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A094424
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Array read by antidiagonals: Solutions to Schmidt's Problem.
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5
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1, 1, 1, 1, 2, 1, 1, 4, 10, 1, 1, 8, 68, 56, 1, 1, 16, 424, 1732, 346, 1, 1, 32, 2576, 48896, 51076, 2252, 1, 1, 64, 15520, 1383568, 6672232, 1657904, 15184, 1, 1, 128, 93248, 39776000, 873960976, 1022309408, 57793316, 104960, 1, 1, 256, 559744, 1159151680, 116758856608, 615833930816, 176808084544, 2117525792, 739162, 1
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OFFSET
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1,5
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COMMENTS
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T(r,k) satisfies sum[k=0,n, C(n,k)^r*C(n+k,k)^r] = sum[k=0,n, C(n,k)*C(n+k,k)*T(r,k)] for all n=0,1,2,3...
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LINKS
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FORMULA
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Zudilin gives a complicated general formula involving binomial coefficients, thus proving that all T(r, k) are integers.
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EXAMPLE
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1 1 1 1 1 1
1 2 10 56 346 2252
1 4 68 1732 51076 1657904
1 8 424 48896 6672232 1022309408
1 16 2576 1383568 873960976 615833930816
1 32 15520 39776000 116758856608 371558588978432
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MATHEMATICA
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eq[r_, n_] := eq[r, n] = Sum[Binomial[n, k]^r*Binomial[n + k, k]^r, {k, 0, n}] == Sum[Binomial[n, k]*Binomial[n + k, k]*t[r, k], {k, 0, n}]; c[r_, k_] := t[r, k] /. Solve[Table[eq[r, n], {n, 0, k}], t[r, k]] // First; lg = 10; m = Table[c[r, k], {r, 1, lg}, {k, 0, lg - 1}];
Flatten[ Table[ Reverse @ Diagonal[ Reverse /@ m, k], {k, lg - 1, -lg + 1, -1}]][[1 ;; 55]] (* Jean-François Alcover, Jul 20 2011 *)
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PROG
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(PARI) A094424row(r, kmax)={ local(nmat, rhs, cv) ; nmat=matrix(kmax+1, kmax+1) ; rhs=matrix(kmax+1, 1) ; for(n=0, kmax, for(k=0, kmax, nmat[n+1, k+1]=binomial(n, k)*binomial(n+k, k) ; ) ; rhs[n+1, 1]=sum(i=0, n, binomial(n, i)^r*binomial(n+i, i)^r) ; ) ; cv=matsolve(nmat, rhs) ; } A094424(nmax)={ local(T, c) ; T=matrix(nmax, nmax) ; for(r=1, nmax, c=A094424row(r, nmax-1) ; for(i=1, nmax, T[r, i]=c[i, 1] ; ) ; ) ; return(T) ; } { rmax=10 ; T=A094424(rmax) ; for(d=0, rmax-1, for(c=0, d, print1(T[d-c+1, c+1], ", ") ; ) ; ) ; } - R. J. Mathar, Oct 06 2006
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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