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A159927 Triangle read by rows: a(1,1) = 1. a(m,m) = sum of all terms in rows 1 through m-1. a(m,n) = a(m-1,n) + (sum of all terms in rows 1 through m-1), for n < m.

%I A159927 #13 Feb 06 2019 00:34:04
%S A159927 1,2,1,6,5,4,25,24,23,19,135,134,133,129,110,886,885,884,880,861,751,
%T A159927 6784,6783,6782,6778,6759,6649,5898,59115,59114,59113,59109,59090,
%U A159927 58980,58229,52331,576527,576526,576525,576521,576502,576392,575641,569743
%N A159927 Triangle read by rows: a(1,1) = 1. a(m,m) = sum of all terms in rows 1 through m-1. a(m,n) = a(m-1,n) + (sum of all terms in rows 1 through m-1), for n < m.
%C A159927 A159928(m) = -A075374(m+4)+A075374(m+3), for m >= 1. -A075374(m+4) = the sum of all terms of triangle A159927 in rows 1 through m. A159928 contains the row-sums of triangle A159927.
%H A159927 Michael De Vlieger, <a href="/A159927/b159927.txt">Table of n, a(n) for n = 1..11476</a> (Rows 1 <= n <= 150).
%e A159927 The triangle starts like this:
%e A159927    1;
%e A159927    2,  1;
%e A159927    6,  5,  4;
%e A159927   25, 24, 23, 19;
%e A159927 The sum of all of these terms is 110. Adding 110 to each term of the 4th row, we get: 25+110=135, 24+110=134, 23+110=133, 19+110=129, 0+110=110. So row 5 is 135,134,133,129,110.
%p A159927 A159927 := proc(n,m) option remember; local rs; if n = 1 then 1; else rs := add(add( procname(i,j),j=1..i),i=1..n-1) ; if n = m then rs; else procname(n-1,m)+rs; fi; fi; end: for n from 1 to 10 do for m from 1 to n do printf("%d,",A159927(n,m)) ; od: od: # _R. J. Mathar_, Apr 28 2009
%t A159927 NestList[{#1 + #2, #2} & @@ {Join[#1, {0}], Total[#1] + #2} & @@ # &, {{1}, 0}, 8][[All, 1]] // Flatten (* _Michael De Vlieger_, Aug 30 2017 *)
%Y A159927 Cf. A159924, A159928, A075374.
%K A159927 nonn,tabl
%O A159927 1,2
%A A159927 _Leroy Quet_, Apr 26 2009
%E A159927 More terms from _R. J. Mathar_, Apr 28 2009