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A193911 Sums of the diagonals of the matrix formed by listing the h-Stohr sequences in increasing order.

%I A193911 #54 Aug 17 2025 21:11:02
%S A193911 1,3,7,14,25,43,69,110,167,255,375,558,805,1179,1681,2438,3451,4975,
%T A193911 7011,10070,14153,20283,28461,40734,57103,81663,114415,163550,229069,
%U A193911 327355,458409,654998,917123,1310319,1834587,2620998,3669553,5242395,7339525,10485230
%N A193911 Sums of the diagonals of the matrix formed by listing the h-Stohr sequences in increasing order.
%H A193911 Vincenzo Librandi, <a href="/A193911/b193911.txt">Table of n, a(n) for n = 1..2000</a>
%H A193911 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/StoehrSequence.html">Stöhr Sequence</a>.
%H A193911 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-6,1,4,-2).
%F A193911 All h-Stohr sequences have formula: h terms 1,2,..,2^(n-1),..,2^(h-1) and then continue (2^h-1)(n-h)+1. - _Henry Bottomley_, Feb 04 2000
%F A193911 So we get the sums from the piecewise function:
%F A193911 for odd n>=1, a(n)=2^((n+1)/2)-n+((n+1)/2)-2+Sum_{i=0..((n+1)/2)-1}(2*i+1)*(2^(((n+1)/2)-i) -1);
%F A193911 for even n>=2, a(n)=2^((n/2)+2)-n-4+Sum_{i=0..(n/2)-1}(2*i+1)*(2^((n/2)-i) -1). - _Jeffrey R. Goodwin_, Aug 09 2011
%F A193911 Let odd m>=3, then a(n)=a(m)-A000295(((m+1)/2)+1), where n>=2 is even. - _Jeffrey R. Goodwin_, Aug 09 2011
%F A193911 Let even m>=2, then a(n)=a(m)-A077802(m/2)=a(m)-A095151(m/2), where n>=1 is odd. - _Jeffrey R. Goodwin_, Aug 09 2011
%F A193911 From _Alexander R. Povolotsky_, Aug 09 2011: (Start)
%F A193911 G.f.: x*(1 + x - x^2)/((-1 + x)^3*(-1 - x + 2*x^2 + 2*x^3)).
%F A193911 a(n+4) = -2*a(n)+3*a(n+2)+n+5.
%F A193911 a(n) = 1/8*(2^(n/2+2)*((10-7*sqrt(2))*(-1)^n+10+7*sqrt(2))-(-1)^n-2*n*(n+12)-79). (End)
%e A193911 Portion of the first three rows:
%e A193911 A033627, 2-Stohr  1  2  4  7
%e A193911 A026474, 3-Stohr  1  2  4  8
%e A193911 A051039, 4-Stohr  1  2  4  8
%e A193911 Thus a(1)=1, a(2)=2+1=3, and a(3)=4+2+1=7.
%t A193911 A193911=t={0,1}; Do[AppendTo[t,t[[-2]]+t[[-1]]]; AppendTo[t,2*t[[-2]]],{n,41}]; Drop[Nest[Accumulate,t,2],1] (* _Vladimir Joseph Stephan Orlovsky_, Jan 27 2012 *)
%t A193911 LinearRecurrence[{2,2,-6,1,4,-2},{1,3,7,14,25,43},40] (* _Harvey P. Dale_, Jun 20 2015 *)
%K A193911 nonn,easy
%O A193911 1,2
%A A193911 _Jeffrey R. Goodwin_, Aug 08 2011