A004984
a(n) = (2^n/n!)*Product_{k=0..n-1} (4*k - 1).
Original entry on oeis.org
1, -2, -6, -28, -154, -924, -5852, -38456, -259578, -1788204, -12517428, -88759944, -636112932, -4599585816, -33511268088, -245749299312, -1812401082426, -13433090375628, -100001895018564, -747382583822952, -5605369378672140, -42173731515723720
Offset: 0
Joe Keane (jgk(AT)jgk.org)
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List([0..25],n->(2^n/Factorial(n))*Product([0..n-1],k->4*k-1)); # Muniru A Asiru, Apr 28 2018
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[1] cat [2^n*(&*[4*k-1: k in [0..n-1]])/Factorial(n): n in [1..30]]; // G. C. Greubel, Aug 22 2019
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seq(-(1/4)*8^n*GAMMA(n-1/4)/(GAMMA(3/4)*n!), n=0..30); # Robert Israel, Sep 29 2014
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Table[8^n*Pochhammer[-1/4, n]/n!, {n,0,30}] (* G. C. Greubel, Aug 22 2019 *)
CoefficientList[Series[Surd[1-8x,4],{x,0,30}],x] (* Harvey P. Dale, Dec 08 2019 *)
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a(n):=-1/n*(if n=1 then (2)^n else sum(sum(binomial(k,j)* binomial(j,n-1-3*k+2*j)*(3/2)^(3*k-n+1-j)*(-1)^(n-1-3*k)*(1/4)^(k-j),j,0,k)*binomial(k+n-1,n-1)*(2)^n,k,1,n-1)); /* Vladimir Kruchinin, Sep 14 2010 */
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a(n):=binomial(1/4,n)*(-8)^n; /* Tani Akinari, Sep 28 2014 */
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for(n=0,28,print1(2^n/n!*prod(k=0,n-1,(4*k-1)),","))
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[8^n*rising_factorial(-1/4, n)/factorial(n) for n in (0..30)] # G. C. Greubel, Aug 22 2019
A025749
4th-order Patalan numbers (generalization of Catalan numbers).
Original entry on oeis.org
1, 1, 6, 56, 616, 7392, 93632, 1230592, 16612992, 228890112, 3204461568, 45445091328, 651379642368, 9419951751168, 137262154088448, 2013178259963904, 29694379334467584, 440175505428578304, 6553724191936610304, 97960930026841964544, 1469413950402629468160
Offset: 0
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seq., Vol. 3 (2000), Article 00.2.4.
- Elżbieta Liszewska and Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.
- Thomas M. Richardson, The Super Patalan Numbers, J. Int. Seq. 18 (2015), Article 15.3.3; arXiv preprint, arXiv:1410.5880 [math.CO], 2014.
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a[n_] := (4^(n-1)*Sum[ Binomial[n+k-1, n-1]*Sum[ Binomial[j, n-3*k+2*j-1] * 4^(j-k) * Binomial[k, j] * 3^(-n+3*k-j+1) * 2^(n-3*k+j-1) * (-1)^(n-3*k+2*j-1), {j, 0, k}], {k, 1, n-1}])/n; a[0] = a[1] = 1; Table[a[n], {n, 0, 17}] (* Jean-François Alcover, Mar 05 2013, after Vladimir Kruchinin *)
a[n_] := 16^(n-1) * Pochhammer[3/4, n-1]/n!; a[0] = 1; Array[a, 21, 0] (* Amiram Eldar, Aug 20 2025 *)
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a(n):=(4^(n-1)*sum(binomial(n+k-1,n-1)*sum(binomial(j,n-3*k+2*j-1)*4^(j-k)*binomial(k,j)*3^(-n+3*k-j+1)*2^(n-3*k+j-1)*(-1)^(n-3*k+2*j-1),j,0,k),k,1,n-1))/n; /* Vladimir Kruchinin, Apr 01 2011 */
A254286
Expansion of (1 - (1-256*x)^(1/4)) / (64*x).
Original entry on oeis.org
1, 96, 14336, 2523136, 484442112, 98180268032, 20645907791872, 4459516083044352, 983075545417777152, 220208922173582082048, 49967406340478261526528, 11459191854083014643417088, 2651480699775516003646046208, 618173786004806016850049630208
Offset: 0
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[Round(2^(8*n)*Gamma(n+3/4)/(Gamma(3/4)*Gamma(n+2))): n in [0..30]]; // G. C. Greubel, Aug 10 2022
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CoefficientList[Series[(1-(1-256*x)^(1/4)) / (64*x),{x,0,20}],x]
CoefficientList[Series[Hypergeometric1F1[3/4,2,256*x],{x,0,20}],x] * Range[0,20]! (* Vaclav Kotesovec, Jan 28 2015 *)
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[2^(8*n)*rising_factorial(3/4,n)/factorial(n+1) for n in (0..30)] # G. C. Greubel, Aug 10 2022
A097179
Triangle read by rows in which row n gives coefficients of polynomial R_n(y) that satisfies R_n(1/2) = 4^n, where R_n(y) forms the initial (n+1) terms of g.f. A077860(y)^(n+1).
Original entry on oeis.org
1, 1, 6, 1, 9, 42, 1, 12, 74, 308, 1, 15, 115, 595, 2310, 1, 18, 165, 1020, 4746, 17556, 1, 21, 224, 1610, 8722, 37730, 134596, 1, 24, 292, 2392, 14778, 73080, 299508, 1038312, 1, 27, 369, 3393, 23535, 130851, 604707, 2376099, 8046918
Offset: 0
Row polynomials evaluated at y=1/2 equals powers of 4:
4^1 = 1 + 6/2;
4^2 = 1 + 9/2 + 42/2^2;
4^3 = 1 + 12/2 + 74/2^2 + 308/2^3;
4^4 = 1 + 15/2 + 115/2^2 + 595/2^3 + 2310/2^4;
where A077860(y)^(n+1) has the same initial terms as the n-th row:
A077860(y) = 1 +3*y +5*y^2 +5*y^3 +1*y^4 -7*y^5 -15*y^6 -15*y^7 +...
A077860(y)^2 = 1 + 6*y +...
A077860(y)^3 = 1 + 9*y + 42*y^2 +...
A077860(y)^4 = 1 + 12*y + 74*y^2 + 308*y^3 +...
A077860(y)^5 = 1 + 15*y + 115*y^2 + 595*y^3 + 2310*y^4 +...
Rows begin with n=0:
1;
1, 6;
1, 9, 42;
1, 12, 74, 308;
1, 15, 115, 595, 2310;
1, 18, 165, 1020, 4746, 17556;
1, 21, 224, 1610, 8722, 37730, 134596;
1, 24, 292, 2392, 14778, 73080, 299508, 1038312;
1, 27, 369, 3393, 23535, 130851, 604707, 2376099, 8046918; ...
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Table[SeriesCoefficient[2*y/((1-8*x*y) +(2*y-1)*(1-8*x*y)^(3/4)), {x, 0, n}, {y,0,k}], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 17 2019 *)
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{T(n,k)=if(n==0,1,if(k==0,1,if(k==n, 2^n*(4^n-sum(j=0,n-1, T(n,j)/2^j)), polcoeff((Ser(vector(n,i,T(n-1,i-1)), x) +x*O(x^k))^((n+1)/n),k,x))))}
Showing 1-4 of 4 results.
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