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The Frobenius Number for Sequences of Binomial Coefficients

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The Frobenius Number for Sequences of Binomial Coefficients The Frobenius number for sequences of binomial coefficients Aureliano M. Robles-Perez´ Universidad de Granada A talk based on a joint work with Jose´ Carlos Rosales (J. Number Theory 186 (2018) 473–492.) INdAM meeting: International meeting on numerical semigroups - Cortona 2018 3-7th September 2018 A. M. Robles-Perez´ (UGR) Frobenius number and binomial sequences IMNS-2018 1 / 29 Frobenius coin problem I Given relatively prime positive integers a1;:::;an, n ≥ 2, find a formula to compute the largest integer that is not representable as a non-negative integer linear combination of a1;:::;an. I F(a1;:::;an) (the Frobenius number of the set fa1;:::;ang) denotes the solution of the previous problem. I F(a1;a2) = a1a2 − a1 − a2: I Frobenius problem is open for n ≥ 3. ∗ Curtis: it is impossible to find a polynomial formula that computes the Frobenius number if n = 3. ∗ Ram´ırez Alfons´ın: the problem is NP-hard for n variables. A. M. Robles-Perez´ (UGR) Frobenius number and binomial sequences IMNS-2018 2 / 29 Particular cases I Arithmetic and almost arithmetic sequences (A. Brauer; M. Lewin; J. B. Roberts; E. S. Selmer), I Fibonacci sequences (J. M. Mar´ın, J. L. Ram´ırez Alfons´ın and M. P. Revuelta), I geometric sequences (D. C. Ong and V. Ponomarenko), I Mersenne, repunit, and Thabit sequences (J. C. Rosales, M. B. Branco, D. Torrao),˜ I squares and cubes sequences (M. Lepilov, J. O’Rourke and I. Swanson; A. Moscariello), I ... A. M. Robles-Perez´ (UGR) Frobenius number and binomial sequences IMNS-2018 3 / 29 Particular cases: motivations (i) I Brauer: n − 2 F(n;n + 1;:::;n + k − 1) = + 1 n − 1: k − 1 I Baker (conjecture): If Tn is the nth triangular (or triangle) number, then 8 3 2 > 6n +18n +12n−8 if n is even; <> 8 F(Tn;Tn+1;Tn+2) = :> 6n3+12n2−6n−20 8 if n is odd. Equivalently, 6n3 + 3(5 + (−1)n)n2 + 3(1 + 3(−1)n)n − (14 − 6(−1)n) F(Tn;T + ;T + ) = : n 1 n 2 8 A. M. Robles-Perez´ (UGR) Frobenius number and binomial sequences IMNS-2018 4 / 29 Particular cases: motivations (ii) I Sequences of binomial coefficients (or combinatorial numbers). nn n+1 n+k−1o ∗ fn;n + 1;:::;n + k − 1g = 1 ; 1 ;:::; 1 nn+1 n+2 n+3o ∗ Tn;Tn+1;Tn+2 = 2 ; 2 ; 2 T1 = 1 T2 = 3 T3 = 6 T4 = 10 T5 = 15 T6 = 21 Figure: First six triangular numbers. A. M. Robles-Perez´ (UGR) Frobenius number and binomial sequences IMNS-2018 5 / 29 Purposes I Prove the Baker’s conjecture. I Find formulas for other binomial coefficients sequences. n+2 ∗ Tetrahedral (or triangular pyramidal) numbers: THn = 3 . Figure: Tetrahedral number TH5 (by layers). ∗ Pentatope (or 4-Hypertetrahedral or 4-dimensional triangular n+3 pyramidal) numbers: Pn = 4 . A. M. Robles-Perez´ (UGR) Frobenius number and binomial sequences IMNS-2018 6 / 29 Tools (i) I Johnson’s formula: if a1;a2;a3 are relatively prime numbers and gcdfa1;a2g = d, then a a F(a ;a ;a ) = d F 1 ; 2 ;a + (d − 1)a : 1 2 3 d d 3 3 I Generalization by Brauer and Shockley: if a1;:::;an are relatively prime numbers and d = gcdfa1;:::;an−1g, then a a − F(a ;:::;a ) = d F 1 ;:::; n 1 ;a + (d − 1)a : 1 n d d n n I Telescopic sequences. A. M. Robles-Perez´ (UGR) Frobenius number and binomial sequences IMNS-2018 7 / 29 Tools (ii) Definition I Let (a1;:::;an) be a sequence of positive integers such that gcdfa1;:::;ang = 1 (where n ≥ 2). I Let di = gcdfa1;:::;aig for i = 1;:::;n. I Then (a1;:::;an) is a telescopic sequence if, for each i = 2;:::;n, ai is representable as a non-negative integer linear combination di of a1 ;:::; ai−1 . di−1 di−1 Remark I If (a1;:::;an) is a telescopic sequence, then a a ∗ 1 ;:::; i is also a telescopic sequence for i = 2;:::;n − 1; di di a1 an−1 ∗ F(a1;:::;an) = dn−1 F ;:::; + (dn−1 − 1)an. dn−1 dn−1 A. M. Robles-Perez´ (UGR) Frobenius number and binomial sequences IMNS-2018 8 / 29 Examples Example: (8,14,19) I d1 = gcdf8g = 8; d2 = gcdf8;14g = 2; d3 = gcdf8;14;19g = 1. 14 8 19 8 14 I = 7 · ; = 3 · + 1 · . 2 8 1 2 2 I (8;14;19) is a telescopic sequence. I 8 14 F(8;14;19) = 2 · F 2 ; 2 + (2 − 1) · 19 = 2 · F(4;7) + 19 = 2 · 17 + 19 = 53: Example: (8,14,17) I d1 = gcdf8g = 8; d2 = gcdf8;14g = 2; d3 = gcdf8;14;17g = 1. 14 8 17 8 14 I = 7 · ; , α1 · + α2 · , for all α1;α 2 2 N = f0;1;2;:::g. 2 8 1 2 2 I (8;14;17) is not a telescopic sequence. I F(8;14;17) =? A. M. Robles-Perez´ (UGR) Frobenius number and binomial sequences IMNS-2018 9 / 29 Triangular numbers sequences ! n + 1 n(n + 1) T = = ; n ≥ 1: n 2 2 Lemma 8 n+1 <> 2 if n is odd; I gcdfTn;Tn+1g = :> n + 1 if n is even. Lemma I gcdfTn;Tn+1;Tn+2g = 1: Proof n o ∗ gcdfTn;Tn+1;Tn+2g = gcd gcdfTn;Tn+1g;gcdfTn+1;Tn+2g : A. M. Robles-Perez´ (UGR) Frobenius number and binomial sequences IMNS-2018 10 / 29 Triangular numbers sequences Proposition I (Tn;Tn+1;Tn+2) and (Tn+2;Tn+1;Tn) are telescopic sequences. Proof I Let n be an odd integer. I n(n+1) n+1 d1 = gcdfTng = 2 ; d2 = gcdfTn;Tn+1g = 2 ; d3 = gcdfTn;Tn+1;Tn+2g = 1. T T T I n+1 · Tn n+2 · Tn n+3 · n+1 n+1 =( n + 2) n(n+1) ; 1 = 0 n+1 + 2 n+1 . 2 2 2 2 I (Tn;Tn+1;Tn+2) is a telescopic sequence if n is odd. I The proofs for the other three cases are similar. A. M. Robles-Perez´ (UGR) Frobenius number and binomial sequences IMNS-2018 11 / 29 Triangular numbers sequences Proposition 8 3 2 > 3n +6n −3n−10 if n is odd; <> 4 I F(Tn;Tn+1;Tn+2) = :> 3n3+9n2+6n−4 4 if n is even. Proposition j k I n F(Tn;Tn+1;Tn+2) = 2 (Tn + Tn+1 + Tn+2 − 1) − 1. A. M. Robles-Perez´ (UGR) Frobenius number and binomial sequences IMNS-2018 12 / 29 Triangular numbers sequences: obstacles I Two consecutive triangular numbers are not relatively prime. I If n = 4 or n ≥ 6, then 0 ! ! ! !1 B n + 1 n + 2 n + 3 n + 4 C B ; ; ; C @B 2 2 2 2 AC is not a telescopic sequence (of four consecutive relatively prime triangular numbers). ∗ In fact, none of its permutations is telescopic. A. M. Robles-Perez´ (UGR) Frobenius number and binomial sequences IMNS-2018 13 / 29 Tetrahedral numbers sequences ! n + 2 n(n + 1)(n + 2) TH = = ; n ≥ 1: n 3 6 Lemma I gcdfTHn;THn+1;THn+2;THn+3g = 1: Proof ∗ (THn+1 − THn;THn+2 − THn+1;THn+3 − THn+2) = (Tn;Tn+1;Tn+2): ∗ Fact • Let (a1;a2;:::;an) be a sequence of positive integers. • Let d1 = gcdfa1;a2;:::;ang and d2 = gcdfa2 − a1;:::;an − an−1g. • Then d1jd2. In particular, if d2 = 1, then d1 = 1. A. M. Robles-Perez´ (UGR) Frobenius number and binomial sequences IMNS-2018 14 / 29 Tetrahedral numbers sequences Lemma 8 > (6k + 1)(3k + 1) if n = 6k; > > (3k + 1)(2k + 1) if n = 6k + 1; > <> (2k + 1)(3k + 2) if n = 6k + 2; I gcd THn;THn+1 = > (3k + 2)(6k + 5) if n = 6k + 3; > > (6k + 5)(k + 1) if n = 6k + 4; > :> (k + 1)(6k + 7) if n = 6k + 5: Lemma 8 > 3k + 1 if n = 6k; > > 2k + 1 if n = 6k + 1; > <> 3k + 2 if n = 6k + 2; I gcd THn;THn+1;THn+2 = > 6k + 5 if n = 6k + 3; > > k + 1 if n = 6k + 4; > :> 6k + 7 if n = 6k + 5: A. M. Robles-Perez´ (UGR) Frobenius number and binomial sequences IMNS-2018 15 / 29 Tetrahedral numbers sequences Proposition I (THn;THn+1;THn+2;THn+3) is telescopic if and only if n ≡ r mod 6 with r 2 f0;1;2;3g. Proposition I (THn+3;THn+2;THn+1;THn) is telescopic if and only if n ≡ r mod 6 with r 2 f4;5g. Proposition 8 n−3 n TH + + nTH + + TH + − TH if n = 6k; > 3 n 1 n 2 2 n 3 n > n−1 n−1 > (n − 1)TH + + TH + + TH + − TH if n = 6k + 1; > n 1 2 n 2 3 n 3 n > n−2 n > (n − 1)TH + + TH + + TH + − TH if n = 6k + 2; I <> n 1 3 n 2 2 n 3 n F(THn;THn+1;THn+2) = > n−3 TH + n−1 TH + (n + 1)TH − TH if n = 6k + 3; > 3 n+1 2 n+2 n+3 n > n+2 n+2 > THn+2 + THn+1 + (n + 2)THn − THn+3 if n = 6k + 4; > 3 2 :> n+1 n+1 (n + 4)THn+2 + 3 THn+1 + 2 THn − THn+3 if n = 6k + 5: A. M. Robles-Perez´ (UGR) Frobenius number and binomial sequences IMNS-2018 16 / 29 Tetrahedral numbers sequences Proposition 8 1 (11n4 + 90n3 + 265n2 + 258n − 36) if n = 6k; > 36 > 1 4 3 2 > (11n + 70n + 133n − 22n − 228) if n = 6k + 1; > 36 > 1 (11n4 + 74n3 + 169n2 + 82n − 132) if n = 6k + 2; I <> 36 F(THn;THn+1;THn+2) = > 1 (11n4 + 102n3 + 373n2 + 570n + 252) if n = 6k + 3; > 36 > > 1 (11n4 + 70n3 + 133n2 − 22n − 228) if n = 6k + 4; > 36 :> 1 4 3 2 36 (11n + 98n + 349n + 526n + 228) if n = 6k + 5: Proposition 8 4 3 2 > 396k + 540k + 265k + 43k − 1 if n = 6k; > > 396k 4 + 684k 3 + 409k 2 + 83k − 1 if n = 6k + 1; > > 396k 4 + 972k 3 + 877k 2 + 333k + 41 if n = 6k + 2; I <> F(THn;THn+1;THn+2) = > 396k 4 + 1404k 3 + 1885k 2 + 1125k + 249 if n = 6k + 3; > > 4 3 2 > 396k + 1476k + 2029k + 1203k + 253 if n = 6k + 4; > :> 396k 4 + 1908k 3 + 3469k 2 + 2811k + 853 if n = 6k + 5: A.
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