Seeing 1, 2, 3, 5, we instinctively answer 8. That is a reasonable answer, but not an inevitable consequence of those four numbers. The aim is not to trick the solver: it is to distinguish the data from the rule used to interpret them. Starting with the supplied document on the sequence 4, 7, 11, 15, 3, we develop a proof and two four-number examples.
- Observed values are not yet a rule
- Lagrange interpolation: fitting the known data
- The source example: 4, 7, 11, 15, 3
- Choosing the sixth term freely
- Four arbitrary numbers: a, b, c, d
- Correcting the fifth term
- First example: even 2, 4, 6, 8 can continue with 42
- Second example: 1, 2, 3, 5 under the Fibonacci rule
- The same data with a polynomial: the fifth term is 9
- The same construction using finite differences
- After 1, 2, 3, 5 we can put N
- What if all terms must be integers?
- The general proof and the meaning of uniqueness
- How to read a sequence puzzle
1. Observed values are not yet a rule
A sequence assigns a value uₙ to each position n. Four known terms specify only four assignments. In the formulas below x is the position, not the value appearing in the sequence: for 1, 2, 3, 5 we require p(1)=1, p(2)=2, p(3)=3 and p(4)=5.
Without further restrictions, any real number can be the fifth term. Requiring integer terms restricts the choice to integers, but does not make it unique. Rather than listing exceptions, we will construct a formula that delivers the number we choose.
2. Lagrange interpolation: fitting the known data
For m values y₁, …, yₘ at distinct positions 1, …, m, there is a unique interpolating polynomial of degree at most m−1. Lagrange uses selector polynomials:
p(x) = Σᵢ yᵢ Lᵢ(x) Lᵢ(x) = ∏ⱼ≠ᵢ (x − j)/(i − j)
Σ means sum and ∏ means product. In Lᵢ, j runs from 1 to m except i. At x=i every ratio equals 1; at another known position one numerator factor is zero. Thus Lᵢ(i)=1 and Lᵢ(j)=0 for j≠i. Multiplying each selector by yᵢ and adding reproduces the data. See the standard formula in NIST DLMF §3.3.
3. The source example: 4, 7, 11, 15, 3
We want p(1)=4, p(2)=7, p(3)=11, p(4)=15 and p(5)=3. The selector denominators are 24, −6, 4, −6, 24. For example, the second is (2−1)(2−3)(2−4)(2−5)=−6:
A(x) = 4(x−2)(x−3)(x−4)(x−5)/24 B(x) = −7(x−1)(x−3)(x−4)(x−5)/6 C(x) = 11(x−1)(x−2)(x−4)(x−5)/4 D(x) = −15(x−1)(x−2)(x−3)(x−5)/6 E(x) = 3(x−1)(x−2)(x−3)(x−4)/24 p(x) = A(x) + B(x) + C(x) + D(x) + E(x)
At the first position A(1)=4; all other terms contain (x−1) and vanish. The same argument checks every other position.
Expand the products and collect their coefficients
P₁ = (x−2)(x−3)(x−4)(x−5) = x⁴−14x³+71x²−154x+120 P₂ = (x−1)(x−3)(x−4)(x−5) = x⁴−13x³+59x²−107x+60 P₃ = (x−1)(x−2)(x−4)(x−5) = x⁴−12x³+49x²−78x+40 P₄ = (x−1)(x−2)(x−3)(x−5) = x⁴−11x³+41x²−61x+30 P₅ = (x−1)(x−2)(x−3)(x−4) = x⁴−10x³+35x²−50x+24 24p(x) = 4P₁ − 28P₂ + 66P₃ − 60P₄ + 3P₅ x⁴: 4−28+66−60+3 = −15 x³: −56+364−792+660−30 = 146 x²: 284−1652+3234−2460+105 = −489 x: −616+2996−5148+3660−150 = 742 1: 480−1680+2640−1800+72 = −288
Using the common denominator 24 gives:
p(x) = (−15x⁴+146x³−489x²+742x−288)/24 p(6) = 4 − 35 + 110 − 150 + 15 = −56
So −56 is the sixth value of the polynomial of degree at most four, not the mandatory sixth value of every possible sequence.
4. Choosing the sixth term freely
We need a correction that leaves the five known values untouched. Q₅ vanishes at those positions but equals 120 at the sixth. Let N be the desired sixth value:
Q₅(x) = (x−1)(x−2)(x−3)(x−4)(x−5) Q₅(1) = … = Q₅(5) = 0 Q₅(6) = 5·4·3·2·1 = 120 f_N(x) = p(x) + k Q₅(x) −56 + 120k = N k = (N+56)/120 f_N(x) = p(x) + (N+56)Q₅(x)/120
At positions 1 through 5 the correction is zero. At position 6 it adds N+56 to −56, giving N. To obtain 100, take k=156/120=13/10: f₁₀₀(6)=−56+156=100. The degree can now reach five: we have enlarged the allowed class of rules.
5. Four arbitrary numbers: a, b, c, d
Place four values at positions 1, 2, 3, 4. The Lagrange denominators are −6, 2, −2, 6. The polynomial of degree at most three is:
p₃(x) = −a(x−2)(x−3)(x−4)/6
+b(x−1)(x−3)(x−4)/2
−c(x−1)(x−2)(x−4)/2
+d(x−1)(x−2)(x−3)/6
p₃(5) = −a + 4b − 6c + 4dWhy the coefficients −1, 4, −6, 4 in the last line? Substitute x=5 into the four selectors: −(3·2·1)/6=−1; (4·2·1)/2=4; −(4·3·1)/2=−6; (4·3·2)/6=4. This shortcut gives the fifth term under the cubic polynomial assumption.
6. Correcting the fifth term
There are now four known positions to preserve, not five. We therefore use four factors and divide by 24 rather than 120:
Q₄(x) = (x−1)(x−2)(x−3)(x−4) Q₄(1) = … = Q₄(4) = 0 Q₄(5) = 4·3·2·1 = 24 g_N(x) = p₃(x) + [N−p₃(5)]Q₄(x)/24 g_N(1)=a, g_N(2)=b, g_N(3)=c, g_N(4)=d g_N(5)=N
The starting polynomial suggests p₃(5). We add precisely the difference N−p₃(5), leaving the first four values unchanged. Here N is the fifth term; in the source example it was the sixth.
7. First example: even 2, 4, 6, 8 can continue with 42
The natural rule “add 2” produces 10. Yet our formula preserves the four data points and produces 42:
2, 4, 6, 8, ? p₃(x) = 2x p₃(5) = −2+4·4−6·6+4·8 = 10 g_N(x) = 2x + (N−10)Q₄(x)/24 N = 42 → g₄₂(x) = 2x + (4/3)Q₄(x) g₄₂(5) = 10 + (4/3)·24 = 42
At positions 1 through 4, Q₄ is zero, so the terms remain 2, 4, 6, 8. This does not make 42 more plausible than 10: it proves that the data alone do not force 10. If the question explicitly specifies an arithmetic progression, however, 10 is required.
8. Second example: 1, 2, 3, 5 under the Fibonacci rule
Choose the rule “from the third term onwards, add the previous two terms”:
u₁=1, u₂=2 uₙ = uₙ₋₁ + uₙ₋₂ (n ≥ 3) 1, 2, 3, 5, 8, 13, 21, …
We check 1+2=3 and 2+3=5. The same rule requires 3+5=8 and then 5+8=13. This is the Fibonacci sequence starting with 1 and 2. 8 is correct under that rule; four values alone do not uniquely identify it.
9. The same data with a polynomial: the fifth term is 9
Start with the function x, which already gives 1, 2, 3 in the first three positions. At the fourth it gives 4, so we must add 1. The product (x−1)(x−2)(x−3) vanishes in the first three positions and equals 6 at the fourth. Divide it by 6:
p₃(x) = x + (x−1)(x−2)(x−3)/6
= (x³−6x²+17x−6)/6
p₃(1)=1, p₃(2)=2, p₃(3)=3
p₃(4)=4+(3·2·1)/6=5
p₃(5)=5+(4·3·2)/6=9
p₃(6)=6+(5·4·3)/6=16This constructs the cubic interpolant without expanding every Lagrange term. The general formula confirms it: −1+4·2−6·3+4·5=9. This is not a mistake in Fibonacci: it is a different rule, continuing with 9 and 16 instead of 8 and 13.
10. The same construction using finite differences
Subtract each term from the next. The four data give first differences 1, 1, 2, second differences 0, 1 and one third difference equal to 1. If we assume constant third differences, the next second difference is 2, the next first difference is 4 and the fifth term is 5+4=9:
u: 1 2 3 5 9 16 Δu: 1 1 2 4 7 Δ²u: 0 1 2 3 Δ³u: 1 1 1
The following step uses second difference 3 and first difference 7, giving 16. Constant third differences describe the cubic model; a single observed third difference does not prove they will stay constant.
11. After 1, 2, 3, 5 we can put N
The cubic predicts 9. To replace it with N without changing earlier values, apply the correction:
g_N(x) = x + (x−1)(x−2)(x−3)/6
+(N−9)(x−1)(x−2)(x−3)(x−4)/24
g_N(5) = 9+(N−9)·24/24 = N
g_N(6) = 16+(N−9)·120/24 = 5N−29| Fifth term N | Coefficient (N−9)/24 | Sixth term of g_N |
|---|---|---|
| 8 | −1/24 | 11 |
| 9 | 0 | 16 |
| 100 | 91/24 | 471 |
Notice the first row: choosing N=8 gives 1, 2, 3, 5, 8, 11, not 13. Matching Fibonacci at the fifth term does not mean matching its rule. The table describes this polynomial family, not every possible sequence.
12. What if all terms must be integers?
Fractional coefficients do not necessarily produce fractional values at integer positions. In our example we can write:
g_N(n) = n + C(n−1,3) + (N−9)C(n−1,4) C(r,k) = r!/[k!(r−k)!] (r ≥ k) C(r,k) = 0 (0 ≤ rC(r,k) is the binomial coefficient: the number of ways to choose k objects from r. For positive integer n both binomial coefficients are integers. Therefore an integer N produces integer values at every positive integer position. This does not promise positive, increasing or bounded values: those would be additional conditions.
13. The general proof and the meaning of uniqueness
With m data points take the interpolating polynomial p of degree at most m−1. The product of m factors vanishes at the known positions and equals m! at the next, where m!=1·2·…·m:
Qₘ(x) = ∏ⱼ₌₁ᵐ (x−j) Qₘ(m+1) = m! F_N(x) = p(x) + [N−p(m+1)]Qₘ(x)/m! H(x) = F_N(x) + λ∏ⱼ₌₁ᵐ⁺¹ (x−j)The first formula assigns any N at position m+1. The second, for any real λ, preserves that value too but changes later terms: infinitely many rules remain even after selecting N.
This does not contradict Lagrange uniqueness. If two polynomials of degree at most m−1 agreed at m distinct points, their difference would have m roots despite degree at most m−1, so it must be zero. Uniqueness holds within the specified degree bound. The corrections use higher degrees. Smoothness alone does not solve the ambiguity: all these polynomials are continuous and differentiable.
14. How to read a sequence puzzle
A puzzle often asks for the simple rule intended by its author. This is useful pattern recognition, but “simple” depends on the language and class of rules considered. A mathematically unique answer needs a stated recurrence, progression or sufficient degree bound.
Check: does 1, 2, 3, 5, 8 force 13 next?
No. Fibonacci requires 13, but the polynomial g₈ above gives 11. Both reproduce the five initial terms.
Check: what follows a, b, c, d if the degree is at most three?
−a+4b−6c+4d. Without that restriction, choose N and use the Q₄ correction.
Conclusion: finding a compatible rule is not proving it is the only rule. The data record what is known; the added rule determines how to continue. Making that choice explicit is part of the educational value of sequences.