cp's OEIS Frontend

A109965 Sum_i {i0) = 1.

%I A109965 #15 Oct 31 2018 17:13:47
%S A109965 1,1,2,3,4,6,8,10,13,16,20,24,28,33,38,44,50,57,64,72,80,88,97,106,
%T A109965 116,126,137,148,160,172,185,198,212,226,241,256,272,288,304,321,338,
%U A109965 356,374,393,412,432,452,473,494,516,538,561,584,608,632,657,682,708,734
%N A109965 Sum_i {i<n} floor(sqrt(a(i))) with a(0) = 1.
%C A109965 The recursion to generate this sequence (excluding the additional extra 1 at the outset) occurs in Chapter 3, Exercise 28, page 97 in Graham, Knuth and Patashnik, Concrete Mathematics, 2nd Edition, Addison Wesley, 1994. A solution is provided on page 509. - Steve Tanny (tanny(AT)math.utoronto.ca), Apr 02 2008
%F A109965 a(n) = a(n-1)+floor(sqrt(a(n-1))) = a(n-1)+A109964(n-1) for n>1.
%F A109965 Contribution from _Paul Weisenhorn_, Jun 26 2010: (Start)
%F A109965 a(2^(j+1)+j+2*k)=2^(2*j)+2^j*(2*k+1)+k*(k-1);
%F A109965 a(2^(j+1)+j+2*k+1)=2^(2*j)+2^j*(2*k+2)+k^2;
%F A109965 a(2^(j+1)+j-1)=2^(2*j); j=0..infinity; k=0..(2^j-1). (End)
%e A109965 a(5) = floor(sqrt(1)) + floor(sqrt(1)) + floor(sqrt(2)) + floor(sqrt(3)) + floor(sqrt(4)) = 1 + 1 + 1 + 1 + 2 = 6.
%e A109965 j=3, k=5: a(29)=172, a(30)=185. [_Paul Weisenhorn_, Jun 26 2010]
%p A109965 a(0):=1: c:=0: for n from 1 to 100 do
%p A109965 a(n):=a(n-1)+c: c:=floor(sqrt(a(n))): end do: # _Paul Weisenhorn_, Jun 22 2010
%p A109965 a(0)=a(1)=b(0)=1;
%p A109965 for n from 1 to 100 do
%p A109965 b(n)=floor(sqrt(a(n))): a(n+1)=a(n)+b(n): end do:
%p A109965 a(n)=A109965(n); b(n)=A109964(n); # _Paul Weisenhorn_, Jun 26 2010
%t A109965 Prepend[RecurrenceTable[{a[n] == a[n - 1] + Floor[a[n - 1]^(1/2)], a[0] == 1}, a, {n, 0, 57}], 1] (* _Geoffrey Critzer_, May 25 2013 *)
%t A109965 Join[{1},NestList[#+Floor[Sqrt[#]]&,1,60]] (* _Harvey P. Dale_, Oct 31 2018 *)
%Y A109965 Essentially the same as A002984.
%Y A109965 Cf. A109964.
%K A109965 nonn
%O A109965 0,3
%A A109965 _Henry Bottomley_, Jul 06 2005