|
%I #45 Sep 08 2022 08:46:19
%S 1,1,3,1,4,1,3,7,1,6,1,3,4,12,1,8,1,3,7,15,1,4,13,1,3,6,18,1,12,1,3,4,
%T 7,12,28,1,14,1,3,8,24,1,4,6,24,1,3,7,15,31,1,18,1,3,4,12,13,39,1,20,
%U 1,3,7,6,18,42,1,4,8,32,1,3,12,36,1,24,1,3,4,7
%N Triangle read by rows in which row n lists the sum of the divisors of each divisor of n.
%C Or, in the triangle A027750(n), replace each element with the sum of its divisors.
%C The row whose index x is a prime power p^m (p prime and m >= 0) is equal to (1, sigma(p), sigma(p^2), ..., sigma(p^(m-1))).
%C We observe the following properties of row n when n is the product of k distinct primes, k = 1,2,...:
%C when n = prime(m), row n = (1, prime(m)+1);
%C when n is the product of two distinct primes p < q, row n = (1, p+1, q+1,(p+1)(q+1));
%C when n is the product of three distinct primes p < q < r, row n = (1, p+1, q+1, r+1, (p+1)(q+1), (p+1)(r+1), (q+1)(r+1), sigma(p*q*r));
%H Michel Marcus, <a href="/A290478/b290478.txt">Table of n, a(n) for n = 1..10006</a> (rows 1 to 1358, flattened).
%F a(n) = sigma(A027750(n)).
%e Row 6 is (a(11), a(12), a(13), a(14)) = (1, 3, 4, 12) because sigma(A027750(11))= sigma(1) = 1, sigma(A027750(12))= sigma(2) = 3, sigma(A027750(13))= sigma(3) = 4 and sigma(A027750(14)) = sigma(6) = 12.
%e Triangle begins:
%e 1;
%e 1, 3;
%e 1, 4;
%e 1, 3, 7;
%e 1, 6;
%e 1, 3, 4, 12;
%e 1, 8;
%e 1, 3, 7, 15;
%e 1, 4, 13;
%e 1, 3, 6, 18;
%e ...
%p with(numtheory):nn:=100:
%p for n from 1 to nn do:
%p d1:=divisors(n):n1:=nops(d1):
%p for i from 1 to n1 do:
%p s:=sigma(d1[i]):
%p printf(`%d, `,s):
%p od:
%p od:
%t Array[DivisorSigma[1, Divisors@ #] &, 24] // Flatten (* _Michael De Vlieger_, Aug 07 2017 *)
%o (PARI) row(n) = apply(sigma, divisors(n)); \\ _Michel Marcus_, Dec 27 2021
%o (Magma) [[SumOfDivisors(d): d in Divisors(n)]: n in [1..20]]; // _Vincenzo Librandi_, Sep 08 2017
%Y Cf. A000203, A027750, A290532.
%Y Cf. A007429 (row sums), A206032 (row products).
%K nonn,tabf
%O 1,3
%A _Michel Lagneau_, Aug 03 2017
|
