In number theory, a perfect digital invariant (PDI) is a number in a given number base (${\displaystyle b}$) that is the sum of its own digits each raised to a given power (${\displaystyle p}$).^[1][2]

Let ${\displaystyle n}$ be a natural number. The perfect digital invariant function (also known as a happy function, from happy numbers) for base ${\displaystyle b>1}$ and power ${\displaystyle p>0}$ ${\displaystyle F_{p,b}:\mathbb {N} \rightarrow \mathbb {N} }$ is defined as:

${\displaystyle F_{p,b}(n)=\sum _{i=0}^{k-1}d_{i}^{p}.}$

where ${\displaystyle k=\lfloor \log _{b}{n}\rfloor +1}$ is the number of digits in the number in base ${\displaystyle b}$, and

${\displaystyle d_{i}={\frac {n{\bmod {b^{i+1}}}-n{\bmod {b}}^{i}}{b^{i}}}}$

is the value of each digit of the number. A natural number ${\displaystyle n}$ is a perfect digital invariant if it is a fixed point for ${\displaystyle F_{p,b}}$, which occurs if ${\displaystyle F_{p,b}(n)=n}$. ${\displaystyle 0}$ and ${\displaystyle 1}$ are trivial perfect digital invariants for all ${\displaystyle b}$ and ${\displaystyle p}$, all other perfect digital invariants are nontrivial perfect digital invariants.

For example, the number 4150 in base ${\displaystyle b=10}$ is a perfect digital invariant with ${\displaystyle p=5}$, because ${\displaystyle 4150=4^{5}+1^{5}+5^{5}+0^{5}}$.

A natural number ${\displaystyle n}$ is a sociable digital invariant if it is a periodic point for ${\displaystyle F_{p,b}}$, where ${\displaystyle F_{p,b}^{k}(n)=n}$ for a positive integer ${\displaystyle k}$ (here ${\displaystyle F_{p,b}^{k}}$ is the ${\displaystyle k}$th iterate of ${\displaystyle F_{p,b}}$), and forms a cycle of period ${\displaystyle k}$. A perfect digital invariant is a sociable digital invariant with ${\displaystyle k=1}$, and a amicable digital invariant is a sociable digital invariant with ${\displaystyle k=2}$.

All natural numbers ${\displaystyle n}$ are preperiodic points for ${\displaystyle F_{p,b}}$, regardless of the base. This is because if ${\displaystyle k\geq p+2}$, ${\displaystyle n\geq b^{k-1}>b^{p}k}$, so any ${\displaystyle n}$ will satisfy ${\displaystyle n>F_{b,p}(n)}$ until ${\displaystyle n<b^{p+1}}$. There are a finite number of natural numbers less than ${\displaystyle b^{p+1}}$, so the number is guaranteed to reach a periodic point or a fixed point less than ${\displaystyle b^{p+1}}$, making it a preperiodic point.

Numbers in base ${\displaystyle b>p}$ lead to fixed or periodic points of numbers ${\displaystyle n\leq (p-2)^{p}+p(b-1)^{p}}$.

The number of iterations ${\displaystyle i}$ needed for ${\displaystyle F_{p,b}^{i}(n)}$ to reach a fixed point is the perfect digital invariant function's persistence of ${\displaystyle n}$, and undefined if it never reaches a fixed point.

${\displaystyle F_{1,b}}$ is the digit sum. The only perfect digital invariants are the single-digit numbers in base ${\displaystyle b}$, and there are no periodic points with prime period greater than 1.

${\displaystyle F_{p,2}}$ reduces to ${\displaystyle F_{1,2}}$, as for any power ${\displaystyle p}$, ${\displaystyle 0^{p}=0}$ and ${\displaystyle 1^{p}=1}$.

For every natural number ${\displaystyle k>1}$, if ${\displaystyle p<b}$, ${\displaystyle (b-1)\equiv 0{\bmod {k}}}$ and ${\displaystyle (p-1)\equiv 0{\bmod {\phi }}(k)}$, then for every natural number ${\displaystyle n}$, if ${\displaystyle n\equiv m{\bmod {k}}}$, then ${\displaystyle F_{p,b}(n)\equiv m{\bmod {k}}}$, where ${\displaystyle \phi (k)}$ is Euler's totient function.

No upper bound can be determined for the size of perfect digital invariants in a given base and arbitrary power, and it is not currently known whether or not the number of perfect digital invariants for an arbitrary base is finite or infinite.^[1]

By definition, any three-digit perfect digital invariant ${\displaystyle n=d_{2}d_{1}d_{0}}$ for ${\displaystyle F_{2,b}}$ with natural number digits ${\displaystyle 0\leq d_{0}<b}$, ${\displaystyle 0\leq d_{1}<b}$, ${\displaystyle 0\leq d_{2}<b}$ has to satisfy the cubic Diophantine equation ${\displaystyle d_{0}^{2}+d_{1}^{2}+d_{2}^{2}=d_{2}b^{2}+d_{1}b+d_{0}}$. ${\displaystyle d_{2}}$ has to be equal to 0 or 1 for any ${\displaystyle b>2}$, because the maximum value ${\displaystyle n}$ can take is ${\displaystyle n=(2-1)^{2}+2(b-1)^{2}=1+2(b-1)^{2}<2b^{2}}$. As a result, there are actually two related quadratic Diophantine equations to solve:

${\displaystyle d_{0}^{2}+d_{1}^{2}=d_{1}b+d_{0}}$ when ${\displaystyle d_{2}=0}$, and

${\displaystyle d_{0}^{2}+d_{1}^{2}+1=b^{2}+d_{1}b+d_{0}}$ when ${\displaystyle d_{2}=1}$.

The two-digit natural number ${\displaystyle n=d_{1}d_{0}}$ is a perfect digital invariant in base

${\displaystyle b=d_{1}+{\frac {d_{0}(d_{0}-1)}{d_{1}}}.}$

This can be proven by taking the first case, where ${\displaystyle d_{2}=0}$, and solving for ${\displaystyle b}$. This means that for some values of ${\displaystyle d_{0}}$ and ${\displaystyle d_{1}}$, ${\displaystyle n}$ is not a perfect digital invariant in any base, as ${\displaystyle d_{1}}$ is not a divisor of ${\displaystyle d_{0}(d_{0}-1)}$. Moreover, ${\displaystyle d_{0}>1}$, because if ${\displaystyle d_{0}=0}$ or ${\displaystyle d_{0}=1}$, then ${\displaystyle b=d_{1}}$, which contradicts the earlier statement that ${\displaystyle 0\leq d_{1}<b}$.

There are no three-digit perfect digital invariants for ${\displaystyle F_{2,b}}$, which can be proven by taking the second case, where ${\displaystyle d_{2}=1}$, and letting ${\displaystyle d_{0}=b-a_{0}}$ and ${\displaystyle d_{1}=b-a_{1}}$. Then the Diophantine equation for the three-digit perfect digital invariant becomes

${\displaystyle (b-a_{0})^{2}+(b-a_{1})^{2}+1=b^{2}+(b-a_{1})b+(b-a_{0})}$

${\displaystyle b^{2}-2a_{0}b+a_{0}^{2}+b^{2}-2a_{1}b+a_{1}^{2}+1=b^{2}+(b-a_{1})b+(b-a_{0})}$

${\displaystyle 2b^{2}-2(a_{0}+a_{1})b+a_{0}^{2}+a_{1}^{2}+1=b^{2}+(b-a_{1})b+(b-a_{0})}$

${\displaystyle b^{2}+(b-2(a_{0}+a_{1}))b+a_{0}^{2}+a_{1}^{2}+1=b^{2}+(b-a_{1})b+(b-a_{0})}$

${\displaystyle 2(a_{0}+a_{1})>a_{1}}$ for all values of ${\displaystyle 0<a_{1}\leq b}$. Thus, there are no solutions to the Diophantine equation, and there are no three-digit perfect digital invariants for ${\displaystyle F_{2,b}}$.

There are just four numbers, after unity, which are the sums of the cubes of their digits:

${\displaystyle 153=1^{3}+5^{3}+3^{3}}$

${\displaystyle 370=3^{3}+7^{3}+0^{3}}$

${\displaystyle 371=3^{3}+7^{3}+1^{3}}$

${\displaystyle 407=4^{3}+0^{3}+7^{3}.}$

These are odd facts, very suitable for puzzle columns and likely to amuse amateurs, but there is nothing in them which appeals to the mathematician. (sequence A046197 in the OEIS)
— G. H. Hardy, A Mathematician's Apology

By definition, any four-digit perfect digital invariant ${\displaystyle n}$ for ${\displaystyle F_{3,b}}$ with natural number digits ${\displaystyle 0\leq d_{0}<b}$, ${\displaystyle 0\leq d_{1}<b}$, ${\displaystyle 0\leq d_{2}<b}$, ${\displaystyle 0\leq d_{3}<b}$ has to satisfy the quartic Diophantine equation ${\displaystyle d_{0}^{3}+d_{1}^{3}+d_{2}^{3}+d_{3}^{3}=d_{3}b^{3}+d_{2}b^{2}+d_{1}b+d_{0}}$. ${\displaystyle d_{3}}$ has to be equal to 0, 1, 2 for any ${\displaystyle b>3}$, because the maximum value ${\displaystyle n}$ can take is ${\displaystyle n=(3-2)^{3}+3(b-1)^{3}=1+3(b-1)^{3}<3b^{3}}$. As a result, there are actually three related cubic Diophantine equations to solve

${\displaystyle d_{0}^{3}+d_{1}^{3}+d_{2}^{3}=d_{2}b^{2}+d_{1}b+d_{0}}$ when ${\displaystyle d_{3}=0}$

${\displaystyle d_{0}^{3}+d_{1}^{3}+d_{2}^{3}+1=b^{3}+d_{2}b^{2}+d_{1}b+d_{0}}$ when ${\displaystyle d_{3}=1}$

${\displaystyle d_{0}^{3}+d_{1}^{3}+d_{2}^{3}+8=2b^{3}+d_{2}b^{2}+d_{1}b+d_{0}}$ when ${\displaystyle d_{3}=2}$

We take the first case, where ${\displaystyle d_{3}=0}$.

b = 3k + 1

Let ${\displaystyle k}$ be a positive integer and the number base ${\displaystyle b=3k+1}$. Then:

• ${\displaystyle n_{1}=kb^{2}+(2k+1)b}$ is a perfect digital invariant for ${\displaystyle F_{3,b}}$ for all ${\displaystyle k}$.

Proof

Let the digits of ${\displaystyle n_{1}=d_{2}b^{2}+d_{1}b+d_{0}}$ be ${\displaystyle d_{2}=k}$, ${\displaystyle d_{1}=2k+1}$, and ${\displaystyle d_{0}=0}$. Then

${\displaystyle {\begin{aligned}F_{3,b}(n_{1})&=d_{0}^{3}+d_{1}^{3}+d_{2}^{3}\\&=k^{3}+(2k+1)^{3}+0^{3}\\&=(k^{2}-k(2k+1)+(2k+1)^{2})(k+(2k+1))\\&=(k^{2}-2k^{2}-k+4k^{2}+4k+1)(3k+1)\\&=(3k^{2}+3k+1)(3k+1)\\&=(3k^{2}+4k+1)(3k+1)-k(3k+1)\\&=(k+1)(3k+1)(3k+1)-k(3k+1)\\&=k(3k+1)(3k+1)+(3k+1)(3k+1)-k(3k+1)\\&=k(3k+1)^{2}+(2k+1)(3k+1)+0\\&=d_{2}b^{2}+d_{1}b+d_{0}\\&=n_{1}\end{aligned}}}$

Thus ${\displaystyle n_{1}}$ is a perfect digital invariant for ${\displaystyle F_{3,b}}$ for all ${\displaystyle k}$.

• ${\displaystyle n_{2}=kb^{2}+(2k+1)b+1}$ is a perfect digital invariant for ${\displaystyle F_{3,b}}$ for all ${\displaystyle k}$.

Proof

Let the digits of ${\displaystyle n_{2}=d_{2}b^{2}+d_{1}b+d_{0}}$ be ${\displaystyle d_{2}=k}$, ${\displaystyle d_{1}=2k+1}$, and ${\displaystyle d_{0}=1}$. Then

${\displaystyle {\begin{aligned}F_{3,b}(n_{2})&=d_{0}^{3}+d_{1}^{3}+d_{2}^{3}\\&=k^{3}+(2k+1)^{3}+1^{3}\\&=(k^{2}-k(2k+1)+(2k+1)^{2})(k+(2k+1))+1\\&=(k^{2}-2k^{2}-k+4k^{2}+4k+1)(3k+1)+1\\&=(3k^{2}+3k+1)(3k+1)+1\\&=(3k^{2}+4k+1)(3k+1)-k(3k+1)+1\\&=(k+1)(3k+1)(3k+1)-k(3k+1)+1\\&=k(3k+1)(3k+1)+(3k+1)(3k+1)-k(3k+1)+1\\&=k(3k+1)^{2}+(2k+1)(3k+1)+1\\&=d_{2}b^{2}+d_{1}b+d_{0}\\&=n_{2}\end{aligned}}}$

Thus ${\displaystyle n_{2}}$ is a perfect digital invariant for ${\displaystyle F_{3,b}}$ for all ${\displaystyle k}$.

• ${\displaystyle n_{3}=(k+1)b^{2}+(2k+1)}$ is a perfect digital invariant for ${\displaystyle F_{3,b}}$ for all ${\displaystyle k}$.

Proof

Let the digits of ${\displaystyle n_{3}=d_{2}b^{2}+d_{1}b+d_{0}}$ be ${\displaystyle d_{2}=k+1}$, ${\displaystyle d_{1}=0}$, and ${\displaystyle d_{0}=2k+1}$. Then

${\displaystyle {\begin{aligned}F_{3,b}(n_{3})&=d_{0}^{3}+d_{1}^{3}+d_{2}^{3}\\&=(k+1)^{3}+0^{3}+(2k+1)^{3}\\&=((k+1)^{2}-(k+1)(2k+1)+(2k+1)^{2})((k+1)+(2k+1))\\&=((k+1)^{2}+k(2k+1)(3k+2)\\&=(k^{2}+2k+1+2k^{2}+k)(3k+2)\\&=(3k^{2}+3k+1)(3k+2)\\&=(3k^{2}+3k)(3k+2)+(3k+2)\\&=3k(k+1)(3k+2)+(3k+2)\\&=(k+1)((3k+1)^{2}-1)+(3k+2)\\&=(k+1)(3k+1)^{2}-(k+1)+(3k+2)\\&=(k+1)(3k+1)^{2}+0(3k+1)+(2k+1)\\&=d_{2}b^{2}+d_{1}b+d_{0}\\&=n_{3}\end{aligned}}}$

Thus ${\displaystyle n_{3}}$ is a perfect digital invariant for ${\displaystyle F_{3,b}}$ for all ${\displaystyle k}$.

More information , ...

Perfect digital invariants

${\displaystyle k}$	${\displaystyle b}$	${\displaystyle n_{1}}$	${\displaystyle n_{2}}$	${\displaystyle n_{3}}$
1	4	130	131	203
2	7	250	251	305
3	10	370	371	407
4	13	490	491	509
5	16	5B0	5B1	60B
6	19	6D0	6D1	70D
7	22	7F0	7F1	80F
8	25	8H0	8H1	90H
9	28	9J0	9J1	A0J

Close

b = 3k + 2

Let ${\displaystyle k}$ be a positive integer and the number base ${\displaystyle b=3k+2}$. Then:

• ${\displaystyle n_{1}=kb^{2}+(2k+1)}$ is a perfect digital invariant for ${\displaystyle F_{3,b}}$ for all ${\displaystyle k}$.

Proof

Let the digits of ${\displaystyle n_{1}=d_{2}b^{2}+d_{1}b+d_{0}}$ be ${\displaystyle d_{2}=k}$, ${\displaystyle d_{1}=2k+1}$, and ${\displaystyle d_{0}=0}$. Then

${\displaystyle F_{3,b}(n_{1})=d_{0}^{3}+d_{1}^{3}+d_{2}^{3}}$

${\displaystyle =k^{3}+0^{3}+(2k+1)^{3}}$

${\displaystyle =(k^{2}-k(2k+1)+(2k+1)^{2})(k+(2k+1))}$

${\displaystyle =(k^{2}-2k^{2}-k+4k^{2}+4k+1)(3k+1)}$

${\displaystyle =(3k^{2}+3k+1)(3k+1)}$

${\displaystyle =(3k^{2}+3k+1)(3k+2)-(3k^{2}+3k+1)}$

${\displaystyle =(3k^{2}+3k+1)(3k+2)-(3k^{2}+2k+k+1)}$

${\displaystyle =(3k^{2}+3k+1)(3k+2)-k(3k+2)-(k+1)}$

${\displaystyle =(3k^{2}+2k+1)(3k+2)-(k+1)}$

${\displaystyle =(3k^{2}+2k)(3k+2)+(3k+2)-(k+1)}$

${\displaystyle =k(3k+2)^{2}+(2k+1)}$

${\displaystyle =d_{2}b^{2}+d_{1}b+d_{0}}$

${\displaystyle =n_{1}}$

Thus ${\displaystyle n_{1}}$ is a perfect digital invariant for ${\displaystyle F_{3,b}}$ for all ${\displaystyle k}$.

More information , ...

Perfect digital invariants

${\displaystyle k}$	${\displaystyle b}$	${\displaystyle n_{1}}$
1	5	103
2	8	205
3	11	307
4	14	409
5	17	50B
6	20	60D
7	23	70F
8	26	80H
9	29	90J

Close

b = 6k + 4

Let ${\displaystyle k}$ be a positive integer and the number base ${\displaystyle b=6k+4}$. Then:

• ${\displaystyle n_{4}=kb^{2}+(3k+2)b+(2k+1)}$ is a perfect digital invariant for ${\displaystyle F_{3,b}}$ for all ${\displaystyle k}$.

Proof

Let the digits of ${\displaystyle n_{4}=d_{2}b^{2}+d_{1}b+d_{0}}$ be ${\displaystyle d_{2}=k+1}$, ${\displaystyle d_{1}=3k+2}$, and ${\displaystyle d_{0}=2k+1}$. Then

${\displaystyle F_{3,b}(n_{3})=d_{0}^{3}+d_{1}^{3}+d_{2}^{3}}$

${\displaystyle =(k)^{3}+(3k+2)^{3}+(2k+1)^{3}}$

${\displaystyle =k^{3}+((3k+2)^{2}-(3k+2)(2k+1)+(2k+1)^{2})((3k+2)+(2k+1))}$

${\displaystyle =k^{3}+((3k+2)(k+1)+(2k+1)^{2})(5k+3)}$

${\displaystyle =k^{3}+(3k^{2}+5k+2+4k^{2}+4k+1)(5k+3)}$

${\displaystyle =k^{3}+(7k^{2}+9k+3)(5k+3)}$

${\displaystyle =k^{3}+5k(7k^{2}+9k+3)+3(7k^{2}+9k+3)}$

${\displaystyle =k^{3}+35k^{3}+45k^{2}+15k+21k^{2}+27k+9}$

${\displaystyle =36k^{3}+66k^{2}+42k+9}$

${\displaystyle =(6k+4)(6k^{2})+42k^{2}+42k+9}$

${\displaystyle =(6k+4)(6k^{2})+(6k+4)(4k^{2})+18k^{2}+26k+9}$

${\displaystyle =(6k+4)(6k^{2}+4k)+18k^{2}+26k+9}$

${\displaystyle =k(6k+4)^{2}+(6k+4)(3k)+14k+9}$

${\displaystyle =k(6k+4)^{2}+(3k+2)(6k+4)+2k+1}$

${\displaystyle =d_{2}b^{2}+d_{1}b+d_{0}}$

${\displaystyle =n_{4}}$

Thus ${\displaystyle n_{4}}$ is a perfect digital invariant for ${\displaystyle F_{3,b}}$ for all ${\displaystyle k}$.

More information , ...

Perfect digital invariants

${\displaystyle k}$	${\displaystyle b}$	${\displaystyle n_{4}}$
0	4	021
1	10	153
2	16	285
3	22	3B7
4	28	4E9

Close

All numbers are represented in base ${\displaystyle b}$.

More information , ...

${\displaystyle p}$	${\displaystyle b}$	Nontrivial perfect digital invariants	Cycles
2	3	12, 22	2 → 11 → 2
	4	${\displaystyle \varnothing }$	${\displaystyle \varnothing }$
	5	23, 33	4 → 31 → 20 → 4
	6	${\displaystyle \varnothing }$	5 → 41 → 25 → 45 → 105 → 42 → 32 → 21 → 5
	7	13, 34, 44, 63	2 → 4 → 22 → 11 → 2 16 → 52 → 41 → 23 → 16
	8	24, 64	4 → 20 → 4 5 → 31 → 12 → 5 15 → 32 → 15
	9	45, 55	58 → 108 → 72 → 58 75 → 82 → 75
	10	${\displaystyle \varnothing }$	4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4
	11	56, 66	5 → 23 → 12 → 5 68 → 91 → 75 → 68
	12	25, A5	5 → 21 → 5 8 → 54 → 35 → 2A → 88 → A8 → 118 → 56 → 51 → 22 → 8 18 → 55 → 42 → 18 68 → 84 → 68
	13	14, 36, 67, 77, A6, C4	28 → 53 → 28 79 → A0 → 79 98 → B2 → 98
	14	${\displaystyle \varnothing }$	1B → 8A → BA → 11B → 8B → D3 → CA → 136 → 34 → 1B 29 → 61 → 29
	15	78, 88	2 → 4 → 11 → 2 8 → 44 → 22 → 8 15 → 1B → 82 → 48 → 55 → 35 → 24 → 15 2B → 85 → 5E → EB → 162 → 2B 4E → E2 → D5 → CE → 17A → A0 → 6A → 91 → 57 → 4E 9A → C1 → 9A D6 → DA → 12E → D6
	16	${\displaystyle \varnothing }$	D → A9 → B5 → 92 → 55 → 32 → D
3	3	122	2 → 22 → 121 → 101 → 2
	4	20, 21, 130, 131, 203, 223, 313, 332	${\displaystyle \varnothing }$
	5	103, 433	14 → 230 → 120 → 14
	6	243, 514, 1055	13 → 44 → 332 → 142 → 201 → 13
	7	12, 22, 250, 251, 305, 505	2 → 11 → 2 13 → 40 → 121 → 13 23 → 50 → 236 → 506 → 665 → 1424 → 254 → 401 → 122 → 23 51 → 240 → 132 → 51 160 → 430 → 160 161 → 431 → 161 466 → 1306 → 466 516 → 666 → 1614 → 552 → 516
	8	134, 205, 463, 660, 661	662 → 670 → 1057 → 725 → 734 → 662
	9	30, 31, 150, 151, 570, 571, 1388	38 → 658 → 1147 → 504 → 230 → 38 152 → 158 → 778 → 1571 → 572 → 578 → 1308 → 660 → 530 → 178 → 1151 → 152 638 → 1028 → 638 818 → 1358 → 818
	10	153, 370, 371, 407	55 → 250 → 133 → 55 136 → 244 → 136 160 → 217 → 352 → 160 919 → 1459 → 919
	11	32, 105, 307, 708, 966, A06, A64	3 → 25 → 111 → 3 9 → 603 → 201 → 9 A → 82A → 1162 → 196 → 790 → 895 → 1032 → 33 → 4A → 888 → 1177 → 576 → 5723 → A3 → 8793 → 1210 → A 25A → 940 → 661 → 364 → 25A 366 → 388 → 876 → 894 → A87 → 1437 → 366 49A → 1390 → 629 → 797 → 1077 → 575 → 49A
	12	577, 668, A83, 11AA
	13	490, 491, 509, B85	13 → 22 → 13
	14	136, 409
	15	C3A, D87
	16	23, 40, 41, 156, 173, 208, 248, 285, 4A5, 580, 581, 60B, 64B, 8C0, 8C1, 99A, AA9, AC3, CA8, E69, EA0, EA1
4	3	${\displaystyle \varnothing }$	121 → 200 → 121 122 → 1020 → 122
	4	1103, 3303	3 → 1101 → 3
	5	2124, 2403, 3134	1234 → 2404 → 4103 → 2323 → 1234 2324 → 2434 → 4414 → 11034 → 2324 3444 → 11344 → 4340 → 4333 → 3444
	6	${\displaystyle \varnothing }$
	7	${\displaystyle \varnothing }$
	8	20, 21, 400, 401, 420, 421
	9	432, 2466
5	3	1020, 1021, 2102, 10121	${\displaystyle \varnothing }$
	4	200	3 → 3303 → 23121 → 10311 → 3312 → 20013 → 10110 → 3 3311 → 13220 → 10310 → 3311

Close

Perfect digital invariants can be extended to the negative integers by use of a signed-digit representation to represent each integer.

A happy number ${\displaystyle n}$ for a given base ${\displaystyle b}$ and a given power ${\displaystyle p}$ is a preperiodic point for the perfect digital invariant function ${\displaystyle F_{p,b}}$ such that the ${\displaystyle m}$-th iteration of ${\displaystyle F_{p,b}}$ is equal to the trivial perfect digital invariant ${\displaystyle 1}$, and an unhappy number is one such that there exists no such ${\displaystyle m}$.

The example below implements the perfect digital invariant function described in the definition above to search for perfect digital invariants and cycles in Python. This can be used to find happy numbers.

def pdif(x: int, p: int, b: int) -> int:
    """Perfect digital invariant function."""
    total = 0
    while x > 0:
        total = total + pow(x % b, p)
        x = x // b
    return total

def pdif_cycle(x: int, p: int, b: int) -> list[int]:
    seen = []
    while x not in seen:
        seen.append(x)
        x = pdif(x, p, b)
    cycle = []
    while x not in cycle:
        cycle.append(x)
        x = pdif(x, p, b)
    return cycle

• Arithmetic dynamics
• Dudeney number
• Factorion
• Happy number
• Kaprekar's constant
• Kaprekar number
• Meertens number
• Narcissistic number
• Perfect digit-to-digit invariant
• Sum-product number