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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. 1

%I #17 Mar 30 2012 17:26:36

%S 0,0,2,14,76,388,1950,9762,48824,244136,1220698,6103510,30517572,

%T 152587884,762939446,3814697258,19073486320,95367431632,476837158194,

%U 2384185791006,11920928955068,59604644775380,298023223876942,1490116119384754,7450580596923816,37252902984619128

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

%C Corresponding formulas for several first primes:

%C p=3, a(n)=(3*3^n+2*n+1)/4 (A047926)

%C p=5, a(n)=(5^n-4n-1)/8 (A185055)

%C p=7, a(n)=(7^n-1)/6

%C p=11, a(n)=(3*11^n+10*n-3)/20

%C p=13, a(n)=(13^n-4*n-1)/8

%C p=17, a(n)=(17^n-1)/8

%C p=19, a(n)=(5*19^n+18n-5)/36

%C p=23, a(n)=3(23^n-1)/22

%C p=29, a(n)=(29^n-4n-1)/8

%C p=31, a(n)=2(31^n-1)/15

%C p=37, a(n)=(37^n-4*n-1)/8

%C p=41, a(n)=(41^n-1)/8

%C p=43, a(n)=(11*43^n+42n-11)/84

%C p=47, a(n)=(3(47^n-1)/23.

%C General formulas for a(n) depend on p mod 8 as follows:

%C p= 1 mod 8 , a(n)=(p^n-1)/8

%C p = 3 mod 8, a(n)=((p + 1)*p^n + 4*(p - 1)*n - (p + 1))/(8*(p - 1))

%C p = 5 mod 8, a(n)=(p^n-4*n-1)/8

%C p = 7 mod 8, a(n)=((p + 1)*(p^n - 1))/(8*(p - 1)).

%F a(n) = (5^n-4n-1)/8.

%e a(2)=2 because 25^2 = 9^2+12^2+20^2 = 12^2+15^2+16^2.

%Y Cf. A047926.

%K nonn

%O 0,3

%A _Zak Seidov_, Mar 02 2012

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