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A113780
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Number of solutions to 24*n+1 = x^2+24*y^2, x a positive integer, y an integer.
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4
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1, 3, 3, 2, 2, 3, 4, 1, 2, 4, 2, 4, 1, 2, 2, 1, 8, 2, 2, 2, 0, 4, 1, 4, 2, 2, 5, 4, 2, 0, 4, 4, 2, 0, 0, 3, 4, 4, 4, 2, 3, 4, 2, 2, 4, 0, 0, 2, 2, 4, 2, 9, 2, 0, 2, 2, 4, 1, 4, 0, 4, 4, 2, 0, 4, 4, 4, 2, 0, 2, 1, 8, 0, 2, 2, 2, 6, 1, 2, 4, 0, 4, 4, 2, 2, 0, 8, 2, 2, 2, 2, 0, 1, 8, 0, 2, 4, 0, 0, 2, 5, 6, 4, 2, 4
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| If 24*n+1 is not a square or if sqrt(24*n+1) == 1 or 11 (mod 12), then A000009(n) == a(n) (mod 4), otherwise A000009(n) == a(n) + 2 (mod 4).
Implied by the arithmetic of Q[sqrt(-6)]: Let 24*n+1 = p_1^e_1 * ... * p_r^e_r * q_1^f_1 * ... * q_s^f_s, where the p_i's are distinct primes == 1, 5, 7, or 11 (mod 24) and the q_i's are distinct primes == 13, 17, 19, or 23 (mod 24). If some f_i is odd, then a(n) = 0. Otherwise, a(n) = (e_1 + 1) * ... * (e_r + 1). a(n) == 2 (mod 4) iff all of the f_i's are even and all but one of the e_i's are even and the one e_i which is odd is == 1 (mod 4). Since A000009(n) and a(n) are both odd if 24*n+1 is a square, we can replace a by A000009 in this.
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EXAMPLE
| If n=51, the solutions (x,y) are: (7,+-7), (19,+-6), (25,+-5), (29,+-4), (35,0) so a(51)=9.
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PROG
| (PARI) {a(n)= if(n<0, 0, n=24*n+1; sumdiv(n, d, kronecker(-12, d)* kronecker(8, n/d)))} /* Michael Somos Mar 11 2007 */
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CROSSREFS
| Cf. A001318 generalized pentagonal numbers, indices of odd values of a(n) and A000009.
A128580(12n)= a(n).
Cf. A114913 = values k such that A000009(k) == 2 (mod 4) and such that a(k) == 2 (mod 4).
Sequence in context: A021305 A075788 A162235 * A007515 A014967 A120992
Adjacent sequences: A113777 A113778 A113779 * A113781 A113782 A113783
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KEYWORD
| nonn
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AUTHOR
| Christian G. Bower (bowerc(AT)usa.net), Jan 20 2006, based on a message from Dean Hickerson (dean.hickerson(AT)yahoo.com).
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