Missing-digit sum

In number theory, a $m$-missing-digit sum in a given number base $b$ is a natural number that is equal to the sum of numbers created by deleting $m$ digits from the original number. For example, the OEIS lists these two integers as 1-missing-digit sums in base ten:

1,729,404 = 729404 (missing 1) + 129404 (missing 7) + 179404 (missing 2) + 172404 + 172904 + 172944 + 172940

1,800,000 = 800000 (missing 1) + 100000 (missing 8) + 180000 (missing first 0) + 180000 + 180000 + 180000 + 180000

Definition

Let $n$ be a natural number. We recursively define the $m$-missing-digit sum function for base $b>1$ $F_{m,b}:\mathbb {N} \rightarrow \mathbb {N}$ to be the following:

$F_{0,b}(n)=n$

$F_{m,b}(n)=\sum _{i=0}^{k-1}F_{m-1,b}\left(n{\bmod {b}}^{i-1}+{\frac {n-(n{\bmod {b}}^{i+1})}{b}}\right)$

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

A natural number $n$ is a $m$-missing-digit sum if it is a fixed point for $F_{m,b}$, which occurs if $F_{m,b}(n)=n$.

For example, the number 121 in base $b=3$ is a missing-digit sum with $m=1$, because $121_{3}=21_{3}+12_{3}+11_{3}$.

A natural number $n$ is a sociable missing-digit sum if it is a periodic point for $F_{m,b}$, where $F_{m,b}^{p}(n)=n$ for a positive integer $p$, and forms a cycle of period $p$. A missing-digit sum is a sociable missing-digit sum with $k=1$, and a amicable missing-digit sum is a missing-digit sum with $k=2$.

The number of iterations $i$ needed for $F_{m,b}^{i}(n)$ to reach a fixed point is the missing-digit sum function's persistence of $n$, and undefined if it never reaches a fixed point.

As a result of the recursive definition of the $m$-missing-digit sum function, when $m$ digits are deleted from a $k$-digit natural number, the number of natural numbers in the sum is equal to the binomial coefficient

${\binom {k}{m}}={\frac {k!}{m!(k-m)!}}$.

For a given base $b$, for every number $n\geq b^{b+1}$, $F_{1,b}(n)>n$ or n is a trivial missing-digit sum, defined below. This means that non-trivial missing-digit sums and cycles of $F_{1,b}(n)>n$ only exist for $n<b^{b+1}$.

Missing-digit sums for $F_{1,b}$

b = k (trivial)

Let $k$ be a positive integer and the number base $b=k$. Then for all integers $0\leq p<k$:

• $n_{1}=pb^{k+1}$ is a trivial missing-digit-sum for $F_{1,b}$ for all $k$, where the digits are $1$ and $k$ 0s.

Proof

Let $n_{1}=pb^{k+1}$. Then

$F_{b}(n_{1})=0+(k)(pb^{k})$

$=pb^{k+1}$

$=n_{1}$

Thus $n_{1}$ is a missing-digit-sum for $F_{b}$ for all $k$.

b = 2k

Let $k$ be a positive integer and the number base $b=2k$. Then:

• $n_{1}=1b^{k+2}+(2k-2)b^{k+1}$ is a missing-digit-sum for $F_{1,b}$ for all $k$, where the digits are $1$, $2k-2$, and $k$ 0s.

Proof

Let $n_{1}=1b^{k+2}+(2k-2)b^{k+1}$. Then

$F_{b}(n_{1})=(2k-2)b^{k+1}+b^{k+1}+(k)(1b^{k+1}+(2k-2)b^{k})$

$=(2k-2)b^{k+1}+b^{k+1}+kb^{k+1}+k(2k-2)b^{k}$

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

$=(3k-1)b^{k+1}+(k-1)b^{k+1}$

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

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

$=n_{1}$

Thus $n_{1}$ is a missing-digit-sum for $F_{b}$ for all $k$.

Searching for missing-digit sums

Searching for 1-missing-digit sums is simplified when one notes that the final two digits of n determine the final digit of its missing-digit sum. One can therefore test simply the final two digits of a given n to determine whether or not it is a potential missing-digit sum. In this way, the search space is considerably reduced. For example, consider the set of seven-digit base-ten numbers ending in ...01. For these numbers, the final digit of the sum is equal to (digit-0 x 1 + digit-1 x 6) modulo 10 = (0 + 6) mod 10 = 6 mod 10 = 6. Therefore, no seven-digit number ending in ...01 is equal to its own missing-digit-sum in base ten.

Now consider the set of seven-digit base-ten numbers ending in ...04. For these numbers, the final digit of the sum is equal to (0 x 1 + 4 x 6) modulo 10 = (0 + 24) mod 10 = 24 mod 10 = 4. This set may therefore contain one or more missing-digit sums. Next consider seven-digit numbers ending ...404. The penultimate (last-but-one) digit of the sum is equal to (2 + 4 x 2 + 0 x 4) modulo 10 = (2 + 8 + 0) mod 10 = 10 mod 10 = 0 (where the 2 is the tens digit of 24 from the sum for the final digit). This set of numbers ending ...404 may therefore contain one or more missing-digit sums. Similar reasoning can be applied to sums in which two, three and more digits are deleted from the original number.

Missing-digit sums and cycles of F_m,b for specific m and b

All numbers are represented in base $b$. This table is currently incomplete.

$m$	$b$	Missing-digit sums	Cycles
1	2	100	$\varnothing$
1	3	112, 121, 1000, 2000	$\varnothing$
1	4	1200, 2223, 10000, 20000, 30000	1332→2121→2031→1332
1	5	33334, 100000, 200000, 300000, 400000
1	6	14000, 444445, 1000000, 2000000, 3000000, 4000000, 5000000
1	7	5555556, 10000000, 20000000, 30000000, 40000000, 50000000, 60000000
1	8	160000, 66666667, 100000000, 200000000, 300000000, 400000000, 500000000, 600000000, 700000000
1	9	777777778, 1000000000, 2000000000, 3000000000, 4000000000, 5000000000, 6000000000, 7000000000, 8000000000
1	10	1800000, 14358846, 14400000, 15000000, 28758846, 28800000, 29358846, 29400000, 1107488889, 1107489042, 1111088889, 1111089042, 3277800000, 3281400000, 4388888889, 4388889042, 4392488889, 4392489042, 4500000000, 5607488889, 5607489042, 5611088889, 5611089042, 7777800000, 7781400000, 8888888889, 8888889042, 8892488889, 8892489042, 10000000000, 20000000000, 30000000000, 40000000000, 50000000000, 60000000000, 70000000000, 80000000000, 90000000000
1	11	9999999999A, 100000000000, 200000000000, 300000000000, 400000000000, 500000000000, 600000000000, 700000000000, 800000000000, 900000000000, A00000000000
1	12	1A000000, AAAAAAAAAAAB, 1000000000000, 2000000000000, 3000000000000, 4000000000000, 5000000000000, 6000000000000, 7000000000000, 8000000000000, 9000000000000, A000000000000, B000000000000
2	10	167564622641, 174977122641, 175543159858, 175543162247, 183477122641, 183518142444, 191500000000, 2779888721787, 2784986175699, 212148288981849, 212148288982006, 315131893491390, 321400000000000, 417586822240846, 417586822241003, 418112649991390, 424299754499265, 424341665637682, 526796569137682, 527322398999265, 533548288981849, 533548288982006, 636493411120423, 636531893491390, 642800000000000, 650000000000000, 738986822240846, 738986822241003, 739474144481849, 739474144482006, 739474144500000, 739512649991390, 745699754499265, 745741665637682, 746186822240846, 746186822241003, 751967555620423, 848722398999265, 849167555620423, 854948288981849, 854948288982006, 855396569137682, 862148288981849, 862148288982006, 957893411120423, 957931893491390, 965131893491390, 971400000000000
4	10	1523163197662495253514, 47989422298181591480943, 423579919359414921365511, 737978887988727574986738

Extension to negative integers

Missing-digit sums numbers can be extended to the negative integers by use of a signed-digit representation to represent each integer.

See also

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

External links

• Miss This — missing-digit sums in base ten and other bases

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