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Euler's Pentagonal Number Theorem

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Euler's Pentagonal Number Theorem Euler's Pentagonal Number Theorem Dan Cranston September 28, 2011 Triangular Numbers:1 ; 3; 6; 10; 15; 21; 28; 36; 45; 55; ::: Square Numbers:1 ; 4; 9; 16; 25; 36; 49; 64; 81; 100; ::: Pentagonal Numbers:1 ; 5; 12; 22; 35; 51; 70; 92; 117; 145; ::: Introduction Square Numbers:1 ; 4; 9; 16; 25; 36; 49; 64; 81; 100; ::: Pentagonal Numbers:1 ; 5; 12; 22; 35; 51; 70; 92; 117; 145; ::: Introduction Triangular Numbers:1 ; 3; 6; 10; 15; 21; 28; 36; 45; 55; ::: Pentagonal Numbers:1 ; 5; 12; 22; 35; 51; 70; 92; 117; 145; ::: Introduction Triangular Numbers:1 ; 3; 6; 10; 15; 21; 28; 36; 45; 55; ::: Square Numbers:1 ; 4; 9; 16; 25; 36; 49; 64; 81; 100; ::: Introduction Triangular Numbers:1 ; 3; 6; 10; 15; 21; 28; 36; 45; 55; ::: Square Numbers:1 ; 4; 9; 16; 25; 36; 49; 64; 81; 100; ::: Pentagonal Numbers:1 ; 5; 12; 22; 35; 51; 70; 92; 117; 145; ::: The kth pentagonal number, P(k), is the kth partial sum of the arithmetic sequence an = 1 + 3(n − 1) = 3n − 2. k X 3k2 − k P(k) = (3n − 2) = 2 n=1 I P(8) = 92, P(500) = 374; 750, etc. and P(0) = 0. I Extend domain, so P(−8) = 100, P(−500) = 375; 250, etc. I fP(0); P(1); P(−1); P(2); P(−2); :::g = f0; 1; 2; 5; 7; :::g is an increasing sequence. Generalized Pentagonal Numbers k X 3k2 − k P(k) = (3n − 2) = 2 n=1 I P(8) = 92, P(500) = 374; 750, etc. and P(0) = 0. I Extend domain, so P(−8) = 100, P(−500) = 375; 250, etc. I fP(0); P(1); P(−1); P(2); P(−2); :::g = f0; 1; 2; 5; 7; :::g is an increasing sequence. Generalized Pentagonal Numbers The kth pentagonal number, P(k), is the kth partial sum of the arithmetic sequence an = 1 + 3(n − 1) = 3n − 2. I P(8) = 92, P(500) = 374; 750, etc. and P(0) = 0. I Extend domain, so P(−8) = 100, P(−500) = 375; 250, etc. I fP(0); P(1); P(−1); P(2); P(−2); :::g = f0; 1; 2; 5; 7; :::g is an increasing sequence. Generalized Pentagonal Numbers The kth pentagonal number, P(k), is the kth partial sum of the arithmetic sequence an = 1 + 3(n − 1) = 3n − 2. k X 3k2 − k P(k) = (3n − 2) = 2 n=1 and P(0) = 0. I Extend domain, so P(−8) = 100, P(−500) = 375; 250, etc. I fP(0); P(1); P(−1); P(2); P(−2); :::g = f0; 1; 2; 5; 7; :::g is an increasing sequence. Generalized Pentagonal Numbers The kth pentagonal number, P(k), is the kth partial sum of the arithmetic sequence an = 1 + 3(n − 1) = 3n − 2. k X 3k2 − k P(k) = (3n − 2) = 2 n=1 I P(8) = 92, P(500) = 374; 750, etc. I Extend domain, so P(−8) = 100, P(−500) = 375; 250, etc. I fP(0); P(1); P(−1); P(2); P(−2); :::g = f0; 1; 2; 5; 7; :::g is an increasing sequence. Generalized Pentagonal Numbers The kth pentagonal number, P(k), is the kth partial sum of the arithmetic sequence an = 1 + 3(n − 1) = 3n − 2. k X 3k2 − k P(k) = (3n − 2) = 2 n=1 I P(8) = 92, P(500) = 374; 750, etc. and P(0) = 0. I fP(0); P(1); P(−1); P(2); P(−2); :::g = f0; 1; 2; 5; 7; :::g is an increasing sequence. Generalized Pentagonal Numbers The kth pentagonal number, P(k), is the kth partial sum of the arithmetic sequence an = 1 + 3(n − 1) = 3n − 2. k X 3k2 − k P(k) = (3n − 2) = 2 n=1 I P(8) = 92, P(500) = 374; 750, etc. and P(0) = 0. I Extend domain, so P(−8) = 100, P(−500) = 375; 250, etc. Generalized Pentagonal Numbers The kth pentagonal number, P(k), is the kth partial sum of the arithmetic sequence an = 1 + 3(n − 1) = 3n − 2. k X 3k2 − k P(k) = (3n − 2) = 2 n=1 I P(8) = 92, P(500) = 374; 750, etc. and P(0) = 0. I Extend domain, so P(−8) = 100, P(−500) = 375; 250, etc. I fP(0); P(1); P(−1); P(2); P(−2); :::g = f0; 1; 2; 5; 7; :::g is an increasing sequence. Let p(n) denote the number of partitions of n. I 3 = 2+1 = 1+1+1, so p(3) = 3. I 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1, so p(4) = 5. I 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p(5) = 7. I 6 = 5+1 = 4+1+1 = 4+2 = 3+1+1+1 = 3+3 = 3+2+1 = 2+1+1+1+1 = 2+2+2 = 2+2+1+1 = 1+1+1+1+1+1, so p(6) = 11. Each summand in a certain partition is called a part. So3 has1 part,2+1 has2 parts, and1+1+1 has3 parts. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers. I 3 = 2+1 = 1+1+1, so p(3) = 3. I 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1, so p(4) = 5. I 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p(5) = 7. I 6 = 5+1 = 4+1+1 = 4+2 = 3+1+1+1 = 3+3 = 3+2+1 = 2+1+1+1+1 = 2+2+2 = 2+2+1+1 = 1+1+1+1+1+1, so p(6) = 11. Each summand in a certain partition is called a part. So3 has1 part,2+1 has2 parts, and1+1+1 has3 parts. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers. Let p(n) denote the number of partitions of n. so p(3) = 3. I 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1, so p(4) = 5. I 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p(5) = 7. I 6 = 5+1 = 4+1+1 = 4+2 = 3+1+1+1 = 3+3 = 3+2+1 = 2+1+1+1+1 = 2+2+2 = 2+2+1+1 = 1+1+1+1+1+1, so p(6) = 11. Each summand in a certain partition is called a part. So3 has1 part,2+1 has2 parts, and1+1+1 has3 parts. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers. Let p(n) denote the number of partitions of n. I 3 = 2+1 = 1+1+1, I 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1, so p(4) = 5. I 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p(5) = 7. I 6 = 5+1 = 4+1+1 = 4+2 = 3+1+1+1 = 3+3 = 3+2+1 = 2+1+1+1+1 = 2+2+2 = 2+2+1+1 = 1+1+1+1+1+1, so p(6) = 11. Each summand in a certain partition is called a part. So3 has1 part,2+1 has2 parts, and1+1+1 has3 parts. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers. Let p(n) denote the number of partitions of n. I 3 = 2+1 = 1+1+1, so p(3) = 3. so p(4) = 5. I 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p(5) = 7. I 6 = 5+1 = 4+1+1 = 4+2 = 3+1+1+1 = 3+3 = 3+2+1 = 2+1+1+1+1 = 2+2+2 = 2+2+1+1 = 1+1+1+1+1+1, so p(6) = 11. Each summand in a certain partition is called a part. So3 has1 part,2+1 has2 parts, and1+1+1 has3 parts. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers. Let p(n) denote the number of partitions of n. I 3 = 2+1 = 1+1+1, so p(3) = 3. I 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1, I 5 = 4+1 = 3+1+1 = 2+2+1 = 2+1+1+1 = 3+2 = 1+1+1+1+1, so p(5) = 7. I 6 = 5+1 = 4+1+1 = 4+2 = 3+1+1+1 = 3+3 = 3+2+1 = 2+1+1+1+1 = 2+2+2 = 2+2+1+1 = 1+1+1+1+1+1, so p(6) = 11. Each summand in a certain partition is called a part. So3 has1 part,2+1 has2 parts, and1+1+1 has3 parts. Partition Numbers A partition of a positive integer n is a way of expressing n as a sum of positive integers. Let p(n) denote the number of partitions of n. I 3 = 2+1 = 1+1+1, so p(3) = 3. I 4 = 3+1 = 2+2 = 2+1+1 = 1+1+1+1, so p(4) = 5.
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