A188628 a(n) = smallest k such that k*2^n - 7 is a square.

7, 4, 2, 1, 1, 1, 2, 1, 11, 11, 32, 16, 8, 4, 2, 1, 4006, 2003, 9284, 4642, 2321, 107566, 53783, 313702, 156851, 1364479, 1493338, 746669, 12145148, 6072574, 3036287, 107186842, 53593421, 323781196, 161890598, 80945299, 3501584548, 1750792274, 875396137

Offset: 0

Author

Michel Lagneau, Apr 06 2011

Keywords

• nonn

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

Crossrefs

Cf. A117619 (n^2+7).

Programs

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 *)

PARI

a(n) = {k=1; while (! issquare(k*2^n - 7), k++); k;} \\ Michel Marcus, Mar 04 2015

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)).

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