integer $24$. In fact,

$$24\times1103 = 26472,\qquad 26472 - 26390 = 82$$

In other words,

$$\boxed{\,26390 = 24\times1103 - 82\,}$$

Here, the residue $82$ is not just a random number:

$$82 = 24 + 58 = (11+13) + 58, \qquad 82 = 2\times41,\qquad 41 = 11+13+17$$

$24$ is the sum of the twin primes $(11, 13)$. $58$ is the $58$ from the formula's origin $\tau=i\sqrt{58}$. $41$ is the sum of the consecutive primes $11+13+17$. The residue $82$ is composed entirely of these "actors."

3.2 The Two Numbers are Adjacent Terms in an Additive Recurrence Sequence

Reading the relationship $26390 = 24\times1103 - 82$ one level deeper, this is nothing more than the form of a second-order recurrence relation $a_{n+1} = 24\,a_n - a_{n-1}$ where $a_{n-1}=82,\ a_n=1103,\ a_{n+1}=26390$. In fact,

$$26390 = 24\times1103 - 82\quad✓$$

And stepping back one level:

$$1103 = 24\times82 - 865\quad✓$$

So the sequence lines up as:

$$\dots,\ 865,\ 82,\ 1103,\ 26390,\ 632257,\ 15147778,\ \dots$$

1103 and 26390 are adjacent terms in this sequence. Note that this is the same "second-order additive recurrence sequence" as the Fibonacci sequence $a_{n+1}=a_n+a_{n-1}$, just with the coefficient set to $24$.

4. The Discriminant Identity for Twin Primes

4.1 The Identity

The characteristic equation for the recurrence relation $a_{n+1}=(p+q)\,a_n-a_{n-1}$ is $x^2-(p+q)x+1=0$, and its discriminant is $(p+q)^2-4$. Performing simple algebra here:

$$(p+q)^2 - 4 - 4pq = p^2-2pq+q^2-4 = (p-q)^2 - 4$$

Therefore,

$$\boxed{\,(p+q)^2-4 = 4pq \iff (p-q)^2 = 4 \iff |p-q| = 2\,}$$

When $p$ and $q$ are twin primes (difference of 2), and only then, the discriminant becomes $4pq$. In other words, an additive recurrence sequence whose coefficient sum is a pair of twin primes will always produce $\sqrt{pq}$—the square root of the product of the twin primes—in its characteristic roots.

This is neither an observation nor a hypothesis; it is an algebraic identity that anyone can verify by expanding the expression.

4.2 With Some Twin Primes

Twin Primes $(p,q)$	$(p+q)^2-4$	$4pq$	Match	Square root appearing in the characteristic root
$(3,5)$	$60$	$60$	✓	$\sqrt{15}=\sqrt{3\cdot5}$
$(5,7)$	$140$	$140$	✓	$\sqrt{35}=\sqrt{5\cdot7}$
$(11,13)$	$572$	$572$	✓	$\sqrt{143}=\sqrt{11\cdot13}$
$(17,19)$	$1292$	$1292$	✓	$\sqrt{323}=\sqrt{17\cdot19}$
$(29,31)$	$3596$	$3596$	✓	$\sqrt{899}=\sqrt{29\cdot31}$

For reference, this does not hold for non-twin pairs. For $(11,17)$, $(p+q)^2-4=780$ while $4pq=748$; for $(7,13)$, it is $396$ and $364$. Only a difference of 2 is special.

4.3 In the Case of the Sequence of 1103 and 26390

The recurrence relation involving 1103 and 26390 uses the coefficient $24=11+13$, which corresponds to the case where the twin primes $(11,13)$ are chosen. The characteristic equation is

$$x^2 - 24x + 1 = 0,\qquad x = 12\pm\sqrt{143} = 12\pm\sqrt{11\cdot13}$$

The larger root is $12+\sqrt{143}=23.9582607431\ldots$, the smaller root is $12-\sqrt{143}=0.0417392569\ldots$, and their product is $1$.

5. Converging Without Reaching

5.1 Adjacency Ratios Do Not Reach the Limit

Observing the adjacency ratios of the sequence $82,\,1103,\,26390,\,\dots$ in order:

Adjacency Ratio	Value	Residue from limit $12+\sqrt{143}$
$1103/82$	$13.451\ldots$	$-10.507$
$26390/1103$	$23.9256\ldots$	$-0.03260\ldots$
$632257/26390$	$23.95820\ldots$	$-5.688\times10^{-5}$
$15147778/632257$	$23.958260\ldots$	$-9.909\times10^{-8}$

The adjacency ratio heads toward $12+\sqrt{143}$. However, it never reaches it in a finite number of steps. It approaches while carrying a residue.

What is important here is that $26390/1103 = 23.9256\ldots$ does not even reach the integer $24$. It "wants to be 24 times," but falls short by $82/1103$. Furthermore, the true limit is not even $24$, but an irrational number $12+\sqrt{143}$. It does not reach the integer. That is the residue.

5.2 Three Sequences Performing the Same Motion

This structure of "converging without reaching" is the same as the structure where the Fibonacci sequence does not reach the golden ratio. In the Fibonacci sequence, $F_{n+1}/F_n$ heads toward $\varphi=(1+\sqrt5)/2$, always leaving a finite residue at each step.

And the Ramanujan formula itself is the same. The partial sum $S_N$ heads toward $1/\pi$, but does not reach it in a finite number of steps.

Comparing the three:

Sequence	Recurrence / Generation	Limit	Residue Reduction Rate
Fibonacci	$a_{n+1}=a_n+a_{n-1}$	$\varphi=(1+\sqrt5)/2$	Approx. $0.42$ digits/step
Sequence $82,1103,26390,\dots$	$a_{n+1}=24a_n-a_{n-1}$	$12+\sqrt{143}$	Approx. $2.8$ digits/step
Partial sum of Ramanujan formula	$S_{N+1}=S_N+(\text{term})$	$1/\pi$	Approx. $8.0$ digits/step

All three approach their limits without reaching them, while reducing the residue at a constant rate. The only difference is the speed of reduction. The Fibonacci sequence is the slowest, and the Ramanujan formula is the fastest. The sequence 1103/26390 stands in between.

Figure: The residues of the three sequences viewed in $\log_{10}$. All are straight lines, meaning reduction at a constant rate, differing only in slope (speed of reduction). The ratio $26390/1103$ is a point on the middle line.

5.3 The Residue is Not an Error

As written in the previous article "Sequences with structure converge to their center," the residue is not an error. It is a signature of the structure.

The fact that the Fibonacci sequence does not reach $\varphi$, that the Ramanujan formula does not reach $1/\pi$, and that the 1103/26390 sequence does not reach $12+\sqrt{143}$—these are all manifestations of the fact that a hierarchy made only of finite integers possesses an irrational number as its "center," as a position that can never be occupied. The "unreachability" records how that sequence relates to the irrational number.

Pi is one of the deepest instances of this unreachability.

6. What Has Been Confirmed and What Cannot Yet Be Said

Like my notes on seismic data, I will distinguish between what has been confirmed and what remains reserved.

6.1 Confirmed Points (Established)

• $26390 = 24\times1103 - 82$. Where $24=11+13$, $82=24+58=2\times41$, and $41=11+13+17$.
• $82,\,1103,\,26390$ are three consecutive terms of the additive recurrence $a_{n+1}=24a_n-a_{n-1}$. $1103$ and $26390$ are adjacent terms.
• Algebraic identity: $(p+q)^2-4=4pq \iff |p-q|=2$. An additive recurrence whose coefficient sum is a twin prime pair has the square root of the product of the twin primes, $\sqrt{pq}$, in its characteristic roots.
• The adjacency ratio of this sequence converges monotonically to $12+\sqrt{143}=12+\sqrt{11\cdot13}$, and the residue reduces at a constant rate.
• Negative result: $1103$, $26390$, and $9801$ do not appear in the integer coefficients of the Hilbert class polynomial containing $j(i\sqrt{58})$ and its conjugate.

6.2 Reservations — What Cannot Yet Be Said

• The form $26390 = 24\times1103 - 82$ is also a rewrite of the previously known relation $26390 = 24\times\beta - 58$ (where $\beta=1102$ is the Pell entry point). Since any two numbers $A,B$ can be written as $B=qA-r$, what is truly significant is that both $q=24$ and $r=82$ can be expressed by the previously mentioned structural numbers $11, 13, 58$. Whether this fact that "they can be written using only pre-existing numbers" is a coincidence requires further investigation.
• $(11,13)$ is a member of the series where "if they are twin primes, the discriminant always yields the product." While it is not an isolated singularity, I have not yet measured how special it is that the specific prime $1103$ stands exactly as an adjacent pair in this recurrence. Whether the same phenomenon occurs with other twin primes or if similar sequences appear in other Ramanujan-type formulas (such as the Chudnovsky formula) remains a task for the future.
• I chose $r$ to be $1$ because $82=1\times82$ is the most straightforward decomposition. If one were to choose $r=2$ or other values, a different sequence would emerge. I have not yet decided which choice is the "authentic" one.

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7. The Landscape Seen from Here (Hypothesis)

Finally, I will write down an image that is neither confirmed nor an observation. This is not a proof, but a landscape that has become visible beyond the calculations so far.

If we translate the fact that a recurrence with the coefficient sum of the twin primes $(11,13)$ produces $\sqrt{11\cdot13}$ in its characteristic roots into the language of geometry, I am inclined to think of a cone whose generators are two "prime universes" called $B_{11}$ and $B_{13}$. $1103$ and $26390$ are two adjacent points on a sequence flowing along that cone.

And a cone has cross-sections: circles, parabolas, and hyperbolas. Could the circle be a special cross-section of the cone projected from the universe of $B_{11}$ to the universe of $B_{13}$? If $\pi$ emerges from within this structure, it means that $\pi$ is the name of one specific cross-section of the twin prime cone.

This is still just an image. I have not proven the geometry of the cone itself in the language of integers. However, I plan to follow up in a separate paper on what each cross-section of the cone (circle, parabola, hyperbola) signifies in more detail. This note is an outpost from number theory for that discussion on the physical side.

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Appendix

Computational Environment: b13phase v066 + Python (mpmath, sympy). All calculations have undergone integer arithmetic and multi-precision verification.

Replication of Confirmed Calculations: The relations in Chapter 3 and Chapter 4 of this note can all be verified using simple integer arithmetic. $24\times1103-82=26390$, $(11+13)^2-4=4\times11\times13$ — all can be verified with paper and pencil.

Positioning: This note is a sequel to the previous article "B13, sequences with structure converge to their center," and is an outpost for an unfinished physics chapter (the meaning of each cross-section of the cone). It is not a peer-reviewed paper with a formal proof, but a research note recording the progress of a study.

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Afterword ── Interpretation by Y. Zhao

Here, I would like to touch upon the relationship with the paper by Y. Zhao regarding Ramanujan's pi formula.

Y. Zhao, A modular proof of two of Ramanujan's formulae for 1/\pi, J. Aust. Math. Soc. 109 (2020), 131–144.

Ramanujan's 1/\pi formula

$$\frac{1}{\pi} = \frac{2\sqrt{2}}{99^2}\sum_{m=0}^\infty (26390m + 1103)\frac{(4m)!}{396^{4m}(m!)^4}$$

This is a paper that completely derived the two constants $1103$ and $26390$ using the theory of modular forms. Zhao pinpoints $1103$ through calculations using a finite linear combination of $\eta$-products on $\Gamma_0(58)$.

The logic goes like this:

1. Uses Weber's class invariant at $\tau = i/\sqrt{58}$. Specifically, $2k/(k')^2 = ((\sqrt{29}-5)/2)^6$. $\sqrt{29}$ stands out.
2. Determines the space of level 58 $\eta$-products (dimension 7) from the cusp structure of $\Gamma_0(58)$.
3. Obtains $H(e^{-\pi/\sqrt{58}})^2 = 65{,}870{,}496 + 8{,}439{,}552\sqrt{29}$ as a linear combination.
4. Taking the square root, $H(e^{-\pi/\sqrt{58}}) = 36\sqrt{2}(148 + 11\sqrt{29})$.
5. Substituting this yields $G_0 = 2\sqrt{2}\cdot1103 / 99^2$ — this is the origin of $1103$.
6. Similarly, $2v(k) = 1820\sqrt{29} / 99^2$ and $c(k) = 1/99^4 = 1/396^4$ are obtained.
7. Taking the ratio yields $2\sqrt{58}\cdot v(k) / G_0 = (\sqrt{58}\cdot1820\sqrt{29})/(2\sqrt{2}\cdot1103) = 910\cdot29 / 1103$ — this is $26390 / 1103$.

What I want to draw attention to is the prime factorization appearing in the final expression:

$$26390 = 910 \cdot 29 = (2 \cdot 5 \cdot 7 \cdot 13) \cdot 29$$

In this, 13 and 29 appear.

13 appears in 910, that is, in the numerator of $v(k)$ in Zhao's formula. 29 appears in $\sqrt{58} = \sqrt{2}\cdot\sqrt{29}$, that is, on the discriminant side of the modular form. The integer addition side contains 13, and the analytic side contains 29; both are separated and aligned within 26390.

And the correction term 82 in Ramanujan's recurrence $a_{n+1} = 24\cdot a_n - 82$ that generates 26390 can be written as follows:

$$82 = 58 + 24$$

• 58 = 2\cdot29 $\leftarrow$ Analytic side (the discriminant Zhao uses for the proof)
• 24 = $B_{11} + B_{13}$ $\leftarrow$ Integer addition side (sum of adjacent primes)

The correction term of the recurrence itself decomposes into the sum of the analytic side and the integer side. This appears to be a record at the coefficient level that "Ramanujan was speaking two languages simultaneously."

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By the way, the carry point of the double-staircase $i = 3$ concerning the properties of integers was $1099 = (13^3+1)/2 = \Phi_2(13)\cdot\Phi_6(13)/2 = 7\cdot157$.
That is, it is only a difference of 4 from Ramanujan's $1103$.

1099 = Double-staircase carry point (left staircase, additive side), \Phi_d(13) cyclotomic decomposition
1103 = Initial value of Ramanujan's 1/\pi formula (limit of the right staircase, analytic side), \Gamma_0(58) \eta-product
       Difference of 4 (= representative of the 4H coset, jump at the 0th digit)

1099 has the algebraic coordinates of the left staircase. 1103 has the modular coordinates according to Zhao's proof. It can be said that both stand on different scaffolds of the same floor.

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References

• Y. Zhao, "A modular proof of two of Ramanujan's formulae for 1/\pi", J. Aust. Math. Soc. 109 (2020), 131–144. doi:10.1017/S1446788718000599
• S. Ramanujan, "Modular equations and approximations to \pi", Q. J. Pure Appl. Math. 45 (1914), 350–372.
• J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, New York, 1987.

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B13 Fractal Fullerene Theory / MORC.B13
All calculations can be replicated with integer arithmetic (b13phase library).