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Irregular array where row n is the positive integers which divide the number of terms in all previous rows. a(1)=1.
1

%I #13 Jul 19 2016 11:08:34

%S 1,1,1,2,1,2,4,1,7,1,3,9,1,2,3,4,6,12,1,2,3,6,9,18,1,2,3,4,6,8,12,24,

%T 1,2,4,8,16,32,1,2,19,38,1,2,3,6,7,14,21,42,1,2,5,10,25,50,1,2,4,7,8,

%U 14,28,56,1,2,4,8,16,32,64,1,71,1,73,1,3,5,15,25,75,1,3,9,27,81,1,2,43,86

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

%e Array begins:

%e 1

%e 1

%e 1,2

%e 1,2,4

%e 1,7

%e 1,3,9

%e 12 terms make up these 6 rows. So row 7 is the divisors of 12, (1,2,3,4,6,12).

%p A119765 := proc(nmax) local a,dvs; a := [1] ; while nops(a) < nmax do dvs := numtheory[divisors](nops(a)) ; a := [op(a),op(dvs) ] ; od ; end: a := A119765(300) ; for i from 1 to nops(a) do printf("%d,",a[i]) ; od ; # _R. J. Mathar_, Jun 23 2006

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

%o (define (A119765 n seq)

%o (cond

%o [(= n 0) seq]

%o [else (A119765 (sub1 n) (append seq (reverse (positive-divisors (length seq)))))]))

%o (A119765 30 (list 1)) ;; _Joshua Zucker_, Jun 21 2006

%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