A030527 A convolution triangle of numbers obtained from A036083.
1, 15, 1, 175, 30, 1, 1750, 575, 45, 1, 15750, 8750, 1200, 60, 1, 131250, 114625, 24375, 2050, 75, 1, 1031250, 1347500, 414750, 52000, 3125, 90, 1, 7734375, 14575000, 6208125, 1084875, 95000, 4425, 105, 1, 55859375, 147468750, 84184375
Offset: 1
Examples
{1}; {15,1}; {175,30,1}; {1750,575,45,1}; {15750,8750,1200,60,1}; ...
Links
- W. Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
Formula
a(n, m) = 5*(5*m+n-1)*a(n-1, m)/n + m*a(n-1, m-1)/n, n >= m >= 1; a(n, m) := 0, n
A033842 Triangle of coefficients of certain polynomials (exponents in decreasing order).
1, 1, 1, 3, 3, 1, 16, 16, 6, 1, 125, 125, 50, 10, 1, 1296, 1296, 540, 120, 15, 1, 16807, 16807, 7203, 1715, 245, 21, 1, 262144, 262144, 114688, 28672, 4480, 448, 28, 1, 4782969, 4782969, 2125764, 551124, 91854, 10206, 756, 36, 1, 100000000
Offset: 0
Comments
See A049323.
Examples
{1}; {1,1}; {3,3,1}; {16,16,6,1}; {125,125,50,10,1}; .... E.g. third row {3,3,1} corresponds to polynomial p(2,x)= 3*x^2+3*x+1.
Links
- W. Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
- Thierry Lévy, The Number of Prefixes of Minimal Factorisations of a Cycle, The Electronic Journal of Combinatorics, 23(3) (2016), #P3.35
Crossrefs
Formula
a(n, m) = binomial(n+1, m)*(n+1)^(n-m-1), n >= m >= 0 else 0.
A053110 Expansion of (-1 + 1/(1-7*x)^7)/(49*x); related to A036226.
1, 28, 588, 10290, 158466, 2218524, 28840812, 353299947, 4121832715, 46164526408, 499416240232, 5243870522436, 53648829191076, 536488291910760, 5257585260725448, 50604258134482437, 479252091744216021
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..400
- Index entries for linear recurrences with constant coefficients, signature (49,-1029,12005,-84035,352947,-823543,823543).
Programs
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Magma
[7^(n-1)*Binomial(n+7, 6): n in [0..30]]; // G. C. Greubel, Aug 16 2018
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Mathematica
CoefficientList[Series[(-1+1/(1-7x)^7)/(49x),{x,0,30}],x] (* or *) LinearRecurrence[{49,-1029,12005,-84035,352947,-823543,823543},{1,28,588,10290,158466,2218524,28840812},30] (* Harvey P. Dale, Jun 03 2015 *) Table[7^(n-1)*Binomial[n+7, 6], {n,0,30}] (* G. C. Greubel, Aug 16 2018 *)
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PARI
vector(30,n,n--; 7^(n-1)*binomial(n+7, 6)) \\ G. C. Greubel, Aug 16 2018
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Sage
[lucas_number2(n, 7, 0)*binomial(n,6)/7^8 for n in range(7, 24)] # Zerinvary Lajos, Mar 13 2009
Formula
a(n) = 7^(n-1)*binomial(n+7, 6);
G.f.: (-1 + (1-7*x)^(-7))/(x*7^2).
A053113 Expansion of (-1 + 1/(1-10*x)^10)/(100*x); related to A053109.
1, 55, 2200, 71500, 2002000, 50050000, 1144000000, 24310000000, 486200000000, 9237800000000, 167960000000000, 2939300000000000, 49742000000000000, 817190000000000000, 13075040000000000000, 204297500000000000000
Offset: 0
Comments
Links
- G. C. Greubel, Table of n, a(n) for n = 0..400
- W. Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
- Index entries for linear recurrences with constant coefficients, signature (100, -4500, 120000, -2100000, 25200000, -210000000, 1200000000, -4500000000, 10000000000, -10000000000).
Programs
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Magma
[10^(n-1)*Binomial(n+10, 9): n in [0..30]]; // G. C. Greubel, Aug 16 2018
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Mathematica
Table[10^(n - 1)*Binomial[n + 10, 9], {n, 0, 30}] (* G. C. Greubel, Aug 16 2018 *) LinearRecurrence[{100,-4500,120000,-2100000,25200000,-210000000,1200000000,-4500000000,10000000000,-10000000000},{1,55,2200,71500,2002000,50050000,1144000000,24310000000,486200000000,9237800000000},20] (* Harvey P. Dale, Jul 30 2025 *)
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PARI
vector(30,n,n--; 10^(n-1)*binomial(n+10, 9)) \\ G. C. Greubel, Aug 16 2018
Formula
a(n) = 10^(n-1)*binomial(n+10, 9).
G.f.: (-1 + (1-10*x)^(-10))/(x*10^2).
A036224 Expansion of (-1+1/(1-6*x)^6)/(36*x); related to A036084.
1, 21, 336, 4536, 54432, 598752, 6158592, 60046272, 560431872, 5043886848, 44019376128, 374164697088, 3108445175808, 25311625003008, 202493000024064, 1594632375189504, 12381851383824384, 94927527275986944
Offset: 0
Links
- Index entries for linear recurrences with constant coefficients, signature (36, -540, 4320, -19440, 46656, -46656).
Programs
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Sage
[lucas_number2(n, 6, 0)*binomial(n,5)/6^7for n in range(6, 24)] # Zerinvary Lajos, Mar 13 2009
Formula
a(n) = 6^(n-1)*binomial(n+6, 5);
g.f. (-1+(1-6*x)^(-6))/(x*6^2).
A049323 Triangle of coefficients of certain polynomials (exponents in increasing order), equivalent to A033842.
1, 1, 1, 1, 3, 3, 1, 6, 16, 16, 1, 10, 50, 125, 125, 1, 15, 120, 540, 1296, 1296, 1, 21, 245, 1715, 7203, 16807, 16807, 1, 28, 448, 4480, 28672, 114688, 262144, 262144, 1, 36, 756, 10206, 91854, 551124, 2125764, 4782969, 4782969, 1, 45, 1200, 21000, 252000
Offset: 0
Comments
These polynomials p(n, x) appear in the W. Lang reference as c1(-(n+1);x), n >= 0 on p.12. The coefficients are given there in eq.(44) on p. 6. - Wolfdieter Lang, Nov 20 2015
Examples
The triangle a(n, m) begins: n\m 0 1 2 3 4 5 6 7 ... 0: 1 1: 1 1 2: 1 3 3 3: 1 6 16 16 4: 1 10 50 125 125 5: 1 15 120 540 1296 1296 6: 1 21 245 1715 7203 16807 16807 7: 1 28 448 4480 28672 114688 262144 262144 ... reformatted. - Wolfdieter Lang, Nov 20 2015 E.g. the third row {1,3,3} corresponds to polynomial p(2,x)= 1 + 3*x + 3*x^2.
Links
- W. Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
Crossrefs
a(n, 0)= A000012 (powers of 1), a(n, 1)= A000217 (triangular numbers), a(n, n)= A000272(n+1), n >= 0 (diagonal), a(n, n-1)= A000272(n+1), n >= 1.
Programs
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Magma
/* As triangle: */ [[Binomial(n+1, k+1)*(n+1)^(k-1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Nov 20 2015
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Maple
seq(seq(binomial(n+1,m+1)*(n+1)^(m-1),m=0..n),n=0..10); # Robert Israel, Oct 19 2015
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Mathematica
Table[Binomial[n + 1, k + 1] (n + 1)^(k - 1), {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 19 2015 *)
Formula
a(n, m) = A033842(n, n-m) = binomial(n+1, m+1)*(n+1)^{m-1}, n >= m >= 0, else 0.
p(k-1, -x)/(1-k*x)^k =(-1+1/(1-k*x)^k)/(x*k^2) is for k=1..5 G.f. for A000012, A001792, A036068, A036070, A036083, respectively.
From Werner Schulte, Oct 19 2015: (Start)
a(2*n,n) = A000108(n)*(2*n+1)^n;
a(3*n,2*n) = A001764(n)*(3*n+1)^(2*n);
a(p*n,(p-1)*n) = binomial(p*n,n)/((p-1)*n+1)*(p*n+1)^((p-1)*n) for p > 0;
Sum_{m=0..n} (m+1)*a(n,m) = (n+2)^n;
Sum_{m=0..n} (-1)^m*(m+1)*a(n,m) = (-n)^n where 0^0 = 1;
p(n,x) = Sum_{m=0..n} a(n,m)*x^m = ((1+(n+1)*x)^(n+1)-1)/((n+1)^2*x).
(End)
Comments