OFFSET
0,1
COMMENTS
From Michel Lagneau, Mar 04 2015: (Start)
The sequence is infinite. Proof by induction:
Given an integer n>0. Suppose there are an infinity of perfect squares of the form k*2^n - 7 = b(n)^2 for some integer k.
For all integers n, there exists an integer b(n) such that b(n)^2 == -7 (mod 2^n). If n <=3, b(n)=1 is appropriate. Suppose there exists b(n) such that b(n)^2 == -7 (mod 2^n) for some n >=3. We prove that b(n+1)^2 == -7 (mod 2^(n+1)). In the first case, we take b(n+1) = b(n) if k is even, and in the second case we take b(n+1) = 2^(n-1)- b(n). Then b(n+1)^2 = b(n)^2 - 2^n*b(n) + 2^(2n-2) == b(n)^2 - 2^n*b(n) (mod 2^(n+1)).
But b(n)^2 = k*2^n-7 => b(n)^2 - 2^n*b(n) = 2^n(k-b(n)) - 7 == - 7 (mod 2^(n+1)) because k - b(n) is even (k is odd and b(n) is odd).
With n>=3 and b(n) odd, the proof is complete.
Finally, we see that the sequence b(n) is unbounded because b(n)^2 >=2^n - 7 for all positive integers n. This completes the proof because, for all p >=n, b(p)^2 == -7 (mod 2^n).
The corresponding squares of the sequence are 1, 1, 1, 3^2, 5^2, 11^2, 11^2, 53^2, 75^2, 181^2, 181^2, 181^2, 181^2, 181^2, 181^2, 16203^2, 16203^2, 49333^2, 49333^2, 49333^2, 474955^2, ...
With the property:
b(2) + b(3) = 1+1 = 2^1;
b(3) + b(4) = 1+3 = 2^2;
b(4) + b(5) = 3+5 = 2^3;
b(5) + b(6) = 5+11 = 2^4;
b(7) + b(8) = 11+53 = 2^6;
b(8) + b(9) = 53+75=2^7;
b(9) + b(10) = 75+181=2^8;
b(15) + b(16) = 181+16203=2^14;
(End)
To get a(n), find the smallest term of A117619 that is divisible by 2^n, and divide it by 2^n. - Michel Marcus, Mar 05 2015
a(n)=1 for n=3, 4, 5, 7, 15; see A060728. - Michel Marcus, Mar 05 2015
LINKS
Michel Lagneau, Proof
FORMULA
b(1) = b(2) = b(3) = 1; b(n+1)= 2^(n-1)-b(n) or b(n+1) = b(n).
k*2^n - 7 = b(n)^2 => a(n) = k = (b(n)^2 + 7)/2^n if b(n) different from b(n-1) or a(n) = a(n-1)/2 if a(n-1) is even (or if b(n) = b(n-1)).
MAPLE
A188628 := proc(n) for k from 1 do if issqr(k*2^n-7) then return k; end if; end do:
end proc:
for n from 0 do printf("%d, \n", A188628(n)) ; end do; # R. J. Mathar, Apr 09 2011
MATHEMATICA
f[n_] := Block[{k = 0}, While[!IntegerQ[Sqrt[k*2^n - 7]], k++]; k]; Table[f[n], {n, 0, 24}] (* Michael De Vlieger, Mar 04 2015 *)
PROG
(PARI) a(n) = {k=1; while (! issquare(k*2^n - 7), k++); k; } \\ Michel Marcus, Mar 04 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Apr 06 2011
EXTENSIONS
a(0) lowered from 8 to 7, a(25) from 2986676 to 1364479 by R. J. Mathar, Apr 09 2011
a(31)-a(38) from Michel Lagneau, Mar 04 2015
STATUS
approved