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A187468 Sum of the squares modulo 2^n of the odd numbers less than 2^n. 2
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
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?
From R. J. Mathar, Aug 26 2013: (Start)
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)
LINKS
FORMULA
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
EXAMPLE
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;
MATHEMATICA
Join[{1, 2}, Table[2^(n - 1) (2^(n - 1) - 3), {n, 3, 20}]]
CROSSREFS
Sequence in context: A184952 A098337 A326483 * A238719 A158213 A012596
KEYWORD
nonn,easy
AUTHOR
J. M. Bergot, Mar 22 2011
EXTENSIONS
Heavily edited by Olivier Gérard, Mar 23 2011
More terms from Colin Barker, Aug 19 2013
STATUS
approved

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