{"author":{"avatar":"https://nguyencungthanh.github.io/animation_avatar%282%29.PNG","name":"Nguyen Cung Thanh","url":"https://nguyencungthanh.github.io/"},"feed_url":"https://nguyencungthanh.github.io/tags/integer-sequence/feed.json","home_page_url":"https://nguyencungthanh.github.io/tags/integer-sequence/","items":[{"content_html":"\n  \u003cdiv class='e-content'\u003e\n    \u003ccenter\u003e\u003cb\u003eAbstract\u003c/b\u003e\u003c/center\u003e\n\u003cp\u003e$\\quad$ \u003ca href=\"https://cs.uwaterloo.ca/journals/JIS/VOL25/Yaqubi/yaq6.pdf\"  rel='noopener noreferrer'\u003eAmirali Fatehizadeh \u0026amp; Daniel Yaqubi\u003c/a\u003e answered the question when does we have the average of first $n$ Fibonacci numbers are integer. In this post, I will generalize that result.\u003c/p\u003e\n\u003ch2 id=\"1-introduction\"\u003e1. Introduction\u003c/h2\u003e\n\u003chr\u003e\n\u003cp\u003eGiven two positive integers $a$ and $m$ such that $a-2$ is not divisible by $4$. Let the sequence $(S_n)_{n \\in \\mathbb{N}}$ defined by $S_0 = 0, S_1 = 1$ and $S_{n+2} = aS_{n+1} - S_n$ for all $n \\in \\mathbb{N}.$\nThen, for each $m$, there exists a positive integer $N$ has $m$ distinct prime divisors such that $$n \\mid S_1 + S_2 + \\ldots + S_N$$\u003c/p\u003e\n\u003ch2 id=\"2-preliminaries\"\u003e2. Preliminaries\u003c/h2\u003e\n\u003chr\u003e\n\u003cp\u003eWe consider the sequence $(T_n)_{n \\in \\mathbb{N}}$ defined by $T_0 = 2, T_1 = a$ and $T_{n+2} = aT_{n+1}-T_n$. As a result, we have\n$$\nS_n = \\dfrac{\\alpha^n-\\beta^n}{\\alpha-\\beta} \\quad \\text{ and } \\quad T_n = \\alpha^n + \\beta^n\n$$\nwhere $\\alpha \\ge \\beta$ are two roots of equation $x^2-ax+1 = 0.$\n\u003cdiv class=\"lemma\"\u003e\n  \u003cb\u003eLemma 1.\u003c/b\u003e \n    \nFrom the general formula of $S_n$ and $T_n$, the following holds for all nonnegative integer $n$\n\u003col class = \"roman-paren\"\u003e\n  \u003cli\u003e$S_{2n} = S_{n} T_n$.\u003c/li\u003e\n  \u003cli\u003e$S_{2n+1}-1 = S_nT_{n+1}$.\u003c/li\u003e\n\u003c/ol\u003e\n\n\u003c/div\u003e \u003c/p\u003e\n\u003cdiv class=\"lemma\"\u003e\n  \u003cb\u003eLemma 2.\u003c/b\u003e \n  \nFor a prime number $p$, we always have $p \\mid S_{p-(\\frac{\\Delta}{p})} $ where $\\Delta = a^2-{4}$ and $\\left(\\frac{\\bullet}{p}\\right)$ is \n\u003ci\u003eLegendre\u003c/i\u003e symbol. \n\n\u003c/div\u003e \n\u003cp\u003e\u003cdiv class=\"lemma\"\u003e\n  \u003cb\u003eLemma 3.\u003c/b\u003e \n  \nFor a prime number $p$ such that $p \\mid S_n $ then $p \\mid \\dfrac{S_{pn}}{S_n}.$ \n\n\u003c/div\u003e \n\u003cdiv class=\"proof\"\u003e\n  \u003ci\u003e Proof. \u003c/i\u003e \n  \nFor a positive integer $n$ such that $p \\mid S_n$, then $$p \\mid \\dfrac{\\alpha^n-\\beta^n}{\\alpha -\\beta}.$$ As a result, there must exist $\\mu \\in \\mathcal{O}_{K}$ where $K = \\mathbb{Q}(\\alpha)$ such that $\\alpha^n-\\beta^n =\\mu p.$ \n\nWe also consider $(p) \\subset \\mathcal{O}_{K}$ be the principal ideal generated by $p$ of $\\mathcal{O}_{K}$ and let $c = \\alpha^n, d = \\beta^n$. From this, we have\n\\begin{align*}\n\\dfrac{S_{pn}}{S_n} \u0026= \\dfrac{c^p-d^p}{c-d} \\\\\\\\ \n\u0026 = \\dfrac{(d+\\mu p)^p-d^p}{\\mu p} \\\\\\\\\n\u0026 = \\dfrac{ \\mathop{\\sum}\\limits_{0\\le j \\le p}\\binom{p}{j}(\\mu p)^jd^{p-j} - d^p}{\\mu p} \\\\\\\\\n\u0026 = \\dfrac{ \\mathop{\\sum}\\limits_{1\\le j \\le p}\\binom{p}{j}(\\mu p)^jd^{p-j}}{\\mu p} \\\\\\\\\n\u0026 =  \\mathop{\\sum}\\limits_{1\\le j \\le p}\\binom{p}{j}(\\mu p)^{j-1}d^{p-j} \\\\\\\\\n\u0026 \\equiv 0 \\pmod{(p)} \n\\end{align*}\nsince $ \\binom{p}{j}$ is divisible by $p$ for all $j = \\overline{1,p-1}.$ \n\n\u003c/div\u003e \u003c/p\u003e\n\u003cp\u003eAs a consequence, we have the following corollary:\u003c/p\u003e\n\u003cp\u003e\u003cstrong\u003eCorollary.\u003c/strong\u003e For a prime $p$ such that $p \\mid S_{k}$ then $p^{n+1} \\mid S_{kp^n}$ for all $n \\in \\mathbb{N}.$\u003c/p\u003e\n\u003ch2 id=\"3-main-result\"\u003e3. Main result\u003c/h2\u003e\n\u003chr\u003e\n\u003cp\u003eConsider the case $m = 1.$ Then, we only need to choose $N = p$ where $p$ is a prime divisor of $\\Delta$ then of course, we have\n$$p \\mid S_{p - (\\frac{\\Delta}{p})} = S_p.$$\u003c/p\u003e\n\u003cp\u003eNow, we consider the case $m \\geq 2.$\u003c/p\u003e\n\u003cp\u003eThen, we denote $p_1,p_2,\\dots, p_m$ are $m$ first primes and we choose $N = 2M$ where $M = (p_1p_2\\cdots p_m)^{p_m}$ which give ours the following result:\n$$\n\\begin{align*}\nS_1+\\ldots+S_n \u0026amp;= \\sum_{i = 1}^n \\dfrac{\\alpha^n-\\beta^n}{\\alpha-\\beta} \\\\\n\u0026amp;= \\dfrac{\\dfrac{ \\alpha^{n+1}-1}{\\alpha-1} - \\dfrac{\\beta^{n+1}-1}{\\beta-1}}{\\alpha-\\beta} \\\\\n\u0026amp;= \\dfrac{(\\beta-1)(\\alpha^{n+1}-1)-(\\alpha-1)(\\beta^{n+1}-1)}{(\\alpha-\\beta)(\\alpha-1)(\\beta-1)} \\\\\n\u0026amp;= \\dfrac{\\alpha^n-\\beta^n+\\alpha-\\beta-(\\alpha^{n+1}-\\beta^{n+1})}{(\\alpha-\\beta)(2-a)} \\\\\n\u0026amp;= \\dfrac{S_{n+1}-1-S_{n}}{a-2} \\\\\n\u0026amp;= \\dfrac{S_M(T_{M+1}-T_M)}{a-2}.\n\\end{align*}\n$$\nNote that $T_0 \\equiv T_1 \\equiv 2 \\pmod{a-2}$ so by induction, $T_n \\equiv 2 \\pmod{a-2}$ for all nonnegative integer $n$.\nTherefore, $S_1+\\ldots+S_n$ is divisible by $n$ is equivalent to $2M \\mid S_M$.\u003c/p\u003e\n\u003cp\u003eNow, consider $p_1 = 2$ then since $a-2$ is not divisible by 4 so $S_2=a$ or $S_3=a^2-1$ is divisible by 4. From this and following lemma 3, we derive that $2^{p_m+1} \\mid S_M$. On the other hand, we fix an index $i \\ge 2$ then there must exist nonnegative integer numbers $\\alpha_{j,i}$ less than $p_m$ for all $j = \\overline{1,i}$ such that $$p_i - \\left( \\frac{\\Delta}{p} \\right) = p_1^{\\alpha_{1,i}} p_2^{\\alpha_{2,i}}\\cdots p_{i-1}^{\\alpha_{i-1,i}} p_i^{\\alpha_{i,i}}$$\nwith note that $\\alpha_{i,i} \\le 1$. Therefore following from lemma, we see that\n$$\np_i \\mid S_{p - \\left(\\frac{\\Delta}{p}\\right)} = S_{p_1^{\\alpha_{1,i}}p_2^{\\alpha_{2,i}}\\cdots p_{i-1}^{\\alpha_{i-1,i}}p_i^{\\alpha_{i,i}}}\n$$\nso\n$$\np_i^{p_m} \\mid S_{p_1^{\\alpha_{1,i}}\\cdot p_2^{\\alpha_{2,i}}\\cdots p_{i-1}^{\\alpha_{i-1,i}}p_i^{\\alpha_{i,i}+p_m-1}} \\mid\nS_{(p_1p_2\\ldots p_m)^{p_m}} = S_K\n$$\u003c/p\u003e\n\u003cp\u003eTherefore, we derive that $$2M = 2(p_1p_2\\cdots p_m)^{p_m} \\mid S_M$$ as desired.\u003c/p\u003e\n\u003ch2 id=\"4-conclusion\"\u003e4. Conclusion\u003c/h2\u003e\n\u003chr\u003e\n\u003cp\u003eI generalized the reference problem by considering the extension of algebraic number field. On the other hand, on the aforementioned paper, \u003ca href=\"https://cs.uwaterloo.ca/journals/JIS/VOL25/Yaqubi/yaq6.pdf\"  rel='noopener noreferrer'\u003eAmirali Fatehizadeh \u0026amp; Daniel Yaqubi\u003c/a\u003e also conjectured that for infinitely many odd numbers $n$ also satisfied the problem. This was proved recently by \u003ca href=\"https://arxiv.org/abs/2504.09938\"  rel='noopener noreferrer'\u003eOisın Flynn-Connolly\u003c/a\u003e. Therefore, if I have time, I will consider to extend the result for odd positive integer in a similar way.\u003c/p\u003e\n\u003ch2 id=\"references\"\u003eReferences\u003c/h2\u003e\n\u003chr\u003e\n\u003cp\u003e[1] Fatehizadeh, A., \u0026amp; Yaqubi, D. (2022). \u003cem\u003eAverage of the Fibonacci Numbers. Journal of Integer Sequences\u003c/em\u003e, 25(2), Article 22.2.6, 10 pp. Retrieved from \u003ca href=\"https://cs.uwaterloo.ca/journals/JIS/VOL25/Yaqubi/yaq6.html\"  rel='noopener noreferrer'\u003ehttps://cs.uwaterloo.ca/journals/JIS/VOL25/Yaqubi/yaq6.html\u003c/a\u003e\u003c/p\u003e\n\u003cp\u003e[2] Flynn-Connolly, O. (2025). \u003cem\u003eOn the divisibility of sums of Fibonacci numbers\u003c/em\u003e (arXiv:2504.09938). arXiv. \u003ca href=\"https://arxiv.org/abs/2504.09938\"  rel='noopener noreferrer'\u003ehttps://arxiv.org/abs/2504.09938\u003c/a\u003e\u003c/p\u003e\n  \u003c/div\u003e\n\u003cp\u003e\n  \u003ca href=\"mailto:nguyencungthanh11@gmail.com?subject=Re: Average%20of%20Generalized%20Fibonacci%20Sequence\" rel='noopener noreferrer'\u003e💬 Reply via e-mail\u003c/a\u003e\n\u003c/p\u003e\n","date_modified":"2026-01-22T00:00:00+00:00","date_published":"2026-01-22T00:00:00+00:00","id":"https://nguyencungthanh.github.io/2026/01/22/average-of-generalized-fibonacci-sequence/","summary":"Abstract $\\quad$ Amirali Fatehizadeh \u0026 Daniel Yaqubi answered the question when does we have the average of first $n$ Fibonacci numbers are integer. In this post, I will generalize that result.\n","tags":["Integer Sequence"],"title":"Average of Generalized Fibonacci Sequence","url":"https://nguyencungthanh.github.io/2026/01/22/average-of-generalized-fibonacci-sequence/"}],"title":"Tag: Integer Sequence - Nguyen Cung Thanh","version":"https://jsonfeed.org/version/1"}