A119763 - OEIS

%I #16 Jul 19 2016 11:08:38

%S 1,1,1,2,1,5,1,11,1,23,1,47,1,5,19,95,1,5,43,215,1,479,1,7,137,959,1,

%T 2063,1,4127,1,5,13,65,127,635,1651,8255,1,83,229,19007,1,38327,1,5,

%U 15331,76655,1,137,1231,168647,1,13,109,239,1417,3107,26051,338663,1,19

%N Irregular array where row n is the positive integers which divide the sum of all previous rows. a(1)=1.

%F a(n,k) | sum(j=1..n-1,l=1,2,...) a(j,l). a(n,k) > a(n,k-1). a(n,1)=1. - _R. J. Mathar_, Jun 23 2006

%e Array begins:

%e 1

%e 1

%e 1,2

%e 1,5

%e 1,11

%e 1,23

%e 1,47

%e 1,5,19,95

%e The sum of these terms is 215.

%e Since the divisors 215 are 1,5,43 and 215, row 9 =(1,5,43,215).

%p A119763 := proc(nmax) local a,dvs; a := [1] ; while nops(a) < nmax do dvs := numtheory[divisors](sum('a[i]',i=1..nops(a))) ; a := [op(a),op(dvs) ] ; od ; end: a := A119763(200) ; for i from 1 to nops(a) do printf("%d,",a[i]) ; od ; # _R. J. Mathar_, Jun 23 2006

%t Module[{arr={1}},Do[AppendTo[arr,Divisors[Total[arr]]];arr=Flatten[arr],{20}];arr] (* _Harvey P. Dale_, May 27 2012 *)

%o (PLT Scheme) ;;positive-divisors gives the list of divisors of n in decreasing order

%o (define (A119763 n seq)

%o (cond

%o [(= n 0) seq]

%o [else (A119763 (sub1 n) (append seq (reverse (positive-divisors (apply + seq)))))]))

%o (A119763 20 (list 1)) ;; _Joshua Zucker_, Jun 21 2006

%Y Cf. A027750.

%K nonn,tabf

%O 1,4

%A _Leroy Quet_, Jun 18 2006

%E More terms from _Joshua Zucker_ and _R. J. Mathar_, Jun 21 2006