A277408 - OEIS

%I #27 May 10 2017 09:38:24

%S 1,1,1,2,3,4,6,12,22,36,24,60,140,300,576,120,360,1020,2700,6576,

%T 14400,720,2520,8400,26460,77952,211680,518400,5040,20160,77280,

%U 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

%N 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.

%H Qiaochu Yuan, <a href="https://math.stackexchange.com/questions/1220519/">Has anyone seen this combinatorial identity involving the Bernoulli and Stirling numbers</a>, Math StackExchange, Apr 05 2015

%F T(n,k) = k!*(n-k)! * Sum_{i=0..n-k+1} (-1)^(n-i+1) * Stirling2(i,n-k+1) * Stirling1(n+1,i)). [From formula in A188881 by _Vladimir Kruchinin_]

%F T(n,k) = k! * A188881(n+1, n-k+1).

%F A003713(n) = Sum_{k=0..n} T(n,k) / k!, where e.g.f. of A003713 is log(1/(1+log(1-x))).

%F Row sums yield A277406.

%e Illustration of initial row polynomials.

%e R_0(y) = 1;

%e R_1(y) = 1 + y;

%e R_2(y) = 2 + 3*y + 4*y^2;

%e R_3(y) = 6 + 12*y + 22*y^2 + 36*y^3;

%e R_4(y) = 24 + 60*y + 140*y^2 + 300*y^3 + 576*y^4;

%e R_5(y) = 120 + 360*y + 1020*y^2 + 2700*y^3 + 6576*y^4 + 14400*y^5;

%e R_6(y) = 720 + 2520*y + 8400*y^2 + 26460*y^3 + 77952*y^4 + 211680*y^5 + 518400*y^6;

%e R_7(y) = 5040 + 20160*y + 77280*y^2 + 282240*y^3 + 974736*y^4 + 3151680*y^5 + 9408960*y^6 + 25401600*y^7;

%e ...

%e Generating Method.

%e R_0(y) = 1, by convention;

%e R_1(y) = Sum_{i=1..1} (1 + i*y);

%e R_2(y) = Sum_{i=1..2, j=1..2, j<>i} (1 + i*y*(1 + j*y));

%e 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)));

%e 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))));

%e etc.

%e This triangle of coefficients begins:

%e 1;

%e 1, 1;

%e 2, 3, 4;

%e 6, 12, 22, 36;

%e 24, 60, 140, 300, 576;

%e 120, 360, 1020, 2700, 6576, 14400;

%e 720, 2520, 8400, 26460, 77952, 211680, 518400;

%e 5040, 20160, 77280, 282240, 974736, 3151680, 9408960, 25401600;

%e 40320, 181440, 786240, 3265920, 12930624, 48444480, 170098560, 552303360, 1625702400;

%e 362880, 1814400, 8769600, 40824000, 182226240, 775656000, 3126297600, 11820816000, 41391544320, 131681894400;

%e 3628800, 19958400, 106444800, 548856000, 2726317440, 12989592000, 59044550400, 254303280000, 1028448368640, 3856920883200, 13168189440000;

%e 39916800, 239500800, 1397088000, 7903526400, 43233886080, 227885011200, 1152535824000, 5563643500800, 25464033745920, 109530230261760, 437429486592000, 1593350922240000; ...

%o (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))}

%o for(n=0,11,for(k=0,n,print1( T(n,k) ,", "));print(""))

%o (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 */

%Y Cf. A277406 (row sums), A277405, A277407, A188881, A003713.

%K nonn,tabl

%O 0,4

%A _Paul D. Hanna_, Oct 16 2016