Odd composites

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The odd composite numbers are the odd positive integers

$N=2m+1,\quad m\geq 4,\,$

of the form

$N=IJ=(2i+1)(2j+1)=4ij+2(i+j)+1,\quad 1\leq j\leq i.\,$

Equivalently, the odd composite numbers are the odd positive integers

$N=2m+1,\quad m\geq 4,\,$

such that

$m=2ij+(i+j),\quad 1\leq j\leq i.\,$

Triangle of 2m+1 = 4ij + 2(i+j) + 1 = (2i+1)(2j+1)

This is a triangle of all the nontrivial factorizations in two factors of any odd composite number. Odd prime numbers, having only the trivial factorization, are thus excluded.

All composite odd numbers $\scriptstyle N\,=\,2m+1,\,m\,\geq \,4,\,$ have one or more entries in the table.

For example,

81 = 2*40+1 = 9*9 = (2*4+1)(2*4+1) = 27*3 = (2*13+1)(2*1+1)

appears twice in the table, e.g. at $\scriptstyle (i,j)\,=\,(4,4)\,$ and $\scriptstyle (i,j)\,=\,(13,1)\,$.

The number of times that $\scriptstyle N\,$ appears in the table is given by the number of nontrivial divisors up to $\scriptstyle {\sqrt {N}}:\ A038548(N)-1\,$.

Number of divisors of n that are at most sqrt(n). (Cf. A038548)

{1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 3, 1, 3, 1, 3, 2, 2, 1, 4, 2, 2, 2, 3, 1, 4, 1, 3, 2, 2, 2, 5, 1, 2, 2, 4, 1, 4, 1, 3, 3, 2, 1, 5, 2, 3, 2, 3, 1, 4, 2, 4, 2, 2, 1, 6, 1, 2, 3, 4, 2, 4, 1, 3, 2, 4, 1, 6, 1, 2, 3, 3, 2, 4, 1, 5, 3, 2, 1, 6, 2, 2, 2, 4, 1, 6, 2, 3, 2, 2, 2, 6, 1, 3, 3, 5, 1, 4, 1, 4, 4, ...}

Triangle of $\scriptstyle 2m+1\,=\,4ij+2(i+j)+1\,=\,(2i+1)(2j+1),\,1\,\leq \,\,j\,\leq \,i.\,$

$\scriptstyle i\,$ = 1	9
2	15	25
3	21	35	49
4	27	45	63	81
5	33	55	77	99	121
6	39	65	91	117	143	169
7	45	75	105	135	165	195	225
8	51	85	119	153	187	221	255	289
9	57	95	133	171	209	247	285	323	361
10	63	105	147	189	231	273	315	357	399	441
11	69	115	161	207	253	299	345	391	437	483	529
12	75	125	175	225	275	325	375	425	475	525	575	625
13	81	135	189	243	297	351	405	459	513	567	621	675	729
14	87	145	203	261	319	377	435	493	551	609	667	725	783	841
15	93	155	217	279	341	403	465	527	589	651	713	775	837	899	961
16	99	165	231	297	363	429	495	561	627	693	759	825	891	957	1023	1089
	$\scriptstyle j\,$ = 1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16

Rows of triangle of 2m+1 = 4ij + 2(i+j) + 1 = (2i+1)(2j+1)

The rows give the infinite sequence of finite sequences

{{9}, {15, 25}, {21, 35, 49}, {27, 45, 63, 81}, {33, 55, 77, 99, 121}, {39, 65, 91, 117, 143, 169}, {45, 75, 105, 135, 165, 195, 225}, {51, 85, 119, 153, 187, 221, 255, 289}, {57, 95, 133, 171, 209, 247, 285, 323, 361}, {63, 105, 147, 189, 231, 273, 315, 357, 399, 441}, ...}

whose concatenation give the sequence (Cf. A172292$\scriptstyle (n),\,n\,\geq \,1\,$)

{9, 15, 25, 21, 35, 49, 27, 45, 63, 81, 33, 55, 77, 99, 121, 39, 65, 91, 117, 143, 169, 45, 75, 105, 135, 165, 195, 225, 51, 85, 119, 153, 187, 221, 255, 289, 57, 95, 133, 171, 209, 247, 285, 323, 361, 63, 105, 147, 189, 231, 273, 315, 357, 399, 441, ...}

given by

$a(n)=N=(2i+1)(2j+1)=(2i+1)(2n+1-(i-1)i)=(2n+1)\,+\,i~(4n+3-2i^{2}+i),\,$

where

$i=\left\lceil {\frac {{\sqrt {1+8n}}-1}{2}}\right\rceil ,\,$

$j=n-t_{i-1}=n-{\frac {(i-1)i}{2}},\,$

giving the formula

$a(n)=N=\left(2\left\lceil {\frac {{\sqrt {1+8n}}-1}{2}}\right\rceil +1\right)\left(2n+1-{\Bigg (}\left\lceil {\frac {{\sqrt {1+8n}}-1}{2}}\right\rceil -1{\Bigg )}\left\lceil {\frac {{\sqrt {1+8n}}-1}{2}}\right\rceil \right),\quad n\geq 1.\,$

Columns of triangle of 2m+1 = 4ij + 2(i+j) + 1 = (2i+1)(2j+1)

The $\scriptstyle j\,$ th column, for $\scriptstyle i\,\geq \,j\,$, corresponds to all the odd multiples $\scriptstyle 2i+1\,$ of $\scriptstyle 2j+1\,$ starting from $\scriptstyle (2j+1)^{2}\,$. So, this triangle is sort of a Sieve of Eratosthenes (for odd numbers) in disguise.

Rising diagonals of triangle of 2m+1 = 4ij + 2(i+j) + 1 = (2i+1)(2j+1)

The $\scriptstyle k\,$ th rising diagonal corresponds to a fixed sum $\scriptstyle i+j\,=\,k+1,\,k\,\geq \,1\,$. This implies a fixed sum $\scriptstyle I+J\,=\,(2i+1)+(2j+1)\,=\,2(k+1)+2\,=\,2(k+2),\,k\,\geq \,1,\,$ of the odd factors of $\scriptstyle N\,=\,I\cdot J\,=\,(2i+1)(2j+1)\,$.

We get the system of two equations in two unknowns

$I\cdot J=N\,$

$I+J=2(k+2)\,$

giving

$I\cdot (2(k+2)-I)-N=0\,$

or

$I^{2}-2(k+2)I+N=0\,$

which gives $\scriptstyle I\,$.

The rising diagonals give the infinite sequence of finite sequences

{{9}, {15}, {21, 25}, {27, 35}, {33, 45, 49}, {39, 55, 63}, {45, 65, 77, 81}, ...}

whose concatenation gives (Cf. A??????)

{9, 15, 21, 25, 27, 35, 33, 45, 49, 39, 55, 63, 45, 65, 77, 81, ...}

Falling diagonals of triangle of 2m+1 = 4ij + 2(i+j) + 1 = (2i+1)(2j+1)

The $\scriptstyle k\,$ th falling diagonal (starting from the right) corresponds to a fixed difference $\scriptstyle i-j\,=\,k-1,\,k\,\geq \,1\,$. This implies a fixed difference $\scriptstyle I-J\,=\,(2i+1)-(2j+1)\,=\,2(k-1),\,k\,\geq \,1,\,$ between the odd factors of $\scriptstyle N\,=\,I\cdot J\,=\,(2i+1)(2j+1)\,$.

We get the system of two equations in two unknowns

$I\cdot J=N\,$

$I-J=2(k-1)\,$

giving

$I\cdot (I-2(k-1))-N=0\,$

or

$I^{2}-2(k-1)I-N=0\,$

which gives $\scriptstyle I\,$.

Triangle of m = 2ij + (i+j)

For all the $\scriptstyle m\,=\,2ij+(i+j),\,1\,\leq \,\,j\,\leq \,i,\,$ entries of the following table, corresponds an odd composite number $\scriptstyle N\,=\,2m+1,\,m\,\geq \,4\,$. All composite odd numbers $\scriptstyle N\,=\,2m+1,\,m\,\geq \,4,\,$ have one or more entries $\scriptstyle m\,=\,2ij+(i+j),\,1\,\leq \,\,j\,\leq \,i,\,$ in the table.

For example,

81 = 2*40+1 = 9*9 = (2*4+1)(2*4+1) = 27*3 = (2*13+1)(2*1+1)

appears twice, as $\scriptstyle m\,=\,40\,$, in the table, e.g. at $\scriptstyle (i,j)\,=\,(4,4)\,$ and $\scriptstyle (i,j)\,=\,(13,1)\,$.

Triangle of $\scriptstyle m\,=\,2ij+(i+j),\,1\,\leq \,\,j\,\leq \,i.\,$

$\scriptstyle i\,$ = 1	4
2	7	12
3	10	17	24
4	13	22	31	40
5	16	27	38	49	60
6	19	32	45	58	71	84
7	22	37	52	67	82	97	112
8	25	42	59	76	93	110	127	144
9	28	47	66	85	104	123	142	161	180
10	31	52	73	94	115	136	157	178	199	220
11	34	57	80	103	126	149	172	195	218	241	264
12	37	62	87	112	137	162	187	212	237	262	287	312
13	40	67	94	121	148	175	202	229	256	283	310	337	364
14	43	72	101	130	159	188	217	246	275	304	333	362	391	420
15	46	77	108	139	170	201	232	263	294	325	356	387	418	449	480
16	49	82	115	148	181	214	247	280	313	346	379	412	445	478	511	544
	$\scriptstyle j\,$ = 1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16

Rows of triangle of m = 2ij + (i+j)

The $\scriptstyle i\,$ th row, for $\scriptstyle j\,\leq \,i\,$, corresponds to all the odd multiples $\scriptstyle 2j+1\,$ of $\scriptstyle 2i+1\,$ ending at $\scriptstyle (2i+1)^{2},\,$ with $\scriptstyle m\,=\,4t_{i}\,=\,2i(i+1)\,$, where $\scriptstyle t_{i}\,$ is the $\scriptstyle i\,$ th triangular number.

The rows give the infinite sequence of finite sequences

{{4}, {7, 12}, {10, 17, 24}, {13, 22, 31, 40}, {16, 27, 38, 49, 60}, {19, 32, 45, 58, 71, 84}, {22, 37, 52, 67, 82, 97, 112}, {25, 42, 59, 76, 93, 110, 127, 144}, {28, 47, 66, 85, 104, 123, 142, 161, 180}, {31, 52, 73, 94, 115, 136, 157, 178, 199, 220}, ...}

whose concatenation gives (Cf. A083487)

{4, 7, 12, 10, 17, 24, 13, 22, 31, 40, 16, 27, 38, 49, 60, 19, 32, 45, 58, 71, 84, 22, 37, 52, 67, 82, 97, 112, 25, 42, 59, 76, 93, 110, 127, 144, 28, 47, 66, 85, 104, 123, 142, 161, 180, 31, 52, 73, 94, 115, 136, 157, 178, 199, 220, ...}

Columns of triangle of m = 2ij + (i+j)

The $\scriptstyle j\,$ th column, for $\scriptstyle i\,\geq \,j\,$, corresponds to all the odd multiples $\scriptstyle 2i+1\,$ of $\scriptstyle 2j+1\,$ starting from $\scriptstyle (2j+1)^{2},\,$ with $\scriptstyle m\,=\,4t_{j}\,=\,2j(j+1)\,$, where $\scriptstyle t_{j}\,$ is the $\scriptstyle j\,$ th triangular number.

Rising diagonals of triangle of m = 2ij + (i+j)

The $\scriptstyle k\,$ th rising diagonal corresponds to a fixed sum $\scriptstyle i+j\,=\,k+1,\,k\,\geq \,1\,$. This implies a fixed sum $\scriptstyle I+J\,=\,(2i+1)+(2j+1)\,=\,2(k+1)+2\,=\,2(k+2),\,k\,\geq \,1,\,$ of the odd factors of $\scriptstyle N\,=\,I\cdot J\,=\,(2i+1)(2j+1)\,$.

We get the system of two equations in two unknowns

$I\cdot J=N\,$

$I+J=2(k+2)\,$

giving

$I\cdot (2(k+2)-I)-N=0\,$

or

$I^{2}-2(k+2)I+N=0\,$

which gives $\scriptstyle I\,$.

The rising diagonals give the infinite sequence of finite sequences

{{4}, {7}, {10, 12}, {13, 17}, {16, 22, 24}, {19, 27, 31}, {22, 32, 38, 40}, {25, 37, 45, 49}, ...}

whose concatenation gives (Cf. A??????)

{4, 7, 10, 12, 13, 17, 16, 22, 24, 19, 27, 31, 22, 32, 38, 40, 25, 37, 45, 49, ...}

Falling diagonals of triangle of m = 2ij + (i+j)

The $\scriptstyle k\,$ th falling diagonal (starting from the right) corresponds to a fixed difference $\scriptstyle i-j\,=\,k-1,\,k\,\geq \,1\,$. This implies a fixed difference $\scriptstyle I-J\,=\,(2i+1)-(2j+1)\,=\,2(k-1),\,k\,\geq \,1,\,$ between the odd factors of $\scriptstyle N\,=\,I\cdot J\,=\,(2i+1)(2j+1)\,$.

We get the system of two equations in two unknowns

$I\cdot J=N\,$

$I-J=2(k-1)\,$

giving

$I\cdot (I-2(k-1))-N=0\,$

or

$I^{2}-2(k-1)I-N=0\,$

which gives $\scriptstyle I\,$.

Sequences

The following sequence, starting from the second term, contains exactly all the entries (without repetition) of the above triangle.

A047845 (n-1)/2, where n runs through odd nonprimes (A014076).

{0, 4, 7, 10, 12, 13, 16, 17, 19, 22, 24, 25, 27, 28, 31, 32, 34, 37, 38, 40, 42, 43, 45, 46, 47, 49, 52, 55, 57, 58, 59, 60, 61, 62, 64, 66, 67, 70, 71, 72, 73, 76, 77, 79, 80, 82, 84, 85, 87, 88, 91, 92, 93, 94, 97, 100, 101, 102, 103, 104, 106, 107, 108, 109, 110, 112, 115, ...}

The following sequence contains the only positive integers which do not appear in the above triangle.

A005097 (Odd primes - 1)/2.

{1, 2, 3, 5, 6, 8, 9, 11, 14, 15, 18, 20, 21, 23, 26, 29, 30, 33, 35, 36, 39, 41, 44, 48, 50, 51, 53, 54, 56, 63, 65, 68, 69, 74, 75, 78, 81, 83, 86, 89, 90, 95, 96, 98, 99, 105, 111, 113, 114, 116, 119, 120, 125, 128, 131, 134, 135, 138, 140, 141, 146, 153, 155, 156, ...}