Fermat's triangle

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Fermat’s triangle

T  (n, m) = m  φ (n) mod n

n																																		∑ n   −  1 m  = 1  m  φ (n) mod n
2																	1																	1
3																1		1																2
4															1		0		1															2
5														1		1		1		1														4
6													1		4		3		4		1													13
7												1		1		1		1		1		1												6
8											1		0		1		0		1		0		1											4
9										1		1		0		1		1		0		1		1										6
10									1		6		1		6		5		6		1		6		1									33
11								1		1		1		1		1		1		1		1		1		1								10
12							1		4		9		4		1		0		1		4		9		4		1							38
13						1		1		1		1		1		1		1		1		1		1		1		1						12
14					1		8		1		8		1		8		7		8		1		8		1		8		1					61
15				1		1		6		1		10		6		1		1		6		10		1		6		1		1				52
16			1		0		1		0		1		0		1		0		1		0		1		0		1		0		1			8
			m = 1		2		3		4		5		6		7		8		9		10		11		12		13		14		15

A066340: Fermat’s triangle:

T  (n, m) = m  φ (n) mod n

, for

n   ≥   2, 1   ≤   m   ≤   n  −  1

. (Concatenated rows of Fermat’s triangle.)

{1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 4, 3, 4, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 6, 1, 6, 5, 6, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 9, 4, 1, 0, 1, 4, 9, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...}

Per Fermat’s little theorem, the rows corresponding to prime

n

consist of all 1s, although the converse is not true: rows corresponding to composite

n

which consist of all 1s are the Carmichael numbers.

Row sums

A?????? Row sums of Fermat’s triangle:

∑ n   −  1 m  = 1  m  φ (n) mod n

, for

n   ≥   2

.

{1, 2, 2, 4, 13, 6, 4, 6, 33, 10, 38, 12, 61, 52, 8, ...}

If

n

is a prime power

p  k, k   ≥   1,

the row sum is

n (   p  −  1  p )

. The converse is not true: if the row sum is

n (  q  −  1  q )

for some

q   ≥   2

, it does not imply that

n

is a prime power

q k, k   ≥   1

.