A007781
a(n) = (n+1)^(n+1) - n^n for n>0, a(0) = 1.
Original entry on oeis.org
1, 3, 23, 229, 2869, 43531, 776887, 15953673, 370643273, 9612579511, 275311670611, 8630788777645, 293959006143997, 10809131718965763, 426781883555301359, 18008850183328692241, 808793517812627212561
Offset: 0
Peter McCormack (peter.mccormack(AT)its.csiro.au)
a(14) = 10809131718965763 = 3 * 61^2 * 968299894201.
- Richard P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see equation (6.7).
- Doug Bell, Table of n, a(n) for n = 0..100
- Andrew Cusumano, Problem H-656, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 45, No. 2 (2007), p. 187; A Sequence Tending To e, Solution to Problem H-656, ibid., Vol. 46-47, No. 3 (2008/2009), pp. 285-287.
- Ronald K. Hoeflin, Mega Test. [Wayback Machine link]
- Eric Weisstein's World of Mathematics, Power Difference Prime.
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[1] cat [(n+1)^(n+1)-n^n: n in [1..20]]; // Vincenzo Librandi, Aug 19 2015
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seq( `if`(n=0,1,(n+1)^(n+1) -n^n), n=0..20); # G. C. Greubel, Mar 05 2020
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Join[{1},Table[(n+1)^(n+1)-n^n,{n,20}]] (* Harvey P. Dale, Feb 09 2011 *)
Differences[Table[n^n,{n,0,20}]] (* Charles R Greathouse IV, Feb 09 2011 *)
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first(m)=vector(m,i,i--;(i+1)^(i+1) - i^i) /* Anders Hellström, Aug 18 2015 */
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[1]+[(n+1)^(n+1) -n^n for n in (1..20)] # G. C. Greubel, Mar 05 2020
A213027
Number A(n,k) of 3n-length k-ary words, either empty or beginning with the first letter of the alphabet, that can be built by repeatedly inserting triples of identical letters into the initially empty word; square array A(n,k), n>=0, k>=0, by antidiagonals.
Original entry on oeis.org
1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 4, 1, 0, 1, 1, 7, 19, 1, 0, 1, 1, 10, 61, 98, 1, 0, 1, 1, 13, 127, 591, 531, 1, 0, 1, 1, 16, 217, 1810, 6101, 2974, 1, 0, 1, 1, 19, 331, 4085, 27631, 65719, 17060, 1, 0, 1, 1, 22, 469, 7746, 82593, 441604, 729933, 99658, 1, 0
Offset: 0
A(0,k) = 1: the empty word.
A(n,1) = 1: (aaa)^n.
A(2,2) = 4: there are 4 words of length 6 over alphabet {a,b}, either empty or beginning with the first letter of the alphabet, that can be built by repeatedly inserting triples of identical letters into the initially empty word: aaaaaa, aaabbb, aabbba, abbbaa.
A(2,3) = 7: aaaaaa, aaabbb, aaaccc, aabbba, aaccca, abbbaa, acccaa.
A(3,2) = 19: aaaaaaaaa, aaaaaabbb, aaaaabbba, aaaabbbaa, aaabaaabb, aaabbaaab, aaabbbaaa, aaabbbbbb, aabaaabba, aabbaaaba, aabbbaaaa, aabbbabbb, aabbbbbba, abaaabbaa, abbaaabaa, abbbaaaaa, abbbaabbb, abbbabbba, abbbbbbaa.
Square array A(n,k) begins:
1, 1, 1, 1, 1, 1, 1, ...
0, 1, 1, 1, 1, 1, 1, ...
0, 1, 4, 7, 10, 13, 16, ...
0, 1, 19, 61, 127, 217, 331, ...
0, 1, 98, 591, 1810, 4085, 7746, ...
0, 1, 531, 6101, 27631, 82593, 195011, ...
0, 1, 2974, 65719, 441604, 1751197, 5153626, ...
Columns k=0-10 give:
A000007,
A000012,
A047099,
A218473,
A218474,
A218475,
A218476,
A218477,
A218478,
A218479,
A218480.
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A:= (n, k)-> `if`(n=0, 1, `if`(k<2, k,
1/n *add(binomial(3*n, j) *(n-j) *(k-1)^j, j=0..n-1))):
seq(seq(A(n, d-n), n=0..d), d=0..12);
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a[0, ] = 1; a[, k_ /; k < 2] := k; a[n_, k_] := 1/n*Sum[Binomial[3*n, j]*(n-j)*(k-1)^j, {j, 0, n-1}]; Table[a[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Dec 11 2013 *)
Showing 1-2 of 2 results.
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