A125714 Alfred Moessner's factorial triangle.

1, 2, 3, 6, 11, 6, 24, 50, 35, 10, 120, 274, 225, 85, 15, 720, 1764, 1624, 735, 175, 21, 5040, 13068, 13132, 6769, 1960, 322, 28, 40320, 109584, 118124, 67284, 22449, 4536, 546, 36, 362880, 1026576, 1172700, 723680, 269325, 63273, 9450, 870, 45, 3628800

Offset: 1

Author

Gary W. Adamson, Dec 01 2006

Keywords

• nonn
• tabl

Comments

Successive numbers arising from the Moessner construction of the factorial numbers. - N. J. A. Sloane, Jul 27 2021

Row sums of the triangle = 1, 5, 23, 119, 719, ...(matching the terms 0, 0, 1, 5, 23, 119, 719, ...; of A033312).

The name of the triangle derives from the fact that A125714(A000124(n)) = A000142(n) for n > 0. Moessner's method uses only additions to compute the factorial n!. - Peter Luschny, Jan 27 2009

If n = (m^2+m+2)/2 then a(n) = (m+1)!. For example, taking m = 3, n = 7, and indeed a(7) = 4! = 24. - N. J. A. Sloane, Jul 27 2021

Examples

			An "x" prefaced before each term will indicate the term following the x being circled.
x1 2 x3 4 5 x6 7 8 9 x10 11 12 13 14 x15 ...
__x2 6 x11 18 26 x35 46 58 71 x85 ...
_____________x6 24 x50 96 154 x225 ...
_________________________x24 120 x274 ...
___________________________________________x120 ...
...
i.e., circle the triangular terms in row 1. In row 2, take partial sums of the uncircled terms and circle the terms offset one place to the left of the triangular terms in row 1. Continue in subsequent rows with analogous operations. The triangle consists of the infinite set of terms prefaced with the x (circled on page 64 of "The Book of Numbers").
		

References

• J. H. Conway and R. K. Guy, "The Book of Numbers", Springer-Verlag, 1996. Sequence can be seen by reading the successive circled numbers in the "factorial" section on page 64 (based on the work of Alfred Moessner).

Links

• Joshua Zucker, Table of n, a(n) for n = 1..66
• G. S. Kazandzidis, On a conjecture of Moessner and a general problem, Bull. Soc. Math. Grèce (N.S.) 2 (1961), 23-30.
• Dexter Kozen and Alexandra Silva, On Moessner's theorem, Amer. Math. Monthly 120(2) (2013), 131-139.
• R. Krebbers, L. Parlant, and A. Silva, Moessner's theorem: an exercise in coinductive reasoning in Coq, Theory and practice of formal methods, 309-324, Lecture Notes in Comput. Sci., 9660, Springer, 2016.
• Calvin T. Long, Strike it out--add it up, Math. Gaz. 66 (438) (1982), 273-277.
• Alfred Moessner, Eine Bemerkung über die Potenzen der natürlichen Zahlen, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss., 29, 1951.
• Ivan Paasche, Ein neuer Beweis des Moessnerschen Satzes S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss. 1952 (1952), 1-5 (1953). [Two years are listed at the beginning of the journal issue.]
• Ivan Paasche, Beweis des Moessnerschen Satzes mittels linearer Transformationen, Arch. Math. (Basel) 6 (1955), 194-199.
• Ivan Paasche, Eine Verallgemeinerung des Moessnerschen Satzes, Compositio Math. 12 (1956), 263-270.
• Hans Salié, Bemerkung zu einem Satz von A. Moessner, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss. 1952 (1952), 7-11 (1953). [Two years are listed at the beginning of the journal issue.]
• Oskar Perron, Beweis des Moessnerschen Satzes, S.-B. Math.-Nat. Kl. Bayer. Akad. Wiss., 31-34, 1951.

Crossrefs

Cf. A000217, A003056, A008275, A033312, A346004-A346007, A346595.

Programs

Maple

a := proc(n) local s,m,k,i; s := array(0..n); s[0] := 1;
for m from 1 to n do s[m] := 0; for k from m by - 1 to 1 do
for i from 1 to k do s[i] := s[i] + s[i - 1] od; lprint(s[k]);
if k = n then return(s[n]) fi od; lprint("-") od end: # Peter Luschny, Jan 27 2009
with(combinat);
s:=stirling1;
A003056 := proc(n) floor((sqrt(1+8*n)-1)/2) ; end proc:
T:=n->n*(n+1)/2; # A000217
g:=proc(n) local i,j,t; global T,A003056;
j:=A003056(n-1)+1;
t:=T(j);
for i from 0 to t-1 do
if ((n+i) mod t) = 0 then return(abs(s(j+1,j-i))); fi;
od;
end;
[seq(g(n),n=1..80)]; # N. J. A. Sloane, Jul 27 2021

Mathematica

n = 10; A125714 = Reap[ ClearAll[s]; s[0] = 1; For[m = 1, m <= n, m++, s[m] = 0; For[k = m, k >= 1, k--, For[i = 1, i <= k, i++, s[i] = s[i] + s[i-1]]; Sow[s[k]]; If[k == n, Print[n, "! = ", s[n]]; Break[]]]]][[2, 1]] (* Jean-François Alcover, Jun 29 2012, after Peter Luschny *)

PARI

T(n, k)={ my( s=vector(n)); for( m=1, n, forstep( j=m,1,-1, s[1]++; for( i=2, j, s[i] += s[i-1]));
k<0 && print(vecextract(s,Str(m"..1"))));
if( k>0,s[n+1-k],vecextract(s,"-1..1"))} /* returns T[n,k], or the whole n-th row if k is not given, prints row 1...n of the triangle if k<0 */ \\ M. F. Hasler, Dec 03 2010

Formula

Starting with the natural numbers, circle each triangular number. Underneath, take partial sums of the uncircled terms and circle the terms in this row which are offset one place to the left of the circled 1, 3, 6, 10, ... in the first row. Repeat with analogous operations in succeeding rows. The circled terms in the infinite set become the triangle.

Given n, let j = A003056(n-1)+1 and set t = j*(j+1)/2. Then, for 0 <= i < t, if n == -i (mod t), a(n) = abs(Stirling_1(j+1,j-i)). - N. J. A. Sloane, Jul 27 2021

Extensions

More terms from Joshua Zucker, Jun 17 2007