|
| |
|
|
A127093
|
|
Triangle read by rows: T(n,k)=k if k is a divisor of n; otherwise, T(n,k)=0 (1 <= k <= n).
|
|
61
|
|
|
|
1, 1, 2, 1, 0, 3, 1, 2, 0, 4, 1, 0, 0, 0, 5, 1, 2, 3, 0, 0, 6, 1, 0, 0, 0, 0, 0, 7, 1, 2, 0, 4, 0, 0, 0, 8, 1, 0, 3, 0, 0, 0, 0, 0, 9, 1, 2, 0, 0, 5, 0, 0, 0, 0, 10, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11, 1, 2, 3, 4, 0, 6, 0, 0, 0, 0, 0, 12, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 13, 1, 2, 0, 0, 0, 0, 7, 0, 0, 0, 0, 0, 0, 14
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,3
|
|
|
COMMENTS
|
Sum of terms in row n = sigma(n) (sum of divisors of n).
Euler's derivation of A127093 in polynomial form is in his proof of the formula for Sigma(n): (let S=Sigma, then Euler proved that S(n) = S(n-1) + S(n-2) - S(n-5) - S(n-7) + S(n-12) + S(n-15) - S(n-22) - S(n-26), ...).
[Young, pp. 365-366], Euler begins, s = (1-x)*(1-x^2)*(1-x^3)*... = 1 - x - x^2 + x^5 + x^7 - x^12 ...; log s = log(1-x) + log(1-x^2) + log(1-x^3) ...; differentiating and then changing signs, Euler has t = x/(1-x) + 2x^2/(1-x^2) + 3x^3/(1-x^3) + 4x^4/(1-x^4) + 5x^5/(1-x^5) + ...
Finally, Euler expands each term of t into a geometric series, getting A127093 in polynomial form: t =
x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + ...
+ 2x^2 + 2x^4 + 2x^6 + 2x^8 + ...
+ 3x^3 + 3x^6 + ...
+ 4x^4 + 4x^8 + ...
+ 5x^5 + ...
+ 6x^6 + ...
+ 7x^7 + ...
+ 8x^8 + ...
T(n,k) is the sum of all the k-th roots of unity each raised to the n-th power. - Geoffrey Critzer, Jan 02 2016
For n > 1, A020639(n) is the leftmost term, other than 0 or 1, in the n-th row of this array. As mentioned in the Formula section, the k-th column is period k: repeat [k, 0, 0, ..., 0], but this also means that it's the characteristic function of the multiples of k multiplied by k. T(n,1) = A000012(n), T(n,2) = 2*A059841(n), T(n,3) = 3*A079978(n), T(n,4) = 4*A121262(n), T(n,5) = 5*A079998(n), and so on.
The terms in the n-th row, other than 0, are the factors of n. If n > 1 and for every k, 1 <= k < n, T(n,k) = 0 or 1, then n is prime. (End)
Row terms of the triangle can be used to calculate E(n) in A002654): (1, 1, 0, 1, 2, 0, 0, 1, 1, 2, ...), and A004018, the number of points in a square lattice on the circle of radius sqrt(n), A004018: (1, 4, 4, 0, 4, 8, 0, 0, 4, ...).
As to row terms in the triangle, let E(n) of even terms = 0,
E(integers of the form 4*k - 1 = (-1), and E(integers of the form 4*k + 1 = 1.
Then E(n) is the sum of the E(n)'s of the factors of n in the triangle rows. Example: E(10) = Sum: ((E(1) + E(2) + E(5) + E(10)) = ((1 + 0 + 1 + 0) = 2, matching A002654(10).
The total numbers of lattice points = 4r^2 = E(1) + ((E(2))/2 + ((E(3))/3 + ((E(4))/4 + ((E(5))/5 + .... Since E(even integers) are zero, E(integers of the form (4*k - 1)) = (-1), and E(integers of the form (4*k + 1)) = (+1); we are left with 4r^2 = 1 - 1/3 + 1/5 - 1/7 + 1/9 - ..., which is approximately equal to Pi(r^2). (End)
T(n,k) is also the number of parts in the partition of n into k equal parts. - Omar E. Pol, May 05 2020
|
|
|
REFERENCES
|
David Wells, "Prime Numbers, the Most Mysterious Figures in Math", John Wiley & Sons, 2005, appendix.
L. Euler, "Discovery of a Most Extraordinary Law of the Numbers Concerning the Sum of Their Divisors"; pp. 358-367 of Robert M. Young, "Excursions in Calculus, An Interplay of the Continuous and the Discrete", MAA, 1992. See p. 366.
|
|
|
LINKS
|
Eric Weisstein's World of Mathematics, Divisor
|
|
|
FORMULA
|
k-th column is composed of "k" interspersed with (k-1) zeros.
Let M = A127093 as an infinite lower triangular matrix and V = the harmonic series as a vector: [1/1, 1/2, 1/3, ...]. then M*V = d(n), A000005: [1, 2, 2, 3, 2, 4, 2, 4, 3, 4, ...]. M^2 * V = A060640: [1, 5, 7, 17, 11, 35, 15, 49, 34, 55, ...]. - Gary W. Adamson, May 10 2007
T(n,k) = ((n-1) mod k) - (n mod k) + 1 (1 <= k <= n). - Mats Granvik, Aug 31 2007
G.f.: Sum_{k>=1} k * x^k * y^k/(1-x^k) = Sum_{m>=1} x^m * y/(1 - x^m*y)^2. - Robert Israel, Aug 08 2016
|
|
|
EXAMPLE
|
T(8,4) = 4 since 4 divides 8.
T(9,3) = 3 since 3 divides 9.
First few rows of the triangle:
1;
1, 2;
1, 0, 3;
1, 2, 0, 4;
1, 0, 0, 0, 5;
1, 2, 3, 0, 0, 6;
1, 0, 0, 0, 0, 0, 7;
1, 2, 0, 4, 0, 0, 0, 8;
1, 0, 3, 0, 0, 0, 0, 0, 9;
...
|
|
|
MAPLE
|
A127093:=proc(n, k) if type(n/k, integer)=true then k else 0 fi end:
for n from 1 to 16 do seq(A127093(n, k), k=1..n) od; # yields sequence in triangular form - Emeric Deutsch, Jan 20 2007
|
|
|
MATHEMATICA
|
t[n_, k_] := k*Boole[Divisible[n, k]]; Table[t[n, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 17 2014 *)
Table[ SeriesCoefficient[k*x^k/(1 - x^k), {x, 0, n}], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, Apr 14 2015 *)
|
|
|
PROG
|
(Excel) mod(row()-1; column()) - mod(row(); column()) + 1 - Mats Granvik, Aug 31 2007
(Haskell)
a127093 n k = a127093_row n !! (k-1)
a127093_row n = zipWith (*) [1..n] $ map ((0 ^) . (mod n)) [1..n]
a127093_tabl = map a127093_row [1..]
(PARI) trianglerows(n) = for(x=1, n, for(k=1, x, if(x%k==0, print1(k, ", "), print1("0, "))); print(""))
/* Print initial 9 rows of triangle as follows: */
|
|
|
CROSSREFS
|
Cf. A127094, A123229, A127096, A127097, A127098, A127099, A000203, A126988, A127013, A127057, A038040, A024916, A060640, A001001.
Cf. A000012 (the first column), A020639, A059841 (the second column when multiplied by 2), A079978 (the third column when multiplied by 2), A079998 (the fifth column when multiplied by 5), A121262 (the fourth column when multiplied by 4).
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
