

A137738


Coefficients of the polynomial giving the nth diagonal of A137743 * n!, read as an upper right triangle.


1



1, 0, 1, 2, 3, 1, 24, 14, 9, 1, 288, 54, 95, 18, 1, 4320, 136, 1110, 315, 30, 1, 80640, 12300, 15064, 5775, 775, 45, 1
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OFFSET

0,4


COMMENTS

Let T(m,n) = number of different strings of length n obtained from "123...m" by iteratively duplicating any substring (A137743).
It can be shown that for given k>=0 and all n>k2, T(n,n+k) = (1/k!) P[k](n), where P[k] is a monic polynomial of degree k with integer coefficients, whose coefficients are the kth column of the upper right triangular array defined by present sequence A137738.
Here rows and columns start at 0 (which also motivated the chosen offset 0), i.e. a(0) = 1 means P[0] = 1, a(1..2) = 0,1 means P[1] = 0 + 1*X, a(3..6) = 2,3,1 means P[2] = 2 + 3*X + 1*X^2, etc.


REFERENCES

M. F. Hasler, in thread "New topic: duplication of substrings" of mailing list seqfan(AT)ext.jussieu.fr, Feb 911, 2008.


LINKS

Table of n, a(n) for n=0..27.
Index entries for doubling substrings


FORMULA

P[k](n) = k! T(n,n+k) for k>=0 and positive n>k2, where T(m,n) is given in A137743.
P[k] = X^k + A045943(k) X^(k1) + O(X^(k2)) for k>=1.
For m>0, T(n,n+m+3) = sum( T(k,k+m+2), k=m+1..n+1)  (1/m!) Q[m](n), where Q[m] is a monic polynomial of degree <= m with integer coefficients (conjectured  see examples).


EXAMPLE

We have the following formulas for T(m,n) as given in A137743:
T(n,n) = 1, T(n,n+1) = n, T(n,n+2) = (n+1)(n+2)/2  2,
T(n,n+3) = A137742 = (1/6)*(n1)*(n+6)*(n+4) for n>1, for n=1 this formula gives 0 instead of 1.
T(n,n+4) = A137741 = (1/24)*(n+3)*(n^3+15*n^2+50*n96) for n>2, for n=2 this gives 15 instead of 16.
T(n,n+5) = A137740 = (1/5!)*(n+4)*(n^2+3*n8)*(n^2+23*n+150)+4 for n>3, for n=3 this gives 137 instead of 138.
T(n,n+6) = A137739 = (1/6!)*(n+9)*(n^5+36*n^4+451*n^3+1716*n^2380*n8880)1 for n>4, for n=4 this gives 1013 instead of 1014.
They satisfy the following relations:
T(n,n+2) = sum( T(k,k+1), k=0..n+1)  2
T(n,n+3) = sum( T(k,k+2), k=1..n+1)  5
T(n,n+4) = sum( T(k,k+3), k=2..n+1)  12  n
T(n,n+5) = sum( T(k,k+4), k=3..n+1)  21  7n/2  n^2/2
T(n,n+6) = sum( T(k,k+5), k=4..n+1) + 49  25n/3  5n^2/2  n^3/6


CROSSREFS

Cf. A137739A137743, A135473, A137744A137748.
Sequence in context: A145643 A323155 A145142 * A009108 A016537 A106385
Adjacent sequences: A137735 A137736 A137737 * A137739 A137740 A137741


KEYWORD

more,tabl,sign


AUTHOR

M. F. Hasler, Mar 18 2008


STATUS

approved



