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 A137738 Coefficients of the polynomial giving the n-th 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 (list; table; graph; refs; listen; history; text; internal format)
 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>k-2, 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 k-th 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 = 1, a(1..2) = 0,1 means P = 0 + 1*X, a(3..6) = -2,3,1 means P = -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 9-11, 2008. LINKS FORMULA P[k](n) = k! T(n,n+k) for k>=0 and positive n>k-2, where T(m,n) is given in A137743. P[k] = X^k + A045943(k) X^(k-1) + O(X^(k-2)) 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)*(n-1)*(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*n-96) for n>2, for n=2 this gives 15 instead of 16. T(n,n+5) = A137740 = (1/5!)*(n+4)*(n^2+3*n-8)*(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^2-380*n-8880)-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. A137739-A137743, A135473, A137744-A137748. Sequence in context: A338208 A323155 A145142 * A009108 A016537 A346381 Adjacent sequences:  A137735 A137736 A137737 * A137739 A137740 A137741 KEYWORD more,tabl,sign AUTHOR M. F. Hasler, Mar 18 2008 STATUS approved

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Last modified July 23 12:19 EDT 2021. Contains 346259 sequences. (Running on oeis4.)