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a(n) is the numerator of the coefficient c_(2n+1) in the expansion Sum_{k=1..j} 1/(k*(k+1)/2)^2 = Sum_{m>=0} c_m/j^m for large values of j.
0

%I #10 Jan 04 2021 07:51:27

%S 0,-4,-116,-340,-356,-1076,-51836,-172,188,-201004,686564,-3423572,

%T 945336244,-34212700,94997798876,-34463365906052,30837284134268,

%U -10310751433852,105261086212083404572,-11719975655366668,1044330873985795459924,-6080390575672283355244

%N a(n) is the numerator of the coefficient c_(2n+1) in the expansion Sum_{k=1..j} 1/(k*(k+1)/2)^2 = Sum_{m>=0} c_m/j^m for large values of j.

%C The infinite sum of the reciprocals of the squares of the positive triangular numbers is Sum_{k>=1} 1/(k*(k+1)/2)^2 = 1/1^2 + 1/3^2 + 1/6^2 + 1/10^2 + ... = 4*Pi^2/3 - 12 (see A340216).

%C For large values of j, the finite sum Sum_{k=1..j} 1/(k*(k+1)/2)^2 = 1/1^2 + 1/3^2 + 1/6^2 + ... + 1/(j*(j+1)/2)^2 approaches 4*Pi^2/3 - 12 - (4/3)/j^3 + 4/j^4 - (116/15)/j^5 + 12/j^6 - (340/21)/j^7 + 20/j^8 - ...; this can be written as c_0 + c_1/j + c_2/j^2 + c_3/j^3 + ... + c_m/j^m + ... where the coefficients are as follows:

%C c_0 = 4*Pi^2/3 - 12

%C c_1 = 0 c_2 = 0

%C c_3 = -4/3 c_4 = 4

%C c_5 = -116/15 c_6 = 12

%C c_7 = -340/21 c_8 = 20

%C c_9 = -356/15 c_10 = 28

%C c_11 = -1076/33 c_12 = 36

%C c_13 = -51836/1365 c_14 = 44

%C c_15 = -172/3 c_16 = 52

%C c_17 = 188/255 c_18 = 60

%C c_19 = -201004/399 c_20 = 68

%C c_21 = 686564/165 c_22 = 76

%C c_23 = -3423572/69 c_24 = 84

%C c_25 = 945336244/1365 c_26 = 92

%C c_27 = -34212700/3 c_28 = 100

%C c_29 = 94997798876/435 c_30 = 108

%C c_31 = -34463365906052/7161 c_32 = 116

%C c_33 = 30837284134268/255 c_34 = 124

%C c_35 = -10310751433852/3 c_36 = 132

%C c_37 = 105261086212083404572/959595 c_38 = 140

%C c_39 = -11719975655366668/3 c_40 = 148

%C c_41 = 1044330873985795459924/6765 c_42 = 156

%C c_43 = -6080390575672283355244/903 c_44 = 164

%C ...

%C For even m > 2, c_m = 4*m - 12; for odd m = 2n+1, c_m appears to be a rational fraction with denominator A001897(n).

%e The sum of the reciprocals of the squares of the first 1000 positive triangular numbers is 1/1^2 + 1/3^2 + 1/6^2 + ... + 1/500500^2 = 1.159472533456470437096484166605...

%e .

%e M | Sum_{m=0..M} c_m/j^m | error

%e ---+--------------------------------+-------------------

%e 0 | 1.1594725347858114917793213... | -1.32934...*10^-09

%e 3 | 1.1594725334524781584459879... | 3.99227...*10^-12

%e 4 | 1.1594725334564781584459879... | -7.72134...*10^-15

%e 5 | 1.1594725334564704251126546... | 1.19838...*10^-17

%e 6 | 1.1594725334564704371126546... | -1.61704...*10^-20

%e 7 | 1.1594725334564704370964641... | 1.99762...*10^-23

%e 8 | 1.1594725334564704370964841... | -2.37053...*10^-26

%Y Cf. A000217, A000537, A001897, A340216.

%K sign,frac

%O 0,2

%A _Jon E. Schoenfield_, Jan 02 2021