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}

 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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