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A232505
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Sum of odd quadratic residues of prime(n).
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2
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1, 1, 1, 1, 18, 13, 38, 50, 26, 83, 66, 137, 224, 242, 147, 303, 509, 395, 578, 364, 714, 563, 965, 1046, 1254, 1155, 1043, 1565, 1323, 1676, 1667, 2440, 2456, 2589, 2563, 2284, 2827, 3362, 2526, 3503, 4408, 3765, 3271, 4902, 4557, 4005, 5829, 5380, 6952, 6093, 7046, 5288, 7626, 8691, 8552, 6871, 8563, 7622, 9007, 10250, 10365, 10233
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OFFSET
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1,5
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COMMENTS
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Seems to have no modular form.
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LINKS
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FORMULA
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EXAMPLE
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a(1), a(2), a(3) and a(4) are all 1, as for the corresponding primes 2, 3, 5 and 7 the quadratic residue sets are {1}, {1}, {1,4} and {1,2,4}, in which all cases, only 1 is an odd residue.
For a(5), which is computed for the 5th prime, 11, we have a set of its quadratic residues (those less than 11) as {1,3,4,5,9}, of which when we sum only the odd residues, 1+3+5+9, we get a(5) = 18.
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MATHEMATICA
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Table[Function[p, Total@ Select[Range[1, p, 2], JacobiSymbol[#, p] == 1 &]]@ Prime@ n, {n, 62}] (* Michael De Vlieger, May 14 2017 *)
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PROG
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(PARI) A232597(n) = {s=0; for(k=1, n, s=s+((k%2)*((1+kronecker(k, n))\2)*k)); return(s); }
(Python)
from sympy.ntheory.residue_ntheory import quadratic_residues as q
from sympy import prime
def a(n): return sum(i for i in q(prime(n)) if i%2)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Missing 1 (as a(1) is value for the first prime, 2) inserted into beginning by Antti Karttunen, Nov 26 2013
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STATUS
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approved
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