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Undid revision 851216462 by Jyoung80 (talk) without a published reference to base this on, it is original research, and in any case it is not clearly explained how these calculations relate to the subject |
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<math>N_k - N_{k-1}=s_{2k-1}: 36 - 1 = 35, 1225 - 36 = 1189,</math> and <math>41616 - 1225 = 40391.</math> In other words, the difference between two consecutive square triangular numbers is the square root of another square triangular number.{{citation needed|date=December 2014}}
Using the square roots of three consecutive triangular square numbers, sum the two most recent square roots, multiply by 5 and subtract the third prior square root:
( 1 + √36 ) * 5 – 0 = 35 '''(then square the answer)''' 35^2 = 1,225,
( √36 + 35 ) * 5 – 1 = 204 '''(then square the answer)''' 204^2 = 41,616,
( 35 + 204 ) * 5 – 6 = 1189 '''(^2)''' = 1,413,721,
( 204 + 1189 ) * 5 – 35 = 6930 '''(^2)''' = 48,024,900,
( 1189 + 6930 ) * 5 – 204 = 40,391 '''(^2)''' = 1,631,432,881,
( 6930 + 40,391 ) * 5 – 1189 = 235,416 '''(^2)''' = 55,420,693,056,
( 40,391 + 235,416 ) * 5 – 6930 = 1,372,105 '''(^2)''' = 1,882,672,131,025,
(235,416 + 1,372,105)*5-40,391 = 7,997,214 '''(^2)''' = 63,955,431,761,796,
(1,372,105+7,997,214)*5-235,416= 46,611,179 '''(^2)''' = 2,172,602,007,770,041,
= 73,804,512,832,419,600,
= 2,507,180,834,294,496,361, etc.
Another method, using the square roots of three consecutive triangular square numbers, is subtracting the newest triangular square radical from the second one back, multiplying by 7, then adding the third prior triangular square radical:
(6 – 1) * 7 + 0 = 35 '''(then square the answer)''' =1225,
(35 – 6) * 7 + 1 = 204 '''^2''' = 41,616,
(204 – 35) * 7 + 6 = 1189 '''^2''' = 1,413,721,
(1189 – 204) * 7 + 35 = 6930 '''^2''' = 48,024,900,
Etc.
The generating function for the square triangular numbers is:<ref>
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