

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
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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(n1).
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(n1) 48*a(n3) +64*a(n4). 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(n1) 324*a(n3) +729*a(n4). 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(n1) 224*a(n2) +512*a(n3).
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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^(n1)]*[2^(n1)  3].
a(n) = 2^(n2)*(2^n6) for n>2. a(n) = 6*a(n1)8*a(n2) for n>4. G.f.: x*(32*x^34*x+1) / ((2*x1)*(4*x1)).  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^(51)]*[2^(51)  3]=16*13=208;


MATHEMATICA

Join[{1, 2}, Table[2^(n  1) (2^(n  1)  3), {n, 3, 20}]]


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



