A168351 -id:A168351 - cp's OEIS frontend

A158497 Triangle T(n,k) formed by the coordination sequences and the number of leaves for trees.

1, 1, 1, 1, 2, 2, 1, 3, 6, 12, 1, 4, 12, 36, 108, 1, 5, 20, 80, 320, 1280, 1, 6, 30, 150, 750, 3750, 18750, 1, 7, 42, 252, 1512, 9072, 54432, 326592, 1, 8, 56, 392, 2744, 19208, 134456, 941192, 6588344, 1, 9, 72, 576, 4608, 36864, 294912, 2359296, 18874368, 150994944, 1, 10, 90, 810, 7290, 65610, 590490, 5314410, 47829690, 430467210, 3874204890

Offset: 0

Thomas Wieder, Mar 20 2009

Consider the k-fold Cartesian products CP(n,k) of the vector A(n) = [1, 2, 3, ..., n].
An element of CP(n,k) is a n-tuple T_t of the form T_t = [i_1, i_2, i_3, ..., i_k] with t=1, .., n^k.
We count members T of CP(n,k) which satisfy some condition delta(T_t), so delta(.) is an indicator function which attains values of 1 or 0 depending on whether T_t is to be counted or not; the summation sum_{CP(n,k)} delta(T_t) over all elements T_t of CP produces the count.
For the triangle here we have delta(T_t) = 0 if for any two i_j, i_(j+1) in T_t one has i_j = i_(j+1): T(n,k) = Sum_{CP(n,k)} delta(T_t) = Sum_{CP(n,k)} delta(i_j = i_(j+1)).
The test on i_j > i_(j+1) generates A158498. One gets the Pascal triangle A007318 if the indicator function tests whether for any two i_j, i_(j+1) in T_t one has i_j >= i_(j+1).
Use of other indicator functions can also calculate the Bell numbers A000110, A000045 or A000108.

			Array, A(n, k) = n*(n-1)^(k-1) for n > 1, A(n, k) = 1 otherwise, begins as:
  1,  1,   1,    1,     1,      1,       1,        1,        1, ... A000012;
  1,  1,   1,    1,     1,      1,       1,        1,        1, ... A000012;
  1,  2,   2,    2,     2,      2,       2,        2,        2, ... A040000;
  1,  3,   6,   12,    24,     48,      96,      192,      384, ... A003945;
  1,  4,  12,   36,   108,    324,     972,     2916,     8748, ... A003946;
  1,  5,  20,   80,   320,   1280,    5120,    20480,    81920, ... A003947;
  1,  6,  30,  150,   750,   3750,   18750,    93750,   468750, ... A003948;
  1,  7,  42,  252,  1512,   9072,   54432,   326592,  1959552, ... A003949;
  1,  8,  56,  392,  2744,  19208,  134456,   941192,  6588344, ... A003950;
  1,  9,  72,  576,  4608,  36864,  294912,  2359296, 18874368, ... A003951;
  1, 10,  90,  810,  7290,  65610,  590490,  5314410, 47829690, ... A003952;
  1, 11, 110, 1100, 11000, 110000, 1100000, 11000000, ............. A003953;
  1, 12, 132, 1452, 15972, 175692, 1932612, 21258732, ............. A003954;
  1, 13, 156, 1872, 22464, 269568, 3234816, 38817792, ............. A170732;
  ... ;
The triangle begins as:
  1
  1, 1;
  1, 2,  2;
  1, 3,  6,  12;
  1, 4, 12,  36,  108;
  1, 5, 20,  80,  320,  1280;
  1, 6, 30, 150,  750,  3750,  18750;
  1, 7, 42, 252, 1512,  9072,  54432, 326592;
  1, 8, 56, 392, 2744, 19208, 134456, 941192, 6588344;
  ...;
T(3,3) = 12 counts the triples (1,2,1), (1,2,3), (1,3,1), (1,3,2), (2,1,2), (2,1,3), (2,3,1), (2,3,2), (3,1,2), (3,1,3), (3,2,1), (3,2,3) out of a total of 3^3 = 27 triples in the CP(3,3).

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Array rows n: A170733 (n=14), ..., A170769 (n=50).
Columns k: A000012(n) (k=0), A000027(n) (k=1), A002378(n-1) (k=2), A011379(n-1) (k=3), A179824(n) (k=4), A101362(n-1) (k=5), 2*A168351(n-1) (k=6), 2*A168526(n-1) (k=7), 2*A168635(n-1) (k=8), 2*A168675(n-1) (k=9), 2*A170783(n-1) (k=10), 2*A170793(n-1) (k=11).
Diagonals k: A055897 (k=n), A055541 (k=n-1), A373395 (k=n-2), A379612 (k=n-3).
Sums: (-1)^n*A065440(n) (signed row).
Cf. A000045, A000108, A000110, A007318, A079901, A158498.

A158497:= func< n,k | k le 1 select n^k else n*(n-1)^(k-1) >;
[A158497(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 18 2025

A158497[n_, k_]:= If[n<2 || k==0, 1, n*(n-1)^(k-1)];
Table[A158497[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 18 2025 *)

def A158497(n,k): return n^k if k<2 else n*(n-1)^(k-1)
print(flatten([[A158497(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Mar 18 2025

T(n, k) = (n-1)^(k-1) + (n-1)^k = n*A079901(n-1,k-1), k > 0.

Sum_{k=0..n} T(n,k) = (n*(n-1)^n - 2)/(n-2), n > 2.

Edited by R. J. Mathar, Mar 31 2009

More terms added by G. C. Greubel, Mar 18 2025

A168432 a(n) = n^5*(n^7 + 1)/2.

Original entry on oeis.org

0, 1, 2064, 265842, 8389120, 122071875, 1088395056, 6920652004, 34359754752, 141214797765, 500000050000, 1569214268886, 4458050348544, 11649042746887, 28346956456560, 64873169325000, 140737488879616, 291311119324809

Offset: 0

N. J. A. Sloane, Dec 11 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).

Cf. A168351.

[n^5*(n^7+1)/2: n in [0..30]]; // Vincenzo Librandi, Aug 29 2011

Table[n^5*(n^7 + 1)/2, {n,0,30}] (* G. C. Greubel, Jul 22 2016 *)
CoefficientList[Series[x (1 +2051 x +239088 x^2 +5093880 x^3 +33159402 x^4 + 81255702 x^5 +81256584 x^6 +33159072 x^7 +5093805 x^8 +239183 x^9 +2032 x^10)/(1-x)^13, {x,0,30}], x] (* Vincenzo Librandi, Jul 23 2016 *)

def A168432(n): return n^5*(n^7+1)//2
print(A168432(n) for n in range(31)) # G. C. Greubel, Mar 20 2025

G.f.: x*(1 + 2051*x + 239088*x^2 + 5093880*x^3 + 33159402*x^4 + 81255702*x^5 + 81256584*x^6 + 33159072*x^7 + 5093805*x^8 + 239183*x^9 + 2032*x^10)/(1-x)^13. - Vincenzo Librandi, Jul 23 2016

E.g.f.: (1/2)*x*(2 + 2062*x + 86551*x^2 + 611511*x^3 + 1379401*x^4 + 1323652*x^5 + 627396*x^6 + 159027*x^7 + 22275*x^8 + 1705*x^9 + 66*x^10 + x^11)*exp(x). - G. C. Greubel, Mar 20 2025

A168462 a(n) = n^5*(n^8 + 1)/2.

Original entry on oeis.org

0, 1, 4112, 797283, 33554944, 610353125, 6530350896, 48444513607, 274877923328, 1270932943689, 5000000050000, 17261356152491, 53496602813952, 151437553481773, 396857386895984, 973097534559375, 2251799814209536

Offset: 0

N. J. A. Sloane, Dec 11 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (14,-91,364,-1001,2002,-3003,3432,-3003,2002,-1001,364,-91,14,-1).

Cf. A168351.

[n^5*(n^8+1)/2: n in [0..30]]; // Vincenzo Librandi, Aug 29 2011

Table[n^5 (n^8+1)/2,{n,0,30}] (* Harvey P. Dale, Jan 10 2013 *)

def A168462(n): return n^5*(n^8+1)//2
print([A168462(n) for n in range(31)]) # G. C. Greubel, Mar 20 2025

From G. C. Greubel, Mar 20 2025: (Start)
G.f.: x*(1 +4098*x +739806*x^2 +22766810*x^3 +211640895*x^4 +752810148*x^5 +1137586884*x^6 +752810148*x^7 +211640895*x^8 +22766810*x^9 +739806*x^10 +4098*x^11 + x^12)/(1-x)^14.
E.g.f.: (1/2)*x*(2 +4110*x +261650*x^2 +2532540*x^3 +7508502*x^4 +9321312*x^5 +5715424*x^6 +1899612*x^7 +359502*x^8 +39325*x^9 +2431*x^10 +78*x^11 +x^12)*exp(x). (End)

A168471 a(n) = n^5*(n^9 + 1)/2.

Original entry on oeis.org

0, 1, 8208, 2391606, 134218240, 3051759375, 39182085936, 339111544828, 2199023271936, 11438396257005, 50000000050000, 189874916872146, 641959232398848, 1968688193035291, 5556003413047920, 14596463013075000

Offset: 0

N. J. A. Sloane, Dec 11 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (15,-105,455,-1365,3003,-5005,6435,-6435,5005,-3003,1365,-455,105,-15,1).

Cf. A168351.

[n^5*(n^9+1)/2: n in [0..30]]; // Vincenzo Librandi, Aug 29 2011

Table[n^5*(n^9 + 1)/2, {n,0,30}] (* G. C. Greubel, Jul 23 2016 *)

def A168471(n): return n^5*(n^9+1)//2
print([A168471(n) for n in range(31)]) # G. C. Greubel, Mar 20 2025

G.f.: x* (1 + 8193*x + 2268591*x^2 + 99205535*x^3 + 1285871130*x^4 + 6421630698*x^5 + 13985589534*x^6 + 13985586558*x^7 + 6421632165*x^8 + 1285871045*x^9 + 99205251*x^10 + 2268723*x^11 + 8176*x^12)/(1 - x)^15. - Vincenzo Librandi, Jul 24 2016

A168507 a(n) = n^5*(n^10 + 1)/2.

Original entry on oeis.org

0, 1, 16400, 7174575, 536871424, 15258790625, 235092496176, 2373780763375, 17592186060800, 102945566076849, 500000000050000, 2088624084788351, 7703510787417600, 25592946507231025, 77784047779175024, 218946945190809375, 576460752303947776

Offset: 0

N. J. A. Sloane, Dec 11 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (16,-120,560,-1820,4368,-8008,11440,-12870,11440,-8008,4368,-1820,560,-120,16,-1).

Cf. A006003, A168351.

[n^5*(n^10+1)/2: n in [0..30]]; // Vincenzo Librandi, Aug 29 2011

Table[n^5*(n^10+1)/2, {n,0,30}] (* G. C. Greubel, Jul 24 2016 *)
LinearRecurrence[{16,-120,560,-1820,4368,-8008,11440,-12870,11440,-8008,4368,-1820,560,-120,16,-1},{0,1,16400,7174575,536871424,15258790625,235092496176, 2373780763375,17592186060800,102945566076849,500000000050000, 2088624084788351,7703510787417600,25592946507231025,77784047779175024, 218946945190809375}, 30] (* Harvey P. Dale, Apr 25 2017 *)

def A168507(n): return n^5*(n^10 +1)//2
print([A168507(n) for n in range(31)]) # G. C. Greubel, Mar 20 2025

G.f.: x*(1 + 16384*x + 6912295*x^2 + 424045664*x^3 + 7520614661*x^4 + 51388498688*x^5 + 155693801427*x^6 + 223769405760*x^7 + 155693801427*x^8 + 51388498688*x^9 + 7520614661*x^10 + 424045664*x^11 + 6912295*x^12 + 16384*x^13 + x^14)/(1 - x)^16. - Vincenzo Librandi, Jul 24 2016

a(n) = A006003(n^5). - G. C. Greubel, Mar 20 2025

cp's OEIS Frontend

A158497 Triangle T(n,k) formed by the coordination sequences and the number of leaves for trees.

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A168432 a(n) = n^5*(n^7 + 1)/2.

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A168462 a(n) = n^5*(n^8 + 1)/2.

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A168471 a(n) = n^5*(n^9 + 1)/2.

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A168507 a(n) = n^5*(n^10 + 1)/2.

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