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%I #20 Mar 12 2021 22:24:43
%S 1,1,0,1,0,0,1,0,0,0,1,0,1,1,0,2,0,0,1,0,0,1,1,0,0,0,0,1,1,0,0,0,0,1,
%T 0,0,2,1,0,1,1,0,1,0,0,1,1,0,1,0,0,1,0,0,0,1,0,2,0,0,0,0,0,0,1,0,1,1,
%U 0,0,0,0,2,1,0,1,0,0,3,0,0,1,1,0,0,0,0,1,0,0,1,2,0,1,0,0,0,0,0,0,1,0,1,1,0
%N Number of representations of n as a sum of a triangular number and twelve times a triangular number.
%C Ramanujan theta functions: f(q) := Prod_{k>=1} (1-(-q)^k) (see A121373), phi(q) := theta_3(q) := Sum_{k=-oo..oo} q^(k^2) (A000122), psi(q) := Sum_{k=0..oo} q^(k*(k+1)/2) (A010054), chi(q) := Prod_{k>=0} (1+q^(2k+1)) (A000700).
%H G. C. Greubel, <a href="/A112607/b112607.txt">Table of n, a(n) for n = 0..1000</a>
%H M. D. Hirschhorn, <a href="http://dx.doi.org/10.1016/j.disc.2004.08.045">The number of representations of a number by various forms</a>, Discrete Mathematics 298 (2005), 205-211
%H Michael Somos, <a href="/A010815/a010815.txt">Introduction to Ramanujan theta functions</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/RamanujanThetaFunctions.html">Ramanujan Theta Functions</a>
%F a(n) = 1/2*( d_{1, 3}(8n+13) - d_{2, 3}(8n+13) ) where d_{a, m}(n) equals the number of divisors of n which are congruent to a mod m.
%F Expansion of q^(-13/8)*(eta(q^2)*eta(q^24))^2/(eta(q)*eta(q^12)) in powers of q. - _Michael Somos_, Sep 29 2006
%F Expansion of psi(q)*psi(q^12) in powers of q where psi() is a Ramanujan theta function. - _Michael Somos_, Sep 29 2006
%F Euler transform of period 24 sequence [ 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, 0, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -2, ...]. - _Michael Somos_, Sep 29 2006
%F a(3n+2)=0. - _Michael Somos_, Sep 29 2006
%e a(15) = 2 since we can write 15 = 15 + 12*0 = 3 + 12*1.
%t a[n_] := DivisorSum[8n+13, KroneckerSymbol[-3, #]&]/2; Table[a[n], {n, 0, 104}] (* _Jean-François Alcover_, Dec 04 2015, adapted from PARI *)
%o (PARI) {a(n)=if(n<0, 0, n=8*n+13; sumdiv(n, d, kronecker(-3,d))/2)} /* _Michael Somos_, Sep 29 2006 */
%o (PARI) {a(n)=local(A); if(n<0, 0, A=x*O(x^n); polcoeff( eta(x^2+A)^2*eta(x^24+A)^2/eta(x+A)/eta(x^12+A), n))} /* _Michael Somos_, Sep 29 2006 */
%Y A123484(24n+15) = 2*a(n). A112609(3n+4) = a(n).
%K nonn
%O 0,16
%A _James A. Sellers_, Dec 21 2005
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