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

%I A188628 #47 Mar 10 2015 02:01:46
%S A188628 7,4,2,1,1,1,2,1,11,11,32,16,8,4,2,1,4006,2003,9284,4642,2321,107566,
%T A188628 53783,313702,156851,1364479,1493338,746669,12145148,6072574,3036287,
%U A188628 107186842,53593421,323781196,161890598,80945299,3501584548,1750792274,875396137
%N A188628 a(n) = smallest k such that k*2^n - 7 is a square.
%C A188628 From _Michel Lagneau_, Mar 04 2015: (Start)
%C A188628 The sequence is infinite. Proof by induction:
%C A188628 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.
%C A188628 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)).
%C A188628 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).
%C A188628 With n>=3 and b(n) odd, the proof is complete.
%C A188628 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).
%C A188628 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, ...
%C A188628 With the property:
%C A188628 b(2) + b(3) = 1+1 = 2^1;
%C A188628 b(3) + b(4) = 1+3 = 2^2;
%C A188628 b(4) + b(5) = 3+5 = 2^3;
%C A188628 b(5) + b(6) = 5+11 = 2^4;
%C A188628 b(7) + b(8) = 11+53 = 2^6;
%C A188628 b(8) + b(9) = 53+75=2^7;
%C A188628 b(9) + b(10) = 75+181=2^8;
%C A188628 b(15) + b(16) = 181+16203=2^14;
%C A188628 (End)
%C A188628 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
%C A188628 a(n)=1 for n=3, 4, 5, 7, 15; see A060728. - _Michel Marcus_, Mar 05 2015
%H A188628 Michel Lagneau, <a href="/A188628/a188628.pdf">Proof</a>
%F A188628 b(1) = b(2) = b(3) = 1; b(n+1)= 2^(n-1)-b(n) or b(n+1) = b(n).
%F A188628 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)).
%p A188628 A188628 := proc(n) for k from 1 do if issqr(k*2^n-7) then return k; end if; end do:
%p A188628 end proc:
%p A188628 for n from 0 do printf("%d,\n",A188628(n)) ; end do; # _R. J. Mathar_, Apr 09 2011
%t A188628 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 *)
%o A188628 (PARI) a(n) = {k=1; while (! issquare(k*2^n - 7), k++); k;} \\ _Michel Marcus_, Mar 04 2015
%Y A188628 Cf. A117619 (n^2+7).
%K A188628 nonn
%O A188628 0,1
%A A188628 _Michel Lagneau_, Apr 06 2011
%E A188628 a(0) lowered from 8 to 7, a(25) from 2986676 to 1364479 by _R. J. Mathar_, Apr 09 2011
%E A188628 a(31)-a(38) from _Michel Lagneau_, Mar 04 2015