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

1, 3, 7, 14, 25, 43, 69, 110, 167, 255, 375, 558, 805, 1179, 1681, 2438, 3451, 4975, 7011, 10070, 14153, 20283, 28461, 40734, 57103, 81663, 114415, 163550, 229069, 327355, 458409, 654998, 917123, 1310319, 1834587, 2620998, 3669553, 5242395, 7339525, 10485230

Offset: 1

Author

Jeffrey R. Goodwin, Aug 08 2011

Keywords

• nonn
• easy

Examples

			Portion of the first three rows:
A033627, 2-Stohr  1  2  4  7
A026474, 3-Stohr  1  2  4  8
A051039, 4-Stohr  1  2  4  8
Thus a(1)=1, a(2)=2+1=3, and a(3)=4+2+1=7.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..2000
• Eric Weisstein's World of Mathematics, Stöhr Sequence.
• Index entries for linear recurrences with constant coefficients, signature (2,2,-6,1,4,-2).

Programs

Mathematica

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 *)
LinearRecurrence[{2,2,-6,1,4,-2},{1,3,7,14,25,43},40] (* Harvey P. Dale, Jun 20 2015 *)

Formula

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

So we get the sums from the piecewise function:

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);

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

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

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

From Alexander R. Povolotsky, Aug 09 2011: (Start)

G.f.: x*(1 + x - x^2)/((-1 + x)^3*(-1 - x + 2*x^2 + 2*x^3)).

a(n+4) = -2*a(n)+3*a(n+2)+n+5.

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)