A277408 Triangle, read by rows, where the g.f. of row n equals the sum of permutations of compositions of functions (1 + k*y*x) for k=1..n with parameter y independent of variable x, as evaluated at x=1.
1, 1, 1, 2, 3, 4, 6, 12, 22, 36, 24, 60, 140, 300, 576, 120, 360, 1020, 2700, 6576, 14400, 720, 2520, 8400, 26460, 77952, 211680, 518400, 5040, 20160, 77280, 282240, 974736, 3151680, 9408960, 25401600, 40320, 181440, 786240, 3265920, 12930624, 48444480, 170098560, 552303360, 1625702400, 362880, 1814400, 8769600, 40824000, 182226240, 775656000, 3126297600, 11820816000, 41391544320, 131681894400, 3628800, 19958400, 106444800, 548856000, 2726317440, 12989592000, 59044550400, 254303280000, 1028448368640, 3856920883200, 13168189440000, 39916800, 239500800, 1397088000, 7903526400, 43233886080, 227885011200, 1152535824000, 5563643500800, 25464033745920, 109530230261760, 437429486592000, 1593350922240000
Offset: 0
Examples
Illustration of initial row polynomials. R_0(y) = 1; R_1(y) = 1 + y; R_2(y) = 2 + 3*y + 4*y^2; R_3(y) = 6 + 12*y + 22*y^2 + 36*y^3; R_4(y) = 24 + 60*y + 140*y^2 + 300*y^3 + 576*y^4; R_5(y) = 120 + 360*y + 1020*y^2 + 2700*y^3 + 6576*y^4 + 14400*y^5; R_6(y) = 720 + 2520*y + 8400*y^2 + 26460*y^3 + 77952*y^4 + 211680*y^5 + 518400*y^6; R_7(y) = 5040 + 20160*y + 77280*y^2 + 282240*y^3 + 974736*y^4 + 3151680*y^5 + 9408960*y^6 + 25401600*y^7; ... Generating method. R_0(y) = 1, by convention; R_1(y) = Sum_{i=1..1} (1 + i*y); R_2(y) = Sum_{i=1..2, j=1..2, j<>i} (1 + i*y*(1 + j*y)); R_3(y) = Sum_{i=1..3, j=1..3, k=1..3, i,j,k distinct} (1 + i*y*(1 + j*y*(1 + k*y))); R_4(y) = Sum_{i=1..4, j=1..4, k=1..4, m=1..4, i,j,k,m distinct} (1 + i*y*(1 + j*y*(1 + k*y*(1 + m*y)))); etc. This triangle of coefficients begins: 1; 1, 1; 2, 3, 4; 6, 12, 22, 36; 24, 60, 140, 300, 576; 120, 360, 1020, 2700, 6576, 14400; 720, 2520, 8400, 26460, 77952, 211680, 518400; 5040, 20160, 77280, 282240, 974736, 3151680, 9408960, 25401600; 40320, 181440, 786240, 3265920, 12930624, 48444480, 170098560, 552303360, 1625702400; ...
Links
- Qiaochu Yuan, Has anyone seen this combinatorial identity involving the Bernoulli and Stirling numbers, Math StackExchange, Apr 05 2015.
Programs
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PARI
{T(n,k) = k!*(n-k)! * sum(i=0,n-k+1, (-1)^(n-i+1) * stirling(i,n-k+1,2) * stirling(n+1,i,1))} for(n=0,11,for(k=0,n,print1( T(n,k) ,", "));print(""))
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PARI
{T(n, k) = if( k<0 || k>n, 0, n! * k! * polcoeff( (x / (1 - exp(-x * (1 + x * O(x^n)))))^(n+1), k))}; /* Michael Somos, May 10 2017 */