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A187468
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Sum of the squares modulo 2^n of the odd numbers less than 2^n.
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2
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1, 2, 4, 40, 208, 928, 3904, 16000, 64768, 260608, 1045504, 4188160, 16764928, 67084288, 268386304, 1073643520, 4294770688, 17179475968, 68718690304, 274876334080, 1099508482048, 4398040219648, 17592173461504, 70368719011840, 281474926379008
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OFFSET
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1,2
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COMMENTS
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There is a simple formula for this case; the sum of the squares of the odd numbers less than 2^n is A016131(n-1).
Can the general case for m^n, m > 2 be calculated with a formula of the same kind?
For n>=3, the sum of the squares of the even numbers less than 2^n (each square mod 2^n) are 8 times the sequence 1, 2, 12, 56, 304, 1376, 6336, 27008 etc. and appear to obey a(n)= +6*a(n-1) -48*a(n-3) +64*a(n-4). For n>=1, the sum of the squares of the odd numbers less than 3^n (modulo 3^n) start as 2 times 1, 12, 144, 1404, 13689, 126360,.. and apparently obey a(n)= +12*a(n-1) -324*a(n-3) +729*a(n-4). For n>=1, the sum of the squares of the odd numbers less than 4^n (modulo 4^n) start as 2 times 1, 28, 688, 13504, 238336,... and seem to obey a(n)= +28*a(n-1) -224*a(n-2) +512*a(n-3).
(End)
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LINKS
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FORMULA
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For n>2 the sum of all r_j = (c_j)^2 mod 2^n for a particular n is given by [2^(n-1)]*[2^(n-1) - 3].
a(n) = 2^(n-2)*(2^n-6) for n>2. a(n) = 6*a(n-1)-8*a(n-2) for n>4. G.f.: x*(32*x^3-4*x+1) / ((2*x-1)*(4*x-1)). - Colin Barker, Aug 19 2013
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EXAMPLE
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For n=5, 2^5=32. The c_j, numbers prime to 32 are the odd numbers
less than 32. The r_j = (c_j)^2 mod 32 are 1,9,25,17,17,25,9,1,1,9,25,17,17,25,9,1=4*52=208.
From the formula, for n=5, [2^(5-1)]*[2^(5-1) - 3]=16*13=208;
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MATHEMATICA
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Join[{1, 2}, Table[2^(n - 1) (2^(n - 1) - 3), {n, 3, 20}]]
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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