Sequences of 4 Consecutive Integers with Falling Divisor Sum

Theorem

The following ordered quadruple of consecutive integers have divisor sums which are strictly decreasing:

$44, 45, 46, 47$

$104, 105, 106, 107$

Proof

\(\ds \map {\sigma_1} {44}\)

\(=\)

\(\ds 84\)

$\sigma_1$ of $44$

\(\ds \map {\sigma_1} {45}\)

\(=\)

\(\ds 78\)

$\sigma_1$ of $45$

\(\ds \map {\sigma_1} {46}\)

\(=\)

\(\ds 72\)

$\sigma_1$ of $46$

\(\ds \map {\sigma_1} {47}\)

\(=\)

\(\ds 48\)

Divisor Sum of Prime Number: $47$ is prime

\(\ds \map {\sigma_1} {104}\)

\(=\)

\(\ds 210\)

$\sigma_1$ of $104$

\(\ds \map {\sigma_1} {105}\)

\(=\)

\(\ds 192\)

$\sigma_1$ of $105$

\(\ds \map {\sigma_1} {106}\)

\(=\)

\(\ds 162\)

$\sigma_1$ of $106$

\(\ds \map {\sigma_1} {107}\)

\(=\)

\(\ds 108\)

Divisor Sum of Prime Number: $107$ is prime

$\blacksquare$

Also see

• Sequences of 4 Consecutive Integers with Rising Divisor Sum

Sources

• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $44$