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A185055 Number of representations of 5^(2n) as a sum a^2 + b^2 + c^2 with 0 < a <= b <= c.

%I A185055 #31 Oct 17 2024 08:27:25
%S A185055 0,0,2,14,76,388,1950,9762,48824,244136,1220698,6103510,30517572,
%T A185055 152587884,762939446,3814697258,19073486320,95367431632,476837158194,
%U A185055 2384185791006,11920928955068,59604644775380,298023223876942,1490116119384754,7450580596923816,37252902984619128
%N A185055 Number of representations of 5^(2n) as a sum a^2 + b^2 + c^2 with 0 < a <= b <= c.
%C A185055 Corresponding formulas for several first primes:
%C A185055 p=3, a(n)=(3*3^n+2*n+1)/4 (A047926)
%C A185055 p=5, a(n)=(5^n-4*n-1)/8 (A185055)
%C A185055 p=7, a(n)=(7^n-1)/6
%C A185055 p=11, a(n)=(3*11^n+10*n-3)/20
%C A185055 p=13, a(n)=(13^n-4*n-1)/8
%C A185055 p=17, a(n)=(17^n-1)/8
%C A185055 p=19, a(n)=(5*19^n+18*n-5)/36
%C A185055 p=23, a(n)=3*(23^n-1)/22
%C A185055 p=29, a(n)=(29^n-4*n-1)/8
%C A185055 p=31, a(n)=2*(31^n-1)/15
%C A185055 p=37, a(n)=(37^n-4*n-1)/8
%C A185055 p=41, a(n)=(41^n-1)/8
%C A185055 p=43, a(n)=(11*43^n+42*n-11)/84
%C A185055 p=47, a(n)=3*(47^n-1)/23.
%C A185055 General formulas for a(n) depend on p mod 8 as follows:
%C A185055 p = 1 mod 8, a(n)=(p^n-1)/8
%C A185055 p = 3 mod 8, a(n)=((p + 1)*p^n + 4*(p - 1)*n - (p + 1))/(8*(p - 1))
%C A185055 p = 5 mod 8, a(n)=(p^n-4*n-1)/8
%C A185055 p = 7 mod 8, a(n)=((p + 1)*(p^n - 1))/(8*(p - 1)).
%H A185055 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-11,5).
%F A185055 a(n) = (5^n-4n-1)/8.
%F A185055 From _Chai Wah Wu_, Jun 07 2024: (Start)
%F A185055 a(n) = 7*a(n-1) - 11*a(n-2) + 5*a(n-3) for n > 2.
%F A185055 G.f.: -2*x^2/((x - 1)^2*(5*x - 1)). (End)
%F A185055 a(n) = 2 * A014827(n-1) for n >= 2. - _Alois P. Heinz_, Jun 07 2024
%e A185055 a(2)=2 because 25^2 = 9^2+12^2+20^2 = 12^2+15^2+16^2.
%Y A185055 Cf. A014827, A047926.
%K A185055 nonn,easy
%O A185055 0,3
%A A185055 _Zak Seidov_, Mar 02 2012