Sums of consecutive pentagonal numbers — Part 1

A number of the form   $T_n \; = \; n \,(n + 1) \,/ \,2$   is called a triangular number.

The n-th pentagonal number is given by the formula    $P_n \; = \; n \, (3 \, n-1) \,/ \,2$

The n-th pentagonal number is one third of the (3n-1)-th triangular number:

$T_{ \,3 \, n-1 \,} \; = \; (3 \, n - 1)(3 \, n) \, \,/2 \; = \; 3/2 \; n \, (3 \, n-1) \; = \; 3 \, P_n$

Let   $N_1 = P_{x}$    and    $N_2 = P_{y-1} + P_{y}$    such that

$N_2 \; - \; N_1 \; = \; 1$

Here are the first few solutions

$5 = P_2$
$6 = P_1 + P_2$

$46376 = P_{176}$
$46377 = P_{124} + P_{125}$

$6906901 = P_{2146}$
$6906902 = P_{1517} + P_{1518}$

$61759818160 = P_{202912}$
$61759818161 = P_{143480} + P_{143481}$

$9198017008005 = P_{2476290}$
$9198017008006 = P_{1751001} + P_{1751002}$

$82246414481329560 = P_{234160080}$
$82246414481329561 = P_{165576180} + P_{165576181}$

$12249128021791713365 = P_{2857636322}$
$12249128021791713366 = P_{2020654021} + P_{2020654022}$

$109528701614527554097376 = P_{270220529216}$
$109528701614527554097377 = P_{191074768624} + P_{191074768625}$

$16312335274403131755399301 = P_{3297709839106}$
$16312335274403131755399302 = P_{2331832989617} + P_{2331832989618}$

$145860905341806700762751902600 = P_{311834256554992}$
$145860905341806700762751902601 = P_{220500117416300} + P_{220500117416301}$

$21723365257604243074488033268245 = P_{3805554296691810}$
$21723365257604243074488033268246 = P_{2690933249364381} + P_{2690933249364382}$