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A101100
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The first summation of row 5 of Euler's triangle - a row that will recursively accumulate to the power of 5.
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5
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1, 27, 93, 119, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120, 120
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OFFSET
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1,2
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REFERENCES
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Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 533.
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LINKS
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FORMULA
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a(n) = 120, n>4.
a(n) = Sum_{j=1..m} Eulerian(m, j-1)*binomial(m+n-j+r, m+r], with m = 5, r = -5.
a(n) = Sum_{j=0..n+1} (-1)^j*binomial(m+1-z, j)*(n-j+1)^n, with m = 5, z = 1.
G.f.: x*(1+26*x+66*x^2+26*x^3+x^4)/(1-x). - Colin Barker, Mar 01 2012
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MATHEMATICA
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MagicNKZ=Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}]; Table[MagicNKZ, {n, 5, 5}, {z, 1, 1}, {k, 0, 34}]
(* or *)
SeriesAtLevelR = Sum[Eulerian[n, i-1]*Binomial[n+x-i+r, n+r], {i, 1, n}]; Table[SeriesAtLevelR, {n, 5, 5}, {r, -5, -5}, {x, 5, 35}]
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PROG
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(PARI) {a(n) = if(n==1, 1, if(n==2, 27, if(n==3, 93, if(n==4, 119, 120))) )}; \\ G. C. Greubel, May 07 2019
(Magma) R<x>:=PowerSeriesRing(Integers(), 40); Coefficients(R!( x*(1+26*x+66*x^2+26*x^3+x^4)/(1-x) )); // G. C. Greubel, May 07 2019
(Sage) a=(x*(1+26*x+66*x^2+26*x^3+x^4)/(1-x)).series(x, 40).coefficients(x, sparse=False); a[1:] # G. C. Greubel, May 07 2019
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CROSSREFS
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Within the "cube" of related sequences with construction based upon MaginNKZ formula, with n downward, k rightward and z backward: Before: this sequence, A101095, A101096, A101098, A022521, A000584, A000539, A101092, A101099. Above: A101104, this sequence.
Within the "cube" of related sequences with construction based upon SeriesAtLevelR formula, with n downward, x rightward and r backward: Before: this sequence, A101095, A101096, A101098, A022521, A000584, A000539, A101092, A101099.
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KEYWORD
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easy,nonn
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AUTHOR
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Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004
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STATUS
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approved
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