A288470
a(n) = Sum_{k=0..n} binomial(n,k)*binomial(2*n,2*k).
Original entry on oeis.org
1, 2, 14, 92, 646, 4652, 34124, 253528, 1901638, 14368844, 109208164, 833981128, 6394017436, 49185717752, 379438594136, 2934361958192, 22741538394694, 176582855512588, 1373431963785332, 10698376362421096, 83447762846703796, 651690159076273192, 5095051571420324264, 39874449115469939152, 312350761370734541596
Offset: 0
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f:= gfun:-rectoproc({n*(2*n-1)*a(n) = (32*(n-2))*(2*n-5)*a(n-3)+(8*(9*n^2-31*n+28))*a(n-2)+(2*(3*n^2+7*n-9))*a(n-1), a(0)=1,a(1)=2, a(2)=14},a(n),remember):
map(f, [$0..30]);
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Table[Sum[Binomial[n, k] Binomial[2 n, 2 k], {k, 0, n}], {n, 0, 24}] (* Michael De Vlieger, Jun 09 2017 *)
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{a(n) = polcoef((-1+(1+x+1/x)^2)^n, 0)} \\ Seiichi Manyama, Nov 21 2019
A168518
Expansion of g.f. (1/2)*( a*(1+x)^n + b*(1-x)^(n+2)*LerchPhi(x, -n-1, 1) + c*2^(n+1)*(1-x)^(n+1)*LerchPhi(x, -n, 1/2) ), where a = -4, b = 2, and c = 2, read by rows.
Original entry on oeis.org
1, 1, 1, 1, 12, 1, 1, 51, 51, 1, 1, 170, 514, 170, 1, 1, 521, 3646, 3646, 521, 1, 1, 1552, 22247, 49472, 22247, 1552, 1, 1, 4591, 125565, 534995, 534995, 125565, 4591, 1, 1, 13590, 677776, 5058698, 9506078, 5058698, 677776, 13590, 1, 1, 40341, 3560448, 43870968, 140136690, 140136690, 43870968, 3560448, 40341, 1
Offset: 0
Triangle begins as:
1;
1, 1;
1, 12, 1;
1, 51, 51, 1;
1, 170, 514, 170, 1;
1, 521, 3646, 3646, 521, 1;
1, 1552, 22247, 49472, 22247, 1552, 1;
1, 4591, 125565, 534995, 534995, 125565, 4591, 1;
1, 13590, 677776, 5058698, 9506078, 5058698, 677776, 13590, 1;
1, 40341, 3560448, 43870968, 140136690, 140136690, 43870968, 3560448, 40341, 1;
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p[x_, n_, a_, b_, c_]= (1/2)*(a*(1+x)^n + b*(1-x)^(n+2)*LerchPhi[x,-n-1,1] + c*2^(n+1)*(1-x)^(n+1)*LerchPhi[x,-n,1/2]);
Table[CoefficientList[p[x,n,-4,2,2], x], {n,0,10}]//Flatten (* modified by G. C. Greubel, Mar 31 2022 *)
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def A168518(n,k,a,b,c): return (1/2)*( a*binomial(n,k) + sum( (-1)^(k-j)*(b*binomial(n+2, k-j)*(j+1)^(n+1) + 2*c*binomial(n+1,k-j)*(2*j+1)^n) for j in (0..k)) )
flatten([[A168518(n,k,-4,2,2) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 31 2022
A155497
Triangle T(n, k) = binomial(n, k)*binomial(n+1, k+1)*binomial(2*n, 2*k)/(n-k+1), read by rows.
Original entry on oeis.org
1, 1, 1, 1, 18, 1, 1, 90, 90, 1, 1, 280, 1400, 280, 1, 1, 675, 10500, 10500, 675, 1, 1, 1386, 51975, 161700, 51975, 1386, 1, 1, 2548, 196196, 1471470, 1471470, 196196, 2548, 1, 1, 4320, 611520, 9417408, 22702680, 9417408, 611520, 4320, 1, 1, 6885, 1652400, 46781280, 231567336, 231567336, 46781280, 1652400, 6885, 1
Offset: 0
Triangle begins as:
1;
1, 1;
1, 18, 1;
1, 90, 90, 1;
1, 280, 1400, 280, 1;
1, 675, 10500, 10500, 675, 1;
1, 1386, 51975, 161700, 51975, 1386, 1;
1, 2548, 196196, 1471470, 1471470, 196196, 2548, 1;
1, 4320, 611520, 9417408, 22702680, 9417408, 611520, 4320, 1;
1, 6885, 1652400, 46781280, 231567336, 231567336, 46781280, 1652400, 6885, 1;
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[Binomial(n, k)*Binomial(n+1, k+1)*Binomial(2*n, 2*k)/(n-k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, May 29 2021
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T[n_, k_]:= Binomial[n,k]*Binomial[n+1,k+1]*Binomial[2*n,2*k]/(n-k+1);
Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* modified by G. C. Greubel, May 29 2021 *)
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flatten([[binomial(n, k)*binomial(n+1, k+1)*binomial(2*n, 2*k)/(n-k+1) for k in (0..12)] for n in (0..12)]) # G. C. Greubel, May 29 2021
A155516
Triangle T(n, k) = binomial(2*n, 2*k)*binomial(2*n+1, 2*k+1)/(2*n-2*k+1), read by rows.
Original entry on oeis.org
1, 1, 1, 1, 20, 1, 1, 105, 105, 1, 1, 336, 1764, 336, 1, 1, 825, 13860, 13860, 825, 1, 1, 1716, 70785, 226512, 70785, 1716, 1, 1, 3185, 273273, 2147145, 2147145, 273273, 3185, 1, 1, 5440, 866320, 14158144, 34763300, 14158144, 866320, 5440, 1, 1, 8721, 2372112, 71954064, 367479684, 367479684, 71954064, 2372112, 8721, 1
Offset: 0
Triangle begins as:
1;
1, 1;
1, 20, 1;
1, 105, 105, 1;
1, 336, 1764, 336, 1;
1, 825, 13860, 13860, 825, 1;
1, 1716, 70785, 226512, 70785, 1716, 1;
1, 3185, 273273, 2147145, 2147145, 273273, 3185, 1;
1, 5440, 866320, 14158144, 34763300, 14158144, 866320, 5440, 1;
1, 8721, 2372112, 71954064, 367479684, 367479684, 71954064, 2372112, 8721, 1;
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[Binomial(2*n, 2*k)*Binomial(2*n+1, 2*k+1)/(2*n-2*k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, May 29 2021
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T[n_, k_]:= Binomial[2*n, 2*k]*Binomial[2*n+1, 2*k+1]/(2*n-2*k+1);
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, May 29 2021 *)
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flatten([[binomial(2*n, 2*k)*binomial(2*n+1, 2*k+1)/(2*n-2*k+1) for k in (0..12)] for n in (0..12)]) # G. C. Greubel, May 29 2021
Showing 1-4 of 4 results.
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