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%I #11 Aug 01 2023 07:41:51
%S 1,17,7,7,56,161,30,95,1020,271,90,802,1452,790,161,7481,309,5882,
%T 8400,980,10260,6850,11152,6081,15903,3589,18221,2023,32166,32805,
%U 9926,110565,18518,4114,29188,4225,83902,47859,70552,89332,37916,93302,34920
%N Group the triangular numbers so that the n-th group sum is a multiple of n. 1, (3, 6, 10, 15), (21), (28), (36, 45, 55, 66, 78), (91, 105, 120, 136, 153, 171, 190), ... Sequence contains n-th group sum divided by n.
%F A114031(n)/n.
%p A000217 := proc(n) option remember ; n*(n+1)/2 ; end: A114031 := proc(n) local strt,su,len,i; strt := 1 ; for su from 1 to n do len := 1; while add( A000217(strt+i),i=0..len-1) mod su <> 0 do len := len+1 ; od ; if su = n then RETURN( add( A000217(strt+i),i=0..len-1) ) ; else strt := strt+len ; fi ; od ; end: A114032 := proc(n) A114031(n)/n ; end: for n from 1 to 80 do printf("%d, ",A114032(n)) ; od ; # _R. J. Mathar_, May 10 2007
%t t[n_] := n(n+1)/2;
%t a[n_] := Module[{strt, su, len}, strt = 1; For[su = 1, True, su++, len = 1; While[Mod[Sum[t[strt + i], {i, 0, len - 1}], su] != 0, len++]; If[su == n, Return[Sum[t[strt + i], {i, 0, len - 1}]/n], strt += len]]];
%t Table[a[n], {n, 1, 43}] (* _Jean-François Alcover_, Aug 01 2023, after _R. J. Mathar_ *)
%Y Cf. A114032, A114033.
%K nonn
%O 1,2
%A _Amarnath Murthy_, Nov 13 2005
%E More terms from _R. J. Mathar_, May 10 2007
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