cp's OEIS Frontend

A350851 Cumulative sums of the first ceiling(n/2)+1 elements of rows 0 to n in Pascal's triangle.

%I A350851 #12 Jan 22 2022 19:44:31
%S A350851 1,3,6,13,24,50,92,191,354,736,1374,2860,5370,11182,21090,43909,83112,
%T A350851 172958,328340,682862,1299528,2700820,5150688,10697070,20437756,
%U A350851 42415272,81170004,168337168,322613196,668607412,1283037084,2657319103,5105342946,10567113352,20323851054
%N A350851 Cumulative sums of the first ceiling(n/2)+1 elements of rows 0 to n in Pascal's triangle.
%e A350851 The first ceiling(n/2)+1 elements from the first four rows of Pascal's are:
%e A350851      1
%e A350851     1 1
%e A350851    1 2
%e A350851   1 3 3
%e A350851 So a(0)=1, a(1)=a(0)+1+1=3, a(2)=a(1)+1+2=6, a(3)=a(2)+1+3+3=13.
%o A350851 (Python)
%o A350851 seq=[];prev=[];total=0
%o A350851 for n in range(30):
%o A350851   row=[1]
%o A350851   last=int(n/2)
%o A350851   for k in range(last):
%o A350851     row.append(prev[k]+prev[k+1])
%o A350851   if n%2==1:
%o A350851     row.append(row[-1])
%o A350851   prev=row
%o A350851   total+=sum(row)
%o A350851   seq.append(total)
%o A350851 print(seq)
%Y A350851 Cf. A007318, A116496 (for n>=2, first differences).
%K A350851 nonn,easy
%O A350851 0,2
%A A350851 _J. Stauduhar_, Jan 18 2022