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Triangle read by rows: T(n, k) is the Sheffer triangle ((1 - 3*x)^(-1/3), (-1/3)*log(1 - 3*x)). A generalized Stirling1 triangle.
+10
13
1, 1, 1, 4, 5, 1, 28, 39, 12, 1, 280, 418, 159, 22, 1, 3640, 5714, 2485, 445, 35, 1, 58240, 95064, 45474, 9605, 1005, 51, 1, 1106560, 1864456, 959070, 227969, 28700, 1974, 70, 1, 24344320, 42124592, 22963996, 5974388, 859369, 72128, 3514, 92, 1, 608608000, 1077459120, 616224492, 172323696, 27458613, 2662569, 159978, 5814, 117, 1, 17041024000, 30777463360, 18331744896, 5441287980, 941164860, 102010545, 7141953, 322770, 9090, 145, 1
OFFSET
0,4
COMMENTS
This is a generalization of the unsigned Stirling1 triangle A132393.
In general the lower triangular Sheffer matrix ((1 - d*x)^(-a/d), (-1/d)*log(1 - d*x)) is called here |S1hat[d,a]|. The signed matrix S1hat[d,a] with elements (-1)^(n-k)*|S1hat[d,a]|(n, k) is the inverse of the generalized Stirling2 Sheffer matrix S2hat[d,a] with elements S2[d,a](n, k)/d^k, where S2[d,a] is Sheffer (exp(a*x), exp(d*x) - 1).
In the Bala link the signed S1hat[d,a] (with row scaled elements S1[d,a](n,k)/d^n where S1[d,a] is the inverse matrix of S2[d,a]) is denoted by s_{(d,0,a)}, and there the notion exponential Riordan array is used for Sheffer array.
In the Luschny link the elements of |S1hat[m,m-1]| are called Stirling-Frobenius cycle numbers SF-C with parameter m.
From Wolfdieter Lang, Aug 09 2017: (Start)
The general row polynomials R(d,a;n,x) = Sum_{k=0..n} T(d,a;n,k)*x^k of the Sheffer triangle |S1hat[d,a]| satisfy, as special polynomials of the Boas-Buck class (see the reference), the identity (we use the notation of Rainville, Theorem 50, p. 141, adapted to an exponential generating function)
(E_x - n*1)*R(d,a;n,x) = -n!*Sum_{k=0..n-1} d^k*(a*1 + d*beta(k)*E_x)*R(d,a;n-1-k,x)/(n-1-k)!, for n >= 0, with E_x = x*d/dx (Euler operator), and beta(k) = A002208(k+1)/A002209(k+1).
This entails a recurrence for the sequence of column k, for n > k >= 0: T(d,a;n,k) = (n!/(n - k))*Sum_{p=k..n-1} d^(n-1-p)*(a + d*k*beta(n-1-p))*T(d,a;p,k)/p!, with input T(d,a;k,k) = 1. For the present [d,a] = [3,1] case see the formula and example sections below. (End)
The inverse of the Sheffer triangular matrix S2[3,1] = A282629 is the Sheffer matrix S1[3,1] = (1/(1 + x)^(1/3), log(1 + x)/3) with rational elements S1[3,1](n, k) = (-1)^(n-m)*T(n, k)/3^n. - Wolfdieter Lang, Nov 15 2018
REFERENCES
Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.
FORMULA
Recurrence: T(n, k) = T(n-1, k-1) + (3*n-2)*T(n-1, k), for n >= 1, k = 0..n, and T(n, -1) = 0, T(0, 0) = 1 and T(n, k) = 0 for n < k.
E.g.f. of row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k (i.e., e.g.f. of the triangle) is (1 - 3*z)^{-(x+1)/3}.
E.g.f. of column k is (1 - 3*x)^(-1/3)*((-1/3)*log(1 - 3*x))^k/k!.
Recurrence for row polynomials is R(n, x) = (x+1)*R(n-1, x+3), with R(0, x) = 1.
Row polynomial R(n, x) = risefac(3,1;x,n) with the rising factorial
risefac(d,a;x,n) := Product_{j=0..n-1} (x + (a + j*d)). (For the signed case see the Bala link, eq. (16)).
T(n, k) = sigma^{(n)}_{n-k}(a_0,a_1,...,a_{n-1}) with the elementary symmetric functions with indeterminates a_j = 1 + 3*j.
T(n, k) = Sum_{j=0..n-k} binomial(n-j, k)*|S1|(n, n-j)*3^j, with the unsigned Stirling1 triangle |S1| = A132393.
Boas-Buck column recurrence (see a comment above): T(n, k) =
(n!/(n - k))*Sum_{p=k..n-1} 3^(n-1-p)*(1 + 3*k*beta(n-1-p))*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, with beta(k) = A002208(k+1)/A002209(k+1). See an example below. - Wolfdieter Lang, Aug 09 2017
EXAMPLE
The triangle T(n, k) begins:
n\k 0 1 2 3 4 5 6 7 8 ...
O: 1
1: 1 1
2: 4 5 1
3: 28 39 12 1
4: 280 418 159 22 1
5: 3640 5714 2485 445 35 1
6: 58240 95064 45474 9605 1005 51 1
7: 1106560 1864456 959070 227969 28700 1974 70 1
8: 24344320 42124592 22963996 5974388 859369 72128 3514 92 1
...
From Wolfdieter Lang, Aug 09 2017: (Start)
Recurrence: T(3, 1) = T(2, 0) + (3*3-2)*T(2, 1) = 4 + 7*5 = 39.
Boas-Buck recurrence for column k = 2 and n = 5:
T(5, 2) = (5!/3)*(3^2*(1 + 6*(3/8))*T(2,2)/2! + 3*(1 + 6*(5/12)*T(3, 2)/3! + (1 + 6*(1/2))* T(4, 2)/4!)) = (5!/3)*(9*(1 + 9/4)/2 + 3*(1 + 15/6)*12/6 + (1 + 3)*159/24) = 2485.
The beta sequence begins: {1/2, 5/12, 3/8, 251/720, 95/288, 19087/60480, ...}.
(End)
MATHEMATICA
T[n_ /; n >= 1, k_] /; 0 <= k <= n := T[n, k] = T[n-1, k-1] + (3*n-2)* T[n-1, k]; T[_, -1] = 0; T[0, 0] = 1; T[n_, k_] /; n<k = 0;
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 20 2018 *)
CROSSREFS
S2[d,a] for [d,a] = [1,0], [2,1], [3,1], [3,2], [4,1] and [4,3] is A048993, A154537, A282629, A225466, A285061 and A225467, respectively.
S2hat[d,a] for these [d,a] values is A048993, A039755, A111577 (offset 0), A225468, A111578 (offset 0) and A225469, respectively.
|S1hat[d,a]| for [d,a] = [1,0], [2,1], [3,2], [4,1] and [4,3] is A132393, A028338, A225470, A290317 and A225471, respectively.
Column sequences for k = 0, 1: A007559, A024216.
Diagonal sequences: A000012, A000326(n+1), A024212(n+1), A024213(n+1).
Row sums: A008544. Alternating row sums: A000007.
Beta sequence: A002208(n+1)/A002209(n+1).
KEYWORD
nonn,easy,tabl
AUTHOR
Wolfdieter Lang, May 18 2017
STATUS
approved
Triple factorial array, read by antidiagonals, where row n+1 is generated from row n by first removing terms in row n at positions {[m*(m+5)/6], m >= 0} and then taking partial sums, starting with all 1's in row 0.
+10
9
1, 1, 1, 4, 2, 1, 28, 10, 3, 1, 280, 80, 18, 4, 1, 3640, 880, 162, 28, 5, 1, 58240, 12320, 1944, 280, 39, 6, 1, 1106560, 209440, 29160, 3640, 418, 52, 7, 1, 24344320, 4188800, 524880, 58240, 5714, 600, 66, 8, 1, 608608000, 96342400, 11022480, 1106560, 95064
OFFSET
0,4
COMMENTS
This is the triple factorial variant of Moessner's factorial array described by A125714 and also of the double factorial array A135876. Another very interesting variant is A136217.
FORMULA
Columns 0, 1 and 2 form the triple factorials A007559, A008544 and A032031, respectively. Column 4 equals A024216; column 6 equals A024395.
EXAMPLE
Square array begins:
(1),(1),(1),1,(1),1,(1),1,(1),1,1,(1),1,1,(1),1,1,(1),1,1,1,(1),1,1,1,...;
(1),(2),(3),4,(5),6,(7),8,(9),10,11,(12),13,14,(15),16,17,(18),19,20,21,..;
(4),(10),(18),28,(39),52,(66),82,(99),118,138,(159),182,206,(231),258,286,..;
(28),(80),(162),280,(418),600,(806),1064,(1350),1696,2074,(2485),2966,3484,..;
(280),(880),(1944),3640,(5714),8680,(12164),16840,(22194),29080,36824,(45474),.;
(3640),(12320),(29160),58240,(95064),151200,(219108),315440,(428652),581680,...;
(58240),(209440),(524880),1106560,(1864456),3082240,...;
where terms in parenthesis are at positions {[m*(m+5)/6], m>=0}
and are removed before taking partial sums to obtain the next row.
To generate the array, start with all 1's in row 0; from then on,
obtain row n+1 from row n by first removing terms in row n at
positions {[m*(m+5)/6], m>=0} and then taking partial sums.
For example, to generate row 2 from row 1:
[(1),(2),(3),4,(5),6,(7),8,(9),10,11,(12),13,14,(15),16,17,(18),...],
remove terms at positions [0,1,2,4,6,8,11,14,17,...] to get:
[4, 6, 8, 10,11, 13,14, 16,17, 19,20,21, 23,24,25, 27,28,29, ...]
then take partial sums to obtain row 2:
[4, 10, 18, 28,39, 52,66, 82,99, 118,138,159, 182,206,231, ...].
Continuing in this way will generate all the rows of this array.
MATHEMATICA
t[n_, k_] := t[n, k] = Module[{a = 0, m = 0, c = 0, d = 0}, If[n == 0, a = 1, While[d <= k, If[c == Quotient[(m*(m + 5)), 6], m += 1, a += t[n - 1, c]; d += 1]; c += 1]]; a]; Table[t[n - k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 06 2013, translated from Pari *)
PROG
(PARI) {T(n, k)=local(A=0, m=0, c=0, d=0); if(n==0, A=1, until(d>k, if(c==(m*(m+5))\6, m+=1, A+=T(n-1, c); d+=1); c+=1)); A}
CROSSREFS
KEYWORD
nice,nonn,tabl
AUTHOR
Paul D. Hanna, Dec 22 2007
STATUS
approved
a(n) = n-th elementary symmetric function of the first n+1 positive integers congruent to 2 mod 3.
+10
5
1, 7, 66, 806, 12164, 219108, 4591600, 109795600, 2951028000, 88084714400, 2891353030400, 103521905491200, 4015191638617600, 167714507921497600, 7506196028811110400, 358368551285791692800, 18180562447078051328000
OFFSET
0,2
COMMENTS
Comment by R. J. Mathar, Oct 01 2016 (Start):
The k-th elementary symmetric functions of the integers 2+j*3, j=0..n-1, form a triangle T(n,k), 0<=k<=n, n>=0:
1
1 2
1 7 10
1 15 66 80
1 26 231 806 880
1 40 595 4040 12164 12320
1 57 1275 14155 80844 219108 209440
1 77 2415 39655 363944 1835988 4591600 4188800
1 100 4186 95200 1276009 10206700 46819324 109795600 96342400
This here is the first subdiagonal. The diagonal seems to be A008544. The first columns are A000012, A005449, A024391, A024392. (End)
LINKS
FORMULA
E.g.f. (for offset 1): -(1/3)*log(1-3*x)/(1-3*x)^(2/3). - Vladeta Jovovic, Sep 26 2003
For n >= 1, a(n-1) = 3^(n-1)*n!*sum(binomial(k-1/3,k)/(n-k), k = 0..n-1). - Milan Janjic, Dec 14 2008, corrected by Peter Bala, Oct 08 2013
a(n) ~ (n+1)! * 3^n * (log(n) + gamma - Pi*sqrt(3)/6 + 3*log(3)/2) / (n^(1/3)*GAMMA(2/3)), where "GAMMA" is the Gamma function and "gamma" is the Euler-Mascheroni constant (A001620). - Vaclav Kotesovec, Oct 07 2013
a(n+1) = (6*n+7) * a(n) - (3*n+2)^2 * a(n-1). - Gheorghe Coserea, Aug 30 2015
a(n) = A225470(n+1, 1), n >= 0. - Wolfdieter Lang, May 29 2017
EXAMPLE
From Gheorghe Coserea, Dec 24 2015: (Start)
For n=1 we have a(1) = 2*5*(1/2 + 1/5) = 7.
For n=2 we have a(2) = 2*5*8*(1/2 + 1/5 + 1/8) = 66.
For n=3 we have a(3) = 2*5*8*11*(1/2 + 1/5 + 1/8 + 1/11) = 806.
(End)
MATHEMATICA
Table[ (-1)^(n+1)*Sum[(-3)^(n - k) k (-1)^(n - k) StirlingS1[n+1, k + 1], {k, 0, n}], {n, 1, 30}]
Join[{1}, Table[Module[{c=NestList[3+#&, 2, n+1]}, Times@@c*Total[1/c]], {n, 0, 20}]] (* Harvey P. Dale, Jul 09 2019 *)
PROG
(PARI)
n = 16; a = vector(n); a[1] = 7; a[2] = 66;
for (k=2, n-1, a[k+1] = (6*k+7) * a[k] - (3*k+2)^2 * a[k-1]);
print(concat(1, a)) \\ Gheorghe Coserea, Aug 30 2015
CROSSREFS
Cf. A024216, A225470 (second column).
KEYWORD
nonn,easy
EXTENSIONS
Formula (see Mathematica line), correction and more terms from Victor Adamchik (adamchik(AT)cs.cmu.edu), Jul 21 2001
STATUS
approved
a(n) = n-th elementary symmetric function of the first n+1 positive integers congruent to 1 mod 4.
+10
3
1, 6, 59, 812, 14389, 312114, 8011695, 237560280, 7990901865, 300659985630, 12511934225955, 570616907588100, 28301322505722525, 1516683700464669450, 87336792132539066775, 5378036128829898836400, 352652348707389385916625, 24533212082483855129037750
OFFSET
0,2
COMMENTS
a(n) is equal to the determinant of the n X n matrix whose (i,j)-entry is KroneckerDelta[i,j]((4*i+2)-1)+1. - John M. Campbell, May 23 2011
From R. J. Mathar, Oct 01 2016: (Start)
The k-th elementary symmetric functions of the integers 1+4*j, j=1..n, form a triangle T(n,k), 0<=k<=n, n>=0:
1
1 1
1 6 5
1 15 59 45
1 28 254 812 585
1 45 730 5130 14389 9945
1 66 1675 20460 122119 312114 208845
1 91 3325 62335 633619 3365089 8011695 5221125
1 120 5964 158760 2441334 21740040 105599276 237560280 151412625
This here is the first subdiagonal. The diagonal seems to be A007696. The 2nd column is A000384, the 3rd A024378, the 4th A024379. (End)
LINKS
FORMULA
a(n) = (8*n-2)*a(n-1) - (4*n-3)^2*a(n-2) for n>1. - Alois P. Heinz, Feb 25 2015
E.g.f.: (4-log(1-4*x))/(4*(1-4*x)^(5/4)). - Gheorghe Coserea, Dec 24 2015
EXAMPLE
For n = 1 we have a(1) = 1*5*(1/1 + 1/5) = 6.
For n = 2 we have a(2) = 1*5*9*(1/1 + 1/5 + 1/9) = 59.
For n = 3 we have a(3) = 1*5*9*13*(1/1 + 1/5 + 1/9 + 1/13) = 812. - Gheorghe Coserea, Dec 24 2015
MAPLE
a:= proc(n) option remember; `if`(n<3, [1, 6, 59][n+1],
(8*n-2)*a(n-1) -(4*n-3)^2*a(n-2))
end;
seq(a(n), n=0..20); # Alois P. Heinz, Feb 25 2015
MATHEMATICA
Table[Det[Array[KroneckerDelta[#1, #2]((4*#1+2)-1)+1&, {k, k}]], {k, 1, 10}] (* John M. Campbell, May 23 2011 *)
RecurrenceTable[{a[0] == 1, a[1] == 6, a[n] == (8 n - 2) a[n - 1] - (4 n - 3)^2 a[n - 2]}, a, {n, 0, 20}] (* Vincenzo Librandi, Dec 26 2015 *)
PROG
(PARI) x = 'x + O('x^33); Vec(serlaplace((4-log(1-4*x))/(4*(1-4*x)^(5/4)))) \\ Gheorghe Coserea, Dec 24 2015
(Magma) I:=[1, 6]; [n le 2 select I[n] else (8*n-10)*Self(n-1)-(4*n-7)^2*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 26 2015
CROSSREFS
Cf. A024216.
KEYWORD
nonn
EXTENSIONS
More terms from Alois P. Heinz, Feb 25 2015
STATUS
approved
Column k=2 of the triangle A286718; Sheffer ((1 - 3*x)^(-1/3), (-1/3)*log(1 - 3*x)).
+10
0
1, 12, 159, 2485, 45474, 959070, 22963996, 616224492, 18331744896, 599061555136, 21339235262784, 823098817664448, 34183157124707200, 1520908498941532800, 72182781516370886400, 3640264913563748243200, 194408478299496756556800, 10961007293837647131724800
OFFSET
0,2
COMMENTS
a(n) is, for n >= 1, the total volume of the binomial(n+2, n) rectangular polytopes (hyper-cuboids) built from n orthogonal vectors with lengths of the sides from the set {1 + 3*j | j=0..n+1}. See the formula a(n) = sigma[3,1]^{(n+2)}_n and an example below.
FORMULA
a(n) = A286718(n+2, 2), n >= 0.
E.g.f.: (d^2/dx^2)((1 - 3*x)^(-1/3)*((-1/3)*log(1 - 3*x))^2/2!) = (2*(log(1-3*x))^2 - 15*log(1-3*x) + 9)/(3^2*(1-3*x)^(7/3)).
a(n) = sigma[3,1]^{(n+2)}_n, n >= 0, with the elementary symmetric function sigma[3,1]^{n+2}_n of degree n of the n+2 numbers 1, 4, 7, ..., (1 + 3*(n+1)).
EXAMPLE
a(2) = 159 because sigma[3,1]^{(4)}_2 = 1*(4 + 7 + 10) + 4*(7 + 10) + 7*10 = 159. There are six rectangles (2D rectangular polytopes) built from two orthogonal vectors of different lengths from the set of {1,4,7,10} of total area 159.
CROSSREFS
Cf. A007559 (k=0), A024216 (k=1), A286718.
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, May 29 2017
STATUS
approved

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