Ramanujan-Style Congruences for primes up to, 23 for the eta-function raised up to the power, 20 By Shalosh B. Ekhad ----------------------------------------------------- Theorem Number, 1 (5 n + 4) For each non-neg. integer n, the coefficients of, q , in infinity --------' ' | | 1 | | ------ | | i | | 1 - q i = 1 is divisible by , 5 Proof: This theorem is not that deep and a Ramanujan-style proof suffices. Let : infinity --------' ' | | i E(q) = | | (1 - q ) | | | | i = 1 Thanks to Jacobi, we have infinity /(n + 1) n\ ----- |---------| 3 \ n \ 2 / E(q) = ) (-1) (2 n + 1) q / ----- n = 0 by considering all the residue classes modulo, 5 (n + 1) n of , ---------, such that, 2 n + 1, is not 0 mod, 5 2 we easily see that only the residue classes, [0, 1], show up so we can write: 3 E(q) = J[0] + J[1] where J[i] consists of those terms in which the power of q is congruent to i mod, 5 Now modulo, 5 1 3 ----, equals , E(q) , to the power , 3 E(q) 5 divided by, E(q ), to the power, 2 that equals 3 (J[0] + J[1]) 5 times a formal power series in powers of, q if we expand, and extract the powers that are congruent to 4, modulo , 5 none of them show up!, QED! ----------------------------------------------------- Theorem Number, 2 (7 n + 5) For each non-neg. integer n, the coefficients of, q , in infinity --------' ' | | 1 | | ------ | | i | | 1 - q i = 1 is divisible by , 7 Proof: This theorem is not that deep and a Ramanujan-style proof suffices. Let : infinity --------' ' | | i E(q) = | | (1 - q ) | | | | i = 1 Thanks to Jacobi, we have infinity /(n + 1) n\ ----- |---------| 3 \ n \ 2 / E(q) = ) (-1) (2 n + 1) q / ----- n = 0 by considering all the residue classes modulo, 7 (n + 1) n of , ---------, such that, 2 n + 1, is not 0 mod, 7 2 we easily see that only the residue classes, [0, 1, 3], show up so we can write: 3 E(q) = J[0] + J[1] + J[3] where J[i] consists of those terms in which the power of q is congruent to i mod, 7 Now modulo, 7 1 3 ----, equals , E(q) , to the power , 2 E(q) 7 divided by, E(q ), to the power, 1 that equals 2 (J[0] + J[1] + J[3]) 7 times a formal power series in powers of, q if we expand, and extract the powers that are congruent to 5, modulo , 7 none of them show up!, QED! ----------------------------------------------------- Theorem Number, 3 (11 n + 6) For each non-neg. integer n, the coefficients of, q , in infinity --------' ' | | 1 | | ------ | | i | | 1 - q i = 1 is divisible by , 11 Proof: We will give a Hirschhorn-style proof. Let : infinity --------' ' | | i E(q) = | | (1 - q ), that equals, thanks to Euler to | | | | i = 1 infinity /(3 n + 1) n\ ----- |-----------| \ n \ 2 / ) (-1) q / ----- n = -infinity by considering all the residue classes modulo, 11 (3 n + 1) n of , ----------- 2 we easily see that only the residue classes, {0, 1, 2, 4, 5, 7}, show up So we can write E(q) = E[0] + E[1] + E[2] + E[4] + E[5] + E[7] where E[i] consists of those terms in which the power of q is congruent to i mod, 11 Thanks to Jacobi, we have infinity /(n + 1) n\ ----- |---------| 3 \ n \ 2 / E(q) = ) (-1) (2 n + 1) q / ----- n = 0 by considering all the residue classes modulo, 11 (n + 1) n of , ---------, such that, 2 n + 1, is not 0 mod, 11 2 we easily see that only the residue classes, [0, 1, 3, 6, 10], show up so we can write: 3 E(q) = J[0] + J[1] + J[3] + J[6] + J[10] where J[i] consists of those terms in which the power of q is congruent to i mod, 11 Now 3 E(q) , to the power, 4, equals 12 E(q) 11 that equals E(q) times, E(q ), to the power, 1 Hence, modulo , 11 4 (J[0] + J[1] + J[3] + J[6] + J[10]) = 11 E(q ) (E[0] + E[1] + E[2] + E[4] + E[5] + E[7]) If we consider the powers of q congruent to, 3, 6, 8, 9, 10 we find that, modulo, 11 3 3 2 2 4 J[0] J[3] + 4 J[0] J[1] + 6 J[1] J[6] + 2 J[0] J[1] J[3] J[10] 2 3 + J[3] J[6] J[10] + 4 J[6] J[10] = 0 3 2 2 3 4 J[0] J[6] + 6 J[0] J[3] + 4 J[1] J[3] + 2 J[0] J[1] J[6] J[10] 2 3 + J[1] J[3] J[10] + 4 J[6] J[10] = 0 2 2 2 3 3 6 J[1] J[3] + J[0] J[1] J[6] + 4 J[1] J[6] + 4 J[3] J[10] 3 + 2 J[0] J[3] J[6] J[10] + 4 J[0] J[10] = 0 3 3 2 2 2 4 J[0] J[3] + 4 J[1] J[6] + J[0] J[3] J[6] + 6 J[0] J[10] 3 + 2 J[1] J[3] J[6] J[10] + 4 J[1] J[10] = 0 3 3 3 4 J[0] J[10] + 4 J[1] J[3] + 2 J[0] J[1] J[3] J[6] + 4 J[3] J[6] 2 2 2 + J[0] J[1] J[10] + 6 J[6] J[10] = 0 Now modulo, 11 1 3 ----, equals , E(q) , to the power , 7 E(q) 11 divided by, E(q ), to the power, 2 that equals 7 (J[0] + J[1] + J[3] + J[6] + J[10]) 11 times a formal power series in powers of, q if we expand, and extract the powers that are congruent to 6, modulo , 11 11 we get a formal power series in , q , times a polynomial,P, of (total) degree, 7, in , [J[0], J[1], J[3], J[6], J[10]] where P equals 3 3 3 3 3 3 8 J[0] J[1] J[3] + 8 J[0] J[6] J[10] + 8 J[1] J[3] J[6] 3 3 3 3 6 + 8 J[0] J[3] J[10] + 8 J[1] J[6] J[10] + 7 J[0] J[6] 5 2 6 5 2 2 5 + 10 J[0] J[3] + 7 J[0] J[1] + 10 J[1] J[6] + 10 J[1] J[3] 6 6 2 5 5 2 + 7 J[3] J[10] + 7 J[3] J[6] + 10 J[0] J[10] + 10 J[6] J[10] 6 4 2 3 + 7 J[1] J[10] + J[1] J[3] J[6] J[10] + 2 J[0] J[3] J[6] J[10] 3 2 2 2 2 + 2 J[0] J[1] J[6] J[10] + 3 J[1] J[3] J[6] J[10] 2 2 2 2 2 2 + 3 J[0] J[1] J[6] J[10] + 3 J[0] J[3] J[6] J[10] 2 2 2 4 4 + 3 J[0] J[1] J[3] J[10] + J[0] J[1] J[3] J[6] + J[0] J[3] J[6] J[10] 3 2 4 + 2 J[1] J[3] J[6] J[10] + J[0] J[1] J[6] J[10] 3 2 2 2 2 + 2 J[0] J[1] J[3] J[10] + 3 J[0] J[1] J[3] J[6] 4 2 3 + J[0] J[1] J[3] J[10] + 2 J[0] J[1] J[3] J[6] Now it is easily verified that, modulo, 11, P equals 3 2 3 3 (10 J[1] + 4 J[0] J[3] + 10 J[1] J[3] J[10]) (4 J[0] J[3] + 4 J[0] J[1] 2 2 2 + 6 J[1] J[6] + 2 J[0] J[1] J[3] J[10] + J[3] J[6] J[10] 3 3 2 + 4 J[6] J[10] ) + (10 J[0] + 10 J[10] J[0] J[1] + 4 J[10] J[6] ) ( 3 2 2 3 4 J[0] J[6] + 6 J[0] J[3] + 4 J[1] J[3] + 2 J[0] J[1] J[6] J[10] 2 3 + J[1] J[3] J[10] + 4 J[6] J[10]) + 3 2 2 2 (10 J[3] + 10 J[0] J[3] J[6] + 4 J[0] J[10] ) (6 J[1] J[3] 2 3 3 + J[0] J[1] J[6] + 4 J[1] J[6] + 4 J[3] J[10] + 2 J[0] J[3] J[6] J[10] 3 2 3 + 4 J[0] J[10] ) + (4 J[1] J[6] + 10 J[3] J[6] J[10] + 10 J[10] ) ( 3 3 2 2 2 4 J[0] J[3] + 4 J[1] J[6] + J[0] J[3] J[6] + 6 J[0] J[10] 3 + 2 J[1] J[3] J[6] J[10] + 4 J[1] J[10] ) + 2 3 3 3 (4 J[1] J[3] + 10 J[0] J[1] J[6] + 10 J[6] ) (4 J[0] J[10] + 4 J[1] J[3] 3 2 2 2 + 2 J[0] J[1] J[3] J[6] + 4 J[3] J[6] + J[0] J[1] J[10] + 6 J[6] J[10] ) that is 0 mod , 11, QED! ----------------------------------------------------- Theorem Number, 4 (5 n + 2) For each non-neg. integer n, the coefficients of, q , in infinity --------' ' | | 1 | | --------- | | i 2 | | (1 - q ) i = 1 is divisible by , 5 Proof: This theorem is not that deep and a Ramanujan-style proof suffices. Let : infinity --------' ' | | i E(q) = | | (1 - q ) | | | | i = 1 Thanks to Jacobi, we have infinity /(n + 1) n\ ----- |---------| 3 \ n \ 2 / E(q) = ) (-1) (2 n + 1) q / ----- n = 0 by considering all the residue classes modulo, 5 (n + 1) n of , ---------, such that, 2 n + 1, is not 0 mod, 5 2 we easily see that only the residue classes, [0, 1], show up so we can write: 3 E(q) = J[0] + J[1] where J[i] consists of those terms in which the power of q is congruent to i mod, 5 Now modulo, 5 1 3 -----, equals , E(q) , to the power , 1 2 E(q) 5 divided by, E(q ), to the power, 1 that equals J[0] + J[1] 5 times a formal power series in powers of, q if we expand, and extract the powers that are congruent to 2, modulo , 5 none of them show up!, QED! ----------------------------------------------------- Theorem Number, 5 (5 n + 3) For each non-neg. integer n, the coefficients of, q , in infinity --------' ' | | 1 | | --------- | | i 2 | | (1 - q ) i = 1 is divisible by , 5 Proof: This theorem is not that deep and a Ramanujan-style proof suffices. Let : infinity --------' ' | | i E(q) = | | (1 - q ) | | | | i = 1 Thanks to Jacobi, we have infinity /(n + 1) n\ ----- |---------| 3 \ n \ 2 / E(q) = ) (-1) (2 n + 1) q / ----- n = 0 by considering all the residue classes modulo, 5 (n + 1) n of , ---------, such that, 2 n + 1, is not 0 mod, 5 2 we easily see that only the residue classes, [0, 1], show up so we can write: 3 E(q) = J[0] + J[1] where J[i] consists of those terms in which the power of q is congruent to i mod, 5 Now modulo, 5 1 3 -----, equals , E(q) , to the power , 1 2 E(q) 5 divided by, E(q ), to the power, 1 that equals J[0] + J[1] 5 times a formal power series in powers of, q if we expand, and extract the powers that are congruent to 3, modulo , 5 none of them show up!, QED! ----------------------------------------------------- Theorem Number, 6 (5 n + 4) For each non-neg. integer n, the coefficients of, q , in infinity --------' ' | | 1 | | --------- | | i 2 | | (1 - q ) i = 1 is divisible by , 5 Proof: This theorem is not that deep and a Ramanujan-style proof suffices. Let : infinity --------' ' | | i E(q) = | | (1 - q ) | | | | i = 1 Thanks to Jacobi, we have infinity /(n + 1) n\ ----- |---------| 3 \ n \ 2 / E(q) = ) (-1) (2 n + 1) q / ----- n = 0 by considering all the residue classes modulo, 5 (n + 1) n of , ---------, such that, 2 n + 1, is not 0 mod, 5 2 we easily see that only the residue classes, [0, 1], show up so we can write: 3 E(q) = J[0] + J[1] where J[i] consists of those terms in which the power of q is congruent to i mod, 5 Now modulo, 5 1 3 -----, equals , E(q) , to the power , 1 2 E(q) 5 divided by, E(q ), to the power, 1 that equals J[0] + J[1] 5 times a formal power series in powers of, q if we expand, and extract the powers that are congruent to 4, modulo , 5 none of them show up!, QED! ----------------------------------------------------- Theorem Number, 7 (11 n + 7) For each non-neg. integer n, the coefficients of, q , in infinity --------' ' | | 1 | | --------- | | i 3 | | (1 - q ) i = 1 is divisible by , 11 Proof: We will give a Hirschhorn-style proof. Let : infinity --------' ' | | i E(q) = | | (1 - q ), that equals, thanks to Euler to | | | | i = 1 infinity /(3 n + 1) n\ ----- |-----------| \ n \ 2 / ) (-1) q / ----- n = -infinity by considering all the residue classes modulo, 11 (3 n + 1) n of , ----------- 2 we easily see that only the residue classes, {0, 1, 2, 4, 5, 7}, show up So we can write E(q) = E[0] + E[1] + E[2] + E[4] + E[5] + E[7] where E[i] consists of those terms in which the power of q is congruent to i mod, 11 Thanks to Jacobi, we have infinity /(n + 1) n\ ----- |---------| 3 \ n \ 2 / E(q) = ) (-1) (2 n + 1) q / ----- n = 0 by considering all the residue classes modulo, 11 (n + 1) n of , ---------, such that, 2 n + 1, is not 0 mod, 11 2 we easily see that only the residue classes, [0, 1, 3, 6, 10], show up so we can write: 3 E(q) = J[0] + J[1] + J[3] + J[6] + J[10] where J[i] consists of those terms in which the power of q is congruent to i mod, 11 Now 3 E(q) , to the power, 4, equals 12 E(q) 11 that equals E(q) times, E(q ), to the power, 1 Hence, modulo , 11 4 (J[0] + J[1] + J[3] + J[6] + J[10]) = 11 E(q ) (E[0] + E[1] + E[2] + E[4] + E[5] + E[7]) If we consider the powers of q congruent to, 3, 6, 8, 9, 10 we find that, modulo, 11 3 3 2 2 4 J[0] J[3] + 4 J[0] J[1] + 6 J[1] J[6] + 2 J[0] J[1] J[3] J[10] 2 3 + J[3] J[6] J[10] + 4 J[6] J[10] = 0 3 2 2 3 2 4 J[0] J[6] + 6 J[0] J[3] + 4 J[1] J[3] + J[1] J[3] J[10] 3 + 2 J[0] J[1] J[6] J[10] + 4 J[6] J[10] = 0 2 2 2 3 3 6 J[1] J[3] + J[0] J[1] J[6] + 4 J[1] J[6] + 4 J[3] J[10] 3 + 2 J[0] J[3] J[6] J[10] + 4 J[0] J[10] = 0 3 3 2 2 2 4 J[0] J[3] + 4 J[1] J[6] + J[0] J[3] J[6] + 6 J[0] J[10] 3 + 2 J[1] J[3] J[6] J[10] + 4 J[1] J[10] = 0 3 3 2 4 J[0] J[10] + 4 J[1] J[3] + 2 J[0] J[1] J[3] J[6] + J[0] J[1] J[10] 3 2 2 + 4 J[3] J[6] + 6 J[6] J[10] = 0 Now modulo, 11 1 3 -----, equals , E(q) , to the power , 10 3 E(q) 11 divided by, E(q ), to the power, 3 that equals 10 (J[0] + J[1] + J[3] + J[6] + J[10]) 11 times a formal power series in powers of, q if we expand, and extract the powers that are congruent to 7, modulo , 11 11 we get a formal power series in , q , times a polynomial,P, of (total) degree, 10, in , [J[0], J[1], J[3], J[6], J[10]] where P equals 2 6 2 6 2 2 5 J[0] J[1] J[6] J[10] + J[1] J[3] J[6] J[10] + 3 J[0] J[3] J[6] J[10] 3 5 3 2 4 + 2 J[1] J[3] J[6] J[10] + 5 J[1] J[3] J[6] J[10] 3 3 3 2 5 2 + 3 J[0] J[3] J[6] J[10] + 3 J[0] J[1] J[6] J[10] 2 3 2 3 3 3 3 + 10 J[0] J[3] J[6] J[10] + 3 J[0] J[1] J[6] J[10] 2 3 4 5 3 + 5 J[0] J[1] J[3] J[10] + 2 J[0] J[3] J[6] J[10] 3 2 2 3 2 6 + 10 J[1] J[3] J[6] J[10] + J[1] J[3] J[6] J[10] 4 3 2 3 4 + 5 J[1] J[3] J[6] J[10] + 10 J[0] J[1] J[3] J[6] J[10] 2 2 4 2 4 2 + 4 J[0] J[1] J[3] J[6] J[10] + 4 J[0] J[1] J[3] J[6] J[10] 3 7 4 6 7 3 3 7 + 10 J[1] J[10] + J[6] J[10] + 10 J[3] J[10] + 10 J[3] J[6] 6 4 4 6 5 4 2 7 + J[0] J[10] + J[1] J[6] + 6 J[3] J[6] J[10] + 8 J[0] J[3] J[10] 8 3 2 5 2 2 6 + 2 J[1] J[3] J[10] + J[0] J[6] J[10] + 6 J[0] J[1] J[10] 4 5 7 2 2 5 3 + 6 J[0] J[3] J[10] + 8 J[0] J[6] J[10] + J[1] J[6] J[10] 2 6 2 4 5 8 + 6 J[3] J[6] J[10] + 6 J[0] J[1] J[10] + 2 J[0] J[3] J[6] 5 4 4 5 2 6 2 + 6 J[1] J[6] J[10] + 6 J[0] J[6] J[10] + 6 J[1] J[3] J[10] 5 4 8 5 3 2 + 6 J[1] J[3] J[6] + 2 J[3] J[6] J[10] + J[0] J[3] J[10] 5 4 7 2 5 2 3 + 6 J[0] J[1] J[6] + 8 J[1] J[6] J[10] + J[1] J[3] J[6] 4 5 2 7 2 6 2 + 6 J[1] J[3] J[10] + 8 J[1] J[3] J[6] + 6 J[0] J[1] J[6] 2 3 5 8 6 2 2 + J[0] J[1] J[3] + 2 J[0] J[1] J[10] + 6 J[0] J[3] J[6] 5 4 8 7 2 + 6 J[0] J[3] J[6] + 2 J[0] J[1] J[6] + 8 J[0] J[1] J[3] 5 4 3 3 3 + 6 J[0] J[1] J[3] + 3 J[0] J[1] J[3] J[10] 3 2 4 2 4 3 + 5 J[0] J[3] J[6] J[10] + 5 J[0] J[1] J[6] J[10] 2 6 2 4 3 + J[0] J[1] J[3] J[6] + 5 J[0] J[3] J[6] J[10] 4 2 3 3 5 + 5 J[0] J[1] J[3] J[10] + 2 J[0] J[1] J[3] J[6] 6 2 5 2 2 + J[0] J[3] J[6] J[10] + 3 J[1] J[3] J[6] J[10] 3 3 3 3 2 3 2 + 3 J[1] J[3] J[6] J[10] + 10 J[0] J[1] J[6] J[10] 6 2 5 3 + J[0] J[3] J[6] J[10] + 2 J[0] J[1] J[6] J[10] 4 3 2 3 3 2 2 + 5 J[0] J[1] J[6] J[10] + 10 J[0] J[1] J[3] J[10] 3 2 4 2 2 3 3 + 5 J[0] J[1] J[3] J[6] + 10 J[0] J[1] J[3] J[6] 2 6 6 2 + J[0] J[1] J[3] J[10] + J[0] J[1] J[3] J[10] 2 5 2 3 5 + 3 J[0] J[1] J[3] J[6] + 2 J[0] J[1] J[3] J[10] 3 3 3 6 2 + 3 J[0] J[1] J[3] J[6] + J[0] J[1] J[3] J[6] 6 2 5 2 2 + J[0] J[1] J[6] J[10] + 3 J[0] J[1] J[3] J[10] 4 3 2 7 3 4 6 6 4 + 5 J[0] J[1] J[3] J[6] + 10 J[0] J[6] + J[0] J[3] + J[1] J[3] 3 7 4 3 + 10 J[0] J[1] + 10 J[0] J[1] J[3] J[6] J[10] 2 2 2 2 2 3 4 + J[0] J[1] J[3] J[6] J[10] + 10 J[0] J[1] J[3] J[6] J[10] 2 4 2 4 2 2 + 4 J[0] J[1] J[3] J[6] J[10] + 4 J[0] J[1] J[3] J[6] J[10] 3 4 2 4 2 + 10 J[0] J[1] J[3] J[6] J[10] + 4 J[0] J[1] J[3] J[6] J[10] 4 3 + 10 J[0] J[1] J[3] J[6] J[10] Now it is easily verified that, modulo, 11, P equals 2 4 4 5 3 2 (8 J[0] J[1] + 4 J[0] J[1] J[3] + J[0] J[3] + 3 J[1] J[3] J[6] 2 3 3 2 3 2 + J[0] J[3] J[6] + 8 J[0] J[1] J[6] + 8 J[0] J[3] J[6] 2 4 5 2 2 + 4 J[1] J[6] + 6 J[10] J[1] + 9 J[10] J[0] J[1] J[3] 3 2 + 4 J[10] J[1] J[3] J[6] + 4 J[10] J[0] J[1] J[3] J[6] 4 2 3 2 2 2 + 9 J[10] J[3] J[6] + 7 J[10] J[1] J[3] + 5 J[10] J[0] J[3] 3 3 3 3 3 + 3 J[10] J[0] J[1] J[6] + 3 J[10] J[6] ) (4 J[0] J[3] + 4 J[0] J[1] 2 2 2 + 6 J[1] J[6] + 2 J[0] J[1] J[3] J[10] + J[3] J[6] J[10] 3 5 3 3 2 4 + 4 J[6] J[10] ) + (6 J[0] J[1] + 6 J[1] J[3] + 9 J[0] J[3] 3 3 2 4 2 + 7 J[0] J[1] J[3] J[6] + 4 J[0] J[3] J[6] + 8 J[0] J[6] 3 2 4 2 + 2 J[10] J[0] J[1] + 8 J[10] J[1] J[3] + 8 J[10] J[0] J[1] J[3] J[6] 2 2 2 3 4 + 4 J[10] J[0] J[1] J[6] + J[10] J[3] J[6] + 4 J[10] J[0] J[6] 2 3 2 3 2 2 2 + 5 J[10] J[0] J[1] + 10 J[10] J[0] J[3] + J[10] J[1] J[6] 3 3 2 2 3 + 6 J[10] J[0] J[1] J[3]) (4 J[0] J[6] + 6 J[0] J[3] + 4 J[1] J[3] 2 3 + J[1] J[3] J[10] + 2 J[0] J[1] J[6] J[10] + 4 J[6] J[10]) + ( 4 2 2 3 4 8 J[1] J[3] + 7 J[0] J[1] J[3] + 2 J[0] J[1] J[6] 3 5 3 2 + 5 J[0] J[1] J[3] J[6] + 6 J[3] J[6] + 8 J[0] J[3] J[6] 3 3 2 3 5 2 3 + 9 J[1] J[6] + 3 J[0] J[3] J[6] + 3 J[10] J[0] + 4 J[10] J[1] J[3] 2 2 3 2 4 + 9 J[10] J[0] J[1] J[3] J[6] + 7 J[10] J[0] J[1] + 8 J[10] J[3] 2 2 2 2 2 3 2 + 6 J[10] J[0] J[3] J[6] + 3 J[10] J[0] J[6] + 6 J[10] J[0] J[1] 4 2 2 2 3 + 4 J[0] J[3] J[10] ) (6 J[1] J[3] + J[0] J[1] J[6] + 4 J[1] J[6] 3 3 + 4 J[3] J[10] + 2 J[0] J[3] J[6] J[10] + 4 J[0] J[10] ) + ( 3 2 3 3 2 3 4 J[0] J[1] J[3] + 5 J[0] J[3] + 3 J[0] J[1] J[6] + 9 J[0] J[3] J[6] 2 2 2 2 3 5 + 3 J[1] J[3] J[6] + 6 J[0] J[1] J[6] + 10 J[1] J[6] 3 4 2 4 + 6 J[10] J[0] J[1] J[3] + 4 J[10] J[6] J[1] + 8 J[1] J[10] 5 3 3 2 + 6 J[3] J[10] ) (4 J[0] J[3] + 4 J[1] J[6] + J[0] J[3] J[6] 2 2 3 + 6 J[0] J[10] + 2 J[1] J[3] J[6] J[10] + 4 J[1] J[10] ) + 5 4 2 4 5 (7 J[1] J[3] + 4 J[1] J[3] J[6] + 8 J[3] J[6] + 6 J[0] J[6] ) ( 3 3 2 4 J[0] J[10] + 4 J[1] J[3] + 2 J[0] J[1] J[3] J[6] + J[0] J[1] J[10] 3 2 2 + 4 J[3] J[6] + 6 J[6] J[10] ) that is 0 mod , 11, QED! ----------------------------------------------------- Conjecture Number, 1 (17 n + 15) The coefficient of, q , of infinity --------' ' | | 1 | | ---------- | | i1 3 | | (1 - q ) i1 = 1 is always divisible by , 17 ----------------------------------------------------- Theorem Number, 8 (11 n + 8) For each non-neg. integer n, the coefficients of, q , in infinity --------' ' | | 1 | | --------- | | i 5 | | (1 - q ) i = 1 is divisible by , 11 Proof: This theorem is not that deep and a Ramanujan-style proof suffices. Let : infinity --------' ' | | i E(q) = | | (1 - q ) | | | | i = 1 Thanks to Jacobi, we have infinity /(n + 1) n\ ----- |---------| 3 \ n \ 2 / E(q) = ) (-1) (2 n + 1) q / ----- n = 0 by considering all the residue classes modulo, 11 (n + 1) n of , ---------, such that, 2 n + 1, is not 0 mod, 11 2 we easily see that only the residue classes, [0, 1, 3, 6, 10], show up so we can write: 3 E(q) = J[0] + J[1] + J[3] + J[6] + J[10] where J[i] consists of those terms in which the power of q is congruent to i mod, 11 Now modulo, 11 1 3 -----, equals , E(q) , to the power , 2 5 E(q) 11 divided by, E(q ), to the power, 1 that equals 2 (J[0] + J[1] + J[3] + J[6] + J[10]) 11 times a formal power series in powers of, q if we expand, and extract the powers that are congruent to 8, modulo , 11 none of them show up!, QED! ----------------------------------------------------- Conjecture Number, 2 (19 n + 9) The coefficient of, q , of infinity --------' ' | | 1 | | ---------- | | i1 7 | | (1 - q ) i1 = 1 is always divisible by , 19 ----------------------------------------------------- Conjecture Number, 3 (19 n + 17) The coefficient of, q , of infinity --------' ' | | 1 | | ---------- | | i1 9 | | (1 - q ) i1 = 1 is always divisible by , 19 ----------------------------------------------------- We discovered, 11, interesting congruences Unfortunately, we were unable to prove, 3, of them. On the other hand, we found, 6, Ramanujan-style relatively easy, proofs, and, 2, deeper Hirschhorn-style proofs To sum up the following 3-element list of lists consisting of pairs [a,[r,p]] for which the statement that for all n>=0, the coeffcient of, q^(n*p+r) in the power series expansion of Product(1/(1-q^i)^a,i=1..infinity) is divisible by p is: (i) probably true, but we were unable to prove it (ii) true and we found a s\ imple Ramanujan-style proof (iii) true and we found a more complicated Hirschhorn-style proof [[[3, [15, 17]], [7, [9, 19]], [9, [17, 19]]], [[1, [4, 5]], [1, [5, 7]], [2, [ 2, 5]], [2, [3, 5]], [2, [4, 5]], [5, [8, 11]]], [[1, [6, 11]], [3, [7, 11]]]] This finishes this gripping article, that took, 441.402, seconds to produce.